1st PUC

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Karnataka 1st PUC Maths Question Bank Chapter 3 Trigonometric Functions

Measurement of Angles:

Question 1.
What is the meaning of trigonometry?
Answer :
The literal meaning of the word trigonometry is the science of measuring the sides and the angles of triangles.

Question 2.
Define an angle.
Answer :
Angle is a measure of rotation of a given ray about its initial point.

Question 3.
Define a degree measure.
Answer :
If a rotation from the initial side to terminal side is \( \left(\frac{1}{360}\right)^{t h} \) of a revolution, the angle is said to have a measure of one degree, written as 1°.

Keen eye:
A degree is divided into 60 minutes and a minute is divided into 60 seconds. One sixtieth of a degree is called a minute, written as 1′ and one sixtieth of a minute is called second, written as 1″
∴ 1 ° – 60′
1’= 60″

Question 4.
Define a radian measure.
Answer :
Angle subtended at the centre by an arc of length 1 unit in a unit circle is said to have a measure of 1 radian.

Keen eye:
\(1 \mathrm{rad}=\frac{180^{\circ}}{\pi}=57^{\circ} 16^{\prime} \)

Degree	30°	45°	60°	90°	180°	270°	360°
Radian	\(\frac{\pi}{6}\)	\(\frac{\pi}{4}\)	\(\frac{\pi}{3}\)	\(\frac{\pi}{2}\)	π	\(\frac{3 \pi}{2}\)	2π

Question 5.
Convert the following into radian measure,
(i) 25°
(ii) 240°
(iii) 520°
(iv) 225°
(v) 150°
(vi) 330°
(vii) – 47° 30′
(viii) 40° 20′
Answer:

Question 6.
Convert the following into degree measure.
(i) 6
(ii) \( \frac{11}{16}\)
(iii) -4
(iv) \( \frac{5 \pi}{3} \)
(v) \( \frac{7 \pi}{6} \)
Answer:

Question 7.
A wheel makes 360° revolutions in one minute. Through how many radians does it turn in one second?
Answer :
Number of revolutions in one minute = 360
Number of revolutions in one second = \( \frac{360}{60}=6  \)
But, in one revolutions angle traced = 360°
= 2π rad
⇒ Angle traced in 6 revolutions
= 6 x 2π= 12π radians
Angle traced in one second = 12π radians

Question 8.
Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm \(\left(\text { use } \pi=\frac{22}{7}\right)\)
Answer :

Question 9.
Find the radius of the circle in which a central angle of 60° intercepts an arc of length \(37.4 \mathrm{cm}\left(\text { use } \pi=\frac{22}{7}\right)\)
Answer :

Question 10.
In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.
Answer :
Given, diameter AB = 40 cm

∴ Radius OB = 20 cm
Chord BC = 20 cm
∴  Δ OBC is an equilateral triangle

Question 11.
The minute hand of a watch is 1.5 cm long. How far does its tip move in 40 minute?
(use π = 3.14).
Answer :
In 60 minutes, the minute hand of a watch completes one revolution.
∴ In 40 minutes, the minute hand turns through  of a revolution

Question 12.
If the arcs of the same lengths in two circles subtend angles 65° and 110° at the centre, find the ratio of their radii.
Answer :

Question 13.
If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.
Answer :
Let r₁ and r₂ be the radii of two circles
\( \text { Given } \quad \theta_{1}=60^{\circ}=\frac{\pi}{3} \mathrm{rad}\)

Question 14.
Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length
(i) 10 cm
(ii) 15 cm
(iii) 21 cm
Answer :

Trigonometric Functions:

Recall:

Graph of trigonometric function:

Question 1.
If \(\cos x=-\frac{3}{5}, x\) lies in the third quadrant, find the values of other five trigonometric functions.
Answer

Question 2.
If \( \cos x=-\frac{1}{2}, x \) lies in the third quadrant, find the values of other five trigonometric functions.
Answer:

Question 3.
If \( \sin x=\frac{3}{5}, x \) lies in the second quadrant, find the values of other five trigonometric functions.
Answer:

Question 4.
If \( \tan x=-\frac{5}{12}, x \)  lies in the second quadrant, find the values of other five trigonometric functions.
Answer:

Question 5.
If \(\cot x=\frac{3}{4}, x \) lies in third quadrant, find the values of other five trigonometric functions.
Answer:

Question 6.
If \(\\cot x=-\frac{5}{12}, x lies\) in second quadrant, find the values of other five trigonometric functions.
Answer:

\(\cot x=-\frac{5}{12}\)

Question 7.
If \(\sec x=\frac{13}{5}, x\) lies in fourth quadrant, find the values of other five trigonometric functions.
Answer:

Measure of angle	sin	cos	tan	cosec	sec	cot
-θ	– sinθ	cosθ	-tanθ	– cosecθ	secθ	– cotθ
\(\frac{\pi}{2}-\theta \)	cosθ	-sinθ	cot θ	secθ	cosecθ	tanθ
\(\frac{\pi}{2}+\theta \)	cosθ	– sinθ	-cotθ	– secθ	cosecθ	-tanθ
π-θ	sinθ	-cosθ	-tanθ	cosec θ	-secθ	-cotθ
π+θ	– sinθ	-cosθ	tanθ	-cosccθ	– secθ	cotθ
\(\frac{3 \pi}{2}-\theta\)	– cosθ	-sinθ	cotθ	– secθ	– cosecθ	tan θ
\(\frac{3 \pi}{2}+\theta\)	-cosθ	sinθ	-cotθ	– secθ	cosecθ	tanθ
2π-θ	-sinθ	cosθ	-tanθ	– cosccθ	secθ	-cotθ
2π+θ	sinθ	cos θ	tanθ	cosecθ	secθ	cosθ

Question 8.
Find the value of
\( \sin \frac{31 \pi}{3} \)
Answer:
We know that values of sinx repeats after an interval of 2π.
\( \sin \frac{31 \pi}{3}=\sin \left(10 \pi+\frac{\pi}{3}\right)=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\)

Question 9.
Find the value of sin 765°.
Answer:
\( \sin 765^{\circ}=\sin \left(720^{\circ}+45^{\circ}\right)=\sin 45^{\circ}=\frac{1}{\sqrt{2}}\)

Question 10.
Find the value of
\( \sin \left(-\frac{11 \pi}{3}\right) \)
Answer:
\(\begin{aligned} \sin \left(-\frac{11 \pi}{3}\right) &=-\sin \left(\frac{11 \pi}{3}\right)=-\sin \left(4 \pi-\frac{\pi}{3}\right) \\ &=\sin \left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2} \end{aligned}\)

Question 11.
Find the value of cos (-1710°)
Answer :
cos (1710°) = cos (1710°)
\( \begin{array}{l}{=\cos \left(1800-90^{\circ}\right)=\cos \left(10 \pi-\frac{\pi}{2}\right)} \\ {=\cos \frac{\pi}{2}=0}\end{array}\)

Question 12.
Find the value of cosec (-1410°)
Answer :
cosec (- 1410°)
= – cosec (1410°)
= – cosec[4(360°) – 30°]
= – cosec (- 30°)
= cosec 30° = 2

Question 13.
Find the value of
\(  \tan \frac{19 \pi}{3}\)
Answer:
\(\tan \frac{19 \pi}{3}=\tan \left(6 \pi+\frac{\pi}{3}\right)=\tan \frac{\pi}{3}=\sqrt{3} \)

Question 14.
Find the value of
\( \cot \left(-\frac{15 \pi}{4}\right)\)
Answer:
\(\begin{aligned} \cot \left(-\frac{15 \pi}{4}\right) &=-\cot \left(\frac{15 \pi}{4}\right)=-\cot \left(4 \pi-\frac{\pi}{4}\right) \\ &=\cot \frac{\pi}{4}=1 \end{aligned}\)

Question 15.
Derive cos(x + y) = cos x cos y – sin x sin y , using unit circle concept.

Answer:

Question 16.
Prove that:
cos(x-y) = cosxcosy+ sinxsiny.
Answer :
We have cos(x + y) = cos xcos y – sin xsin y
Replace y by – y, we get
cos[x + (-y)] = cos xcos(-y) – sin xsin(-y)
cos(x – y) = cos x cos y + sin x sin y

Question 17.
Prove that
\( \cos \left(\frac{\pi}{2}-x\right)=\sin x \)
Answer:

Question 18.
Prove that
\( \sin \left(\frac{\pi}{2}-x\right)=\cos x\)
Answer:

Question 19.
Prove that sin(x + y) = sinx cosy + cosx siny .
Answer :

Question 20.
Prove that sin( x – y) = sin x cos y – cos x sin y.
Answer :
sin(x – y) = sin[x + (-y)]
= sin x cos(-y) + cos x  sin(-y)
= sin x cos y-cos x sin y .

Question 21.
Prove the following
(i) \(\cos \left(\frac{\pi}{2}+x\right)=-\sin x \)
(ii) \( \sin \left(\frac{\pi}{2}+x\right)=\cos x \)
(iii) cos(π – x) = – cosx
(iv) sin(π – x) =sinx
(v) cos(π + x) = – cosx
(vi) sin(π + x) = – sinx
(vii) cos(2π – x) =cosx
(viii) sin(2π – x) = – sin x.
Answer:

Question 22.
If none of the angles ac, y and x + y is an odd multiple of \(\frac{\pi}{2} \)then prove that
\(\tan (x \pm y)=\frac{\tan x \pm \tan y}{1 \mp \tan x \tan y} \)
Answer:

Question 23.
If none of the angles x,y and (x+y) is a multiple of π, then prove that
\( \cot (x \pm y)=\frac{\cot x \cot y \pm 1}{\cot y \pm \cot x}\)
Answer :
As x, y and x+y are not the multiple of π, so Sinx, sin y and sin(x+ y) are non zero.

Question 24.
Prove that
\(\cos 2 x=\left\{\begin{array}{l}{\cos ^{2} x-\sin ^{2} x} \\ {2 \cos ^{2} x-1} \\ {1-2 \sin ^{2} x} \\ {\frac{1-\tan ^{2} x}{1+\tan ^{2} x}}\end{array}\right.\)
Answer:
We have cos(x + y) = cos xcos y – sin xsin y
Replacing y by x, we get
cos(x + x) = cosx cosx-sinx sinx

Question 25.
Prove that
\( \sin 2 x=\left\{\begin{array}{c}{2 \sin x \cos x} \\ {\frac{2 \tan x}{1+\tan ^{2} x}}\end{array}\right.\)
Answer:
We have sin(x + y) = sin xcos y + cos xsin y
Replacing y by x we get
sin 2x = 2sinx cosx

Question 26.
Prove that
\(\tan 2 x=\frac{2 \tan x}{1-\tan ^{2} x} \)
Answer:

Question 27.
Prove that sin3x = 3sinx – 4sin³ x .
Answer :
sin 3x = sin(2x + x)
= sin 2x cos x + cos 2x sin x
= 2sin xcos x cos x + (1 – 2sin² x)sin x
= 2 sin x(1- sin² x) + sin x – 2 sin³ x
= 3sinx-4sin³ x

Question 28.
Prove that cos3x = 4cos³ x -3cosx .
Answer :
cos3x = cos(2x + x)
= cos 2x cos x – sin 2x sin x
= (2 cos² x -1) cos x – 2 sin x cos x sin x
= 2cos³x-cosx-2cosx(1-cos²x)
= 4cos³ x-3cosx

Question 29.
Prove that
\(\tan 3 x=\frac{3 \tan x-\tan ^{3} x}{1-3 \tan ^{2} x}\)
Answer:

Question 30.
Prove the following
(i) \(\cos x+\cos y=2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}\)
(ii) \( \cos x-\cos y=-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2} \)
(iii) \( \sin x+\sin y=2 \sin \frac{x+y}{2} \cos \frac{x-y}{2} \)
(iv) \( \sin x-\sin y=2 \cos \frac{x+y}{2} \sin \frac{x-y}{2} \)
Answer:
We have
cos(A + B) = cosAcosB-sinAsinB …(1)
cos(A – B) = cos AcosB + sin Asin B … (2)
Adding and subtracting (1) and (2) we get
cos( A + B) + cos( A -B) = 2 cos A cos B … (3)
cos( A + B) – cos (A -B) = – 2 sin A sin B … (4)
Further
sin(A + B) = sinAcosB + cosAsinB         …(5)
sin(A-B) = sinAcosB-cosAsinB …(6)
Adding and subtracting (5) and (6) we get
sin( A + B) + sin( A – B) = 2 sin A cos B … (7)
sin( A + B) – sin(A – B) = 2cosA sinB … (8)
Let A + B = x and A-B = y


Keen eye;

• 2 cos xcos y = cos(x + y) + cos(x – y)
• -2 sin x sin y = cos(x + y) – cos(x – y)
• 2sin x cos y = sin(x + y) 4-sin(x- y)
• 2cosx siny = sin(x+ y)-sin(x-y)

Question 31.
Prove the following
(i) \( 3 \sin \frac{\pi}{6} \sec \frac{\pi}{3}-4 \sin \frac{5 \pi}{6} \cot \frac{\pi}{4}=1\)
(ii) \( \sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}=-\frac{1}{2}\)
(iii) \( 2 \sin ^{2} \frac{\pi}{6}+\csc ^{2} \frac{7 \pi}{6} \cos ^{2} \frac{\pi}{3}=\frac{3}{2} \)
(iv) \( 2 \sin ^{2} \frac{3 \pi}{4}+2 \cos ^{2} \frac{\pi}{4}+2 \sec ^{2} \frac{\pi}{3}=10 \)
(v) \( \cot ^{2} \frac{\pi}{6}+\csc \frac{5 \pi}{6}+3 \tan ^{2} \frac{\pi}{6}=6 \)
Answer:

Question 32.
Find the Value of
(i) sin 75°
(ii) cos 75°
(iii) tan 75°
(iv) sin 15°
(v) cos 15°
(vi) tan 15°
(vii) sin 105°
(viii) cos 105°.
Answer:

Question 33.
Find the value of \( \tan \frac{13 \pi}{12} \)
Answer:

Question 34.
Prove that
\( \frac{\sin (x+y)}{\sin (x-y)}=\frac{\tan x+\tan y}{\tan x-\tan y} \)
Answer:

Question 35.
Prove that
\(\begin{array}{l}{\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)} \\ {=\sin (x+y)}\end{array} \)
Answer:
\( \begin{aligned} L H S &=\cos \left[\frac{\pi}{4}-x+\frac{\pi}{4}-y\right] \\ &=\cos \left[\frac{\pi}{2}-(x+y)\right] \\ &=\sin (x+y)=R H S \end{aligned} \)

Question 36.
Prove that
sin[(n + l)x]sin[(n + 2)x] +cos[(n+1)x]cos[(n + 2)x] = cosx
Answer :
LHS = cos[(n + 2)x – (n + 1)x]
= cos x = RHS .

Question 37.
Prove that
(sin3x + sinx)sinx + (cos3x -cosx)cosx = 0
Answer :
LHS = sin 3x  sin x+sin² x + cos 3xcos x – cos² x
= [cos 3x cos x + sin 3x sin x] – [cos² x – sin² x]
= cos(3x – x) – cos 2x
= 0
= RHS

Question 38.
Prove that
\(\frac{\tan \left(\frac{\pi}{4}+x\right)}{\tan \left(\frac{\pi}{4} x\right)}=\left(\frac{1+\tan x}{1-\tan x}\right)^{2} \)
Answer:
\( L H S=\frac{\frac{1+\tan x}{1-\tan x}}{\frac{1-\tan x}{1+\tan x}}=\left(\frac{1+\tan x}{1-\tan x}\right)^{2}=R H S\)

Question 39.
Prove that
\( \frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}=\cot ^{2} x\)
Answer:
\(L H S=\frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)}=\cot ^{2} x=R H S \)

Question 40.
Prove that
\( \begin{array}{l}{\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)} \\ {\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]=1}\end{array}\)
Answer:

Question 41.
Prove that
\(\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x\)
Answer:

Question 42.
Prove that sin²6x-sin²4x=sin2xsinl0x.
Answer :
LHS = sin(6x + 4x)sin(6x – 4x)
∵ sin( A + B) sin(A – B)  = sin² A – sin² B
LHS = sin 10x . sin2x
= RHS.

Question 43.
Prove that cos² 2x-cos²6x =sin4x sin8x .
Answer :
LHS = (1 – sin² 2x) – (1 – sin² 6x)
= sin² 6x – sin² 2x = sin(6x + 2x) sin(6x – 2x)
= sin8x sin4x
= RHS

Question 44.
Prove that
\( \cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)=\sqrt{2} \cos x \)
Answer:
\(\begin{aligned} L H S &=\cos \frac{\pi}{4} \cos x-\sin \frac{\pi}{4} \sin x+\cos \frac{\pi}{4} \cos x+\sin \frac{\pi}{4} \sin x \\ &=2 \cos \frac{\pi}{4} \cos x=2\left(\frac{1}{\sqrt{2}}\right) \cos x=\sqrt{2} \cos x \\ &=R H S \end{aligned} \)

Question 45.
Prove that
tan3x tan2x tanx = tan3x-tan2x-tanx.
Answer :
Consider: tan 3x = tan(2x + x) tan 2x + tan x
\( \tan 3 x=\frac{\tan 2 x+\tan x}{1-\tan 2 x \tan x} \)
⇒ tan 3x(1 – tan 2x tan x) = tan 2x + tan x
⇒ tan3x-tan3x tan2x tanx = tan2x + tanx
∴ tan 3x – tan 2x – tan x = tan 3x tan 2x tan x

Question 46.
Prove that
cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1.
Answer :
Consider cot 3x = cot(2x + x)
\( =\frac{\cot 2 x \cot x-1}{\cot 2 x+\cot x} \)
⇒ cot 3x(cot 2x+cot x) – cot 2x cot x -1
⇒ 1 = cot 2x cot x – cot 3x cot 2x – cot 3x cot x

Question 47.
Prove that
\(\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}\)
Answer:

Question 48.
Prove that
\(\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x \)
Answer:

Question 49.
Prove that
\(\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x\)
Answer:

Question 50.
Prove that
\(\frac{\cos 7 x+\cos 5 x}{\sin 7 x-\sin 5 x}=\cot x\)
Answer:

Question 51.
Prove that
\(\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \left(\frac{x-y}{2}\right) \)
Answer:

Question 52.
Prove that
\(\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x \)
Answer:

Question 53.
Prove that
\(\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}=\cot 3 x \)
Answer:

Question 54.
Prove that
\(\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x\)
Answer:

Question 55.
Prove that
\(\frac{\sin 9 x+\sin 7 x+\sin 5 x+\sin 3 x}{\cos 9 x+\cos 7 x+\cos 5 x+\cos 3 x}=\tan 6 x\)
Answer:

Question 56.
Prove that
\((\cos x+\cos y)^{2}+(\sin x-\sin y)^{2}=4 \cos ^{2}\left(\frac{x+y}{2}\right)\)
Answer:

Question 57.
Prove that
\((\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}=4 \sin ^{2}\left(\frac{x-y}{2}\right)\)
Answer:

Question 58.
Prove that
sinx + sin3x + sin5x + sin7x
= 4 cos x cos 2x sin 4x
Answer :
LHS = (sin 7x + sin x) + (sin 5x + sin 3x)
= 2 sin 4x cos 3x + 2 sin 4x cos x
= 2 sin 4x(cos 3x + cos x)
= 2 sin 4x 2 cos 2x cos x
= 4cosx cos2x sin4x
= RHS.

Question 59.
Prove that
\(\sin 3 x+\sin 2 x-\sin x=4 \sin x \cos \frac{x}{2} \cos \frac{3 x}{2} \)
Answer:

Question 60.
Prove that
\(2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0\)
Answer:

Question 61.
Prove that
\(\tan 4 x=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1-6 \tan ^{2} x+\tan ^{4} x}\)
Answer:

Question 62.
Prove that cos4x = 1 – 8sin² x cos² x.
Answer :
LHS = cos2(2x)
= 1-2sin² 2x = 1-2(2sinx cosx)²
= 1-8sin²xcos²x
= RHS

Question 63.
Prove that
cos6x = 32cos⁶ x – 48cos⁴ x + 18cos² x -1.
Answer :
LHS = cos3(2x)
= 4cos³ 2x-3cos2x
= 4(2 cos² x -1)³ – 3(2 cos² x -1)
= 4[8 cos⁶ x -1 – 3(2cos² x)(2cos² x -1)] -6cos²x + 3
= 32cos⁶x-4-24cos²x(2cos²x-1) -6cos²x + 3
= 32 cos⁶ x – 48 cos⁴ x +18 cos² x -1
= RHS.

Question 64.
If \(\sin x=\frac{3}{5}, \cos y=-\frac{12}{13} \)
Answer:

 

Question 65.
Prove that
\(\cos 2 x \cos \frac{x}{2}-\cos 3 x \cos \frac{9 x}{2}=\sin 5 x \sin \frac{5 x}{2}\)
Answer:

Question 66.
Find the value of  \( \tan \frac{\pi}{8}\)
Answer:

Question 67.
Prove that
\( \cos ^{2} x+\cos ^{2}\left(x+\frac{\pi}{3}\right)+\cos ^{2}\left(x-\frac{\pi}{3}\right)=\frac{3}{2}\)
Answer:

Question 68.
\(\begin{aligned} &\text { If } \tan x=\frac{3}{4}, \pi<x<\frac{3 \pi}{2}, \text { find the value of }\\ &\sin \frac{x}{2}, \cos \frac{x}{2} \text { and } \tan \frac{x}{2} \end{aligned}\)
Answer:

Question 69.
Find \(\sin \frac{x}{2}, \cos \frac{x}{2} \text { and } \tan \frac{x}{2} \)
(i) \(\tan x=-\frac{4}{3}, \) x in quadrant II
(ii) \( \cos x=-\frac{1}{3}, \) x in quadrant III
(iii) \( \sin x=\frac{1}{4}, x \)in quadrant II
Answer:

Trigonometric Equations

Question 1.
Define trigonometric equations.
Answer :
Equations involving trigonometric functions of a variable are called trigonometric equations.

Question 2.
Define principal solutions of trigonometric equations.
Answer :
The solutions of a trigonometric equation for which 0 ≤ x ≤ 2π are called principal solutions.

Question 3.
Define general solution of trigonometric equation.
Answer :
The expression involving integer ‘n’ which gives all solutions of a trigonometric equation is called the general solution.

Question 4.
Find the principal solution of the following equations
(i) \(\sin x=\frac{\sqrt{3}}{2}\)
(ii) \(\tan x=-\frac{1}{\sqrt{3}}\)
(iii) \( \tan x=\sqrt{3}\)
(iv) sec x=2
(v) \( \cot x=\sqrt{3}\)
(vi) cosec x=-2
Answer:

Question 5.
Prove that, for any real numbers x and y, sin x = sin y implies x =nπ + (-1)ⁿ y , where n∈Z
Answer:

Question 6.
Prove that, for any real numbers x and y, cosx=cosy, implies x = 2nπ±y,where n∈Z
Answer:

Question 7.
Prove that, if x and y are not odd multiple of \(\frac{\pi}{2} \)
Answer:
Given: tan x=tan y

Question 8.
Find the solution of  \(\sin x=-\frac{\sqrt{3}}{2}\)
Answer:

Note:
\(\frac{4 \pi}{3}\) is one such value of x for which \(\sin x=-\frac{\sqrt{3}}{2} \)
One may take any other value of x for which \(\sin x=-\frac{\sqrt{3}}{2}\)
The solutions obtained will be the same although these may apparently look different.

Question 9.
Solve: \(\cos x=\frac{1}{2}\)
Answer:
\(\begin{array}{l}{\text { We have, } \cos x=\frac{1}{2}=\cos \frac{\pi}{3}} \\ {x=2 n \pi \pm \frac{\pi}{3}, n \in Z}\end{array}\)

Question 10.
Solve:
\(\tan 2 x=-\cot \left(x+\frac{\pi}{3}\right)\)
Answer:

Question 11.
Find the general solution for which cos4x = cos2x.
Answer:

Question 12.
Solve: sin 2x – sin4x + sin 6x = 0
Answer :

Question 13.
Solve: cos3x + cosjc-cos2jc = 0.
Answer :
Given: (cos 3x + cosx) – cos 2x = 0
⇒ 2cos2x cosac-cos 2x = 0
cos 2x(2cos x -1) = 0

Question 14.
Solve: sin2x+cosx = 0.
Answer :
Given sin 2x + cosx = 0
⇒ 2sinxcosx + cosx = 0
⇒ cosx(2sin x+1)=0

Question 15.
Solve: sinx + sin3x+sin5x = 0.
Answer :
Given: (sin5x + sinx) + sin3x = 0
⇒2sin3xcos2x + sin3x = 0
∴ sin3x(2cos2x + 1) = 0

Question 16.
Solve: 2cos²x + 3sinx = 0
Answer :
Given: 2(1 – sin² x) + 3sin x = 0
⇒ 2sin² x-3sinx-2 = 0
⇒ (2sin x+1)(sin x-2)=0

Question 17.
Solve: sec²2x = 1-tan2x.
Answer :

Sine and cosine formulae

Question 1.
State and prove sine formula.
Answer :
Statement: In any triangle, sides are proportional to the sines of the opposite angles. That is, in a triangle ABC
\(\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\)
Proof : Let ABC be either acute angled triangle or obtuse angled triangle.

The altitude h is drawn from the vertex B to meet the side AC in point D.
From the right angled triangle ABD, we have

Question 2.
State and prove cosine formula.
Answer :
Statement: Let A, B and C be angles of a triangle and a, b and c be lengths of sides opposite to angles A, B and C respectively, then
a² =b² + c²-2bc cos A
b² =c² +a²-2ca cos B
c² =a² +b² – 2ab cos C
Proof:
Let ABC be triangle as given below

Referring to second triangle ,we have
BC² = BD² + DC² = BD² + (AC – AD)²
= BD² + AD² + AC² – 2AC AD
= AB² + AC² -2AC AB cos A
= a² =b² +c² -2bc cos A
Similarly we can obtain
b² =c² + a² – 2ca cosB
and c² = a² +b² – 2ab cos C

Keen eye:
\(\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}\)
\(\cos B=\frac{c^{2}+a^{2}-b^{2}}{2 a c} \)
\(\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}\)

Question 3.
(Napier Analogies) In triangle ABC, .prove that
(i)\(\tan \frac{B-C}{2}=\frac{b-c}{b+c} \cot \frac{A}{2}\)
(ii) \(\tan \frac{B-C}{2}=\frac{b-c}{b+c} \cot \frac{A}{2}\)
(iii) \(\tan \frac{B-C}{2}=\frac{b-c}{b+c} \cot \frac{A}{2}\)
Answer:

Question 4.
In any triangle ABC , if a = 18, b = 24, c = 30, find
(i) cosA, cosB, cosC
(ii) sinA, sinB, sinC
Answer :
(i) We have

Question 5.
For any triangle ABC,Prove that
\(\frac{a+b}{c}=\frac{\cos \left(\frac{A-B}{2}\right)}{\sin \frac{C}{2}}\)
Answer:

Question 6.
For any triangle ABC, Prove that
\(\frac{a-b}{c}=\frac{\sin \left(\frac{A-B}{2}\right)}{\cos \frac{C}{2}}\)
Answer:

Question 7.
For any triangle ABC, Prove that
\(\sin \left(\frac{B-C}{2}\right)=\frac{b-c}{a} \cos \frac{A}{2}\)
Answer:

Question 8.
In any ΔABC, prove that a(bcosC-c cos B) = b²-c².
Answer :

Question 9.
In any ΔABC, prove that
\( a(\cos C-\cos B)=2(b-c) \cos ^{2} \frac{A}{2}\)
Answer:

Question 10.
In any ΔABC, prove that
\(\frac{\sin (B-C)}{\sin (B+C)}=\frac{b^{2}-c^{2}}{a^{2}} \)
Answer:

Question 11.
In any ΔABC, prove that
\((b+c) \cos \frac{B+C}{2}=a \cos \frac{B-C}{2}\)
Answer:

Question 12.
In any ΔABC, prove that
acosA+bcosB + ccosC = 2asin BsinC.
Answer :
LHS – k sin A cos A + k sin B cos B + k sin C cos C

Question 13.
In any ΔABC, prove that
\(\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{a^{2}+b^{2}+c^{2}}{2 a b c}\)
Answer:

Question 14.
In any ΔABC, prove that
\( \left(b^{2}-c^{2}\right) \cot A+\left(c^{2}-a^{2}\right) \cot B+\left(a^{2}-b^{2}\right) \cot C=0\)
Answer:

Question 15.
In any ΔABC, prove that
\( \frac{b^{2}-c^{2}}{a^{2}} \sin 2 A+\frac{c^{2}-a^{2}}{b^{2}} \sin 2 B+\frac{a^{2}-b^{2}}{c^{2}} \sin 2 C=0\)
Answer:

Question 16.
In any ΔABC, prove that
asin(B -C) + bsin(C – A) + csin(A-B) = 0
Answer :
LHS = k sin A sin(B -C) + ksinB sin(C – A) +k sin C sin( A – B)
= k[sin(B + C) sin(B – C)+sin(C + A) sin(C – A) +sin( A + B) sin( A – B)]
= K[sin² B – sin² C + sin² C -sin² A + sin² A – sin² B]
= k x 0
=0=RHS.

Question 17.
A tree stands vertically on a hill side which makes an angle of 15° with the horizontal. From a point on the ground 35 m down the hill from the base of the tree, the angle of elevation of the top of the tree is 60°. Find the height of the tree.
Answer :
In ΔOBT,
\( \frac{h}{\sin 45^{\circ}}=\frac{35}{\sin 30^{\circ}}\)

Question 18.
Two ships leave a port at the same time. One goes 24 km per hour in the direction N45°E and other travels 32 km per hour in the direction S75°E. Find the distance between the ships at the end of 3 hours.
Answer :
Let C be the port A and B are the positions of the two ships after 3 hours.

Question 19.
Two trees, A and B are on the same side of a river. From a point C in the river the distance of the trees A and B is 250 m and 300 m, respectively. If the angle C is 45°, find the distance of the trees.
Answer :
Given AC = 250 m
BC = 300 m

Question 20.
The angle of elevation of the top point P of the vertical tower PQ of height h from a point A is 45° and from a point B, the angle of elevation is 60°, where B is a point at a distance ‘d’ from the point A measured along the line AB which makes an angle 30° with AQ. Prove that \(d=h(\sqrt{3}-1) \)
Answer:

Question 21.
A lamp past is situated at the middle point M of the side AC of a triangular plot ABC with BC = 7 m, CA = 8 m and AB = 9 m. Lamp post subtends an angle 15° at the point B. Determine the height of the lamp post.
Answer:

You can Download Chapter 6 Analysis of Bivariate Correlation and Regression Questions and Answers, Notes, 1st PUC Statistics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Statistics Question Bank Chapter 6 Analysis of Bivariate Correlation and Regression

1st PUC Analysis of Bivariate Correlation and Regression Two Marks Questions and Answers

Question 1.
What is correlation?
Answer:
Correlation is said to exists between two variables if they vary directly or indirectly together.

Question 2.
Mention different types of correlation.
Answer:

1. Positive correlation
2. Negative correlation
3. zero correlation

Question 3.
Give examples for different types of correlation.
Answer:

• Demand and supply, Height and weight -positive correlation
• Price and demand, price and sales-negative correlation
• Height or Weight of a student and marks obtained in an exam-zero correlation

Question 4.
Define the co-efficient of correlation
Answer:
The degree or the extent of the correlation that exists between the variables is called coefficient of correlation, denoted by gamma (γ).

Question 5.
What is scatter diagram?
Answer:
When the values of the two variables are plotted on graph on an xy-plane, the spread or the scattered ness of the plotted points gives the types of correlation and the degree of the correlation that exists between the variables, the diagrams is called scattered diagrams.

Question 6.
Define positive correlation and negative correlations.
Answer:
The two variables are said to be positively correlated if they vary in the same direction, eg., demand and supply and, the two variables are said to be negatively correlated if they vary in the opposite direction, eg., price and sales.

Question 7.
What is causation?
Answer:
The cause and effect relationship between the two correlated variables is called causation.

Question 8.
What is spurious correlation?
Answer:
The absence of causation between two variables even though they vary together is called spurious correlation.

Question 9.
Write down any two properties of coefficient of correlation.
Answer:

• It is independent of units of measurement
• It is independent of origin and the scale.

Question 10.
What is the range of correlation co-efficient?
Answer:
γ = ±1 ie., it can take the least value -1 and the highest value +1

Question 11.
What is the value of co-efficient correlation for any two independent variables?
Answer:
γ = 0

Question 12.
Define Rank correlation.
Answer:
The ordinary co-efficient of correlation between the two ranking variables OR the product moment co-efficient correlation between the two ranking variables is called rank correlation, denoted by ‘p’/R

Question 13.
Write any difference between correlation and Rank correlation.
Answer:
Correlation is the measure of relationship between two quantitative characteristics called variables, such as ht. & wt, age & wt, of persons, ages of husbands and wives etc.,Where as the Rank correlation is the measure of relationship of two qualitative characteristics called attributes, such as the aesthetic senses as IQ, Beauty, Intelligence, Taste etc between the persons.

Question 14.
Define the term regression.
Answer:
Regress means go back to the origin, regression means estimation of unknown value of a given variable by knowing all values of the other variable when the variables are correlated

Question 15.
What are regression equations? And write down the two regression equations.
Answer:
The functionai relation developed between the two correlated variables are called regression equations.
The regression equation of x on y is: (X – X̄) = b_xy (Y – Ȳ) where bxy-the regression coefficient of x on y.
The regression equation of y on x is: (Y – Ȳ) = b_xy (X – X̄) where byx-the regression coefficient of y on x.

Question 16.
What are regression lines?
Answer:
The graph of regression equations are called regression lines.

17.
What are regressions co-efficient?
Sol.
b_xy and b_yx are called regression co-efficients.

18.
Write the properties of regression lines.
Sol.

• The two regression lines coincides at \((\bar{x}, \bar{y})\)
• The two regression lines intersect when there is a perfect correlation.
• The two regression lines are perpendicular to each other when there is no correlation.

19.
Write the properties of regressions co-efficients.
Sol.

• The regression co-efficients are independent of the origin but dependent on the scale.
• The geometric mean of the two regression co-efficients is numerically equal to the co-efficient of correlation, ie., \(r=\pm \sqrt{b_{x y} \times b_{y x}}\)
• Since the r cannot be greater than 1, so the product of the two regression co-efficients never exceeds 1.
• If the value of one of regression co-efficient is greater than 1 and the other should be less than 1.

1st PUC Analysis of Bivariate Correlation and Regression Problems

Question 1.
Draw a slatter diagram or the following data ant common on correlation.

Answer:

From the diagram, the scatterd points shows raising straight line. There exists a high degree positive correlation between price and supply.

Question 2.
Draw a scatter diagram for the following data and interpret it.

Answer:

From the scattered diagram, the spraded for the points show down ward diagram. Hence x and y are negatively correlated.

Question 3.
The following data gives the age of a motor cycle and the annual maintenance cost over a period. Find the product moment coefficient of correlation and interpret the same.

Answer:
Let x and y be the age and maintenance cost, the data is small in size, so use direct method.

∴ There exists a high degree positive correlation.

Question 4.
Calculate karl pearson’s coefficient of correlation between x and y.

Answer:
The data is large in size, so use step deviation or short cut method.

where A, B assumed means and C – common factors from x and y.

Here γ > 0.5. The exists high degree or close to perfect positive correlation.

Question 5.
Find K.P.C.C of the following data

Answer:
Let X and Y be ages of husbands and wifes. Here by observation method, Arithmetic means are Whole numbers/ integers. 50, it is convenient to use actual mean method

Question 6.
Calculate the coefficient fo correlation between the number of children and age of mothers from the following bivariate frequency table.

Answer:
Let x and y be the ages of mothers and number of children.
The data is discrete nature for y and with a common difference in x:

There exists a high degree negative correlation between ages of mothers and number of children.

Question 7.
Calculate the karl pearson’s coefficient of correlation between ages of students and marks obtained in a certain test and interpret.

Answer:
Let x and y be the marks and age

Here, there exists high degree negative correlation between Age of students and their marks.

Question 8.
Calculate K.P.C.C from the following date.

Answer:
Both variables are continuous so, use step deviation method.

Question 9.
If the covariance between x and y variables is 12.5 and variance of x and y are respectively 16.4 and 13.8. Find the coefficient of correlation between them.
Answer:

Question 10.
Given COV (x,.y) = -30, Var (x) = 225 and Var (y) = 9, find karl pearson’s coefficient of correlation.
Answer:

Question 11.
In a bivariate data, if standard deviations of x and y are 5 and 9 respectively. If the COV (x, y) = 8, find the coefficient of correlation.
Answer:

Question 12.
In a bivariate Σ(x – x̄)² = 35, Σ(x – x̄)² = 60, Σ(x – x̄)(y – ȳ) = 42. Find the coefficient of  correlation
Answer:

Question 13.
In a bivariate data Σxy = 500, Σx = 15, Σy = 200, Σx² = 98, Σy² = 23250 and n = 10. Find the coefficient of correlation.
Answer:

Question 14.
The coefficient of correlation between two variables x and y is 0.65, their covariance is 2.32. The variance of x is 18. Find standard deviation of y.
Answer:

Question 15.
If the coefficient of correlation between x and y is ‘o’ comment.
Answer:
If γ = 0. the two variables x and y are independent, i.e., Non correlated.

Question 16.
Define spurious correlation.
Answer:
Absence of causation between the variables is called sparious correlation.

Question 17.
Calculate spearman’s rank correlation for the following data regarding sports and studies of 8 students:

Answer:
Ranks are given. Let R₁ and R₂ be the ranks in sports and studies.

There exists low degree negative correlation between sports and studies.

Question 18.
Ten competitors in a pointing competition are ranked by two judges. Using rank correlation find out whether the two judges have common table in painting:

Answer:
Ranks are given. Let R₁ and R₂ be the ranks of judges I and II

the two Judges have high degree of similar taste in beauty of paintings.

Question 19.
The corresponding values of 2 series are given below.

Find the coefficient of rank correlation of the above series.
Answer:
Ranks are not given. We have to assign rank. Assign ranks R₁ and R₂ for the variables x and y in descending order as highest value as 1, 2, upto 7.

There exsits a high degree positive correlation

Question 20.
Calculate rank correlation from the following marks in Accountancy and statistics

Answer:
Ranks are not given. Assign ranks as R₁ and R₂ for the marks in Accountancy and statistics as 1^st who got highest marks and 2^nd to next less marks.

There exists low degree correlation between marks in Accountancy and Statistics.

Question 21.
Calculate rank correlation coefficient from the following data:

Answer:
Here the values of the variables are repeated,
So use, the formula:

m₁ = 3 (65 repeated 3 times)
Ranks are averaged as = \(\frac{6+7+8}{3}=7\)

m₂ = 2(13 repeated 2 times)
Ranks are averages as = \(\frac{4+5}{2}=4.5\)


There exists high degree positive correlation between x and y.

Question 22.
Calculate coefficient of correlation undere rank difference method for the following data

Answer:
Repeated values in the variables, so use correation factor formula.

m – available repeated number of times
Let R₁ and R₂ be the ranks of variables x and y.

There exsits a low degree positive correlation between the variables.

Question 23.
Mention the uses of Rank correlation.
Ans.
Spearman’s rank correlation useful

• To find the correlation between two qualitative characteristics, such as Honesty, Intelligence, Taste etc.
• This method is easy and simple as compared to karl pearsons method when the data has no repetitions.

Question 24.
When do you use Rank method of correlation?
Ans.
Rank method of correlation is used to find correlation between two qualitative characteristics.

Question 25.
The coefficiently rank correlation is 0.75 and the sum of the squares of the difference in rank is 30, then find the value of V.
Answer:
Given: p = 0.75, Σd² = 30
∴ \(r=1-\frac{6 \Sigma d^{2}}{n^{3}-n}\) \(0.76=1-\frac{6 \times 30}{n^{3}-n}\)
\(\frac{180}{n^{3}-n}=1-0.75=0.25\)
\(\frac { 180 }{ 0.25 }\) = n³ – n 720 = n³ – n
9 × 80 = n(n₂ – 1) 9(81 – 1) = n(n² – 1)
9(9² – 1) = n(n² – 1) ∴ n= 9

Question 26.
If n = 10 and Σd² = 200, find the coefficient of rank correlation.
Answer:

Question 27.
If Σd² = 182 and r = -0.1030 find ‘n’
Answer:
\(\mathrm{r}=1-\frac{6 \Sigma d^{2}}{n^{3}-n} \quad-0.1030=1-\frac{6 \times 182}{n^{3}-n}\)
\(\frac{1092}{n^{3}-n}=1+0.1030=1.103 ; \quad n^{3}-n=\frac{1092}{1.103}\)
n³ – n = 990 = 10 × 99; n(n² – 1) = 10(100 – 1)
n(n² – 1) = 10(10² – 1); ∴ n = 10

Regression 

Question 28.
From the following data:
(a) Calcualte the regression equation of x on y.
(b) Estimate the value of x when y = 40.

Answer:
Regression equation x on y is :
(x – x̄) = b_xy (y – ȳ)
Here, x and y variables have their Arithmetic means with whole number / integers. So, it is convenient to use actual mean method.

\(n=5 ; \bar{x}=\frac{140}{5}, \bar{y}=\frac{80}{5}=16\)
b_xy = \(\frac { 160 }{ 160 }\) = 1

(a) Regression equation of x on y is
x – 28 = 10-16)
x = y- 16 + 28 ∴ x = y + 12

(b) Estimation of x when y = 40
x = 40 + 12 = 52

Question 29.
From the following data of the age of Husband (x) and the age of the wife (x) obtain the regression equation of y on x and calculate the wife’s age (y). When husband’s age(x0’s 25 years.

Answer:
Regression equation y on x is (y – ȳ) = b_yx (x – x̄)
Here x and y variables Arithmetic means are not in whole numbers/ integers. So, it is better to use shortcut method.

Regression equation of y on x is:
y – 24.8 = 1.35 (x – 32)
y -24.8 = 1.35x – 1.35 × 32
y = 1.35x- 43.2 + 24.8
∴ y = 1.35x – 18.4
Estimation of (y) wife’s age when husbands age (x) is 25:
y = 1.35 (25) – 18.4 = 33.75 – 18.4
y = 15.35 years.

Question 30.
Construct two regression equations for the following data and estimate the value of x when y = 70 and y when x = 650.

Answer:
The two regression equations are:
Regression equation of x on y : (x – x̄) = b_xy (y – ȳ)
Regression equation of y on x : (y – ȳ) = b_xy (x – x̄)
Both variables having common differences, so use step deviation methods.

Here i_x common width in x = 100
i_y common width in v = 10 and n = 7;

Regression equation of x on y:
x – 400 = 6.05 (y-80) = 6.05y-605 × 80
x = 6.05y – 48.4+ 400

Regression equation of x on y : (x – x̄) = b_xy (y – ȳ)
x – 14 = 1(y – 8)
x = y – 8 + 14
∴ x = y + 6

and regression equation y on x is : (y – ȳ) = b_xy (x – x̄)
y – 8 = 0.8695 (x – 14)
y = 0.8695x – 12.173 + 8
∴ y = 0.8695x -4.173
Estimation of x when y = 20.
∴ From the Regression equation of x on y:
x = 20 + 6 = 26
We know that the coeffiicent of correlation:
\(\gamma=\pm \sqrt{b_{x y} \cdot b_{y x}}=\sqrt{1 \times 0.8695}=0.9325\)

Question 31.
Eight observations on each of two variables x and y are given below:

1. obtain two regression equation and
2. Find γ_xy
Answer:

∴ x = 6.05y – 84
pregression equation of y on x:
y – 80 = 0.1643(x – 400)
= 0.1643x – 0.1643 × 400
= 0.1643x – 65.72+80
∴ y = 0.1643x + 14.28
Equation y when x = 70
x = 605 (70) – 84 = 423.5 – 84
∴ x = 339.5
Equation y when x = 650
y = 0.1643(650) + 14.28 = 106.795 + 14.28
y = 121.075

Question 32.
Calculating the regression equation of x on y and y on x from the following data and estimate x when y = 20. Also determine the value of correlation coefficient.

Answer:
The values of x and y varibles are small in size so, use direct method for easy calculation.

Regression coefficient of x on y: (x – x̄) = b_xy (y – ȳ)
x – 2.5 =0.4762 (y – 4.5)
= 0.4762y – 0.4762x
x = 0.4762y – 2.1429 + 2.5
∴ x = 0.4762j + 0.3571

Regression equation of y on x = (y – ȳ) = b_xy (x – x̄)
y – 4.5 = 2(x – 2.5)
y = 2x – 2 × 2.5 + 4.5
y = 2x – 5 + 4.5
∴ = 2x – 0.5

We know the relation between coefficient of correlation and the regression coefficients:
\(\gamma=\pm \sqrt{b_{x y} \cdot b_{y x}}=\sqrt{0.4762 \times 2}\)
γ =0.9759

Question 33.
Calculate the two regression equations from the following bivariate table and determine y.

Answer:
For Bivariate, both x andy variables having common differences. So use step deviation method:

Regression equation of x on y : (x – x̄) = b_xy (y – ȳ)
x – 25 = 0.256 (y – 19.8)
= 0.256y – 0.256 × 19.8x
=0.256y – 5.0688 + 25
∴ x = 0.256y + 19.9312

Regression equation of y on x = (y – ȳ) = b_xy (x – x̄)
y – 19.8 = 0.667(x – 25)
∴ = 0.6667x – 0.667 × 25
y = 0.6667x – 16.675 + 19.8
∴ y = 0.667x + 3.125
and \(\gamma=\pm \sqrt{b_{x y} \cdot b_{y x}}=\sqrt{0.256 \times 0.667}\)
γ_xy = 0.4132

Question 34.
Obtain the regression equation of y on x and hence estimate y when x = 32
Answer:


Choose A and B the assumed means as : A = 25, and B = 27.5 from the respective mid points of the variables x and y width x: ix = 10 and width y : iy = 5


Regression equation of y on x
4 – 34.5 = 0.723 (x – 38.667)
= 0.723 x – 27.96 + 34.5
y = 0.723 x – 27.956+ 34.5
y = 0723x +6.544
Estimation of y when x = 32
y = 0.723 (32)+ 6.544
= 23.136 + 6.544 = 29.68

Question 35.
Find the most probable value of y if x is 70 and the most probable value of x when y 1890 give that

	X series	Y series
Mean	18	100
Standard deviation	14	20

co-efficient of correlation between x and y is 0.8
Answer:
Given:
x̄ = 18 ȳ = 100
σ_x = 14, σ_y = 20 and γ _xy = 0.8
To find y when x = 70 we need regression equation of y on x:

= 1.1429(52)
y = 59.43 + 100
∴ y= 159.43
To find x when y = 90, we need regression equation of x on y.

Question 36.
The following data relates to marks obtained by 250 students in accountancy and statistics is 0.8. (i) Esitimate the marks obtained by a student in statistics in II PUC. Examination

Subject	Arithmetic mean	Variance
Accountancy (x)	48	16
Statistics (y)	55	25

Co-efficient of correlation between mavasin accountancy and statistics is 0.8
1. Estimate the marks obtained by a student instatistics who secured 50 marks in accountancy and
2. Estimate the marks obtained in accountancy when marks in statistics is 65.
Answer:
Given:
\(\bar{x}=48, \bar{y}=55\) σ_x² = 16, ∴ σ_x = 4
σ²₄ = 25 ∴ σ_y = 5 and γ = 0.8

1. Estimation y when x = 50 regression equation of yon x : (y – ȳ) = b_xy (x – x̄)

y – 55 = 0.8 \(\frac { 5 }{ 4 }\) (50-48)
y = 1 (2) + 55
∴ y = 2 + 55 = 57

2. Estimation g x when y = 65, use regression equation g x on y : (x – x̄) = b_xy (y – ȳ)

x – 48 = 0.8 ×\(\frac { 4 }{ 5 }\)(65-55)
∴ x = 64+48 = 54.4

Question 37.
Find two regression equations when x̄ = 68.2, ȳ = 99, \(\frac{\sigma_{y}}{\sigma_{x}}\) = 0.44 and g = 0.76. Also
estimate (a) the most probable value g x when y is 12 (b) the most probable value of 4 when value g x is 16.
Answer:
Given : \(\frac{\sigma_{y}}{\sigma_{x}}\) = 0.44, Also \(\frac{\sigma_{y}}{\sigma_{x}}\) = \(\frac { 1 }{ 0.44 }\) = 2.272
The regression equation of x on y is

x – 682 = 0.76 × 2.272 (y – 9.9) = 1.7267 y – 1.7267 × 9.9
x= 1.7267 y – 17.0943 + 68.2
∴ x = 17267y + 51.1057

(a) Estimation g x when y = 12
x= 1.7267 (12) + 51.1057 x = 71.8261
The regression equation of yon x is :

y – 9.9 = 0.76 × 0.44 (x – 68.2) = 0.3344 x – 0.3344 × 68.2 = 0.3344 x – 22.806 + 9.9
y = 0.3344x – 12.906

(b) Estimation of y when x = 16
y = 0.3344 (16) – 2.906 .
y = 5.3504 – 12.906 = – 7.5556

Question 38.
If the regression lines x and y are given by 2x – 3y = 0 and 47 – 5x – 8 = 0. Find the means of JC and y, also the coefficient of correlation.
Answer:
2x – 3y = 0 and 4y – 5x – 8 = 0
Solving both equations we get x and y. And so x̄,ȳ

Put y = – 2.286 in 2x – 3y = 0
2x – 3(-2.286) = 0
2x = 6.8568
∴ x = 3.4284
x̄ =3.4284, ȳ = -2.286
Consider, 2x – 3y = 0. be the regression of y on x :
3y = 2x + 0
y = \(\frac { 2 }{ 3 }\)x + 0
∴ b_yx = \(\frac { 2 }{ 3 }\)
And 4y – 5x = 8 be the regression equation g x on J18
5x = 44 – 8
x = \(\frac { 4 }{ 5 }\)y – \(\frac { 8 }{ 5 }\)x
∴ b_xy = \(\frac { 4 }{ 5 }\)
The coefficient of correlation is : \(\gamma=\pm \sqrt{\mathrm{b}_{\mathrm{xy}} \cdot \mathrm{b}_{\mathrm{yx}}}\)
\(\gamma=\sqrt{\frac{4}{5} \times \frac{2}{3}}=0.7303\)

Question 39.
Calculate the two regression coefficients when γ = 0.9, σ_x = 10, and σ_y = 1.5
Answer:
Regression coefficient g x ony : \(\mathrm{b}_{\mathrm{xy}}=\frac{\gamma \sigma_{\mathrm{x}}}{\sigma_{\mathrm{y}}}\)
\(b_{x y}=0.9 \times \frac{10}{1.5}=6\)

Regression coefficient of yon x is :\(\mathrm{b}_{\mathrm{yx}}=\frac{\gamma \sigma_{\mathrm{y}}}{\sigma_{\mathrm{x}}}\)
\(b_{y x}=0.9 \times \frac{1.5}{10}=0.135\)

Question 40.
If b_xy = 0.8 and b_yx = 0.6. Find γ_xy
Answer:
\(\gamma_{x y}=\pm \sqrt{b_{x y} \cdot b_{y x}}=\sqrt{0.8 \times 0.6}=0.6928\)

Question 41.
Find y when the two regression coefficients are 0.6 and – 1.4
Answer:
\(\gamma=\pm \sqrt{\mathrm{b}_{\mathrm{xy}} \cdot \mathrm{b}_{\mathrm{yx}}}=\pm \sqrt{-0.6 \times-1.4}\)
∴ γ = -0.9165 (Based on properties of regression coefficients and correlation coefficient all are negative or positive)

Question 42.
A student calculated the regression co-efficients g xony as – 1.25 abd regression co¬efficient g yon x is -2.40 commenton his calculation.
Answer:
Given b_xy =-1.25 and b_yx =-2.4
It is the property of the two regression coefficient is that, the geometic mean of two regression coefficients is numbericaliy equi to the coefficient of correlation. And coeficient of correlation
taks a value between + 1 and – 1. i.e., \(\gamma=\pm \sqrt{\mathrm{b}_{\mathrm{xy}} \times \mathrm{b}_{\mathrm{yx}}}\)
Here both regression coefficients are more than 1 and so their product is also greater than 1. The students calculation is wrong.

Question 43.
If one of the regression coefficient is 1.5 and y = 0.55, find the value of other regression coefficient.
Answer:
we know that \(\gamma=\pm \sqrt{\mathrm{b}_{\mathrm{xy}} \cdot \mathrm{b}_{\mathrm{yx}}}\)
\(\begin{array}{l}{0.55=\sqrt{\mathrm{b}_{\mathrm{xy}} \times 1.5}} \\ {(0.55)^{2}=\mathrm{b}_{\mathrm{xy}} \times 1.5} \\ {\frac{0.3025}{1.5}=\mathrm{b}_{\mathrm{xy}}}\end{array}\)
∴ b_xy =0.2017

You can Download Chapter 7 Associates of Attributes Questions and Answers, Notes, 1st PUC Statistics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Statistics Question Bank Chapter 7 Associates of Attributes

1st PUC Associates of Attributes Question and Answers

Question 1.
What is meant by Association of Attribute? How does it differ from correlation?
Answer:
Association of attributes measures the degree of relationship between the two attributes; such as sex and literacy, literacy and employment, intelligence and employment, Beauty and Intelligence etc.

Whereas Correlation coefficient measures the degree of relationship between the variables; such as height and weight of persons, ages of husbands and wives, demand and supply of items etc.

Question 2.
How do you interpret regarding the Yule’s coefficient of association (Q)?
Answer:

• The value of this coefficient lies between ±1 (i.e., – i < Q < l).
• If Q = +1 there is perfect positive association between the attributes.
• If Q = -l there is perfect negative association (perfect dissociation) between the attributes and,
• If Q = zero the two attributes are independent.

Question 3.
What is the difference between coefficient of correlation and association of attributes?
Answer:
As we know that correlation coefficient measures the degree of relationship between the ‘ variables; such as height and weight of persons, ages of husbands and wives, demand and supply of items etc. Whereas the method of association of attributes measures the degree of relationship between the attributes; such as sex and literacy, literacy and employment, intelligence and employment etc.

Question 4.
Write the formula of Yule’s coefficient of Association.
Answer:

Question 5.
What are frequencies of first order and second orders? And mention them with their notations.
Answer:
A class having one attribute is known as the class of the first order, class of the combination two attributes as class of the second order. The total number of observations is denoted by N.
Here, N is frequency of zero order, because it has no attribute to indicate.
(A), (B), (α), (β) are called frequencies of the first order, and
(AB), (αB), (Aβ), (αβ) are called frequencies of the second order.

Question 6.
Calculate Yule’s coefficient of Association between marriage and failure of students from the following data pertaining to 525 students.

	Passed	Failed
Married	90	65
Unmarried	260	110

Answer:
Let A and B denotes Passed and Marriage, a and a denotes Failed and Unmarried. The 2 × 2 contingency table can be prepared as:

Note : In the above table figure in bold letter are adjusted figures.
Yule’s Coefficient of Association:

There exists a low degree of negative association between pass (A) and (B) marriage.

Question 7.
Eighty eight residents of a city were interviewed during sample survey are classified below, according to their smoking and tea drinking habits. Calculate Yule’s coefficient of Association and comment on its value.

	Smokers	Non -smokers
Tea drinkers	35	33
Non tea drinkers	8	12

Answer:
Let A and B denotes smokers and Tea drinkers, a and a denotes Failed and Unmarried. The 2 × 2 contingency table can be prepared as:

Note: In the above table figure in bold letter are adjusted figures.
Yule’s Coefficient of Association:

There exists a low degree of positive association between smoking (A) and tea drinking (B)

Question 8.
Compute Yule’s coefficient of Association from the following data.
(AB)=150, N=1000, (A)=200, (B)= 300.
Answer:
By putting the known values in the nine square tables find the unknown values to determine the Yule’s Coefficient of Association.

Note : In the above table figure in bold letter are adjusted figures.
Yule’s Coefficient of Association:


There is a high degree +ve association between the attributes.

Question 9.
Compute Yule’s co-efficient of Association from the following data.
N = 250, (Aβ) = 70, (A) = 100, (B) = 50. Ans: 0.4717
Answer:
By putting the known values in the nine square tables find the unknown values to determine the Yule’s Coefficient of Association.

Note : In the above table figure in bold letter are adjusted figures.
Yule’s Coefficient of Association:

There is a low degree +ve association between the attributes

Question 10.
Given N=500, (αβ)=280, (A)=160 and (B)=200. Calculate Yule’s coefficient of Association.
Answer:
By putting the known values in the nine square tables find the unknown values to determine the Yule’s Coefficient of Association.

Note: In the above table figure in bold letter are adjusted figures.
Yule’s Coefficient of Association:

There is a high degree +ve association between the attributes.

Question 11.
Given N=250, (AB)=40, (α)=210 and (β)=90
Calculate Yule’s coefficient of Association.  Ans: 1
Answer:
By putting the known values in the nine square tables find the unknown values to determine the Yule’s Coefficient of Association.

Note : In the above table figure in bold letter are adjusted figures.
Yule’s Coefficient of Association:

Students can Download Maths Chapter 1 Sets Questions and Answers, Notes Pdf, 1st PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Maths Question Bank Chapter 1 Sets

Question 1.
Define a set.
Answer :
A collection of well-defined objects is called a set. The objects in a set are called its members or elements. We denote sets by capital letters A, B, C, X. Y, Z, etc. If ‘a’ is an element of a set A, we write, a ∈ A, which means that a belongs to A or that a is an element of If ‘a ’ does not belong to A, we write, a ∉ A

Question 2.
How to describe a set?
Answer :
There are two methods of describing a set.
1. Roster or tabular form: In the roster form, we list all the members of the set within brackets { } and separate them by commas.

2. Set-Builder form: In the set-builder form, we list the property or properties satisfied by all the elements of the set.

• N: The set of all-natural numbers.
• Z: The set of all integers.
• Q: The set of all rational numbers.
• M: The set of real numbers.
• Z⁺: The set of positive integers.
• Q⁺: Set of positive rational numbers.
• R⁺: The set of positive real numbers.

Question 3.
Which of the following are sets? Justify your answer.
(i) The collection of all the months of a year beginning with the letter.
(ii) The collection of ten most talented writers of India.
(iii) A team of eleven best-cricket batsmen of the world.
(iv) The collection of all boys in your class.
(v) The collection of all natural numbers less than 100.
(vi) A collection of novels written by the writer Munshi Prem Chand.
(vii) The collection of all even integers.
(viii) The collection of questions in this chapter.
(ix) A collection of most dangerous animals of the world.
Answer :
(i) The given collection is {January, June, July}, which consists of definite objects. So, it is a set.
(ii) The term most talented is vague. So, this collection is not a set.
(iii) The term best is vague. So, this collection is not a set.
(iv) This collection is well defined and hence is a set.
(v) This collection is well defined and hence it is a set.
(vi) The collection of novels written by Munshi Prem Chand is well defined, so it is a set.
(vii) {0,±2,±4,±6,—} is the collection of well-defined objects, so it is a set.
(viii) The collection of questions in this chapter is well defined, so it is a set.
(ix) The term most dangerous is vague. So, this collection is not a set.

Question 4.
Let A = {1, 2, 3,4,5, 6}. Insert the appropriate symbol 6 or g in the blank spaces:
(i) 5 ……………. A
(ii) 8…………….. A
(iii) O……………. A
(iv) A ……………. A
(v) 2……………. A
(vi) 10 ……………. A
Answer :
(i) 5 ∈ A
(ii) 8 ∉ A
(iii) 0 ∉ A
(iv) 4 ∈ A
(v) 2 ∈ A
(vi) 10 ∉A

Question 5.
Write the following sets in roster form:
(i) A = {x : x is an integer and – 3 < x < 7}
(ii) B = {x : x is a natural number less than 6}
(iii) C = {x : x is a two-digit natural number such that the sum of its digits is 8}
(iv) D =  {x : x is a prime number which is divisor of 60}
(v) E = The set of all letters in the word TRIGONOMETRY.
(vi) F = The set of all letters in the word BETTER.
(vii) G = The solution set of the equation x² + x-2 = 0.
(viii) H = {x : x is a positive integer and x² < 40}
Answer :
(i) A = {-2,-1,0,1,2,3,4,5,6}
(ii) B = {1,2,3,4,5}
(iii) C = {17,26,35,44,53,62,71,80}
(iv) D = {2,3,5}
(v) E = {T,R,I,G,0,N,M,E,Y}
(vi) F = {B,E,T,R}
Given equation is x² + x – 2 = 0
⇒ (x – 1)(x + 2) = 0
i.e., = 1,-2
∴ G = { 1,-2}
(viii) H ={1,2,3,4,5,6}

Question 6.
Write the following sets in the set-builder form
(i) {3,6,9,12}
(ii) {2,4,8,16,32}
(iii) {5,25,125,625}
(iv) {2,4,6,… }
(v) {1,4,9,…, 100}
(vi) {1,4,9,16,…}
(vii) \(\frac { 1 }{ 2 } ,\frac { 2 }{ 3 } ,\frac { 3 }{ 4 } ,\frac { 4 }{ 5 } ,\frac { 5 }{ 6 } \frac { 6 }{ 7 } \)
Answer :
The given sets in set – builder form are
(i) {x : x = 3n, n∈N,n<5}
(ii) {x : x = 2ⁿ,n∈N,n<6}
(iii) {x : x = 5ⁿ,n∈ N, n<5}
(iv) {x : x is an even natural number}
(v) {x : x = n²,n∈ N,n≤10}
(vi) {x : x = n²,n∈ N}
(vii)
\(\left\{x: x=\frac{n}{n+1}, n \in N, n<7\right\}\)

Question 7.
List all the elements of the following sets:
(i) A = {x : x is an odd natural number}
(ii) B = { x : x is an integer, \(-\frac { 1 }{ 2 } <x<\frac { 9 }{ 2 } \)
(iii) C = {x : x is an integer, x² ≤ 4}
(iv) D = {x : x is a letter in the word “LOYAL”}
(v) E = { x : x is a month of a year not having 31 days}
(vi) F =  {x : x is a consonant in the English alphabet which precedes k}
Answer :
(i) A = {1,3,5,.. }
(ii) B = {0,1,2,3,4}
(iii) C = {-2,-1,0,1,2}
(iv) D = {L,0,Y,A}
(v) E = {February, April, June, September,November}
(vi) F = {b,c,d,f,g,h,j}

Question 8.
Match each of the set on the left in the roster form with the same set on the right described in set-builder form:

(i)	(1,2,3,6}	(a)	(x : x is a prime number and ‘ a divisor of 6}
(ii)	{2,3}	(b)	{x : x is an odd natural number less than 10}
(iii)	{M, A, T, H, E, I,C,S}	(c)	{x : x is a natural number and divisor of 6}
(iv)	11,3,5,7,9}	(d)	{x : x is a letter of the  word MATHEMATICS}

Answer:
(i) →(c), (ii) → (a), (iii) → (d), (iv) → (b).

Question 9.
Match each of the set of on the left described in the roster form with the same set on the right described in the set – builder form.

Answer :
(i) → (d), (ii) → (c), (iii) → (a), (iv) → (b).

Question 10.
Define an empty set.
Answer :
A set which does not contain any element is called the empty set or the null set or the void set and it is denoted by { } or φ.

Question 11.
Define a finite set.
Answer :
A set which is empty or consists of a definite number of elements is called finite set.

Question 12.
Define an infinite set.
Answer :
A set having infinite elements is called an infinite set.

Question 13.
Which of the following are examples of the null set.
(i) Set of odd natural numbers divisible by 2.
(ii) Set of even prime numbers.
(iii) {x : x is a natural numbers, x < 5 and x >7}
(iv) { y: y is a point common to any two parallel lines}.
Answer :
(i) There is no odd natural number divisible by 2. Therefore, given set is an empty set / nullset.
(ii) 2 is the even prime number.
∴ Given set = {2} ≠φ
(iii) There is no natural number x, which is simultaneously less than 5 and greater than 7. So, given set is null set.
(iv) We know that two parallel lines have no common point and hence the given set is null set.

Question 14.
Which of the following sets are finite or infinite?
(i) The set of months of a year,
(ii) {1, 2, 3, ….}
(iii) {1, 2, 3, …….99, 100}
(iv) The set of positive integers greater than 100
(v) The set of prime numbers less than 99.
Answer :
(i) Finite set
(ii) Infinite set
(iii) Finite set
(iv) Infinite set
(v) Finite set.

Question 15.
State which of the following sets are finite or infinite.
(i) {x : x ∈ N and (x – 1)(x – 2) = 0}
(ii) {x : x ∈ N and x² = 4}
(iii) {x : x ∈ N and 2x-1 = 0}
(iv) {x : x ∈ N and x is prime}
(v) {x : x ∈ N and x is odd}
(vi) The set of lines which are parallel to the x-axis.
(vii) The set of letters in the English alphabet.
(viii) The set of numbers which are multiple of 5.
(ix) The set of animals on the earth
(x) The set of circles passing through the origin (0,0).
Answer :
(i) Given set = {1,2}. Hence, it is finite.
(ii) Given set = {2}. Hence, it is finite.
(iii) Given set = φ. Hence, it is finite.
(iv) We know that there are infinitely many primes. Hence, given set is infinite.
(v) There are infinitely many odd numbers. Hence, given set is infinite set.
(vi) There are infinitely many parallel lines to the X-axis. Hence, given set is infinite set.
(vii) Finite sets
(viii) Infinite set
(ix) Finite set
(x) Infinite set.

Question 16.
Define equal sets.
Answer :
Two sets A and B are said to be equal if they have exactly the same elements and we write A = B.

Question 17.
In the following, state whether A = B or not.
(i) A = {a,b,c,d}, B={d,c,b,a}.
(ii) A = {4,8,12,16}, B = {8,4,16,18}.
(iii) A = {2,4,6,8,10}, B ={x : x is positive even integer and x ≤ 10}.
(iv) A = {x : x is a multiple of 10}, B={10,15,20,25,30,-}.
(v) A = {x: x is a prime numbers ≤ 6},
B = {x: x is a prime factors of 30}.
Answer :
(i) A = B because A and B have same elements though in different order which is immaterial.
(ii) A≠B ∵ 12 ∈ A but 12 ∉ B
(iii) A = B ∵ A and B have same elements
(iv) A≠B ∵ 15 ∈ B
(v) A = {2,3,5}, B = {2,3,5} ∴ A = B.

Question 18.
Are the following pairs of sets equal? Give reasons.
(i) A = {2,3}, B = {x : x is solution of x²+5x + 6 = 0}.
(ii) A = {x : x is a letter in the word FOLLOW}.
B={y: y is a letter in the word WOLF}.
Answer :
(i) x² + 5x + 6 = 0
⇒ (x + 3)(x + 2) = 0
∴ x = -3,-2
∴B = {-3,-2}- But A = {2,3}
∴ A≠B(ii) Given sets are A = {F, O, L, W}
B = {W,0,L,F}
∴ A = B
∵ A and B have same elements.

Question 19.
From the sets given below, select equal sets.
A = {2,4,8,12}
B = {1,2,3,4}
C ={4,8,12,14}
D ={3,1,4,2}
E = {-1,1}
F={0,a}
G = {1,-1}
H ={0,1}.
Answer :
Here, B = D and E = G.

Question 20.
Define subset of a set.
Answer :
A set A is said to be subset of set B if every element of A is also an element of B, and we write, A ⊆ B

• If A ⊆ B then B is called a superset of A, and we write, B ⊇ A
• If A ⊆ B and A ≠ B then A is called a proper subset of B.
• If there exists even a single element in A which is not in B, then A is not a subset of B and we write, A⊄B
• Every set A is a subset of itself
• Empty set φ is a subset of every set.

Question 21.
Consider the sets φ, A={1,3},B{1,5,3},C ={1,3,5,7,9}. Inset the symbol ⊂ or ⊄ between each of the following pair of sets:
(i) φ……… B
(ii) A … B
(iii) A… C
(iv) B… C.
Answer:
(i) φ ⊂ B, as empty set is a subset of every set
(ii) A ⊄ B, ∵ 3∈  A ⇒ but 3∉ B
(iii) A⊂C, as 1,3∈ A ⇒1,3 ∈ C
(iv) B⊂C, as 1,5,9∈ A ⇒ 1,5,9 ∈ C.

Question 22.
Make correct statements by filling in the symbols ⊂ or ⊄ in the blank spaces:
(i) {2,3,4} … {1,2,3,4,5}.
(ii) {a, b, c} … {b,c,d}.
(iii) {x : x is a circle in the plane } … {x : x is a circle in the same plane with radius 1 unit}.
(iv) {x : x is a student of I PUC of your college} … {x : x is student of your college}.
(v) {x : x is a triangle in a plane } … {x: x is a rectangle in the plane}.
(vi) {x : x is an equilateral triangle in a plane} … { x : x is a triangle in the same plane}.
(vii) {x : x is an even natural number} … {x : x is an integer}.
Answer :
(i) {2,3,4} ⊂ {1,2,3,4,5} ∵ 2,3,4 ∈{1,2,3,4,5}

(ii) {a,b,c}⊄ {b,c,d} ∵ a ⊄ {b,c,d}

(iii) {x: x is a circle in the plane} ⊄ {x: x is a circle in the same plane with radius 1 unit}. Because there exists a circle having a radius other than 1 unit.

(iv) {x: x is a student of 1 PUC of your college} ⊄ {x: x is a student of your college}
∵ Every student of I PUC is a student of the college.

(v) {x: x is a triangle in a plane} ⊄ {x: x is a rectangle in the plane.
∵ Triangle is not a rectangle.

(vi) {x: x is an equilateral triangle in a plane} ⊂ {x: x is a triangle in the same plane}.
∵ Every equilateral triangle in a plane is a triangle in the same plane.

(vii) {x : x is an even natural number} ⊂ {x : x is an integer}
∵ Every even natural number is an integer.

Question 23.
Examine whether the following statements are true or false:
(i) {a, b} ⊄ {b,c,a}
(ii) {a,e} ⊂ {x:x is a vowel in the English alphabet}
(iii) {1,2,3} ⊂ {1,3,5}
(iv) {a} ⊂ {a,b,c}
(v) {a}∈{a,b,c}
(vi) {x: x is an even natural number less than 6} ⊂ {x : x is a natural number which divides 36}.
Answer:
(i) False ∵ a,b∈ {b,c,a}
(ii) True ∵ a,e∈ {a,e,i,o,u}
(iii) False ∵ 2∉{1,3,5}
(iv) True ∵ a∈ {a,b,c}
(v) False ∵ {a} is a set but not in [a, b,c}
(vi) True ∵ {2,4} ⊂ {1,2,3,4,6,9,12,18,36}

Question 24.
Let A = {1, 2, {3,4}, 5}. Which of the following statements are incorrect and why?
(i) {3,4} ⊂ A
(ii) {3,4} ∈ A
(iii) {{3,4}} ⊂ A
(iv) 1 ∈ A
(v) 1⊂ A
(vi) {1,2,5} ⊂ A
(vii) {1,2,5} ∈ A
(viii) {1,2,3} ⊂ A
(ix) φ ∈ A
(xi) {φ} ⊂ A
(x) φ ⊂A
Answer :
(i) Incorrect, ∵ 3∈ {3,4} but 3∉ A.
(ii) Correct, ∵ {3,4} is a member of A.
(iii) Correct, ∵ {3,4}∈ A.
(iv) Correct, ∵  1 is a member of A.
(v) Incorrect, ∵ 1 is  A but not 1⊂ A.
(vi) Correct, ∵ 1, 2, 5, ∈ A.
(vii) Incorrect, ∵ {1,2,5}∉A.
(viii) Incorrect, ∵ 3∈ {1,2,3} but 3∉A
(ix) Incorrect, ∵ φ ∉ A
(x) Correct, ∵ φ is a subset of every set.
(xi) Incorrect, ∵ φ ∈  {φ} and φ ∈ A.

Question 25.
Define a power set of set.
Answer :
The collection of all subsets of a set A is called the power set of A and is denoted by
P (A). If n(A) = m, then n[P(A)] = 2^m.Question 26.
Define the universal set.
Answer :
If there are some sets under consideration, then there happen to be a set which is a super set of each one of the given sets. Such a set is known as the universal set for those sets. The universal set is denoted by U.Question 27.
Write down all the subsets of the following sets:
(i) {a}
(ii) {a, b}
(iii) {1,2,3}
(iv) (-1,0,1}
Answer :
(i) Subsets of {a} are {a}, φ
(ii) Subsets of {a, b} are {a, b},φ,{a},{b}
(iii) Subsets of {1, 2, 3} are {1, 2, 3}, are {1}, {2}, {3}, {1,2}, {2, 3}, {3,1}
(iv) Subsets of {-1, 0, 1} are {-1, 0, 1}, φ, {-1}, {0}, {1}, {-1,0}, {0,1}, {-1,1}

Question 28.
How many elements has P(A), if A =φ?
Answer :
P(A) has only one element, namely φ.
∴ PW = {φ}

Question 29.
What universal set (s) would you propose for each of the following:
(i) The set of right triangles.
(ii) The set of isosceles triangles.
Answer :
(i) Given set is the set of right triangles, so our proposed universal set is the set of all triangles.
(ii) Given set is the set of isosceles triangles, so our proposed universal set is the set of all triangles.

Question 30.
Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of the following may be considered as universal set (s) for the three sets A, B, and C.
(i) {0, 1, 2, 3, 4, 5, 6}
(ii) ϕ {0,1, 2,3,4,5, 6, 7, 8, 9,10}
(iii) {0,1,2,3,4,5,6 7, 8, 9,10}
(iv) {1, 2,3, 4, 5,6,7,8}
Answer :
Here universal set

Question 31.
Write the following as intervals:
(i) {x: x ∈ R,-4< x ≤6},
(ii) {x:x ∈ R,-12<x <-10}
(iii) {x:x∈ R,0≤x<7}
(iv) {x:x∈ R,3≤x≤4}.
Answer :
(i) (-4,6]
(ii) (-12,-10)
(iii) (0,7)
(iv) [3,4]

Question 32.
Write the following intervals in set – builder form:                                 ,
(i) (-3,0)
(ii) [6,12]
(iii) (6,12]
(iv) [-23,5).
Answer :
(i) (-3,0)={x: x ∈ R, -3 < x < 0}
(ii) [6,12]={x:x ∈ R,6<x≤12]
(iii) (6,12]={x:x ∈ R,6<x≤12}
(iv) [-23,5)=[x: x ∈ R,-23≤x< 5}.

Question 33.
Decide, among the following sets, which sets are subsets of one and another:
A = {x: x ∈ R and satisfy x² – 8x +12 = 0},
B ={2,4,6}
C={2,4,6,8,-}
D = {6}.
Answer :
x²-8x + 12 = 0
⇒(x-6)(x-2) = 0
⇒ x = 2,6
∴ A = {2,6}
∴ A ⊂ B and A ⊂ C; B ⊂ C, D ⊂ B and D ⊂ C.

Question 34.
In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.
(i) If x ∈ A and A∈B, then x∈B
(ii) If A ⊂ B and BC, then A ∈ C
(iii) If A⊂ B and B⊂ C, then A ⊂ C
(iv) If A ⊄ B and B ⊄C,then A ⊄C
(v) If x ∈A and A⊄B, then x ∈ B
(vi) If A ⊂ B and x∈B, then x ∉A
Answer :
(i) False: For example, let A = {1, 2} and
B = (2, 3, {1, 2}}. As 1∈A and A∈ B but 1∉ B

(ii) False: For example, let A = {1,2},B = {1,2,3} and C = {3,4,{1,2,3}}
Now clearly, A ⊂ B and B⊂C but A∉C,

(iii) True: Given A ⊂ B and
Now we have to prove A ⊂ C
∴ x∈C ∵ B∈C.
∀ X∈A ⇒X∈C∴ A∈C.(iv) False: For example, let
A=(1,2), B={2,3),C=(1,2,4)
Clearly A⊄B and B⊄C but A⊂C(v) False: For example, let
A=(l,2), B={2,3,4,5}
Now 1 ∈A and A ∉ B but 1 ∉ B(vi) True: Suppose x∈A then X∈ B A⊂B.

Question 35.
If A⊂B, then show that C-B⊂C-A
Answer :
Let x∈C-B ⇒x∈C but x ∉ B
⇒x∈C but x∉A ∵ A⊂B
∴ x∈C- A
Thus, C-B⊂C-A

Question 36.
Assume that P(A) P(B). Show that A – B
Answer :
Let x∈A∴(X)∈P(A)
={x}∈P(B) ∵ P(A)=P(B)
∴ x∈ B.
∴ A⊂B
Let y∈ B ∴(y)∈ P(B)
(y)∈P(A) ∵ P(A)=P(B)
y∈A
B⊂A
From(1) and (2),we get A=B

Question 37.
Define union of sets.
Answer :
The union of two sets A and B denoted by A∪B, is the set of all those elements which arc either in A or in B or in both A and B.
Thus, AB=(x: x∈A or x∈B)
∴ x∈A∪B x∈A or x∈B
x∉A∪B=x∉A and z∉B

• A∪B=B∩A (Commutative Law)
• (A∪B)∪C=A∪(B∪C) (Associative law)
• A∪φ=A
• A∪A=A
• U∪A=U

Question 38.
Define intersection of sets.
Answer :
The intersection of sets A and B is the set of all elements which are common to both A and B and is denoted by A∩B. Thus, A∩B = {x;x∈ A and x∈ B}


Question 39.
Define disjoint sets.
Answer :
If A and B are two sets such that A∩B =φ, then A and B are called disjoint sets.

Question 40.
Define difference of sets.
Answer :
The difference of the sets A and B i.e., A – B. is the set of all the elements of A but not in B.

Question 41.
Let A = {2,4, 6, 8} and B = (6, 8,10,12}. Find A∪B
Answer :
A∪B = {2,4,6,8,10,12}.

Question 42.
Let A = {a, e, i, o, u} and B = {a, i, u}. Show that A∪B = A.
Answer :
A∪B = {a,c,i,o,u} = A
∴ A∪B = AQuestion 43.
Let X = {Ram, Geeta, Akbar} be the set of students of class XI, who are in school hockey team.
Let Y = {Geeta, Divid, Ashok} be the set of students from Class XI who are in school foot ball team. Find X∪Y and interpret the set.
Answer :
X∪Y= {Ram, Geeta, Akbar, Divid, Ashok}
This is the set of students from Class XI who are in the hockey team or the foot ball team or both.

Question 44.
Find the union and the intersection of each of the following pairs of sets:
(i) X ={1,3,5}, Y ={1,2,3}
(ii) A = {a,e,i,o,u},B={a,b,c}
(iii) A = {x : x is a natural number and multiple of 3}
B = {x : x is a natural number less than 6}.
(iv) A = {x:x is a natural number and 1<x≤6}
B ={x:x is a natural number and 6 < x< 10}
(v) A = {1,2,3},B = φ
Answer :
(i) X∪Y = {1,2,3,5}
X∩Y = {1,3}(ii) A∪B = {a,b,c,e,i,o,u}
A∩B = {a}(iii) A = {3,6,9,12,…….. }
B = { 1,2,3,4,5,6}
∴ A∪B = {l,2,3,4,5,6,9,12, ………….. }(iv) A = {2,3,4,5,6}
B = {7,8,9}
A∪B = {2,3,4,5,6,7,8,9}
A∩B = {2,3,4,5,6,7,8,9}(v) A∪B = {1, 2,3} = A
A∩B = φ = B

Question 45.
Let A = {a,b), B ={a,b,c}- Is A⊂B? What is A∪B?
Answer :
Given sets are A = {a,b} and B = {a,b,c}.
Clearly, every element of A is the element of B, therefore A⊄B- A∪B = {a,b,c].

Question 46.
If A and B are two sets such that A‎⊂B, then what is A∪B?
Answer :
Given A⊂B i.e., every element of A is contained in the set B and hence A∪B = B

Question 47.
If A ={1,2,3,4}, B ={3,4,5,6}, C ={5,6,7,8} and D = {7,8,9,10} ; find
(i) A∪B
(ii) A∪C
(iii) B∪C
(iv) B∪D
(v) A∪B∪C
(vi) A∪B∪D
(vii) B∪C∪D .
Answer :
(i) A∪B = {1,2,3,4,5,6}
(ii) A∪C = {1,2,3,4,5,6,7,8}
(iii) B∪C = {3,4,5,6,7,8}
(iv) B∪D = {3,4,5,6,7,8,9,10}
(v) A∪B∪C = {1,2,3,4,5,6,7,8}
(vi) A∪B∪D = {1,2,3,4,5,6,7,8,9,10}
(vii) B∪C∪O = {3,4,5,6,7,8,9,10}Question 48.
If A = {3,5,7,9,11}, B ={7,9,11,13}, C = {11,13,15} and D = {15,17}; find
(i) A∩B
(ii) B∩C
(iii) A∩C∩D
(iv) A∩C
(v) B∩D
(vi) A∩(B∪C)
(vii) A∩D
(vii) A∩(B∪D)
(ix) (A∩B)∩(B∪C)
(x) (A∪D)∩(B∪C).
Answer :
(i) A∩B = {7,9,11}
(ii) B∩C = {11,13}
(iii) A∩C∩D = φ
(iv) A∩C ={11}
(v) B∩D ={ } =φ
(vi ) A∩{B∪C) = {3,5,7,9,11}∩{7,9,11,13,15}= {7,9,11}
A∩D = φ
(viii) A∩(B∪D) = {3,5,7,9,11} ∩ {7,9,11,13,15,17} = {7,9,11}
(A∩B)∩(B∪C) = {7,9,11}∩{7,9,11,13,15} = {7,9,11}
(A∪D)∩(B∪C)
= {3,5,7,9,11,15,17} ∩ {7,9,11,13,15}.
= {7,9,11,15}

Question 49.
If A = {x : x is a natural number}, B={x:x is even natural number}, C-{x : x is an odd natural number} and D = { x: x is a prime number},find
(i) A∩B
(ii) A∩C
(iii) A∩D
(iv) B∩C
(v) B∩D
(vi) C∩D
Answer :
Given A = {1,2,3,4,………………}
B = {2,4,6,8, …………..}
C = {1,3,5,7,……… }
D = {2,3,5,7,11,13, …………..}.
(i) A∩B = {2,4,6,8,……….. } = B
(ii) A∩C = {1,3,5,7,……….. } = C
(iii) A∩D = {2,3,5,7,11,13,………….}
(iv) B∩C = φ
(v) B∩D = {2}
(vi) C∩D = {3,5,7,11,13,…………}

Question 50.
Show that A∪B = A∩B implies A = B
Answer :
Given A∪B = A∪B
Let x∈A.
Then x∈A∪B
x∈A∩B∴ A∩B = A∪B
⇒ y∈B
∴A ⊂B
Similarly, let y∈B .
Then y∈ A∪B
y∈A∩B ∴ A∩B = A∪B
= y∈A
∴ B⊂A
Thus, A = B.

Question 51.
Using properties of sets, show that
(i) A∪(A∩B) = A
(ii) A∩(A∪B) = A
Answer :
(i) A∪(A∩B) = (A∪A)∩(A∪B) ∵ distributive law
= A∩(A∪B) = A
(ii) A∩(A∪B)
= (A∩A)∪(A∩B) ∵ distributive law
= A∪(A∩B) = A.

Question 52.
Show that A∩B = A∩C need not imply B = C
Answer :
Let A = {1, 2}, B = {1, 3}, C = {1,4}.
A∩B = {1} and A∩C = {1}
A∩B = A∩C
But B≠C.

Question 53.
Let A and B be sets. If A∩X =B∩X = φ and A∪X =B∪X for some set X, show that A = B.
Answer :

Question 54.
Find sets A, B and C such that A∩B,B∩C and A∩C are non-empty sets and A∩B∩C =φ.
Answer:
Let A = {a,b}, B = {b,c} and C = {a,c}.
∴ A∩B = {b}, B∩C = {c}, A∩C = {a}
⇒A∩B, B∩C and A∩C are non-empty sets Now A∩B∩C = A∩(5∩C)
= {a,b}∩{c} = φQuestion 55.
For any sets A and B, show that P(A∩B) = P(A)∩P(B)
Answer:

Question 56.
Is it true for any sets A and B,
P(A)∪P(B) = P(A∪B)? Justify your answer.
Answer:
Let A = {φ{a}}, and B = {b}
∴ A∪B = {a,b}
P(A) = {φ,{a}}, P(B) = {φ,{b}}
and P(A∪B) = {φ,{a},{b} {a,b}}………………. (1)
and P(A)∪P(B) = {φ,{a},{b}} ………………. (2)
From (1) and (2), we get, P(A∪B) ≠P(A)∪P{B)

Question 57.
Which of the following pairs of sets are disjoint
(i) {1, 2, 3, 4} and {x : x is a natural number and 4 < x < 6}
(ii) {a,e,i,o,u} and {c,d,e,f}
(iii) {x: x is an even integer} and {x : x is an odd integer}.
Answer :
(i) Given sets are {1, 2, 3,4} and {4, 5, 6}
∴ 4 is a common element.
∴ Given sets are not disjoint sets.

(ii) Given sets are {a,e,i,o,u} and {c,d,e,f}
∴ ‘e’ is a common element.
∴ Given sets are not disjoint sets.

(iii) Given sets are {0,±2,±4,- -} and {±1,+3±5,—}
∴ There is no common element.
∴ Given sets are disjoint sets.

Question 58.
State whether each of the following statement is true or false. Justify your answer.
(i) {2,3,4,5} and {3,6} are disjoint sets.
(ii) {a,e,i,o,u} and {a,b,c,d} are disjoint sets.
(iii) {2,6,10,14} and {3,7,11,15} are disjoint sets
(iv) {2,6,10} and {3,7,11} are disjoint sets.
Answer :
(i) Let A = {2,3,4,5} and B = {3,6}
∴ A∩B = {3} ∴ A∩B≠φ
∴Given sets are disjoint sets.(ii) Let A = {a,e,i,o,u} and B = {a,b,c,d}
A∩B = {a}.
Given sets are disjoint sets.

(iii) Given sets are A = {2,6,10,14} and B = {3,7,11,15}.
∴ A∩B =φ(iv) Given sets are disjoint sets.
Let A = {2,6,10} and B = {3, 7,11}
∴ A∩B = φ
∴ Given sets are disjoint sets.

Question 59.
Let A, B and C be the sets such that
A∪B = A∪C and A∩B = A∩C . Show, that B = C
Answer :
Given A∪B=A∪C
⇒(A∪B)∩C = (A∪C)∩C
⇒(A∩C)∪(B∩C) = C
⇒(A∩B)∪(B∩C) = C ∵ A∩B = A∩C
Now, A∪B=A∪C
⇒(A∪B)∩B = (A∪C)∩B
⇒(A∩B)u(B∩B) = (A∩B)∪(C∩B)
⇒B = (A∩B)∪(B∩C)
B = C

Question 60.
If A = {3,6,9,12,15,18,21}, B = {4,8,12,16,20}, C = {2,4,6,8,10,12,14,16}, £>={5,10,15,20}. Find
(i) A-B
(ii) A-C
(iii) A-D
(iv) B – A
(v) C – A
(vi) D – A
(vii) B – C
(viii) B – D
(ix) C – B
(z) D – B
(xi) C – D
(xii) D – C.
Answer :
(î) A – B={3,6,9,15,18,21)
(ii) A – C=(3,9,15,18,21)
(iii) A – D=(3,6,9,12,18,21)
(iv) B – A={4,8,16,20)
(v) C – A={2,4,8,10,14,16}
(vi) D – A=(5,10,20}
(vii) B – C={20)
(viii) B – D = (4,8,12,16)
(ix) C – B={2,6,10,14)
(x) D – B={5,10,15)
(xi) C – D={2,4,6,812,14,16}
(xii) D – C={5,15,20).

Question 61.
IV X = {a,b,c,d) and Y = (f,b,d,g), find
(i) X – Y
(ii) y – x
(iii) X∩Y.
Answer :
(j) X – Y={a,c)
(ii) Y- X=(f,g)
(iii) X∩Y={b,d}

Question 62.
If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?
Answer :
R – Q = set of irrational numbers.

Question 63.
Show that the following four conditions are equivalent
(i) A⊂B
(ii) A – B=φ
(iii) A ∪ B = B
(iv) A ∩ B A.
Answer:
(i) ⇔ (ii):
A⊂B ⇔ All elements of A are in B
⇔ A – B = φ(ii) ⇔ (iii):
A-B = φ⇔ All elements of A are in B
⇔ A∪B = B(iii) ⇔(iv)
A∪B = B ⇔ All elements of A are in B
⇔ All elements of A are common in A and B.
⇔ A∩B = A
∴ All the four given conditions are equivalent.

Question 64.
Show that for any sets A and B,
A = (A∩B)∪(A – B) and A∪(B-A) = A∪B.
Answer :

Question 65.
Define complement of a set.
Answer :
Let U be the universal set and A a subset of U. Then the complement of A is the set of all elements of U which are not the elements of A and is denoted by A’.
Properties:

• A∪A’
• A∩A’=φ
• (A’)’=A
• (A∪B)’= A’∩B’
• (A∩B)’ -A’∪B’

Question 66.
Let U = {1,2,3,4,5,6,7,8,9,10} and A = {1,3,5,7,9}. Find A’.
Answer :
A’ = U -A = {2,4,6,8,10}Question 67.
Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}. Find A’, B’, A’∩B’,(A∪B) and hence show that (A∪B)’ = A’∩B’.
Answer :
A’ = ∪ -A = { 1,4,5,6}
B’ = U-B = { 1,2,6}
A’∩B’ = {1,6}
A∩B = {2,3,4,5}
∴ (A∪B)’ = {1,6} = A’∩B’

Question 67.
If ∪={a,b,c,d,e,f ,g,h] find the complements of the following sets
(i) A={a,b,c}
(ii) B={d,e,f,g}
(iii) C ={a,c,e,g}
(iv) D = {f,g,h,a}.
Answer :
(i) A’= {d,e,f,g,h}
(ii) B’ = {a,b,c,h}
(iii) C’ = {b,d,f,h]
(iv) D’ = {b,c,d,e}.

Question 68.
If U={a,b,c,d,e,f ,g,h] find the complements of the following sets
(i) A={a,b,c}
(ii) B={d,e,f,g}
(iii) C ={a,c,e,g}
(iv) D = {f,g,h,a}.
Answer :
(i) A’= {d,e,f,g,h}
(ii) B’ = {a,b,c,h}
(iii) C = {b,d,f,h]
(iv) D’ = {b,c,d,e}.

Question 68.
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2,3,5,7}. Verify that
(i) (A∪B)’ = A’∩B’
(ii) (A∩B)’ = A’∪B’.
Answer :
(i) A’ = {1, 3, 5, 7, 9},
(ii) B’ = {1,4, 6, 8, 9}
(iii) C’ = {b,d,f,h}
(iv) D’={b,c.d,e}

Question 69.
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A’ = {2, 4, 6, 8} and B’ = {2,3,5,7}. Verify that
(i) (A∪B)’ = A’∩B’
(ii) (A∩B)’ = A’∪B’.
Answer:

Question 70.
Taking the set of natural numbers as the universal set, write down the complements of the following sets:
(i) {x: x is an even natural number}
(ii) {x : x is an odd natural number}
(iii) {x : x is positive multiple of 3}
(iv) {x : x is a prime number}
(v) {x : x is a natural number divisible by 3 and 5}
(vi) {x : x is a perfect square}
(vii) {x : x is a perfect cube}
(viii) {x : x + 5 = 8}
(ix) {x : 2x + 5 = 9}
(x) {x : x > 7}
(xi) {x : x ∈N and 2x + 1 > 10}.
Answer :
(i) {x : x is an odd natural numbers}
(ii) {x : x is an even natural numbers}
(iii) {x: x is a natural number and not multiple of 3}
(iv) {x : x is a positive composite number and x = 1}
(x) {x : x∈ N and x is not divisible by 3 and 5}
(xi) {x: x∈N and x is not a perfect square}
(xii) {x:x∈N and x is not a perfect cube}
(xiii) {x:x∈2V and x≠3}
(ix) {x:x∈2V and x≠2}
(x) {1,2,3,4,5,6}
(xi) {x:x∈N and 2x + 1≤10} = {1,2,3,4}.

Question 71.
Draw appropriate Venn diagram for each of the following:
(i) (A∪B)’
(ii) A’∩B’
(iii) (A∩B’)
(iv) A’∪B’.
Answer:

Question 72.
Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60°, what is A’?
Answer :
A’ = set of equilateral triangles.

Question 73.
Fill in the blanks to make each of the following a true statement:
(i) A∪A’ = —
(ii) φ’∩A = —
(iii) A∩A’ = —
(iv) U’∩A = —
Answer :
(i) A∪A’ = U
(ii) (ii) φ’∩A = A
(iii) A∩A’ = φ
(iv) U’∩A = φ.

Question 74.
In X and Y are two sets such that X∪Y has 50 elements, X has 28 elements and Y has 32 elements, how many elements does X∩Y have?
Answer :
Given n(X∪Y) = 50; n(X) = 28; n(Y) = 32
But n(X∪Y) = n(x) + n(y)-n(X∩Y)
∴ n(X∩Y) = 28 + 32-50 = 10

Question 75.
If X and Y are two sets such that n(X) = 17, n(Y) = 23 and n(X∪Y) = 38 find n(X∩Y).
Answer :
Given n(X) = 17, n(Y) = 23, n(X∪Y) = 38
But n(X∪Y) = n(X) + n(Y)-n(X∩Y)∩(X∩Y)
= 17 + 23-38 = 2

Question 76.
If X and Y are two sets such that X∪Y has 18 elements X has 8 elements and Y has 15 elements; how many elements does X∩Y have?
Answer :
Given n(X∪Y) = 18, n(X) = 8, n(Y) = 15
But n(X∪Y) = n(X) + n(Y) – n(X ∩Y) n(X∩Y)
= 8 + 15-18 = 5.

Question 77.
In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English?
Answer :
Let H = set of people speaking Hindi;
E = set of people speaking English.
Then n(H∪E) = 400; n(H) = 250; n(E) = 200
∴ n(H∩E) = n(H) = n(E)-n(H∪E)
= 250 + 200 – 400 = 50
Hence, 50 people can speak both Hindi and English.

Question 78.
If S and T are two sets such that S has 21 elements, T has 32 elements and S∩T has 11 elements, how many elements does S∪T have?
Answer :
Given n(5) = 21; n(T) = 32; n(S∩T) = 11
n(S∪T) = n(S) + n(T) – n(S∩T)
= 21 + 32-11 =42.

Question 79.
If X and Y are two sets such that X has 40 elements, X∪Y has 60 elements and X∩Y has 10 elements, how many elements does Y have?
Answer :
Given n(X) = 40; n(X∪Y) = 60; n(X∩Y) = 10
n(Y) = n(X∪Y) + n(X∩Y)-n(X)
= 60 + 10 – 40 = 30Question 80.
In a group of 70 people, 37 like coffee, 52 like tea and each person likes atleast one of the two drinks. How many people like both coffee and tea?
Answer :
Let C = set of people who like coffee
T = set of people who like tea
Then n(C∪T) = 70; n(C) = 37; n(T) = 52 n(C∩T)
= n(C) + n(J)-n(C∪T)
= 37 + 52-70 =19

Question 81.
In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?
Answer :
Let C = set of people who like cricket
T = set of people who like tennis.
Then n(C∪T) = 65; n(C) = 40; n(C∩T) = 10
∴ n(T) = n{C∪T) + n(C∩T)-n(C)
= 65 + 10-40 = 35
Number of people who like tennis only
n(T-C) = n(T)-n(C∩T)
= 35-10 = 25

Question 82.
In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many speak atleast one of these two languages?
Answer :
Let F = set of people speaking French
S = set of people speaking Spanish.
Then n(S) = 20, n(F) = 50 n(F∩S) = 10
∴ n(F∪S) = n(S) + n(F)-n(F∩S)
= 20 + 50-10 = 60
∴ 60 people speak French or Spanish or both.Question 83.
In a school there are 20 teachers who teach Mathematics or Physics. Of these, 12 teach Mathematics and 4 teach both Physics and Mathematics. How many teach Physics?
Answer :
Let M = set of teachers who teach Mathematics
P = set of teachers who teach Physics
Then, n(M∪P) = 20, n(M) = 12, n(M∩P) = 4
n(P) = n(M∪P) + n(M∩P)- n(M)
= 20 + 4-12 = 12.

Question 84.
In a class of 35 students, 24 like to play cricket and 16 like to play football. Also, each student likes to play at least one of the two games. How many students like to play both cricket and football?
Answer :
Let C = set of students who like cricket F = set of students who like football.
Then n(C uf) = 35, n(C) = 24, n(F)
= 16 n(C n F) = n(C) + n(F) – n(C u F)
= 24 + 16-35 = 5
∴ 5 Students like to play both games.

Question 85.
In a survey of 400 students in a school, 100 were listed as taking apple juice, 150 as taking orange juice and 75 were listed as taking both apple as well as orange juice. Find how many students were taking neither apple juice nor orange juice.
Answer :
Let U = set of surveyed students
A = set of students taking apple juice
B = set of students taking orange juice.
Then n(∪) = 400; n(A) = 100; n(B) = 150;
and n(A∩B) = 75
Now n(A’∩B’) = n(A∪B)’
= n(∪) – n(A∪B)
= n(∪)-n(A)-n(B) + n(A∩S)
= 400-100-150 + 75 = 225
∴ 225 students were taking neither apple juice nor orange juice.

Question 86.
In a survey of 600 students in a school. 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?
Answer :
Let U = set of students in a school.
A = set of students taking tea
B = set of students taking coffee
Then n(∪) = 600; n(A) = 150; n(B) = 225; n(A∩B) = 100
∴ n(Au B) = n(A) + n(B)-n(A∩B)
= 150 + 225-100 =275
Now n(A’∩B’) = n(A∪B)
= n(∪)-n(A∪5)
= 600-275 = 325
∴ 325 students were taking neither tea nor coffee.

Question 87.
In a group of students, 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?
Answer :
Let H = set of students, who know Hindi
E = set of students, who know English
Then n(H) = 100; n(E) = 50 and n(H∩E) = 25.
∴ n(H∪E) = n(H) + n(E)-n(H∩E)
= 100 + 50-25 = 125
Hence, 125 students are there in the group.

Question 88.
There are 200 individuals with a skin disorder, 120 had been exposed to the chemical C₁, 50 to chemical C₂ and 30 to both the chemicals C₁ and C₂. Find the number of individuals exposed to
(i) Chemical C₁ but not chemical C₂
(ii) Chemical C₂ but not chemical C₁
(iii) Chemical C₁ or chemical C₂.
Answer :
Let U = set of individuals suffering from the skin disorder.
A = set of individuals exposed to the chemical C₁.
B = set of individuals exposed to the chemical C₂.
Then, n(u) = 200, n(A) = 120, n(B) = 50 and n(A∩B) = 30

(i) From the Venn diagram, we have
A = (A-B)∪(A∩B)
∴ n(A) = n(A -B) + n(A∩B)
A – B and A∩B are disjoint
∴ n(A -B) = n(A) – n(A∩B)
= 120-30 = 90
Hence, 90 individuals exposed to chemical C₁ but not to chemical C₂.

(ii) Similarly, B = (B – A) ∪ (A∩B)
n(B) = n(B – A) + n(A∩B)
∵ (B-A)∩(A∩B) = φ
∴ n(B-A) = n(B)-n(A∩B)
= 50-30 = 20
Hence, 20 individuals exposed to chemical C₂ but not to chemical C₁.

(iii) We have, n(A∪5) = n(A) + n(B) – n(A∩B)
= 120 + 50-30 = 140
Hence, 140 individuals exposed either chemical C₁ or chemical C₂.

Question 89.
A market research group conducted a survey of 1000 consumers and reported that 720 consumers like product A and 450 consumers like product B, what is the least number that must have liked both products?
Answer :
Let U = set of consumers questioned
S = set of consumers who like product A.
T = set of consumers who like the product B.
Then, n(u) = 1000, n(S) = 720, n(T) = 450
∴ n(S∪T) = n(S) + n(T)-n(S∩T)
= 720 + 450 – n(S∩T)
= 1170-n(S∩T).
Therefore, n(S∪T) is maximum when n(S∩T) is least.
But S∪T⊂U =n(S∪T)≤n(U) = 1000
So, maximum value of n(S∪T) is 1000. Thus, least value of n(S∩T) is 170.
Hence, the least number of consumers who liked both products is 170.

Question 90.
Out of 500 car owners investigated, 400 owned car A and 200 owned car B, 50 owned both A and B cars. Is this data correct?
Answer :
Let U = set of car owners investigated
M = set of persons who owned car A
S = set of persons who owned car B.
Then n(u) = 500, n(M) = 400, n(S) = 200 and n(S∩M) = 50
∴ n(S∪M) = n(S) + n(M)-n(S∩M)
= 200 + 400-50 = 550
But S∪M⊂U⇒n(5∪M)≤n(U)
This is contradiction. So, the given data is incorrect.

Question 91.
A college awarded 38 medals in football, 15 in basket ball and 20 in cricket If these medals went to a total of 58 men and only three men got medals in all the three sports, how many received medals in exactly two of the three sports?
Answer :
Let F = set of men who received medals in football
B = set of men who received medals in basket ball
C = set of men who received medals in cricket.

Question 92.
In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T and I, 3 read all three news papers. Find:
(i) the number of people who read at least one of the newspapers
(ii) the number of people who read exactly one newspaper. ‘
Answer :
Let H = set of people who read paper H.
I = set of people who read newspaper I
T = set of people who read newspaper T
Then, n(H) = 25, n(T) = 26, n(I) = 26, n(H∩I) = 9
n(H∩T) = 11, n(T∩I) = 8, n(H∩T∩I) = 3(i) Number of people who read atleast one newspaper
n(H∪T∪T) = n(H) + n(T) + n(I)
-n(H∩T)-n(T∩I)-n(I∩H) + n(H∩T∩I)
= 25 + 26 + 26-9-11-8 + 3 =52(ii) Number of people who read exactly one paper

Question 93.
In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.
Answer :
Given n(A) = 21, n(B) = 26, n(C) = 29
n(A∩B) = 14, n(A∩C) = 12, n(B∩C) = 14
n(A∩B∩C) = 8

From the figure, number of persons liking product C
only = n(C) – n(A∩C) – n(B∩C) + n(A∩B∩C)
= 29-12-14 + 8 = 11.

(i)	{P, R, I, N, C, A, L}	(a)	{x : x is a positive integer and is a divisor of 18}
(ii)	{0}	(b)	{ x : x is a integer and x2 – 9 = 0}
(iii)	(1,2,3, 6, 9,18}	(c)	{x : x is an integer and x + 1 = 1}
(iv)	{3,-3}	(d)	{x : x is a letter of the word PRINCIPAL}

You can Download Chapter 8 Interpolation and Extrapolation Questions and Answers, Notes, 1st PUC Statistics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Statistics Question Bank Chapter 8 Interpolation and Extrapolation

1st PUC Interpolation and Extrapolation Question and Answers

Question 1.
What is Interpolation?
Answer:
‘Interpolation is a procedure of estimating the unknown value of dependent variable for a given value of independent variable which is within the limits or the range of the independent variable’

Question 2.
What is extrapolation?
Answer:
Extrapolation is a procedure of estimating the unknown value of dependent variable for a given value of independent variable which is outside the limits or the range of the independent variable’

Question 3.
Mention the situations where the technique of interpolation is used.
Answer:
The procedure of estimating the missing value of y for a given value of x, where x is within the limits x₀ and x_n we use the technique Interpolation.

Question 4.
What are the assumptions made in interpolation?
Answer:
In making use of the techniques of interpolation the following assumptions are made

• There are no sudden jumps in the values of independent variable from one period to another.
• The rate of change of figures from one period to another is uniform.

Question 5.
Using Binomial expansion method expand (y – 1)5 = 0.
Answer:
The equation (y – 1 )⁵ = 0 is:
y₅ – 5y-₄ + 10y₃ -10y₂ + 5y₁ -y₀ = 0

Question 6.
If in a Binomial expansion y₄ – 4y₃ + 6y₂ – 4y₁ + y₀ = 0, if y₃=157, y₂=124, y₁ = 107 and  y₀=100, then find y₄
Answer:
Substituting the known values in the equation, we obtain,
y₄ – 4(157) + 6(124) – 4(107)+ 100 = 0
y₄ – 1056 + 844= 0
y₄ – 212= 0
∴ y₄ = 212

Question 7.
Distinguish between interpolation and extrapolation.
Answer:
The procedure of estimating the missing value of y for a given value of x, where x is within the limits x₀ and x_n we use Interpolation. Here “Interpolation is a procedure of estimating the unknown value of dependent variable for a given value of independent variable which is within the limits or the range of the independent variable”

But if the value of y is to be estimated for a value of x which is outside the limits x₀ and x_n then procedure Extrapolation is used. “Extrapolation is a procedure of estimating the unknown value of dependent variable for a given value of independent variable which is outside the limits or the range of the independent variable”

Question 8.
Interpolate the export of handlooms during 2008 from the following data

Answer:
Let X and Y be years and Exports

(years)X	(Export in crores Rs. ) Y
1998 (x0)	10 (y0)
2000 (x1)	13 (y1)
2002 (x2)	15 (y2)
2004 (x3)	23 (y3) 18
2006 (x4)	26 (y4) 23
2008 (x5)	– (y5)
2010 (x6)	32 (y6)

Using binomial expansion method: Here we have to estimate y₅ Since there are 6 known values of y, the formula of estimation is based on the expansion of (y-1)₆= 0
We get the equation,
y₆ – 6y₅ + 15y₄ – 20y₃ + 15y₂ – 6y₁ + y₀= 0
Substituting the known values in the equation, we get,
i.e.32 – 6y₅+ 15 x 23 -20 x 18 + 15 x15 -6 x 13 + 10 = 0
i.e.32 – 6y₅+ 345 – 360 + 225 – 78 + 10 = 0 ;
-6y₅ = -174
⇒ y₅ = 29
Hence the probable export for the year 2008 is 29 (crores Rs.)

Question 9.
Interpolate the missing figure.

Answer:
Let X and Y be years and sales.

(years)X	(sales ) Y
2001 (x0)	100 (y0)
2002 (x1)	120 (y1)
2003 (x2)	(y2)
2004 (x3)	180 (y3)
2005 (x4)	210 (y4)

Here we have to estimate y2 Since there are 4 known values of y, the formula of estimation is based on the expansion of (y – 1)⁴= 0
We get the equation,
Y₄ -4 y₃ + 6y₂ – 4y₁ + y₀=o
Substituting the known values in the equation, we obtain,
i.e.210 – 4 x 180 + 6y₂ – 4 x 120 +100 = 0
i.e.210 – 720 + 6y₂ – 480 +100 = 0;
6y₅= 890
⇒ y₂ = 148.33
Hence the probable sales for the 2003 is 148.33

Question 10.
Interpolate the index for 2008 from the following data.

Answer:
Using Binomial expansion method – let X and Y be year and Index number.

(Years) x	(Index No.) y
2006 (x0)	278 (y0)
2007 (x1)	281 (y1)
2008 (x2)	– (y2)
2009 (x3)	313 (y3)
2010 (x4)	322 (y4)

We have to estimate y₂ Since there are 4 known values of y, the formula of estimation is • based on the expansion of (y – 1)⁴=0
We get the equation,
y₄ – 4y₃ + 6y₂ – 4y₁ y_o = 0
Substituting the known values in the equation, we get,
i.e. 322 – 4(313) + 6y₂ – 4(281) + 278 = 0
i.e. 322 – 1252 + 6y₂ – 1124 + 278 = 0
6y₂= 1776
⇒ y₂ = 296
Hence the probable value of the index number of 2008 is 296.

Question 11.
From the following data interpolate the production of cement in 2007.

Answer:
Let X and Y be the year and production

X	Y
2005 (x0)	44 (y0)
2006 (x1)	90 (y1)
2007 (x2)	(y2)
2008 (x3)	160 (y3)
2009 (x4)	210 (y4)
2010 (x5)	290 (y5)

Using Binomial expansion method We have to estimate y₂ Since there are 5 known values of y, the formula of estimation is, the expansion of (y-1)⁵ = 0 We get the equation,
Y₅ – 5y₄ + 10y₃ – 10y₂ + 5y₁ – y₀= 0.
Substituting the known values in the equation we obtain,
290 – 5(210)+ 10 x 160 – 10y₂+ 5(90) – 44 =0
290 – 1050+ 1600 – 10y₂ + 450 – 44 =0
-10y₂ – 1246 =0
– 10y₂ =1246
∴ y₂ = \(\frac { 1246 }{ 10 }\)
⇒ y₂ = 127 is 124.6.

Students can Download Maths Chapter 2 Relations and Functions Questions and Answers, Notes Pdf, 1st PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Maths Question Bank Chapter 2 Relations and Functions

Question 1.
Define ordered pair.
Answer :
Two numbers a and b listed in specific order and enclosed in parentheses form is called an ordered pair (a, b).
Keen Eye: Equality of two ordered pairs:
We have {a, b)-(c,d)⇔a-c and b – d.

Question 2.
Define a Cartesian product of two sets.
Answer :
Let A and B two non-empty sets. Then, the Cartesian product of A and B is the set denoted by Ax B, consisting of all ordered pairs (a, b) such that a e A and be B.
A x B= {(a, b): a ∈ A, b∈B}
Keen Eye:

• If n(A) = p and n(B) = q, then n (A x B) = pq and n (B x A) = pq
• If at least one of A and B is infinite then AxB is infinite and B x A is infinite,
• In general, A x B ≠ B x A
• A x A x A = {(a, b, c) : a, b, c ∈A}. Here (a, b, c) is called an ordered triplet.

Question 3.
If (x + 1, y – 2) = (3,1), find the values of x
Answer :
Given (x + 1, y – 2) = (3,1)
⇒ x+1=3  ∴x=2
y-2=1  ∴ y=3

Question 4.
If \( \left(\frac{x}{3}+1, y-\frac{2}{3}\right)=\left(\frac{5}{3}, \frac{1}{3}\right)\)
Answer:
Given \( \left(\frac{x}{3}+1, y-\frac{2}{3}\right)=\left(\frac{5}{3}, \frac{1}{3}\right)\)

Question 5.
If P={a,b,c}and Q={r},from P×Q and Q x P.Are these two products equal?
Answer:
PxQ = {(a,r),(b,r)(c,r)}
QxP = {(r, a), (r, b), (r, c)}
Clearly PxQ≠QxP

Question 6.
If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in
Answer :
Given n(A) = 3; n(B) = 3.
∴ n(AxB) = 3×3 = 9

Question 7.
If G=(7, 8} and H={5, 4, 2), find G x II and
Answer :
GxH = {(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)}
HxG = {(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)}

Question 8.
State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
(i) If P={m, n} and Q = {n, m}, then P x Q = {(m, n), (n, m)}.
(ii) If A and B are non-empty sets, then A x B is a non-empty set of ordered pairs (at, y) such that x∈B and y∈A
(iii) If A = {1,2}, B = {3,4}, then A x {B∩φ ) = φ
Answer :
(i) Given statement is false:
Correct statement:
PxQ={(m, n), (m, m), (n, n), (n, m)}.

(ii) Given statement is false:
Correct statement:
AxB = {(x, y) :x∈A, y ∈B}.                                 ‘

(iii) True statement,

Question 9.
If A x B = {(a, x), (a, y), (b, x), (b, y)}. Find A and
Answer :
A = {a, b} and B – {x, y}

Question 10.
If A x B = {(p, q), (p, r), (m, q), (m, r)}, find A and
Answer :
A = set of first elements = {p, m}
B = set of second elements = {q, r}

Question 11.
Let A = (1, 2}, B = [1, 2, 3, 4}, C = { 5, 6} and D = (5,6,7,8}. Verify that
(i) A x (B∩C) = (A x B)∩(A x C).
(ii) A x C is a subset
Answer :
(i) B∩C = { }
∴ Ax(B∩C)=φ ………….. (1)
A x B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2,4)}
A x C = {(1, 5), (1,6), (2, 5) (2,6)}
∴ (A x B) ∩ (A x C) = φ ………………. (2)
From (1) and (2), we get
A x (B∩C) = (A x B)∩(A x C)

(ii) A x C = {(1, 5), (1,6), (2, 5), (2, 6)}
B x D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}.
Clearly every elements of A x C is an element of B x D.
A x C ⊂B x D.

Question 12.
Let A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}. Find
(i) A x (B ∩ C)
(ii) (A x B) ∩ (A x C)
(iii) A x (B∪C)
(iv) (A x B)∪(A x C)
Answer :
(i) B∩C={4}
A x (B∩C) = (1,4), (2, 4), (3,4)}

(ii) A x B = {(1, 3), (1,4), (2, 3), (2, 4), (3, 3), (3, 4)}
A x C = {(1, 4), (1, 5) (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(A x B)∩(A x C)= {(1, 4), (2, 4), (3, 4)}

(iii) B ∪ C={3,4, 5, 6}
∴ Ax(B∪C)  = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}

(iv) A x B = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
A x C = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(A x B)∪(A x C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6) (3, 3), (3, 4), (3, 5), (3,6)}.

Question 13.
Let A = {1, 2} and B = {3, 4}. Write A x B. How many subsets will A x B have? List them.
Answer :
Given A = {1, 2} and B = {3,4}
A x B= {(1, 3), (1,4), (2, 3), (2, 4)}
∴n (A x B) = 4
Number of subsets of A x B = 2⁴=16
Subsets of A x B are: A x B, φ, {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3), (1, 4)}, {(1, 3), (2, 3)}, {(1, 3), (2,4)}, {(1,4), (2, 3)}, {(1, 4), (2, 4)} {(2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1,4), (2, 3), (2, 4)}, {(2, 3), (2,4), (1, 3)}.

Question 14.
Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A x B, find A and B, where x, y, z are distinct elements.
Answer :
A = {x, y, z} and B = {1, 2}.

Question 15.
The Cartesian product A x A has 9 elements along which are found (-1, 0) and (0,1). Find the set A and the remaining elements Ax A.
Answer :
Given n(A x A) = 9 = 3²
⇒n(A) = 3
But (-1, 0) and (0, 1) are in A x A
∴ A= {-1,0,1}.
Remaining elements of A x A: (-1, -1), (-1, 1),
(0,-1), (0,0), (1,-1), (1,0), (1,1).

Question 16.
If P = {1,2}, form the set
Answer :
P x P x P = {(1, 1, 1), (1, 1, 2), (1, 2, 1),
(1, 2,2), (2, 1,1), (2, 1, 2), (2, 2,1), (2, 2, 2)}

Question 17.
If A = {-1,1}, find A x A x A.
Answer :
A x A x A = {(-1, -1, -1), (-1, -I, 1), (-1, 1, -1), (-1, 1, 1), (1, -1, -1), (1, -1, 1), (1,1,-1), (1,1,1)}

Question 18.
If R is the set of all real numbers, what do the Cartesian products R x R and R x R x R represent?
Answer :
We have R x R = {(x, y) : x, y ∈ R } which represents the coordinates of all the points in two dimensional space and R x R x R = {(x, y, z)  x,y,z ∈ R } which represents the coordinates of all the points in three-dimensional space.

Question 19.
Define a relation.
Answer :
A relation R from a non-empty set A to non empty set B is a subset of the Cartesian product A x B.

Question 20.
Define domain of a relation.
Answer :
The set of all first elements of the ordered pairs in a relation R from a set A to a set B is called the domain of the relation R.

Question 21.
Define range of a relation.
Answer :
The set of all second elements in a relation R from a set A to a set B is called the range of the relation R. The whole set B is called the co-domain of the relation R.

Question 22.
Let A = {1, 2, 3, 4, 5, 6}. Define a relation R from A to A by R = {(x,y): y = x + 1}
(i) Depict this relation using an arrow diagram
(ii) Write down the domain, condomain and range of
Answer :
Given R = {(x, y): y = x + 1}
= {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

Domain = {1, 2, 3,4, 5,}; Co-domain = A
Range = {2, 3,4, 5, 6}.

Question 23.
Let A = {1, 2, 3, ………….14}. Define a relation R from A to A by R = {(x, y): 3x – y = 0, x,
y ∈ A}. Write down its domain, co-domain and range.
Answer :
Given R = {(x, y): 3x -y = 0, x, y ∈ A}
= {(1, 3), (2, 6), (3, 9), (4,12)}
Domain = {1, 2, 3,4}
Co-domain = A Range = {3, 6, 9,12}

Question 24.
Define a relation R on the set N of natural numbers by R – {(x, y) : y = x + 5, x is a natural number less than 4; x, y ∈ . N}. Depict this relationship using roster form. Write down the domain and the range.
Answer :
Given R = {(x, y): x, y ∈ N and y = x + 5, x < 4}
= {(1,6), (2,7), (3, 8)}
Domain = {1, 2, 3}
Range = {6, 7, 8}

Question 25.
A = {1, 2, 3, 5} and B = (4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ A, y∈ B}. Write R in roster form.
Answer :
Given A = {1, 2, 3, 5} and B = {4, 6, 9}
R = {(x, y): the difference between x and y is odd; x∈ A,y∈B}
= {(1,4), (1, 6), (2, 9), (3, 4), (3, 6), (5,4), (5, 6)}

Question 26.
The figure shows a relationship between the sets P and Write this relation
(i) in set- builder-form
(ii) roster form. What is its domain and range?

Answer:
Given P={5,6,7} and Q={3,4,5}
(i) Set builder form
R= {(x,y):x-y = 2; x∈P,y ∈Q)

(ii) Roster form
R = {(5, 3), (6,4), (7, 5)}
Domain of R = P
Range of R = Q.

Question 27.
Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by {(a, b): a, b ∈A, b is exactly divisible by a},
(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R
Answer :
Given A = { 1, 2, 3,4, 6}
R- {(a, b),a,b∈A,bis exactly divisible by a}

(i) Roster form:
R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4,4), (6, 6)}

(ii) Domain of R = {1, 2, 3,4, 6} = A

(iii) Range of R = {1, 2, 3,4, 6} = A

Question 28.
Determine the domain and range of the relation R .defined by R = {(x, x + 5): x e {0,1, 2,3,4,5}}.
Answer :
Given R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}.
Domain of R = {0, 1, 2, 3,4, 5}
Range of R = {5, 6, 7, 8, 9, 10}

Question 29.
Write the relation R = {(x, x³) : x is a prime number less than 10} in roster form.
Answer :
Given R = {(x, x³): x g {2, 3, 5, 7}}
= {(2, 8), (3, 27), (5, 125), (7, 343)}

Question 30.
Let A – {x, y, z} and B = (1, 2}. Find the number of relations from A to
Answer :
Given n(A) = 3 and n(B) = 2
∴ n (A x B) = 3 x 2 = 6
Number of relations from A to B = 2^{n (A x B)} = 2⁶ = 64

Question 31.
Let R be the relation on Z defined by R = {(a, b): a, b ∈ Z, a – b is an integer}. Find the domain and range of
Answer :
Given: R = {(a, b): a, b ∈Z,a-b is an integer} Domain of R-Z Range of R = Z

Question 32.
Let R be a relation from Q to Q defined by R = {(a, b)\ a,b ∈Q and a – b ∈ Z}. Show that (a, a) g R, for all a ∈Q
(ii) (a, b) ∈ R implies that (b, a) ∈R
(iii) (a, b) ∈ R and (b, c) ∈R implies that (a, c) ∈ R.
Answer :
Given R = {{a, b): a,b∈Q and a-b ∈ Z)
(i) ∀ a ∈ Q, a – a = 0 ∈ Z ⇒ (a, a)∈R

(ii) Let (a, b) ∈ R⇒ a- b ∈ Z
b – a ∈ Z ⇒ (b, a) ∈ R

(iii) Let (a, b) and (b, c) g R ⇒ a – b ∈Z and
b – c ∈ Z
a- c = (a-b) + (b – c) ∈Z
∴ (a, c)∈ R

Question 33.
Let R be a relation from A to A defined by R = {(a, b): a,b ∈ N and a = b²} Are the following true?
(i) (a, a) ∈ R for all a ∈ A
(ii) (a, b) ∈ R, implies (b, a) ∈ R
(iii) (a, b) ∈ R, (b, c) ∈ R, implies (a, c) ∈ R.
Justify your answer in each case.
Answer :
Given R= {(a, b): a,b∈N and a = b²}
= {(1,1), (2,4), (3, 9), (4, 16),…}
(i) (a, a)∈ R for all a ∈ N is not true because (2, 2) ∉ R.
(ii) (a, b) ∈ R implies (b, a) ∈ R is not true, because  (2,4) ∈ R but (4,2) ∉ R
(iii) (a, b) ∈ R, (b, c) ∈ R implies (a,c) ∈ R is not true because (2,4) and (4,16) ∈ R but (2,16) ∉ R.

Question 34.
Define a function.
Answer :
A relation/from a set A to a set B is said to be a function if every element of set A has one and only one image in set B, and
we write f : A → B

Question 35.
Define
(i) a real valued function
(ii) a real function
Answer :
A function which has either R or one of its subsets as its range is called a real valued function. A function f: A → B is said to be a real function if both A and B are subsets of R.

Question 36.
Let N be the set of natural numbers and the relation R be defined on N such that
R = {(x,y):y = 2x,x,y ∈ N} What is the domain, co-domain and range of R? Is this relation a function?
Answer :
Given R = {x, y): y = 2x; x, y ∈ N}
Domain of R = set of natural numbers Co-domain of
R = set of natural number
Range of R = set of even natural numbers
Clearly, every natural number is related to unique image, so this relation is a function.

Question 37.
Examine each of the following relations given below and state in each case, giving reasons whether it is a function or not?
(i) R = {(2,1), (3,1), (4,2)}
(ii) R = {(2,2), (2,4), (3,3), (4,4)}
(iii) R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6,7)}
Answer :
(i) Given R= {(2,1), (3,1), (4, 2)}
Here every element of domain is related to unique element of co-domain, so it is a function.

(ii) Given R = {(2, 2), (2,4), (3, 3), (4, 4)}
Since the element 2 has two images namely 2 and 4, so it is not a function.

(iii) Given R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7)}
Since every element of domain is related to unique

Question 38.
Let N be the set of natural numbers. Define a real valued function f: N → N by
f(x) = 2x + 1. Using this definition, complete the table given below:

X	1	2	3	4	5	6	7
y	F : (1) =	F : ( 2) =	F : (3) =	F : (4) =	F : (5) =	F : (6) =	F : (7) =

Answer:
Given function is f(x) = 2x + 1.

X	1	2	3	4	5	6	7
y	F:(1)=3	F:(2)=5	F:(3)=7	F:(4)=9	F:(5)=11	F:(6)=13	F:(7)=15

Question 39.
Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
(i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}
(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14,7)}
(iii) {(1,3), (1,5), (2,5)}.
Answer :
(i) Clearly, every element of domain is related to unique element of co-domain, so it is a function.
Domain = {2, 5, 8, 11, 14, 17}
Range = {1}

(ii) Clearly, every element of domain is related to unique element of co-domain, so it is a function.
Domain = {2,4, 6, 8,10,12,14}
Range = {1,2, 3,4, 5, 6,7}

(iii) 1 is related to two elements of co-domain, namely 3 and 5, so it is not a function.

Question 40.
Let A={1,2,3,4), B={1,5,9,11,15,16} and f = {(1,5),(2,9), (3,1), (4,5),(2, 11)). Are the
following true?
(i) f is a relation from A to B
(ii) f is a function from A to B. Justify your answer in each case.
Answer :
(i) Every element off is an element of A x B, so f is a relation.
(ii) ‘f’ is not a function the element 2 has two images.

Question 41.
Let f be the subset of Z x Z defined by f ={(ab,a +b):a,b ∈z) Is f a function from Z to Z? Justify your answer.
Answer :
Given f={(ab,a+b):a,b∈Z)
If a = 1 and b = 4 = ab = 4 and a+b=5
∴ (4,5) ∈ fIf a=2 and b=2⇒ab=4 and a+b=4
∴ (4,4) ∈ f∴ The element 4 has two images, so f is not a function.

Question 42.
The relation f is defined by
\( f(x)=\left\{\begin{array}{ll}{x^{2},} & {0 \leq x \leq 3} \\ {3 x,} & {3 \leq x \leq 10}\end{array}\right.\)
The relation g is defined by
\( g(x)=\left\{\begin{array}{ll}{x^{2},} & {0 \leq x \leq 2} \\ {3 x,} & {2 \leq x \leq 10}\end{array}\right.\) .
Show that/is a function and g is not a function.
Answer :
Since f(x) is unique for 0 ≤ x ≤ 10.
f(x) is a function. g(2) = 2² =4 and g(2) = 3(2) = 6
∴ z has two images under g.
∴ g is not a function.

Question 43.
Let f= {(1,1),(2,3),(0,-1),(-1,-3)} be a linear function from Z into Z. Find f(x) ).
Answer :
Since/is a linear function.
∴f{x) = ax + b
But (0,-1) ∈ f . f(0) ∴ a(0) + b ⇒ -1 = b
Similarly, (1,1) ∈ f ∴ f(1) = a(1) +a(1)+b
⇒ 1=a+b ∴ a=2 ∴ f(x) = 2x -1

Question 44.
A function f is defined by f(x) = 2x-5. Write down the values of (i) f(0) (ii) f(17) (iii) f(-3).
Answer :
Given: f(x) = 2x-5

• f(0) = 2(0) – 5 = -5
• f(17) = 2(17)-5 = 29
• f(-3) = 2(-3) – 5 = -11

Question 45.
The function ‘t’ which maps temperature in degree Celsius into temperature in degree. Fahrenheit is defined by \( t(c)=\frac{9 c}{5}+32 \). Find
(i) t (0)
(ii) t (28)
(iii) t (-10)
(iv) The value of c, when t(c) – 212
Answer:

Question 46.
\(\text { If } f(x)=x^{2}, \text { find } \frac{f(1 \cdot 1)-f(1)}{1 \cdot 1-1}\)
Answer:

Question 47.
Find the range of each of the following functions:
(i) f{x) = 2-3x, x∈R,x>0
(ii) f(x) = x² + 2, x is a real number
(iii) f(x) = x, x is a real number.
Answer :
(i) Given f(x) = 2-3x, x∈R,x>0
For x > 0,f(x) = 2 – 3x < 2
∴ Range of f= (-∞, 2)

(ii) Given f(x) = x² + 2, x ∈ R
For x ∈ R, f(x) = x² + 2 ≥ 2
Range of f = [2, ∞)

(iii) Given f(x) = x, x∈ R For r∈ E, f(x) = x∈R
Range of f = R

Question 48.
Let A = (9, 10, 11, 12, 13} and let f:A→N be defined by f(n) = the highest prime factor of n. Find the range of f.
Answer :
Given f(n) = the highest prime factor of n.
f(9) = the highest prime factor of 9 = 3
f(10) = the highest prime factor of 10 = 5
f(11) = the highest prime factor of 11 = 11
f(12) = the highest prime factor of 12 = 3
f(13) = the highest prime factor of 13 = 13
Range of f ={3,5,11,13}

Question 49.
\(\text { Let } f=\left\{\left(x, \frac{x^{2}}{1+x^{2}}\right): x \in \mathbb{R}\right\}\) be a function from R to R .Determine the range of f.
Answer:

Question 50.
Find the domain of the function \(f(x)=\frac{x^{2}+3 x+5}{x^{2}-5 x+4}\)
Answer:
Given \(f(x)=\frac{x^{2}+3 x+5}{x^{2}-5 x+4}\)
f(x) is defined for all real numbers except x² – 5x + 4 = 0
But x² -5x + 4 = (x-4) (x-1)
Domain of f= R-{1,4}

Question 51.
Find the domain of the function
\( f(x)=\frac{x^{2}+2 x+1}{x^{2}-8 x+12}\)
Answer:
\(f(x)=\frac{x^{2}+2 x+1}{x^{2}-8 x+12} \)
is not defined when
x² – 8+12 = 0.
∴ (x-6)(x-2) = 0
∴ Domain of f = R-{2,6}

Question 52.
Find the domain and range of the following real functions:
(i) \( f(x)=-|x| \)
(ii) \(f(x)=\sqrt{9-x^{2}}\)
(iii) \(f(x)=\sqrt{x-1}\)
(iv) \(f(x)=|x-1|\)
Answer:

Remark: Operations of functions:
(i) Addition of two real functions:
Let f: X → R and g : X → R Then.
(f + g):X → R; (f + g)(x) = f(x) + g(x) for all x ∈ X

(ii) Difference of two real functions:
Let f : X → R and g : X → R Then.
(f – g): X → R; (f- g)(x) = f(x) – g(x) for all x ∈ X

(iii) Scalar multiplication of a function:
Let f: X → R and let ‘a’ he a scalar. Then,
(αf):X → R; (fg)(x) = f(x) for all x ∈ X

(iv) Multiplication of two real functions:
Let f: X → R and g : X → R .Then
(fg): X → R; (fg)(x)= f(x)g(x), for all x ∈ X

(v) Quotient of two real functions:
Let f: X → R and  g : X → R for all x for which g(x) ≠ 0. Then
\( \left(\frac{f}{g}\right): X \rightarrow \mathbb{R} ;\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)} \)

Question 53.
Let f(x) = x² and g(x) = 2x + 1 be two real functions. Find (f+g)(x),(f-g)(x)
\((f g)(x),\left(\frac{f}{g}\right)(x)\)
Answer:

Question 54.
Let \( f(x)=\sqrt{x} \text { and } g(x)=x \) be two functions defined over the set of non-negative real numbers. Find (f+ g)(x), (f – g)(x) \((f g)(x) \text { and }\left(\frac{f}{g}\right)(x)\)
Answer:

Question 55.
Let f,g: R→R be defined, respectively by f(x) = x + 1,g(x) = 2x-3- Find f + g,f-g and \( \frac{f}{g} \)
Answer:
Given: f(x) = x + 1,g(x) = 2x-3
(f + g)(x) = f (x) + g(x) = x + 1 + 2x-3 = 3x-2
(f – g)(x) = fix) – g(x) = x + 1-2x + 3 = -x + 4
\( \left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x+1}{2 x-3}, x \neq \frac{3}{2}\)

Question 56.
Define an identity function and draw its graph also find its domain and range.
Answer :
The function f: M → R; f(x) = x for a II x∈R is called an identity function on R.

Question 57.
Define a constant function and draw its graph also find its domain and range.
Answer :
Let c be a fixed real number. Then, the function f : R→R, f(x) = c for all x∈R is called the constant function.

f(x) is defined for all real number,
∴ Domain = M
Range = { c }

Question 58.
Define a polynomial function.
Answer :
A function f : M → R is said to be polynomial function if for each x in IR,
y = f (x) = a₀ + a_xx + a₂x² + ……………. + a_nxⁿ
where n is a non-negative integer and a₀,a₁,a_{2………………….}a_n ∈ R .

Question 59.
Draw the graph of the function f (x) = x² and write its domain and range.
Answer:
Given function: f(x)= x²

Domain = R
Range = set of non-negative reals.

Question 60.
Draw the graph of the function f: R → R defined by f(x) = x³ Find its domain and range.
Answer :
Let f : R → R: f(x) = x³, ∀ ∈ R
Then, domain of f = R and range of f = R . we have

Question 61.
Define a rational function
Answer:
The functions of the type \(\frac{f(x)}{g(x)}\) where f(x) g(x) and g(x) are polynomial functions of x, defined in a domain, where g(x) ≠ 0

Question 62.
Let f : R – {0} → R defined by \( f(x)=\frac{1}{x}, \forall x \in \mathbb{R}-\{0\}\) . Find its domain and x range. Also, draw its graph.
Answer :
Given function is f : R – {0} → R defined by \( f(x)=\frac{1}{x} \)
∴ Domain = R-{0} and range =R-{0}

X	-4	-2	-1	-0-5	-0-25	0-25	0-5	1	2
f(x)=1/x	-0-25	-0-5	-1	-2	-4	4	2	1	0-5

Question 63.
Define a modulus function. Find its domain and range. Also, draw its graph.
Answer :
Let f: R → R defined by f(x) =1 x I, for each x ∈ R, is called modulus function.
\( \text { i.e., } f(x)=|x|=\left\{\begin{array}{ll}{x,} & {\text { if } x \geq 0} \\ {-x,} & {\text { if } x<0}\end{array}\right.\)

Domain = R
Range = set of non negative real numbers

Question 64.
Define Signum function. Draw its graph and find its domain and range.
Answer :
The function f: R → R defined by
\( f(x)=\left\{\begin{array}{lll}{1,} & {\text { if }} & {x>0} \\ {0,} & {\text { if }} & {x=0} \\ {-1,} & {\text { if }} & {x<0}\end{array}\right.\) is called signum function
We have

Domain = R
Range = {-1,0,1}

Question 65.
Define a greatest integer function. Draw its graph and find its domain and range.
Answer:

The function f : R → R define by f(x) = [x], x∈ R assumes the value of the greatest integer, less than or equal to x. Such a’ function is called the greatest integer function or step function. We have
[x] = -2 for – 2 ≤ x < -1
[x] = -1 for -1≤x<0
[x] = 0 for 0≤x≤1
[x] = 1 for 1 ≤ x < 2
[x] = 2 for 2 ≤ x < 3.
Hence, domain of f = R and range = Z.

Question 66.
Define a linear function.
Answer :
The function f : R → R defined by f(x) = mx + c, x ∈ R is called linear function, where m and c are constant.

Question 67.
Let R be the set of real numbers. Define the real function f : R →R by f(x) = x + 10 and sketch the graph of this function.
Answer :
Given f(x) = x +10
We have

Question 68.
The function f is defined by
\( f(x)=\left\{\begin{array}{cl}{1-x,} & {x<0} \\ {1,} & {x=0} \\ {x+1,} & {x>0}\end{array}\right.\).
Draw the graph of f(x)
Answer:
We have

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Karnataka 1st PUC Computer Science Question Bank Chapter 7 Introduction to C++

1st PUC Computer Science Introduction to C++ One Mark Questions and Answers

Question 1.
Who is the creator of the C++ programming language?
Answer:
The creator of C++ is Bjarne Stroustrup.

Question 2.
What is C++?
Answer:
C++is an object-oriented programming language.

Question 3.
What was the earlier name of C++?
Answer:
The earlier name of C++ was ‘C with classes’.

Question 4.
Who gave the name ‘C++’?
Answer:
The name C++ was given by Rick Mascitti.

Question 5.
Name any two characteristics of C++.
Answer:
Object-oriented programming and portability are two characteristics of C++.

Question 6.
Describe tokens.
Answer:
The smallest individual units in a program are known as tokens, on lexical units.

Question 7.
Mention a few tokens.
Answer:
A few tokens are keywords, identifiers, constants, strings, and operators.

Question 8.
What is a keyword?
Answer:
All keywords (reserved words) are basically the sequences of characters that have one or more fixed meanings.

Question 9.
What are identifiers?
Answer:
Identifiers are names given to the program elements such as variables, arrays, and functions.

Question 10.
What are constants?
Answer:
The program elements, which not change during the execution of a program, are known as constants.

Question 11.
Define a character set.
Answer:
It is a set of symbols that a programming language identifies and is used in writing data and instructions in a programming language.

Question 12.
What is an integer constant?
Answer:
Integer constants are whole numbers without any fractional part.

Question 13.
What are floating-point constants?
Answer:
Floating-point constants are numeric values that contain a decimal point and can also contain exponents.

Question 14.
Define octal constant.
Answer:
It consists of a sequence of digits starting with 0 (zero).

Question 15.
Define hexadecimal constants.
Answer:
It consists of a sequence of digits from 0 to 9 and A, B, C, D, E, F alphabet symbols that represent decimal numbers 10, 11, 12, 13,14, and 15 respectively, preceded by ox or OX.

Question 16.
What are the character constants?
Answer:
A character constant in C++ must contain one or more characters and must be enclosed in single quotation marks.

Question 17.
What are punctuators?
Answer:
Punctuators are symbols other than alphabets and numbers that are used in C++.

Question 18.
Define string constants.
Answer:
A sequence of characters enclosed within double quotes is called a string constant or literal.

Question 19.
What is the operator?
Answer:
An operator is a symbol that tells the compiler to performs specific operations and give a value as the result.

Question 20.
What are the arithmetic operators?
Answer:
The symbols that perform addition, subtraction, multiplication, modulus, and division are called arithmetic operators.

Question 21.
Give the difference between / and % arithmetic operators.
Answer:

• The arithmetic operator / performs division and gives out the quotient as a result.
• The arithmetic operator % performs division and gives out the remainder as a result.

Question 22.
What is an operand?
Answer:
The operand is a data item on which the operator performs some activity.

Question 23.
What are unary operators?
Answer:
An operator that takes only one operand to perform some operation is called a unary operator.

Question 24.
What are binary operators?
Answer:
The operators which take two operands to perform some operation is called a binary operator.

Question 25.
What are the relational operators?
Answer:
The operators which perform an operation of the relation between two operands are called relational operators.

Question 26.
What are the logical operators?
Answer:
The operator which perform combine or negate the expressions that contain relational operators are called logical operators.

Question 27.
What is the function of the bitwise operator?
Answer:
Bit manipulation operators manipulate individual bits within a variable. Bitwise operators modify variables considering the bit patterns that represent the values they store.

Question 28.
What is meant by shorthands in C++?
Answer:
In C++ short hands means writing certain types of assignment statements in a simplified manner.

Question 29.
What is a ternary operator?
Answer:
The operator that operates on three or more operands is called a ternary operator.

Question 30.
What is an expression?
Answer:
An expression is a combination of constants, variables, operators and function calls which produces a particular value to be used in some other context.

Question 31.
What is a statement?
Answer:
The statement is an instruction to the computer telling it what to do for instance assigning an expression value to a variable because it produces an instruction tells the computer to assign a value to something.

Question 32.
Give an example for assignment operators.
Answer:
Example for assignment operators is a = 20;

Question 33.
Give an example for sizeof operator.
Answer:
If k is integer variable, the size of (k) returns 2.

Question 34.
What is operator precedence?
Answer:
The order in which the arithmetic operators (+,-,*,/,%) are used in a given expression is called the order of precedence.

Question 35.
What is typecast?
Answer:
Typecasting is making a variable of one type, such as an int act like another type, a char, for anyone single operation.

Question 36.
Mention the types of typecasting.
Answer:
The two types of typecasting are implicit conversion and explicit conversion.

Question 37.
What is the use of iostream.h in C++?
Answer:
iostream stands for input/output stream. It is a header file which we include in our programs, to perform basic input-output operations.

Question 38.
Mention any two ctype.h functions.
Answer:
Toupper() and tolower() are the two examples for ctype.h functions.

Question 39.
Mention any two string.h functions.
Answer:
Strupr() and strlwr()

Question 40.
Mention any two functions of stdio.h
Answer:
Printf() and scanf() are two functions of stdio.h

Question 41.
Mention any two functions of stdlib.h
Answer:
Atoi() and itoa() are two functions for stdlib.h

Question 42.
What is a preprocessor directive?
Answer:
A preprocessor directive is a command that is considered for execution before the processor executes the program.

1st PUC Computer Science Introduction to C++ Two/Three Marks Questions and Answers

Question 1.
What is C++?
Answer:
C++ is an object-oriented programming language. It is also known as C with classes. It is a combination of C and Simula 67.
It was invented by Bjarne Stroustrup at AT &T Bell Lab in New Jersey, USA. in the early 1980’s.

Question 2.
Write any two characteristics of C++.
Answer:
Object-oriented programming and portability are important two characteristics of C++.

Question 3.
What is meant by a character set? Give an example.
Answer:
Character set is a set of valid characters that a language can recognize.

• Letters A – Z, a – z
• Digits 0 – 9
• Special Characters + – * / ˆ \ () [] {} = != <> ‘ “ $ , ; : % ! & ? _ # <= >= @

Question 4.
What are the different tokens available in C++?
Answer:
A token is a group of characters that logically belong together. C++ uses the following types of tokens. Keywords, Identifiers, Literals, Punctuators, and Operators.

Question 5.
What is the function of the keywords? Write any four keywords.
Answer:
These are some reserved words in C++ which have a predefined meaning to the compiler, called keywords. Some commonly used keywords are:
int, float, char, and case

Question 6.
Write any two rules for naming the identifier.
Answer:
Two rules for the formation of an identifier are that it can consist of alphabets, digits and/or underscores and it must not start with a digit.

Question 7.
Mention different types of literals available in C++.
Answer:
The following types of literals are available in C++.

• Integer-constants
• Character-constants
• Floating-constants
• Strings-constants

Question 8.
Write any four punctuators of C++.
Answer:
Four punctuator of C++are (),{}, ; , and :

Question 9.
How many types of operators are available in C++? Mention any two.
Answer:
C++ provides six types of operators. Arithmetical operators and Relational operators are two such operators.

Question 10.
Write the use of arithmetical operators.
Answer:
Arithmetical operators +, -, *, /, and % are used to perform an arithmetic (numeric) operation.

Question 11.
Why is a relational operator used?
Answer:
Relational operators are used to test the relation between two values. A relational expression returns zero when the relation is false and a non-zero when it is true.

Question 12.
Why are logical operators are used? Give an example.
Answer:
Logical operators are used to combine one or more relational expressions. Examples of logical operators are || (OR), && (AND).

Question 13.
How many unary operators are available in C++? Give an example.
Answer:
C++ provides two unary operators for which only one variable is required.
For Example a = – 50;

Question 14.
Explain a ternary operator with an example.
Answer:
The operator which uses three or more expressions is called a ternary operator. It is also called a conditional operator. For example, conditional Expression? expression1: expression2; if the conditional Expression is true, expression 1 executes, otherwise if the conditional Expression is false, expression 2 executes.

Question 15.
Explain shorthand operators with an example.
Answer:
In C++ shorthand operators mean writing certain type of assignment statements in a simplified manner. For example, if x = 5 and x + = 10 means x = x +10 i.e., x = 5 + 10 = 15.

Question 16.
Mention different bitwise operators.
Answer:

Question 17.
Mention any two special operators and their meaning.
Answer:
Sizeof() – it returns the size of a variable.
. (dot) – member operator used to reference member of a structure.

Question 18.
What is the use of assignment operator? Give an example.
Answer:
The assignment operator ‘=’ is used for assigning a variable to a value. This operator takes the expression on its right-hand-side and places it into the variable on its left-hand-side.
For example: m = 5;

Question 19.
What do you mean by the precedence of operators? Give an example.
Answer:
The order in which the Arithmetic operators (+,-,*,/,%) are used in a given expression is called the order of precedence. The following table shows the order of precedence.

Question 20.
What is implicit conversion? Give an example.
Answer:
When two operands of different types are encountered in the same expression, the lower type variable is converted to the higher type variable which is called implicit conversion.
For example, float = int + float

Question 21.
Explain the use of explicit conversion with an example.
Answer:
This is also called as typecasting. It temporarily changes a variable data type from its declared data type to a new one.
For example, T_Pay = double (salary) + bonus;

Question 22.
Write a simple C++program.
Answer:
#include
main()
{
cout<< “Hello World”; // prints Hello World
return 0;
}

Question 23.
How are comments included in C++? Give an example.
Answer:
Comments are included using//
For example,
main()
{
cout<< “Hello World”; // prints Hello World
return 0;
}

1st PUC Computer Science Introduction to C++ Five Marks Questions and Answers

Question 1.
Write a note on C++.
Answer:

• C++ is a typed, compiled, general-purpose, case-sensitive, free-form programming language that supports procedural, object-oriented, and generic programming.
• C++ is regarded as a middle-level language, as it comprises a combination of both high-level and low-level language features.
• C++ was developed by Bjarne Stroustrup starting in 1979 at Bell Labs in Murray Hill, New Jersey as an enhancement to the C language and was originally named C with Classes but later it was renamed C++ in 1983. by Rick Masatti.
• C++ is a superset of C, and virtually, any legal C program is a legal C++ program.

Question 2.
Write the applications of C++.
Answer:

• It is a versatile language for handling very large programs
• It is suitable for virtually any programming task, including development of editors, compilers, databases, communication systems and any complex real life application systems
• It allows us to create hierarchy-related objects, so we can build special object-oriented libraries which can be used later by other programmers.
• While C++ is able to map the real-world problem properly, the C part of C++ gives the language the ability to get close to the machine-level details.
• C++ programs are easily maintainable and expandable.

Question 3.
Write a short note on the keywords of C++.
Answer:
These are some reserved words in C++ which have some predefined meanings to the compiler and are called keywords. Some commonly used Keywords are given below:

Question 4.
Write the rules to be followed, while naming the identifier.
Answer:
The identifier is a sequence of characters taken from the C++ character set. The rules for the formation of an identifier are:

• It can consist of alphabets, digits and/or underscores.
• It must not start with a digit
• C++ is case sensitive, that is upper and lower case letters are considered as different from each other.
• It should not be a reserved keyword.

Question 5.
Write a short note on literals.
Answer:
The five arithmetical operations supported by the C++ language are:
+ addition
– subtraction
* multiplication
/ division
% modulus
Operations of addition, subtraction, multiplication, and division literally correspond with their respective mathematical operators.
For example,
sum = 5 + 10;
difference = 10 – 5;
product = 2 * 5;
quotient = 10 / 2;
remainder = 11 % 3;

Question 6.
Write a short note on the precedence of operators.
Answer:
The following table shows the order of precedence.

The following table shows the precedence of operators.

Question 7.
Explain the Structure of a C++ program (with an example)
Answer:
#include
// main()
int main()
{
cout<< “Hello World”; // prints Hello World
return 0;
}
The various parts of the above program:

• Headers, which contain information that is either necessary or useful to the program. For this program, the header is needed.
• The next line // main() is where the program execution begins. It is a single-line comment available in C++. Single-line comments begin with // and stop at the end of the line.
• The line int main() is the main function where program execution begins.
• The pair of {} indicates the body of the main function.
• The next line cout<< “Hello World.”; causes the message “Hello World” to be displayed on the screen.
• The next line return 0; terminates main( )function and causes it to return the value 0 to the calling process.

You can Download Chapter 9 Elements of Probability Theory Questions and Answers, Notes, 1st PUC Statistics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Statistics Question Bank Chapter 9 Elements of Probability Theory

1st PUC Elements of Probability Theory Question and Answers

Question 1.
What do you mean by ‘probability’?
Answer:
Measurement of chances is called ‘Probability.’

Question 2.
Define Random experiment
Answer:
An experiment in which the outcomes of that experiment cannot be predictable, is called Random experiment.Eg., tossing a coin, throwing a die.

Question 3.
Define Trial and Event
Answer:
Conducting an experiment is called Trial, and getting an outcome is an Event.
Eg. Tossing a coin is a trial and getting outcome Head Or Tail are Events.

Question 4.
What is sample space?
Answer:
The set of all possible outcomes of a random experiment. Denoted by S
S = {H,T}, S = {1,2,3,4,5,6}

Question 5.
Define Exhaustive events
Answer:
Total number of all possible outcomes of a random experiment is called Exhaustive events.denoted by ‘n’. in the above examples, n=2 & n=6.

Question 6.
Define Favourable events
Answer:
The number of outcomes which entail as an outcome, or number of outcomes which favours the happening of an event, denoted by ‘m’
Eg., in the above example to get H, m = 1, & to get ‘Even’ no. S={2,4,6}, m=3.

Question 7.
What are Mutually Exclusive events?
Answer:
Two or more events are said to be mutually exclusive, if no two events can occur at a time in the same trial. Eg, When a coin is tossed getting Head in a trial is mutually exclusive of getting Tail in the same trial.

Question 8.
What are Equally likely events?
Answer:
Outcomes of a random experiment are called equally likely if they have equal chance of occurrence.Eg. When a die is thrown, all outcomes to get 1,2,.. .6 are having equal chance to get as an outcome.

Question 9.
What are Independent events?
Answer:
The outcomes of a random experiment are said to be independent if the happening of an Event in the trial is independent of getting the same (or other) Event in the subsequent trials.Eg., Getting Head in the first Toss is independent of getting Head in the second toss of a coin.

Question 10.
What are Complementary events
Answer:
Let S be the sample space, A be an event, then A‘ is called complementary event of A is it contains elements of S, but it does’t contain any elements of event
Eg., S = {1,2,3,4,5,6} if A = {1,2}, then A = {3,4,5,6}

Question 11.
Give the Classical/Priori/Mathematical definition of probability.
Answer:
Let a random experiment have n possible outcomes which are equally likely, mutually exclusive and exhaustive. Let m of these outcomes be favourable to an event A. Then, probability of A

Question 12.
Write down Posterior/Statistical definition of Probability
Answer:
Let a random experiment be repeated n times essentially under identical conditions. Let m of these repetitions results in the occurrence of an event A. Then, the probability of event A is the limiting value of the ratio m/n as n increases indefinitely.

Here, it is assumed that a unique limit exists.

Question 13.
Write down the Axiomatic definition of Probability.
Answer:
Let A and B be the events of a sample space S. Let P(A) and P(B) are the real numbers (probabilities) assigned to these events. Then, P(A) is the probability of A if, the following axioms are satisfied.
Axiom (i) : P(A) ≥ 0 (non-negativity condition)
Axiom (ii) : P(S) = 1. ‘S’ being the sure event.
Axiom(iii): For any two disjoint events A and B,P(A + B) = P(A) + P(B)

Question 14.
Prove that 0 ≤ P[A] ≤ 1
Answer:
statement: 0 ≤ P(A) ≤ 1. That is, P(A) is the value between 0 and 1
(The limits of probability are 0 and 1).
Proof: Here, by definition of probability of an event A is P(A) = \(\frac { m }{ n }\)
The least and highest possible values of m are zero and n.
That is, 0 ≤ m ≤ n dividing by n
\(\frac{\mathrm{o}}{\mathrm{n}} \leq \frac{\mathrm{m}}{\mathrm{n}} \leq \frac{\mathrm{n}}{\mathrm{n}}\)
i.e., 0 ≤ P(A) ≤ 1

Question 15.
Show P[A]+P[A¹]=1
Answer:
P(A) + P(A²) = 1. That is, the sum of probabilities of complementary events is 1.
Proof: We knowthat, Out of ‘n’ outcomes, if ‘m’ outcomes are favourableto event A, then the remaining (n-m) outcomes are favourable to event A¹.
By definition P(A) = \(\frac { m }{ n }\) and P(A²) = \(\frac { n-m }{ n }\)
\(\therefore \mathrm{P}(\mathrm{A})+\mathrm{P}\left(\mathrm{A}^{2}\right)=\frac{\mathrm{m}}{\mathrm{n}}+\frac{\mathrm{n}-\mathrm{m}}{\mathrm{n}}=\frac{\mathrm{m}+\mathrm{n}-\mathrm{m}}{\mathrm{n}}=\frac{\mathrm{n}}{\mathrm{n}}=1\)

Question 16.
StatetheAdditiontheoremforanytwoevents.
Answer:
Statement: Let A and B be two events with respective probabilities P(A) and P(B). Then, the probability of occurrence of at least one of these two events is
P (A ∪ B) = P(A) + P(B) – P(A ∩ B) or
P(A+B) = P(A) + P(B) – P(AB)

Question 17.
State the Addition theorem for any two mutually exclusive events.
Answer:
Let A and B be two mutually exclusive events with respective probabilities P (A) and P(B). Then, the probability of occurrence of at least one of these two events is
P(A ∪ B) = P(A) + P(B) or
P(A+B) = P(A) + P(B)

Question 18.
State the Multiplication theorem for any two Independent events.
Answer:
Statement: Let A and B be two independent events with respective probabilities P(A) and P(B). Then, the probability of simultaneous occurrence of Aand B is
P(A ∩ B)= P(A) × P(B).

Question 19.
Define conditional probability.
Answer:
The probability of occurrence of one event under the condition that another event has already occurred is known as conditional probability.
The probability that B will occur under the condition that Alias already occurred is denoted by

Similarly, The probability that A will occur under the condition that B has already occurred is denoted by P(A/B).

Question 20.
If P[A]=l/4, what is P[A’]?
Answer:
P(A) + P(A²) = 1 or P(A²) = 1- P(A);
P(A²) = 1-1/4 ; P(A²) = 3/4

Question 21.
Find the probability of getting head when a coin is tossed.
Answer:
when a coin is tossed the sample space S ={Head, Tail};
n(s) = 2; n(H)=l;

Question 22.
P(H) = \(\frac { 1 }{ 2 }\). If P[A]=2/5 and P[B]=l/5. Find P[A∪B], when A and B are Mutually exclusive events
Answer:
we know that, P(A ∪B) = P(A) + P(B);
P(A ∪B) = 2/5 + 1/5
∴ P(A ∪ B) = 3/5

Question 23.
P[A]=3/4, P[B]=l/2 and P[A∪B]=l/4. Find P[A)”B] and P[A/B]
Answer:
we know that; P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
1/4 = 3/4 + 1/2 – P(A ∩ B)
P(A ∩ B) = 3/4 + 1/2 – 1/4
∴ P(A∩B) = 1/2 = 0.5

Question 24.

Answer:
we know that, P (A ∪ B) =P(A)+P(B)-P(A∩B)
= 1/3 + 2/3 – 3/4
=0.33 + 0.67 – 0.75
P(A∪B) = 0.25

Question 25.
IfP[ABl=l/3 and P[B]=2/3, find P[A/B]
Answer:
we know that P(B/A) – p(A) =

Question 26.
IfP[A ∩ B]=l/2, and P[B]=2/3, find P[A/B]
Answer:
We know that P(A/B) =

Question 27.
If P[B]=3/5 and P[A/B]=l/3 find P[B ∩ A]
Answer:

Question 28.
IfP[A]=l/2, P[B]=l/3 and P(A ∩ B)=l/4, find P[B/A] and P[AUB].
Answer:
we know that; P(A∪B) = P(A) + P(B) – P(A ∩ B)
P(A∪B)= 1/2+ 1/3 – 1/4 = 0.5833

Question 29.
IfP[A∪B]=l/3, P[A ∩ B]=l/12 and P[A]=l/6, find P[B].
Answer:
P(A∪B) = P(A) + P(B)- P(A∩B)
1/3 = 1/6 + P(B) – 1/12;  1/3 – 1/6 + 1/12 = P(B)
∴ P(B) = 0.25

Question 30.
IfP[A]=0.5, P[B]=0.1 and P[A ∪B]=0.7, find P[A/B].
Answer:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
0. 7 =0.5 + 0.1 -P(A ∩B);  ∴ P(A ∩B) = 0.7 – 0.5 – 0.1 =0.1

Question 31.
If the sample space S={1,2,3,4,5,6} and A={3,4}, find P[A}
Answer:
Here n= 6 and m = 2 ; ∴ P(A) = m/n = 2/6 =0.33

Question 32.
If P[A]=l/8, P[B]=l/6 and P[A∪B]=1/10, find P[A ∩ B].
Answer:
P(A∪B) = P(A) + P(B) – P(A∩B)
1/10= 1/8+ l/16 – P(A∩B)
P(A∩B)= 1/8 + 1/16 – 1/10 = 0.0875

Question 33.
If P[B/A]=2/5 and P[A]=5/6, find P(B ∩ A)
Answer:

Question 34.
A bag contains 4 white, 2 pink and 3 Black balls. A ball is drawn at random, What is the probability that it is
1. a white
2. a pink
3. either pink or black ball.
Answer:
Here total bal Is 4(W) + 3 (B) + 2(P) = 9 and n = 9
1. m Favorable ways to draw a white ball = 4
P(drawing a white ball) = P(B) = \(\frac { m }{ n }\) = \(\frac { 4 }{ 9 }\)

2. m = pink balls = 2
P(pink) = \(\frac { 2 }{ 9 }\)

3. m = either pink or black ball = 2 + 3 = 5
∴ P(pink or black) = \(\frac { 5 }{ 9 }\)

Question 35.
A card is drawn from a wll su filed pack of placing cards. Find the probability that the card draw is
(a) a King
(b) a club
(c) a Red
(d) a King or Queen
(e) an Ace or Spade.
Answer:
n=Total cards=52
(a) m = No. of Kings = 4
∴ P (drawing an king card) = \(\frac { m }{ n }\)= \(\frac { 4 }{ 52 }\)

(b) m = clubs = 13
P(club) = \(\frac { 13 }{ 52 }\)

(c) m = Red cards = 13(Hearts) + 13(Diamonds) = 26
P(Red card) = \(\frac { 26 }{ 52 }\)

(d) m : King or Queen = 4 + 4 = 8
P(K or Q) = \(\frac { 8 }{ 52 }\)

(e) m = An ace or a club card = 4+13-1 = 16
(Since there is a card, which is a Ace and a Club)
P(Ace or Club) = \(\frac { 16 }{ 52 }\)

Question 36.
Adie is thrown. Find the probability that the face of a die results in
1. multiple of 2
2. multiple of 3
3. an odd number.
Answer:
S= {1,1, 3,4,5,6}: n = 6
1. m = multiple 2 : (2, 4, 6) = 3
P(multiple of 2) = \(\frac{m}{n}=\frac{3}{6}=\frac{1}{2}\)

2. m = multiple of 3 : (3, 6) = 2
P(multiple of 3) = \(\frac{2}{6}=\frac{1}{3}\)

3. m = odd number : (1, 3, 5) = 3
P(odd no.) = \(\frac{3}{6}=\frac{1}{2}\)

Question 37.
Two coins are tossed, find the probabiliti getting
1. both coins heads
2. exactly one head
3. no heads
4. atleast one head.
Answer:
S = {HH, HT, TH, TT} : n = 22 = 4
1. m = both heads : (HH) =1
P(2H) = \(\frac{\mathrm{m}}{\mathrm{n}}=\frac{1}{4}\)

2. m = exactly one H: (HT, TH) = 2
P(one heads) = \(\frac{2}{4}=\frac{1}{2}\)

3. m : no heads : (TT) =1
P(no head) = p(2T) = \(\frac { 1 }{ 4 }\)

4. m = atleast one H : (HT, TH, HH) = 3
P(atleast one H) = \(\frac { 3 }{ 4 }\)

Question 38.
Three coins are tossed, find the probability of getting
1. Exactly one Head
2. Exactly 2H
3. No ehads/only tails
4. Atleast one Tail.
Answer:
S = {HHH, HHT, HTT, TTT, TTH, THH, THT, HTH} : n = 23 = 8
1. m = Getting exactly one Head : (HTT, TTH, THT) = 3 .
P(One H) = \(\frac{\mathrm{m}}{\mathrm{n}}=\frac{3}{8}\)

2. m = Exactly 2 H : (HHT, THH, HTH) = 3
P(2H) = \(\frac { 3 }{ 8 }\)

3. m = no heads/ Tails only : (TTT) =1
P(No heads/Tails only) = \(\frac{m}{n}=\frac{1}{8}\)

4. m = Atleast one Tail: Except (HHH) = 7
P(Atleast one Tail) = \(\frac { 7 }{ 8 }\) OR
P(A) = p(getting atleast one T) = 1 – P (Not getting No Tail) .
= l- P(A’)=l – \(\frac { 1 }{ 8 }\) = \(\frac { 7 }{ 8 }\)

Question 39.
Two dice are rolled once. Find the probability of getting
1. both with number 5
2. first die with no.
3. sum on both the dice is 7
4. sum is 10 or more
5. both with same no.
Answer:
n = 6² = 36
1. m = both with no. 5 = (5, 5) = 1
P(both 5) = \(\frac{\mathrm{m}}{\mathrm{n}}=\frac{1}{36}\)

2. m = first dice with no. 1 : (1, 1) (1, 2) (1, 3) ……………………. (1, 6) = 6
P(First die with no. 1) = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)

3. m = sum is 7 : (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) = 6
P(sum is 7) = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)

4. m = sum is 10 or more : sum (10) or (11) or (12)
(4, 6), (5, 5), (6, 4) + (6, 5), (5, 6) + (6, 6) = 6
P(sum is 10 or more) = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)

5. m = Both with same number: (1, 1) (2, 2) (3, 3)
P(both dice with same no.) = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)

Question 40.
Two coins are tossed, find the probabiliti getting
1. both coins heads
2. exactly one head
3. no heads
4. atleast one head.
Answer:
S = {HH, HT, TH, TT} : n = 2² = 4
1. m = 2 white marbles = ³C₂ = 3

2. m = 2 Red marbles = ⁵C₂ = 10

3. m = 2 marbles of same colour: (2W) OR (2R)
= ³C₂ OR ⁵C₂ = 3 + 10 = 13 ways

4. m =2 marbles of different colour = 1W and 1R
= ³C₁ × ⁵C₁ = 3 × 5 = 15 ways.

OR P(A) = p(same colour) = 1 – P(A’) = 1 – P (different colour)
∴ P(different colour) = 1 – P(same colour) = \(1-\frac{13}{28}=\frac{15}{28}\)

Question 41.
Two cards are drawn from a peck of 52 playing cards. Find the probability that they are of
1. Kings
2. Clubs
3. Red cards
4. a king and a Ace
5. a Heart and a spade.
Answer:
n = Two cards can bedrawn from 52 cards = ⁵²C₂
n = 1326 ways
1. m = 2 kings from 4 kings can be draw = ⁴C₂ = 6 ways

2. m = 2 clubs can be drawn from 13 = ¹³C₂ = 78

3. m = 2 Red cards can be drawn from 26 = ²⁶C₂ = 325
(i.e., out of 13 diamonds and 13 hearts)

4. m = a king and an Ace can be drawn = ⁴C₁ × ⁴C₁ = 4 × 4 = 16

5. m = a Hearts and a spade can be drawn from 13 hearts and 13 spades
= ¹³C₁ × ¹³C₁ = 13 × 13 = 169

Question 42.
A book rack at home has 4 Novels, 5 magezines and 3 Cockery books. Two books are chosen at raondom. Find the rpboability that the selected books are
1. of the same type
2. of different type
3. only Novels
4. one is a Novel and the ohter is a cokcery.
Answer:
n – (4N + 5m + 3C) = ¹²C₂ =66 :
1. m = 2 books of same type can be drawn
= 2 N or 5 m or 3C = ⁴C₂ + ⁵C₂ + ³C₂
∴ m = 6+ 10 + 3 = 19

Let P(A) = \(\frac { 19 }{ 16 }\)

2. Let P(A’) = p(selecting different type of books)
P(A’)= 1 – P(A) = 1 – \(\frac { 19 }{ 16 }\) = \(\frac { 47 }{ 66 }\)

3. m = selecting z novels = ⁴C₂

4. m = selecting one Novel other cockery
= ⁴C₁ × ³C₁ =4 × 3 = 12

Question 43.
The chance of winning a cricket match by India against Australia is 5/7. Find the probability that India loses against Australia.
Answer:
Let P(A) = p(Indian wins against Australia) = \(\frac { 5 }{ 7 }\)
Then P(A’) = p(India loses agaisnt Australia)
P(A’) = 1 – P(A)
P(A’) = l – \(\frac { 5 }{ 7 }\) = \(\frac { 2 }{ 7 }\)

Question 44.
In a city out of 1200 New born babies in that month, 580 were girls. Find the probability that a new born baby is a girl.
Answer:
Here n = 1200 Total births, m = 580 No. of girls born
P(New born baby is a girl) = \(\frac { m }{ n }\) = \(\frac { 580 }{ 1200 }\) = 0.4833

Question 45.
State and prove addition theorem of propability of two not mutually exclusive events.
Answer:
Statement: Let Aand B be any two events (subsets of sample sapce S) with respective probabilite is P(A) and P(B). Then, the probability of occurrence of atleast one of these events is:
P(A ∪B) = P(A) + P(B) – P(A∩B)
Here the event (AnB) is the simultaneous occurrence of A and B.
Proof: A random experiment results ‘n’ exhaustive out comes, of which’m₁ ‘ outcomes are favorable to event A ‘m₂‘ outcomes are favorable to event B and ‘p’ outcomes are common to both A and B.
Consider the venn diagram
‘

The event of occarrence of atleast one of A or B is (A∪B) has (m₁ + m₂ – P) favorable outcomes

P(A∪B) = P(A) + P(B) – P(A∩B), Hence the proof.

Question 46.
State and prove addition theorem of probability of two mutually exclusive events.
Answer:
Statement: Let A and B be two mutually exclusive events (subsets of sample space S) with respective prababilities P(A) and P(B). Then, the probability of occurrence of atleast one of these events is
P(A∪B) = P(A) + P(B)
Proof: A random experiment results ‘n’ exhaustive outcomes, of which’m’ outcomes are favourable to event A and ‘m₂‘ outcome of are favorable to event B.
Consider the venn diagram.


The event of occurrence of atleast oen of A or B is (A∪B) has (m₁ + m₂) favorable out comes.

P(A∪B) = P(A) + P(B). Hence the proof.

Question 47.
State and prove multiplication theorem of probability for any two dependent events.
Answer:
Statement: If A and B be any two events in evently in a random experiment. then the probability of occurrence of events A and B is,
P(A∩B) = P(A). P(B/A)
Where the events P(B/A) is the conditional probability of occurrence of event B, known that the event Alias already occurrence.
Proof: Suppose, a random experiment results ‘n’ outcomes, of are favorable events A and B and ‘P’ outcomes are favorable to both the events A and B.

Suppose, event A has occurred, which has’m,’ favorable outcomes to A.
If event ‘B’ occurred, it has out’m’ outcomes, ‘P’ outcomes are favorable to B
∴ Probability of happening of an event ‘B’ known that A has already happened is:

Then, the probability of occurrence of events A and B together is :

P(A∩B) = P(A) × P(B/A) from the results (1) & (2)
Hence the proof.
Note: The conditional probabilities can also be extend to prove : P(B∩A) = P(B). P(A/B)

Question 48.
State and prove the multiplication theorem of probabilities for two independent events:
Answer:
Statement: Let A and B be any two independent events with respective probabilities P(A) and P(B). Then, the probability of occurrence of events A and B is P(A∩B) = P(A). P(B)
Proof: Suppose, a random experiment, results ‘n’ outcomes, of which’m’, outcomes are favorable
k to events A and another random experiment results n₂ outcomes of which m₂ outcomes are favorable to event B.

The occurrence of the events A and B together m₁ and m₂ i.e., m₁ × m₂ favorable events out of n₁, and n₂ i.e., n₁ × n₂ outcomes.
∴ Probability of ocurrence of events A and Btogther is.

Hence the proof.

Question 49.
A card are drawn from a pack of playing cards. Find the probability that the card drawn is
1. King or a Red
2. Heart or a Red.
Answer:
n = A card can be drawn from 52 cards in 52 ways.
1. Let A = drawing a King = 4
B = drawing a Red card = 26
Here (A∪B) = King and a Red card = 1
P(A∪B) = P(A) + P(B) – P(A∪B) =\(\frac{4}{52}+\frac{26}{52}-\frac{1}{52}=\frac{29}{52}\)

2. Let A – Card drawn is a Heart = 13
B – Card drawn is a Red = 26
and (A∩B) – Heart and Red cards =13
P(A∪B) = P(A) + P(B) – P(A∩B) = \(\frac{13}{52}+\frac{26}{52}-\frac{13}{52}=\frac{26}{52}\)

Question 50.
A Die is trolled once. Find the probability of getting a face with a multiple 2 or multiple of 3.
Answer:
S = {1,2, 3,4, 5,6} ; n = 6
Let A= Multiple of 2 : (2, 4, 6) = 3
B-Multiple of 3 : (3,6) = 3 and (A∩B) common in A and B : (6)= 1
∴ P(A∩B) = P(A) + P(B). P(A∩B) = \(\frac{3}{6}+\frac{3}{6}-\frac{1}{6}=\frac{5}{6}\)

Question 51.
Probability of hitting a target is \(\frac { 1 }{ 3 }\) and that of B is \(\frac { 1 }{ 4 }\). If both attempt to hit the target, what is the probability that
1. both hit
2. the raged is hit.
Answer:
Let P(A) – P(Hittign target by (A) = \(\frac { 1 }{ 3 }\)
P(B) – P(Hitting targed by B) = \(\frac { 1 }{ 4 }\)

1. Since Hitting tanget by A and B are independent
∴ P(A∩B) = P(Both hit) = P(A) x P(B) = \(\frac{1}{3} \times \frac{1}{4}=\frac{1}{12}\)

2. P(the target is hit) = P(Atleast one hit)
= P(A∩B) = P(A) + P(B) – P(A∩B) =\(\frac{1}{3}+\frac{1}{4}-\frac{1}{12}=\frac{6}{12}=\frac{1}{2}=0.5\)

Question 52.
Probability that a student A can solve a problem is \(\frac { 3 }{ 5 }\) and that of B can solve is \(\frac { 5 }{ 7 }\) Find
the probability that
1. both solve
2. atleast one solve
3. None solve the problem.
Answer:
Let P(A) = \(\frac { 3 }{ 5 }\) , P(B) = \(\frac { 5 }{ 7 }\)
1. P(Both solve) = P(A∩B) = P(A). P(B). Since A and B are independent.
\(=\frac{3}{5} \times \frac{5}{7}=\frac{3}{7}\)

2. P(atleast one solve the problem) = P(A∪B) = P(A) + P(B) – P(A∩B)
\(=\frac{3}{5}+\frac{5}{7}-\frac{3}{7}=\frac{31}{35}\)

3. P(None solve the problem)

Question 53.
A can hit a target 2 times with 5 shots. B can hit it 3 times with 4 shots and can hit it 5 times with 8 shots. If they fire at a volley, what is the probability that atleast oen of them hits it?
Answer:
Let P(A) = (A can hit a target)= \(\frac { 2 }{ 5 }\)
Similarly P(B) = \(\frac { 3 }{ 4 }\), P(C) = \(\frac { 5 }{ 8 }\)
P(atleast one of them hit) = P(A∪B∪C) = 1 – P(None hit the target) = 1 – P(A’∪B’∪C’)
Here P(A’) = 1 – P(A) = 1 – \(\frac { 2 }{ 5 }\) = \(\frac { 3 }{ 5 }\), p(B’) = \(\frac { 1 }{ 4 }\), P(c’) = \(\frac { 3 }{ 8 }\)

∴ p(atleast one of them hit) = 1- P(A’). P(B’) P(c’)
\(=1-\left(\frac{3}{5} \times \frac{1}{4} \times \frac{3}{8}\right)=1-\frac{9}{160}=\frac{151}{160}=0.944\)

Question 54.
In a class of 20 girls and 40 boys, half of the girls and half of the boys use cell phones. Find the probability that a student chosen at random is
1. a girl or using cell phone,
2. a boy or a girl
3. a boy or has a cellphone)
Answer:
Let A: Student is a boy = 40,
B : Student is a girl = 20
C : Boy using cell phones = 20,
D: Girls using cell phones =10,
E: Students using cell phone = 30.
1. P(Chosen student is a girl or using cell phone)
= P(B∪E) = P(B) + P(E) – P(A∩E) = \(\frac{20}{60}+\frac{30}{60}-\frac{10}{60}=\frac{40}{60}=\frac{2}{3}=0.67\)

2. P(a boy or a girl) = P(A∪B) = P(A) + P(B)= \(\frac{40}{60}+\frac{20}{60}=\frac{60}{60}=1\)

3. P(a boy or has a cell phone)
P(A∪E) – P(A) + P(E) – P(A∩E) = \(\frac{40}{60}+\frac{30}{60}-\frac{20}{60}=\frac{50}{60}=0.833\)

Question 55.
A machine has two parts A and B and it fails to work either of the parts tails. The probability of part a fails is 0.25 and the probability of part B fails 0.12. Find the probability that the machine fails.
Answer:
Let P(A) : P(part A fails) = 0.25 and P(B): P(part B fails) = 0.12
P (machine fails) =P(partSfails or B fails) = P(A∪B) = P(A) + P(B) – P(A∩B)
= P(A) + P(B) – P(A).P(B)
Since parts A and B are independent
P(A∪B) = 0.25 + 0.12-0.25 × 0.12 = 0.37 – 0.03 = 0.34

Question 56.
An corn contains 9 red balls and 6 yellow balls. If Deeput chooses 2 balls one after the other with replacement. Find the probaiblity that
1. both are red
2. one of each colour.
Answer:
Let A: First ball drawn is red, B : Second ball drawn is red – C: The ball drawn is yellow. Since A and B are independent events
1. P(Drawing 2 red balls one after the other with replacement) = P(A∩B) = P(A). P(B); by multiplication theorem.
\(=\frac{9}{15} \times \frac{9}{15}=\frac{9}{25}=0.36\)

2. P (drawing one of each colour ball)
= P(A∩C) or P(C∩A) = \(\frac{9}{15} \times \frac{6}{15}=\frac{6}{25}=0.24\)

Question 57.
A bowl contains 4 Red and 3 blue marbles. Another bowl contians 3 Red and 5 blue marbles. One of the bowl is randomly chosen and a marble is drawn, what is the probability that it is a blue marble?
Answer:
Let A: First bowl is selected, B : Second bowl is selected, C : Marble drawn is blue.
Here P(A) = P(B) = \(\frac { 1 }{ 2 }\)
P(drawing a blue marble) = P(First bowl is selected and 9 blue marble is drawn from it) Or P(Second bowl is selected and a blue marble is drawn from it)
= P(A∩C) ∩P(B∩C) = P(A). P(C/A) + P(B). P(C/B)
\(=\frac{1}{2} \cdot \frac{3}{7}+\frac{1}{2} \cdot \frac{5}{8}=\frac{3}{14}+\frac{5}{16}=0.527\)

Question 58.
A box has 4 blue and 3 green balls. Another box has 3 blue and 5 green balls A ball is selected from each boxes, find the probability both balls drawn are green.
Answer:
Let A: drawing a green ball from I box
B : drawing a green ball from II box
Since drawing balls from each box sare independent.
∴ P(drawing green balls from both boxes)
= P(A∩B) = P(A).P(B)=\(=\frac{3}{4} \times \frac{5}{8}=\frac{15}{40}=0.375\)

Question 59.
What is the probability that there will be 53 Sundays in a random 4 selected
1. Non-leap year and
2. Leap year.
Answer:
1. ANon-leap year has 365 days, with 52 weeks has 364 days. The remaining one day may be
: Mon, Tue, Wed, Thu, Fri, Sat and San = 7

∴ Probability of 53 Sundays in a non-leap year = \(\frac { 1 }{ 7 }\)

2. A Leap year has 366 days with 52 weeks, has 364 day. The remaining = days may be. (Mon, Tue), (The, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun) and (Sun, Mon) =
7 possible combination of days.
∴ Probability 53 Sundays in a leep year = \(\frac { 2 }{ 7 }\) ;
i. e., may be of (Sat, Sun), (Sun, Mon)

Question 60.
A study shows that a person randomly chosen will go out to a mall with probability 0.74, and the probability that a person will go to get some chats is 0.45 during a weekend. And the probability that the person will get chats given that he will go to a mall.
Answer:
Let A: Person gets chats, B: Person goes out to a mall.
P(a person gets chats given that he goes to a mall)

Question 61.
An urn contains 8 black balls and 4 white balls. Two balls are taken from the urn without replacement. Compute the probability that both balls are white.
Answer:
Let A: First balls’s white, B: Second ball is white
P(drawing 2 white balls one after the other without replacements) = P(A∩B) = P(A). P(B/A)
\(=\frac{4}{12} \cdot \frac{3}{11}=\frac{1}{11}=0.0909\)

Question 62
In a bag, there are 8 blue balls and 6 red, 2 balls are picked up one after two other without replacement. So, what is the probability of picking red ball in 2nd attempt knowing that a blue balls has already been picked.
Answer:
A: I ball drawn l’s blue B: II ball drawn is red.
P(A∩B) =P(A).P(B/A) = \(\frac{8}{14} \times \frac{6}{13}\)
Here first ball l’s not replaced.
∴ P(A/B) = \(\frac { 6 }{ 13 }\) is the required

You can Download Chapter 4 Diagrammatic and Graphical Presentation of Data Questions and Answers, Notes, 1st PUC Statistics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Statistics Question Bank Chapter 4 Diagrammatic and Graphical Presentation of Data

1st PUC Statistics Diagrammatic and Graphical Presentation of Data Two Marks Questions and Answers

Question 1.
What are diagrams and graphs?
Solution:
Frequency distributions when they are presented in Tabular form becomes dull and un¬interesting, moreover they require close reading of the figures to understand what is represented. Diagrams and graphs are the means of making the data easy to understand even by layman at a glance.

Question 2.
What is the Need/Objectives/Purpose for graphic presentation of statistical data ? Or Mention the uses of diagrammatic /graphical presentation, or How diagrams and graphs are useful in representing statistical data?
Solution:

• They are attractive and hence in News papers and magazines, diagrams and graphs are commonly used in advertisements.
• They gives Birds-eye view of the entire data at glance.
• They can be easily understood by common man.
• They can be remembered for longer period of time.
• They facilitates comparison.

Question 3.
What is one dimensional diagram?
Solution:
In one dimensional diagram only the height (length) is considered. They are mostly bar diagrams. Here, only one characteristic is considered to represent the data.

Question 4.
Mention the various types of one dimensional diagram?
Solution:
Some of the one dimensional diagrams are:

• Simple Bar diagram
• Multiple Bar diagram
• Component/Subdivided Bar diagram
• Percentage Bar diagram.

Question 5.
Name different graphs used for presentation of frequency distribution?
Solution:
Some commonly used graphs are :

• Histogram
• Frequency Polygon
• Frequency Curve
• Ogives.

1st PUC Statistics Diagrammatic and Graphical Presentation of Data Five Marks Questions and Answers

Question 1.
What are the general rules for drawing a diagram?
Solution:
General Rules for constructing Diagrams and Graphs are: –

• Every diagrams and should have a suitable Title and is written above it.
• The proper scale according with the size of the paper should be selected.
• It should be neat and clean .
• It should not be overloaded with more information
• To indicate different parts suitable shadings, colours, crossings should made use of.
• An Index indicating different shades colours, crossing etc. used should be clearly shown
• It should be complete in all respects
• It should be simple and self explanatory.

Question 2.
Write short notes on (a) simple bar diagram (b) sub divided bar diagram.
Solution:
(a) Simple bars or thick lines aredrawn to represent the items of the data. The length of the bar is taken in proportion to the magnitude of the item in the data. The proper width of the bar is taken merely for attraction, but’it is nothing to do with the data. It represents only one character of the data.
Ex- The figures of Imports, Exports, and Population etc. for few years are to be represented by simple bars.

(b) In subdivided bar diagrams, where in some problems it is required to represent more than one or two variables of the same kind, we use component/sub-divided bar diagrams. In this case each bar is subdivided to in to various components and different shades, crossings, colours are used, and an Index is given to that effect. These are useful in comparing the total magnitudes, along with the components.
Ex: The figures of expenditure of family, Exports or Imports of certain commodities, population according to sex,

Question 3.
What is a pie diagram? or What are Pie-charts?/Explain the construction procedure
Solution:
As in bar diagram, a bar is sub-divided to represent its components. Where as in the pie- diagram a circle is subdivided into sectors by subtending the angle at the circle. The total of the components are equated to 360° and each component is expressed in degrees. The area of the sector formed by the angle measured by the degrees of the component is proportional to the magnitude of that component. To distinguish between the different components shades, coloures are used. It is so called because, it resembles a pie (cake) and components resemble slices of the pie. It is also known as angular diagram or sector graphs.

Question 4.
Write short notes on
(a) histogram
(b) frequency polygon
(c) frequency curve
(d) Ogives.
Solution:
Histogram: – “A Histogram is a pictorial representation of graphs of frequency distribution by means of adjacent rectangles, whose areas are proportional to the frequencies represented” The Histogram can be constructed by taking variable (class intervals) on x-axis and class frequency (f) along y-axis. On each of the class intervals rectangles are erected. The width and height of the rectangles are proportional to the length of the class and class frequencies respectively.

The graph formed by series of such rectangles adjacent to one another is called Histogram. From the Histogram Mode(Z) can be obtained by joining the end point of the highest rectangle to the diagonal end point of the adjacent rectangles, and a perpendicular drawn to intersection of these lines to the x-axis, which gives the value of mode.

On the basis of Histogram, Frequency polygon and Frequency curve can be constructed. Frequency distribution with Inclusive class intervals should be converted into Exclusive and for unequal width of the class interval. Histogram is constructed with, width of the class interval against Frequency density (f/w).

Frequency Polygon: – This graph is preferred when two or more frequency distributions are required to compare on the same graph. It is so called because of its resemblance with the plane geometrical figure polygon (many angled) representing frequency distribution. We can construct polygon in two ways- By drawing first Histogram and then joining the. mid-points of the upper horizontal lines of each rectangles. Thus obtained polygons ends are extended to touch the base line at a distance of half class interval.

Another method of drawing a polygon is that by taking all the mid-points of the class intervals and the corresponding frequencies areplotted. Thus obtained end points are joined by straight lines. And end points are extended to reach the base line at a distance of half class interval.

Frequency Curve: – A frequency polygon obtained from the Histogram or direct by midpoints of the various classes, is not a smooth curve. Its boundaries are made up of straight lines and it has sharp corners, these sharp corners can be removed by a free hand curve drawn along the frequency polygon.

Ogives/cumulative frequency curves: – Sometimes it is required to plot a graph of variables which is less than some value or more than a value. So, in such cases we are required to add up the frequencies lying below or above a given point of variable. Thus added frequencies are called cumulative frequency. The curve obtained by plotting cumulative frequency and the respective variable is called cumulative frequency curves or Ogives.
There are two types of Ogives (i) Less than Ogive (ii) More than Ogive

• ‘If a curve is drawn for the cumulative frequency added from the top (l.c.f) and the upper limits of the class is called Less Than Ogive ’.
• Similarly ‘If a curve is drawn for the cumulative frequency from below (m.c.f) and lower limits of the classes then curve is called More Than Ogive’’
• Here the variable is taken along x-axis and l.c.f / m.c.f. along y-axis. The corresponding points are joined by a smooth curve. The resulting graph is Less/More than Ogive

Question 5.
What is a frequency polygon? How is it constructed?
Solution:
Frequency Polygon: – This graph is preferred when two or more frequency distributions are required to compare on the same graph. It is so called because of its resemblance with the plane geometrical figure polygon (many angled) representing frequency distribution. We can construct polygon in two ways- By drawing first Histogram and then joining the. mid-points of the upper horizontal lines of each rectangles. Thus obtained polygons ends are extended to touch the base line at a distance of half class interval.

Another method of drawing a polygon is that by taking all the mid-points of the class intervals and the corresponding frequencies areplotted. Thus obtained end points are joined by straight lines. And end points are extended to reach the base line at a distance of half class interval.

Question 6.
Distinguish between histogram and frequency polygon.
Solution:
A Histogram is a pictorial representation of graphs of frequency distribution by means of adjacent rectangles, whose areas are proportional to the frequencies represented”

Where as frequency polygon is preferred when two or more frequency distributions are required to compare on the same graph. It is so called because of its resemblance with the plane geometrical figure polygon (many angled) representing frequency distribution. We can construct polygon in two ways- By drawing first Histogram and then joining the mid-points of the upper horizontal lines of each rectangles.

Question 7.
Write a note on false base line.
Solution:
Generally X and Y-axes begin from zero ie, the origin is at zero. When lowest value to be plotted is high and detailed scale needed to study all the variations in the data taking zero at the origin becomes impractical. Hence, if the fluctuations in the variable are too small, or if the lowest value of the variable is large, the False Base Line should be used, Here False Base Line can be shown by vertical wavy line between zero and first scale or the axes can be broken by kinked line.

Question 8.
From which diagram Mode can be obtained and how do you locate Mode graphically?
Solution:
Mode can be obtained from Histogram, and is obtained by joining the upper end points of the highest rectangle to the diagonal end points of adjacent rectangles, the intersection of diagonal end points a perpendicular drawn to the x-axis, which gives the mode.

Question 9.
From which curves Median can be obtained graphically/How do you locate Median graphically?
Solution:
Usually the two Ogives are drawn together with common axes. The point of intersection of two Ogives correspond to Median at the x-axis of the distribution.

Question 10.
What are limitations of diagrams and graphs?
Solution:
Graphs /diagrams are -only visual aids, and they cannot be considered as alternatives to numerical data.

• Graphs/diagrams are not accurate and gives only a rough idea of,the data.
• They cannot be used for further analysis of data.
• They can be easily misled and can create wrong impression about the data.

Question 11.
Write down any two Comparison of diagrams and graphs.
Solution:

1. Diagrams give only an approximate idea where as graphs give more accurate information.
2. Diagrams are drawn on plain paper where as graphs need graph paper.
3. Diagrams can be easily understood than graphs.
4. Time series and frequency distributions can be represented by graphs but not by diagrams.
5. Diagrams are more suitable for publicity and advertisement, where as graphs are suitable for statistical analysis.

Question 12.
Represent the following data regarding the production of paddy (in ’000 tons) by . simple bar diagram.

Solution:
Following is the simple bar diagram representing the production of paddy

Question 13.
The incomes of the different categories of persons in Bangalore are given below:

Represent the data by suitable diagram.
Solution:

Businessmen Teachers Lecturers Call centre IT/BT Employees

Question 14.
Draw the multiple bar diagram showing the working of men, women and children in the Bangalore
Table showing the sex wise population of Mangalore and Mysore.

Solution:
Multiple Bar diagram showing the sex and children wise distribution population of Bangalore, Mangalore and Mysore.

Question 15.
Present the following data in Multiple Bars of results of II PUC statistics examination in 2005, 2006, and 2007.

Solution:
Multiple bar diagram showing the result statistics of II puc students for the year 2005-07

Question 16.
Following is the data regarding the strength of students of a university during 2008-10. constructs a percentage bar diagram.

Solution:
percentage of a value can be calculated using the following formula:
% of a value =(individual value/Total value) × 100
For the year 2008: Arts = (200/450) × 100 = 44.44, similarly calculate for the others.

Footnote:Cum-cumulative percentage
Percentage bar diagram showing student’s strength of a university

Question 17.
The following data relates to the. monthly expenditure (in Rs.) of two families A and B. Represent the data by a rectangular diagram on percentage basis.

Solution:

Percentage bar diagram showing the expenditure of families A and B

Question 18.
For the following data regarding the income of the government from different sources, draw a pie and component diagram :

Solution:
The angles for the components can be calculated as below:
The angle of an individual component = \(\frac{\text { Individual value }}{\text { Total value }} \times 360^{\circ}\)
For Customs: (80/540) × 360 = 53.33 ≅ 53 ; similarly calculate for the others also.

Question 19.
Percentage breakup of the cost of construction of a house in Bangalore (Excluding land cost) is given below :
Labour: 20% , Bricks:12%, Cement:20%, Steel:15%, Wood:13%,
Supervision:15%. Other expenses: 5% . Construct a pie diagram.
Solution:
The angles for the components can be calculated as below:
Angle of an individual component = \(\frac{\text { Individual value }}{\text { Total value }} \times 360^{\circ}\)
Angle for Labour = (20/100) × 360 = 72; similarly others can be calculated.

Question 20.
Draw a Histogram from the following data.

Solution:
Draw Histogram by taking the width of class intervals as 10 each and height with resect to – class frequencies.

Question 21.
Draw a Histogram from the following data and locate the mode from the Histtogram:

Solution:
Convert the mid points as class intervals: the common difference between the mid points is 7, half of it is 3.5. Subtract 3.5 in all midpoints to get lower limits and add 3.5 to all upper limits to get upper limits.

To get mode (Z) join the upper vertices of the tallest rectangle to their opposite upper vertices of the rectangles adjacent it. Then from the point of intersection of these diagonals draw a perpendicular line to the x-axis, The point at which the perpendicular line touches the x-axis, gives the value of mode.

Question 22.
Draw a frequency polygon to the following frequency distribution.

Solution:
Here drawing frequency polygon for a continuous frequency distribution is same as of discrete frequency distribution. Simply plot the mid points against frequencies and join them by straight line and extend the end points to reach x-axis at a distance of half class interval.

Question 23.
Prepare a frequency polygon from the following data :

Solution:
To get polygon plot the points against micl points against the class frequencies then join the plotted points by a straight, extend the end points to reach the x-axis at a distance of half a class interval.

Question 24.
Draw the histogram, frequency polygon and frequency curve for the following data.

Solution:
First draw Histogram, obtain the polygon and then draw a smooth line along the polygon, which we get a frequency curve

Question 25.
Draw frequency curve from the following frequency distribution.

Solution:
First plot the points mid values against the class frequencies and these points are joined by a free hand curve, which gives the frequency curve.

Question 26.
Draw Ogives for the following distribution and locate the median from the graph.

Solution:
To get the less than ogive and more than, the upper class limits (lower class limits) are taken along X-axis and the corresponding less (more) than cumulative frequencies along Y-axis with appropriate scale . The plotted points on the graph are joined by smooth curve. The obtained curve is less (more) than Ogive.

The point of intesection of ogives corresponds to Median of the distribution. From the point of intersection of two Ogives the perpendicular line (projection line ) is drawn to x-axis. The point on the x-axis gives the value of the median . From the ogives Median = 79.

Question 27.
Draw an Ogive and locate the median, lower quartile and upper quartile from the graph.

Solution:

Median = (N/2)th term
(80/2) = 40th term
From the ogive M = 35 Marks
FirstlLowcr quartile Q₁ (N/4)th term
(80/4)=2Oth term
∴ From the ogive Q₁ =26Marks
ThirdlUppcr quartile Q₃ (N/4)th term
= 3(80/4) 60th term
∴ From the ogive Q₃ = 42Marks

You can Download Chapter 8 Data Types Questions and Answers, Notes, 1st PUC Computer Science Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Computer Science Question Bank Chapter 8 Data Types

1st PUC Computer Science Data Types One Mark Questions and Answers

Question 1.
What is meant by data types?
Answer:
The set of values along with the operations that can be performed on these values are called as data types.

Question 2.
Give the classification of data types.
Answer:
Data types are classified into built-in or basic data types and user-defined or derived data types.

Question 3.
Name the built-in data types of C++.
Answer:
The built-in or basic data types supported by C++ are integer, floating-point, and character types.

Question 4.
How is derived data type created?
Answer:
The derived data type is created by using basic data types.

Question 5.
What is linear data structure?
Answer:
When data elements are arranged in a sequential manner it is called as linear data structure.

Question 6.
What is a non-linear data structure?
Answer:
When the data elements are arranged non-sequentially it is called as non linear data structure.

Question 7.
Give the range of values that int data type can store.
Answer:
The range of values that int data type can store is from -32768 to 32767.

Question 8.
Give the ‘char’ data type range of values.
Answer:
The ‘char’ can store 256 characters in the range of-128 to 127.

Question 9.
Give the storage size of the float data type.
Answer:
The float data type needs 4 bytes of memory for each number with a fractional part.

Question 10.
Define modifiers.
Answer:
The modifiers change the meanings of the predefined built-in data types and expand them to a much larger set.

Question 11.
Write the different modifiers.
Answer:
The different modifiers are long, short, signed and unsigned.

Question 12.
What is the range of values that unsigned int can store?
Answer:
The range of values that unsigned int stores is 0 to 65535.

Question 13.
Give the unsigned char range of values.
Answer:
The unsigned char can store the value in the range of 0 to 255.

Question 14.
Mention a few user-defined data types.
Answer:
A few user-defined data types are structure, union, class and enumerated data types.

Question 15.
What is an enumerated data type?
Answer:
An enumerated type is a data type where every possible value is defined as a symbolic constant (called an enumerator).

1st PUC Computer Science Data Types Two/Three Marks Questions and Answers

Question 1.
What are the variables? Give an example.
Answer:
It is a location in the computer memory which can store data and is given a symbolic name for easy reference.
For example, Total = 20.00; In this statement, a value 20.00 has been stored in a memory location Total.

Question 2.
Give the declaration syntax of a variable and example.
Answer:
The syntax for declaring a variable is

1. datatype variablename;
2.  example, on

Question 3.
variable initialized? Give an example.
Answer:
The variable is initialized during declaration of the variable with initial value.
For example, int a = 20;

Question 4.
Explain lvalue and rvalue. Give a examples.
Answer:
Lvalue is the location value and r value is the data value of a variable.
int age =16;
lvalue is the location value of age and 16 is the r-value.

Question 5.
What is dynamic initialization? Give its advantage.
Answer:
It is a method of initialization C++ adopts. In this method, C++ declares a variable during the run time of the program. The advantage is variables can be initialized anywhere in the program before they are used.

Question 6.
Write the features of int data type.
Answer:
Integers are those values which have no decimal part and they can be positive or negative, like 12 or-12.

1. int keyword is used for integers.
2. It takes two bytes in memory.

There are two more types of int data type

1. signed int or short int
2. unsigned int or unsigned short int.

Question 7.
Give the features of char data type.
Answer:
C++ offers a predefined data type that is one byte in size, which can hold exactly one character such as ‘a’ or ‘A’.
To declare a variable of type char,
char ch;

Question 8.
Write the features of the float data type.
Answer:
C++ defines the type float data as representing numbers that have a fractional part. For example, 12.55. Floating-point variables can either be small or large. A variable with float type occupies 4 bytes in size and can hold numbers from 10^-308 to 10⁺³⁰⁸ with about 15 digits of precision. There is a long double, also available, that can hold numbers from 10^-4932 to 10⁺⁴⁹³².

Question 9.
What is a void? Give the usage.
Answer:
The data type void has no values and no operations means it is empty. It plays the role of generic data type and can represent any of the other standard types. It is also used in functions which do not return any value.

Question 10.
Explain bool data type.
Answer:
The bool data type means Boolean, that has the logical value true or false. It can be used to manipulate logical expressions.

Question 11.
What is named constant? Give the declaration.
Answer:
The named constant is a memory location, whose value cannot be changed during the execution of a program.
Declaration syntax:
const datatype variable = initial value;

Question 12.
What are the conversion rules for enum type?
Answer:
There is an implicit conversion from any enum type to int. It does not support for implicit conversion from into to enum type.

1st PUC Computer Science Data Types Five Marks Questions and Answers

Question 1.
Write a short note on variables.
Answer:
“A variable is a temporary container to store information, it is a named location in computer memory where varying data, like numbers and characters, can be stored and manipulated during the execution of the program”.
A variable must be declared before it is used.
For example, int my_age;
A few naming conventions must be taken in consideration.

1. It must start with an english alphabet character, either lowercase or uppercase, including underscore (not a hyphen). It may not start with a digit. The rest is optional. It can either be a letter or a digit (0-9).
2. C++ keywords like main, case, class, if, else, do, while, for, tyedef, etc cannot be used as a variable names.
3. It must be unique within the scope.

Question 2.
Explain the basic data types in details.
Answer:
In computer programming, information is stored in a computer memory with different data types. We must know what is to be stored in a computer memory, whether it is a simple number, a letter or a very large number.
Basic Data types in C++
1. character:
C++ offers a predefined data type that is one byte in size, which can hold exactly one character such as ‘a’ or ‘A’. To declare a variable of type char, we have
char ch;
Suppose we want to store a character value ‘a’, in a char data type eh, it is enclosed within a single quote.
ch = ‘a’;
Only a single character can be stored in a variable of type char.

2. integer:
On most machines, the size of int type is 2 bytes. C++ defines this type as consisting of the values ranging from -32768 to 32767. This range is for the small integer. If a long integer is needed, the type long or long int can be used. The range of long int is too big that is from -2147483648 to 2147483647, which occupies 4 bytes in memory.

3. float:
C++ defines the data type float, as representing numbers that have a fractional part. For example, 12.55 as opposed to integers which have no fractional part. Floating-point variables can either be small or large. A variable with type float occupies 4 bytes in size and can hold numbers from 10^-308 to 10⁺³⁰⁸ with about 15 digits of precision. There is a long double, also available, that can hold numbers from 10^-4932 to 10⁺⁴⁹³².

4. Bool:
It is an additional data type for representing a Boolean value. A variable associated with a bool data type may be assigned an integer value 1 to the literal true and a value 0 to the literal false.

Question 3.
Write a short note on modifiers.
Answer:
C++ allows the char, int, and double data types to have modifiers preceding them. A modifier is used to alter the meaning of the base type so that it more accurately fits the needs of various situations.
The data type modifiers are listed here:

1. signed
2. unsigned
3. long
4. short

The modifiers signed, unsigned, long and short can be applied to integer base types. In addition, signed and unsigned can be applied to char, and long can be applied to double. The modifiers signed and unsigned can also be used as prefix to long or short modifiers. For example, unsigned long int.

C++ allows a shorthand notation for declaring unsigned, short, or long integers. You can simply use the word unsigned, short or long, without the int. The int is implied. For example, the following two statements both declare unsigned integer variables, unsigned x; unsigned int y;

Question 4.
Write the different data types, memory size in bytes, minimum value, maximum value
Answer:
Data types in C++