## Karnataka 1st PUC Model Question Papers with Answers 2019-2020 Science Commerce New Syllabus

We hope the given New Syllabus Karnataka 1st PUC Class 11 Model Question Papers with Answers 2019-2020 Pdf Free Download of 1st PUC Model Question Papers PCMB with Answers, PUC 1st Year Model Question Papers, 1st PUC Commerce Model Question Papers with Answers, 1st PUC Model Question Papers Science Arts, Karnataka 1st PUC Previous Year Question Papers with Answers in English Medium and Kannada Medium will help you. If you have any queries regarding Karnataka State Board NCERT Syllabus First PUC Class 11 Model Question Papers for 1st PUC Science Commerce Arts with Answers 2019-2020 Pdf, drop a comment below and we will get back to you at the earliest.

## 1st PUC Question Banks with Answers Karnataka

Expert Teachers at KSEEBSolutions.com has created Karnataka 1st PUC Question Banks with Answers Solutions, Notes, Guide Pdf Free Download of 1st PUC Textbook Questions and Answers, Model Question Papers with Answers, Study Material 2019-20 in English Medium and Kannada Medium are part of KSEEB Solutions. Here We have given the Department of Pre University Education (PUE) Karnataka State Board NCERT Syllabus 1st Year PUC Question Banks with Answers Pdf. Students can also read 1st PUC Model Question Papers with Answers hope will definitely help for your board exams.

## Karnataka 1st PUC Question Banks with Answers

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## 1st PUC Basic Maths Question Bank Chapter 17 Straight Lines

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## Karnataka 1st PUC Basic Maths Question Bank Chapter 17 Straight Lines

Question 1.
Find the slope of the line joining the points (3, -4) and (-7,3).

Question 2.
Show that the line joining points (2, -3) and (-5,1) is parallel to the line joining the points (7, -1) and (0,3) and perpendicular to the line joining the points (4,5) and (0, -2).
Let m1 be the slope of the lien l1 joining points (2, -3) and (0, 3)

Let m2 be the slope of the line l2 going (7, -1) and (0,3)

⇒ m1 = m2
∴ L1 and L2 are parallel to each other.

Question 3.
L1 and L3 or l2 l3 are perpendicular to each other. The slope of a line is double the
slope of another line. If the tangent of the angle between the is $\frac { 1 }{ 3 }$. Find the slope of the lines.
If slope of one line is m, then the slope of the other is 2m.
Let the angle between them θ then tan θ = $\frac { 1 }{ 3 }$

$\frac { 1 }{ 3 }$ = $\frac{m}{1+2 m^{2}}$
⇒ 1 + 2m2 = 3m
⇒ 2m2 – 3m + x = 0
⇒ (2m – 1)(m – 1 ) = 0
∴ 2m – 1 = 0 and m – 1 = 0
m = $\frac { 1 }{ 2 }$ m = 1
∴ Slope are $\frac { 1 }{ 2 }$, and 1.

On standard form of straight lines:

Question 1.
Find the equation of line passing through (-3, 5) with slope $\frac { -1 }{ 5 }$
Let (x1 y1) = (-3, 5). slope m = $\frac { -1 }{ 5 }$
The equation of line is y – y1 = m(x – x1)
= y – 5 = $\frac { -1 }{ 5 }$ (x + 3)
⇒ 5y – 25 = -x – 3
⇒ x + 5y – 22 = 0

Question 2.
Find the equation line passing through (-3, 2) and (11, -1).
Let (x1, y1) = (-3,2), (x2, y2) = (11,-1)

$\frac{y-2}{x+3}=\frac{-3}{14}$
⇒ 14y – 28 = -3x – 9
⇒ 3x + 14y – 19 = 0 is the equation

Question 3.
Find the ratio in which the line join (1, 2) and (4, 3) is divided by the line joining the points (2,3) and (4, 1).
Let A = (2,3) B = (4, 1)
∴ the equation of line is a given by $\frac{y-3}{x-2}=\frac{1-3}{4-2}$
$\frac{y-3}{x-2}=\frac{-2}{2}$ = -1
⇒ y – 3 = -x + 2
∴ x + y – 5 = 0 is the equation of line
Let c = (1,2) and D(4,3) cut the line AB at P(x ,y) is the ratio 2 : 1 then

⇒ 4x + 1 + 3r + 2 – 5(r + 1) = 0
⇒ 2r – 2 = 0
⇒ r = 1
∴ the required ratio is 1 : 1

Question 4.
Express 9x – 4y – 13 = 0 in the intercept from
Consider 9x – 4y = 13 = $\frac{9}{13} x-\frac{4}{13} y$ = 1

∴ x – intercept = $\frac{13}{9}$
y – intercept = $\frac{-13}{9}$

Question 5.
If P ¡s the length of the perpendicular from the origin on a line, whose x and y intercepts are respectively a and b then show that = $\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$
Let the line any the axis at A and y – axis at B
Then by data OA = a and OB = b from the figure

Area of OAB = $\frac { 1 }{ 2 }$ (O.A) (O.B) = $\frac { 1 }{ 2 }$ ab
or = $\frac { 1 }{ 2 }$ (AB).P
$\frac { 1 }{ 2 }$ ab = $\frac { 1 }{ 2 }$ (AB).P ⇒ a2b2 = (AB)2. p2
Also AB2 = OA2 + OB2 ⇒ AB2 = a2 + b2 (∴ because a2b2 = (a2 + b2) . p2

Intersection of two lines and Concurrency of Lines

Question 1.
Find the equation of line parallel to the y-axis and drawn through the point of intersection of x – 7y + 5 = 0 and 3x + y – 7 = 0.
The equation of line through the point of intersection of the given lines is of the form x – 7y + 5 + k(3x + y – 7) = 0
∴ (1 + 3k)x + (k -7)y + 5 – 7k = 0 (1)
Since the lines are parallel to y-axis, coefficient of y = 0
i.e., k – 7 =0
∴ k = 7 substitute in (1) we get
22x – 44 = 0
∴ x – 2 = 0 is required equation

Question 2.
Find the equation of the line through the intersection of 5x – 3y = 1 and 2x + 3y – 23 = 0 and perpendicular to the line whose equation is 5x – 3y -1 = 0 and perpendicular to the line whose equation is 5x – 3y – 1 = 0.
We have 5x – 3y – 1 = 0 ….. (1)
and 2x + 3y – 23 = 0 ….. (2) .
∴ the equation will be in the form
(5x – 3y – 1) + k(2x + 3y – 23) = 0
⇒ (5 + 2k)x + 3(k – 1)y – (23k + 1) = 0 …… (3)

∴ Slope of line = $\frac{5+2 k}{3(k-1)}$
Also 5x – 3y – 1 = 0
⇒ y = $\frac{5}{3} x \frac{1}{3}$ …….. (1)
∴slope = $\frac{5}{3}$
Given (4) and (5) are perpendicular

⇒ k = -34. Substituting the value of k in (3), we get
(5x – 3y – 1) -34 (2x + 3y – 25) = 0
⇒ 5x – 68x – 3y – 102y – 1 + 782 = 0
⇒ 63x + 105y – 781 = 0

Question 3.
If lines whose equation are y = m1 + c1y = m2x + c2 and y = m5x + c3 meet is a point then prove that m1( (c2 – c3) + m2 (c3 – c1) + m3(c1 – c2) = 0
The equation of the given lines are
m1x – y + c1 = 0 ……. (1)
m2x – y + c2 = 0 …… (2)
m3x – y + c3 = 0 .. (3)

The three lines will be concurrent if the point of intersection of (1) and (2) lies on (3)

⇒ m3(c1 – c2) – (m2c1 – m1c2) + c3 (m2 – m1) =0, simplify,
⇒ m1(c2 – c3) + m2(c3 – c1) + m3(c1 – c2) =0

Question 4.
For what values of k are the lines x – 2y + 1 = 0, 2x – 5y + 3 = 0 and 5x – 9y + k = 0 are concurrent.
consider
x – 2y + 1 = 0 ……. (1)
2x – 5y + 3 = 0 ….. (2)
5x – 9y + k = 0 …….. (3)
Solve (1) and (2) $\frac{x}{-6+5}=\frac{-y}{3-2}=1$
$\frac{x}{-1}=\frac{-y}{-1}=\frac{1}{-1}$
⇒ x = 1, y = 1
point of intersection of (1) and (2) is (1,1),
⇒ (1, 1) lies on (3) = 5 – 9 + k = 0 = k = 4

Angle between the lines, Length of perpendicular

Question 1.
Find the angle between the lines 2x + 3y – 4 = 0 and 3x – 2y + 5 =0
By data
2x + 3y – 4 = 0 … (1)
3x – 2y + 5 = 0 …… (2)
∴ Slope of (1) = $\frac{-a}{b}=\frac{-2}{3}$ = m1

∴ the angle between the lines is 90°

Question 2.
Determine the position of the points (2, 1) and (-1, 1) w.r.t the line 4x – ly + 1 = 0.
4(2) -7(1) = -7 + 1 = 2 > 0
4( -1) – 7(1) = – 4 – 7 + 2 = -10 < 0
Since the two points are opposite in sign the two points lie an either sides of the given line 4x – 7y + 1 = 0

Question 3.
If points (9,8) and (-3,3) are equidistant from the line 5x + 2y + 7 = 0 find ‘a’.
Distance from (a, 8) to the line 5x + 2y + 7 = 0

Given that the distance are equal

⇒ 5a + 23 = 2
or
5a + 23 = -2
⇒ a = $\frac{-21}{5}$ and a = -5

## 1st PUC Basic Maths Question Bank Chapter 16 Locus and its Equations

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## Karnataka 1st PUC Basic Maths Question Bank Chapter 16 Locus and its Equations

Question 1.
Find the equation of the locus of a point which moves such that its distance from the point (-5,7) is twice its distance from (-3,1).
Let A = (-5, 7) and B = (-3, 1) and P(y, b) be any point of the locus
By data PA = 2PB = PA2 = 4PB2
⇒ (x + 5)5 + (y – 7)2
= 4[(x + 3)2 + (y – 1)2]
⇒ x2 + 10x + 25 + y2 + 49 – 14y
= 4(x2 + 9 + 6x + y2 + 1 – 2y)
= 4x2 + 36 + 24x + 4y2 + 4 – 8y
= 3x2 + 3y2 + 14x + 6y – 34 = 0
which is the equation as to the locus.

Question 2.
Find the equation to the locus of the perpendicular bisector of the line joining A(3, 2) and B(4,1).
The perpendicular bisector of the line joining A and B is the locus of the point which moves such that it is equidistant from A and B.
By data we have A = (3, -2), B = (4, 1)
Let P(x, y) be any point on the perpendicular bisector
Thus we have PA = PB = PA2 = PB2
=> (x – 3)2 + (y + 2)2 = (x – 4)2 + (y – 1 )2

⇒ -6x + 4y + 13
= -8x – 2y + 17
⇒ 2x + 6y – 4 = 0.
⇒ x + 3y – 2 = 0 is the equation of the locus of a point.

Question 3.
Find the equation of the locus of the point which moves such that the sum of its distances from (0, 5) and (0, -4) is 10 units.
Let A = (0, 6) and B = (0, -6).
Let P(x, y) be the locus of the point; By data PA + PB = 10
$\sqrt{x^{2}+(y-6)^{2}}+\sqrt{x^{2}+(y+6)^{2}}$ = 0
$\sqrt{x^{2}+(y-6)^{2}}$ = 10 – $\sqrt{x^{2}+(y+6)^{2}}$
Squaring both sides

⇒ 10y – 12y – 100 = -20$\sqrt{x^{2}+(y+6)^{2}}$
⇒ -22y + – 100 = -20$\sqrt{x^{2}+(y+6)^{2}}$ + by – 2
⇒ 11y + 50 = 10$\sqrt{x^{2}+(y+6)^{2}}$
squaring again we get
⇒ 121y2 + 2500 + 110y
= 100(x2 + y2 + 36 + 12y)
= 100x2 + 10y2 + 3600 + 122y
⇒ 100x2 – 21y2 + 100y + 1100 = 0 is the locus of the equation.

## 1st PUC Basic Maths Question Bank Chapter 15 Co-ordinate System in a plane

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## Karnataka 1st PUC Basic Maths Question Bank Chapter 15 Co-ordinate System in a plane

Question 1.
Find the reflection of (-4, 3) through x-axis, y-axis and through the centre.
Reflection through x – axis → (-4, – 3)
Reflection through y – axis → (4,3)
Reflection through center → (4, -3)

Question 2.
Find ‘a’ if the distance between the points (a, 2) and (3, 4) is V8 units.
Let P = (a, 2) and Q = (3, 4)
By data |PQ| = √8
⇒ √8 = $\sqrt{(3-a)^{2}+(4-2)^{2}}$
⇒ 8 = (3 – a)2 + 22 = 8
⇒ 9 + a2 – 6a + 4
⇒ a2 – 6a + 5 = 0
⇒ (a – 5)(a – 1) = 0
⇒ a = 5 or a = 1

Question 3.
Show that the points (2, 2) (6, 3) and (4,11) form a right angled triangle.
Let A = (2, 2), B = (6, 3), C = (4, 1)
Consider

By pythagorus theorem AC2 = AB2 + BC2

⇒ 85 = 17 + 68 = 85
∴ the points form a right angled triangle.

Question 4.
The points (x, 2) is equidistant from (8, -2) and (2, -2). Find the value of x.
Let A = (2,2) B = (8,-2) C = (2,-2)
By data AB = AC = AB2 = AC2
⇒ (8 – x)2 + (-2 -2)2 = (x – 2)2 + (2 + 2)2

-16x + 4x = 4 -64 ⇒ – 12x = -60
∴ x = 5

Question 5.
If two vertices of an equilateral triangle are (3,4) and (-2,3). Find the co-ordinates of third vertex.
Let A = (3, 4), B = (-2, 3) and C = (x, y)
bydata AB = BC = CA
⇒ AB2 = BC2 = CA2
consider BC2 = CA2
⇒ (x + 2)2 + (y – 3)2 = (x – 3)2 + (y – 4)2

4x + 6x – 6y + 8y = 25 – 13
∴ 10x + 2y =12
5x + y = 6 …… (1)
Again AB = BC
⇒ AB2 = BC2
∴ 5(3 + 2)2 + (4 – 3)2 = (x + 2)2 + (y – 3)2 ………. (2)
from(1) y = 6 – 5x
∴ substitute (1) is (2)
(x + 2)2 + (6 – 5x – 3)2 = 26
x2 + 4 + 4x + 9 + 25x2 – 30x = 26
26x2 – 26x – 13 = 0
⇒ 2x2 – 2x – 1 = 0

Section Formula

Question 1.
Find the co-ordinates of the point which divide the line joining the points
(i) (1, -3) and (-3, 9) internally in the ratio 1 : 3 and the line joining the points
(ii) (2, -6) (4, 3) externally is the ratio 3 : 2.
(i) Let A = (1, -3) B = (-3 9), Ratio is 1 : 3

(ii) Let A = (2, -6) B = (4, 3) ratio 3 : 2

Question 2.
Find the ratio in which (2, 7) divides the line joining the point (8, 9) and (-7, 4).
Let
A = (x1 y1) = (8, 9)
B = (x2 y2) = (-7, 4)
p = (x1 y2) = (2, 7)
The ratio l : m in which p divides AB, is given by

⇒ The ratio is l : m = 2 : 3
∴ the points (2,7) divides internally

Question 3.
Determine the ratio in which the line 3x + y – 9 = 0 divide the segment joining the points (1, 3) and (2, 7).
Let the of division k : 1 as and point then the coordinates of the points are $\left(\frac{2 k+1}{k+1}, \frac{7 k+3}{k+1}\right)$
But the point ‘C’ lies on 3x + y – 9 = 0

⇒ 4x – 3 = 0
⇒ k = $\frac { 3 }{ 4 }$
∴ The required ratio is 3 : 4 internally

Question 4.
Find the lengths of the medians of a triangle whose vertices are (3, 5) (5, 3) and (7, 7).
Median is a tine joining the vertex of a triangle to the middle points of the opposite side.
Let A(3, 5) B(5, 3) C(7, 7) the vertices
Let D, E, F be the mid points of BC, CA, AB respectively

## 1st PUC Basic Maths Question Bank Chapter 14 Standard Angles of Allied Angles

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## Karnataka 1st PUC Basic Maths Question Bank Chapter 14 Standard Angles of Allied Angles

Question 1.
Find the value of
(i) cosec (-765°)
(ii) sec (-1305°)
(iii) sin 585°
(i) -cosec 765°
= -cosec (360x2 + 45°)
= -cosec 45°
= – √2

(ii) sec (1305°)
= sec (360.3 + 225)
= sec 225°
= sec (180 + 45°)
= -sec 45°
= – √2

(iii) sin (360 + 225°)
= sin 225°
= sin (180 + 45°)
= – sin 45°
= – √2

Question 2.
Evaluate: $\frac { 4 }{ 3 }$tan2 120° + 3 sin2 300° – 2cosec2240°- $\frac { 3 }{ 4 }$cot2 60°
Now tan 120° = tan (180° – 60°)
= -tan 60°
= -√3
sin 300° = sin (360° – 60°)
= – sin 60° = $\frac{-\sqrt{3}}{2}$
cosec 240° = cosec (180° + 60°)
= – cosec 60° = $\frac{-2}{\sqrt{3}}$

Question 3.

## 1st PUC Basic Maths Question Bank Chapter 13 Angles and Trigonometric Ratios

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## Karnataka 1st PUC Basic Maths Question Bank Chapter 13 Angles and Trigonometric Ratios

Question 1.
Find the length of the arc and the area of the sector formed by an arc of a circle of radius 8 cm subtending an angle 25° as the centre.

Question 2.
Find the angle between minute hand and the hourhand of a clock when the time is 5.40.
Angle described by the hourhand is 12 hours = 360°
∴ per hour = $\frac{360}{12}$ = 30°

Now the angle described by minute hand is 60° minutes = 360°
i.e., Angle in 40° minutes = 40 × 6° = 240°,

Question 3.
A circular wire of radius 6 cm is cut and bent so as to lie along the circumference of a loop whose radius in 66 cm. Find in degrees the angle which subtended at the centre of the loop.
The length of the arc subtended is the circumference of the circle of radius 6cm.
∴ S = 2πr = 2π × 6 = 127 cm
radius of the loop = 66cm

Question 4.
The angles of a triangle are in A.P and the greatest is double the least. Express the angles in degrees and radians.
Let the angle, be (a – d)° a° and (a + d)°, (a > d > 0°)
∴ (a – d) + a + (a + d) = 180° ⇒ 3a = 180° ⇒ a = 60°
i. e., (a – d) is the least and (a + d) is the greatest,
given (a + d) = 2 (a – d) = a + d = 2a – 2d
⇒ a – 3d = 0 ⇒ 60° – 3d = 0 ⇒ d = 20°

Question 5.
Express 25°. 30′ 10″ in radians.

Trigonometric Ratios of Acute Angle

Question 1.
Prove that tan α$\sqrt{1-\sin ^{2} \alpha}$ = sin α
LHS : tan α . $\sqrt{1-\sin ^{2} \alpha}$ = tan α . $\sqrt{\cos ^{2} \alpha}$
= $\frac{\sin \alpha}{\cos \alpha}$ × cos α = sin α = RHS

Question 2.
Prove that $\frac{1-\cos \theta}{\sin \theta}=\frac{\sin \theta}{1+\cos \theta}$

Question 3.
Prove that $\sqrt{\frac{1-\sin A}{1+\sin A}}$ = sec A – tan A.

Question 4.
Show that $\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}$ = 1 + sec A cosec A.

Question 5.
If x = a cos θ + sin θ. y = a sin θ – b cos θ. Show that x2 + y2 = a2 + b2
Consider x2 = (a cos θ + b sin θ)2
⇒ x2 = a2cos2θ + 2ab cos θ sin θ + b2 sin2θ …….(1)
Again consider
y2 = (a sin θ – b cos θ)2 => y2 = a2 sin2θ – ab cos θ sin θ + b2 cos2 θ …….. (2)
(1) + (2) ⇒ x2 + y2 = a2 + b2

Question 6.
If sin θ = $\frac{3}{5}$ and θ is acute, find the value of
By data sin θ = $\frac{3}{5}$
cos θ = $\frac{3}{5}$ . tan θ = $\frac{3}{4}$, cot = $\frac{4}{3}$

Trigonometric Ratios of some Standard Angles

Question 1.
Find the value of cos $\frac{\pi}{3}$ – sin $\frac{\pi}{6}$ – tan3 $\frac{\pi}{4}$

Question 2.
Find value of 3 tan230° + $\frac{4}{3}$sin260° – $\frac{1}{2}$cosec230° – $\frac{1}{3}$cos2 45°

Question 3.
Find x, if $\frac{x \csc ^{2} 30^{\circ} \cdot \sec ^{2} 45^{\circ}}{6 \cos ^{2} 45^{\circ} \sin 30^{\circ}}$ = tan2 45° – tan2 60°

Question 4.
Show that cos2 $\frac{\pi}{4}$ – cos4 $\frac{\pi}{6}$ + sin4 $\frac{\pi}{6}$ + sin4 $\frac{\pi}{3}$ = $\frac{9}{16}$

Question 5.

Heights and Distances

Question 1.
A person is at the top of tower 75 feet high from there he observes a vertical pole and finds the angles of depression of the top and the bottom of the pole which are 30° and 60° respectively. Find the height of the pole.
Let AB be the tower = 75 feet,
CD be the pole = x
AX is ∥lel to MC
∴ AĈM = XÂC = 30°
AX ∥lel BD
Hence AD̂B = XÂD = 60°

Now from the right triangle ADB, we have
tan 60° = $\frac{\mathrm{AB}}{\mathrm{BD}}$
⇒ BD = $\frac{\mathrm{AB}}{\tan 60^{\circ}}$ = BD = $\frac{75}{\sqrt{3}}$
tan 30° = $\frac{\mathrm{AM}}{\mathrm{MC}}$
⇒ AM = MC . tan 30°
⇒ AM = BD $\frac{1}{\sqrt{3}}=\frac{75}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}}$ = 25
Now x = CD = MB
⇒ x = AB – AM = 75 – 25 = 50 feet

Question 2.
A person standing on the bank of a river observe that the angle of elevation of the top of a tree on the opposite bank is 60°, when he returns 60 feet from the bank, he finds the angle to be 30°. Find the height of the tree and breadth of the river.
PQ represent tree A and B be the points of observation. BQ is the breadth of the river. From the figure

tan 60° = $\frac{P Q}{B Q}$
⇒ PQ = BQ . √3
Again from the figure we have
tan 30° = $\frac{P Q}{A Q}$
⇒ AQ = $\frac{P Q}{\tan 30^{\circ}}$ = AQ = √3PQ
⇒ AQ = √3 . (√3BQ) = 3BQ. Now BQ ⇒ AQ – AB
⇒ BQ = 3BQ – AB
⇒ 2BQ = 60
∴ BQ = 30° is the breadth of the river
∴ height of the tree = PQ = √3.BQ = √3.30 = 30√3

Question 3.
The angle of elevation of a stationary cloud from a point 3000 cm above the lake is 30° and the angle of depression of its reflection in the lake is 60°. What is the height of the cloud above the lake.
D – the image of the cloud C.
AB the surface of the lake
Also CB = BD
Let P be the position of observation.
Thus by data AP = 3000 cm
By data, EP̂C = 30°, EP̂D = 60°

⇒ PE = √3 . CE ….. (1)

⇒ PE = $\frac{\mathrm{ED}}{\sqrt{3}}$ ….. (2)
From (1) and (2) we get
⇒ √3CE = $\frac{1}{\sqrt{3}}$ED = 3CE = ED
⇒ 3(BC – BE) = EB + BD
⇒ 3(BC – 3000) = 3000 + BC (∵ BD = BC and BE PA = 3000)
⇒ 3BC – 9000 = 3000 + BC
⇒ 2BC = 12,000 = BC = 6000 cm
∴ height of the cloud = 600 cms

## 1st PUC Basic Maths Question Bank Chapter 12 Linear Functions

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## Karnataka 1st PUC Basic Maths Question Bank Chapter 12 Linear Functions

Question 1.
A manufacturer produces and sells pens ’10 per unit. His fixed costs are ‘ 600 and variable cost per pen is ‘ 3.50. Calculate
(i) Revenue function
(ii) Cost function
(iii) Profit function
(iv) Breakeven point.
(i) Revenue function R(x) = p.x. = lQx
(ii) Cost function C(x) = ax + b
(iii) Profit function p(x) = Rx – C(x) = 8x – [3.50x + 6500]
(iv) Breakeven point at BEP ⇒ TR = TC ⇒ R(x) = R(x) ⇒ P(x) = 0
∴ 4.50x – 6500 = 0 ⇒ x = $\frac{6500}{4.50}$ = 1445. units
Breakeven point revenue is RS = 1445 × 10 = Rs. 14450

Question 2.
The daily cost of production ‘C’ in R for ‘x unit of an assembly is C(x) = 12.5x + 6400. If each unit is sold for ’25/- then determine the minimum number of units that should be produced and sold to ensure no loss. If the selling price is reduced by Rs. 2.5 unit, What would be the break even point.
By date C(x) = 12.5x + 6400, S(x) = 25x
Breakeven point ⇒ C(x) = S(x)
⇒ 12.5x + 6400 = 25x ⇒ 6400 = 12.5x
⇒ x = $\frac{6400}{12.5}$ = 512
Selling price is reduced by 2.5 unit
S(x) = (25 – 2.5) = (22.5)x
Breakeven point ⇒ C(x) = S(x)
12.5x + 640 = 22.5x
⇒ 6400 = 10x = x = $\frac{6400}{10}$ = 640

## 1st PUC Basic Maths Question Bank Chapter 11 Percentages of Profit and Loss Sales tax, Vat

Students can Download Basic Maths Chapter 11 Percentages of Profit and Loss Sales tax, Vat Questions and Answers, Notes Pdf, 1st PUC Basic 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 Basic Maths Question Bank Chapter 11 Percentages of Profit and Loss Sales tax, Vat

Question 1.
Ravi purchased a pair of shoes costing ‘950. Calculate the total amount to be paid by him, if the rate of sales tax is 7%.
Sales price = Rs. 950

Total amount paid by Ravi = ‘950 + ’57 = ‘1007.

Question 2.
Sneha purchased confectionery goods costing 165 on which the rate of sales tax is 6% and some tooth paste, cold cream, soap etc. costing ‘ 550 which the rate of sales tax is 10%. If she gives 1000 note to the shop keeper, What money will be return to Mrs Sneha?
Price of confectionary goods with sales tax = ’65 + 6% + 165 = ‘174.90
Price of toothpaste, cold cream soap with tax = ‘550 + 10% of 550 = 550 + 55 = 605
∴ the average to be paid = 174.90 + 605 = 774.90
Amount to be returned by shop keeper = 1000 – 774.90 = ‘225.90 paise.

Question 3.
The price of an article inclusive of sales tax of 15% is 3450. Find its marked price if the sales tax is reduced to 6%. How much less does the customer pay for the article?
Let the marked price = ‘x
∴ x + 15% of x = 3450
$\frac{115 x}{100}$ = 3050 ⇒ x = 3000
∴ Market price of the article = ‘3000
Since new sales tax = 60%
Now the customer will pay = ‘3000 + 6% of ‘3000 = $\frac{106}{100}$ × ‘3000 = 3180
∴ customer will pay for the article = 3450 – 3180 = 270 less

Problems on Overhead charges, Discount

Question 1.
A shop keeper buys an article for Rs. 1500 and spends20% of the cost on its packing transportation etc. Then he marked price on it. If he sells the article for ‘2,452.50 including 9% sales tax on the price marked. Find its profit as percent.
Let marked price of the article be ‘x.
Sales tax price = 9% of x = $\frac{9 x}{100}$
According to given statement n + $\frac{9 x}{100}$ = 2452.50
$\frac{100 x+9 x}{100}$ = 2452.50 ⇒ x = 2452.50 × $\frac{100}{109}$ = 2,250
∴ Marked price = ‘2,250 = selling price
Since the shopkeeper buys for ‘1500 and spends 20% of the cost as overhead.
∴ Total cost price of the article = ‘1500 + 20% ‘1500 = ‘1500 + ‘300= ‘1800
Profit = Selling price – total cost price = 2250 – 1800 = 450
∴ Profit % = $\frac{450}{1800}$ × 100 = 22%

Question 2.
Prabha bought an article ‘374. Which included a discount of 15% on the market price and a sales tax of 10% on the reduced price. Find the market price of the article.
Let the market price of the article be ‘x
∴ Discount 15% of x = $\frac{15}{100} \times x=\frac{3 x}{20}$
⇒ Remaining cost = x – $\frac{3 x}{20}=\frac{17 x}{20}$
Sales tax = 10%
∴ price paid by prabha = $\frac{110}{100} \times \frac{17 x}{20}=\frac{187 x}{200}$
Given $\frac{187 x}{200}$ = 374 = x = 374 = $\frac{200}{187}$ = 400
∴ Market price = ‘400

Question 3.
A Shopkeeper buys an article at a rebate of 20% on the printed price. He spends ” 40 on transportation of article. After charging a sales tax of 7% on the printed price he sells the article for ‘1,070. Find his gain on percent.
Let printed price of the article be ‘ x
⇒ Sales tax on it = 7% of ‘x = $\frac{7 x}{100}$
By data x$\frac{7 x}{100}$ = 1070 ⇒ x = 1000
∴ Printed price = ‘1000
Again the shop keeper buy the article at 20% rebate.
∴ The cost price to the shop keeper = ‘1000 – 20% of 1000 = ‘ 800
Since he spends ’40 on the transportation of the article
⇒ Total cost price = ‘800 + ’40 = ‘840
The selling price = printed price = ‘1000
⇒ Profit = ‘100 – ‘840 =’160
And profit = $\frac{160}{840}$ × 100 = 19 $\frac{1}{21} \%$

Question 1.
A Shopkeeper sales an article at its marked price ‘7,500 and charges sales tax at the rate of 12% from the customer. If the shopkeeper pays a VAT of ‘180, calculate the price (inclusive of tax) paid by the shopkeeper.
Since the shopkeeper sells the article for ‘7,500 and changes sales tax at the rate of 12%
∴ tax charged by the shopkeeper = 12% of ‘7500
= $\frac{12}{100}$ x 7500 = 900
VAT = Tax charge – tax paid ,
⇒ ‘180 = ‘900 – Taxpaid
Tax paid by the shopkeeper = ‘ 900 – ‘180 =’ 720
If the shopkeeper buys the article for ‘ x,
Tax on it = 12% of ‘x = ‘720
⇒ x = ‘6,000 ∴ The price (inclusive of tax)
paid by the shopkeeper = ‘6000 + ‘720 = ‘6720.

Question 2.
During the financial year a shopkeeper purchased goods worth ‘4,15,000 and paid a total tax of ‘38,000. His sales during the period consisted of a taxable turnover of ‘50,000 for goods taxable at 5% and ‘3,20,000 for goods takable at 12%. He also sold tax exempted goods worth ‘ 45,000. Calculate his tax liability for the financial year.
Goods taxable at 5% turover = ‘ 50,000
Tax = 50% of ‘50,000 = ‘ 2,500
of goods taxable at 125, turover = ‘3,20,000
tax = 12% of ‘3,20,000 = ‘38,400
Given tax exampled sales = ‘45,000
Total tax = ‘2,500 + ‘38,400 = ‘40,900
Tax paid = ‘38,000
‘ Net tax payable = Total tax changed – tax paid
= ‘40,900 – ‘38,000
=’2,900
∴ Tax liability (under VAT) = ‘2,900

Question 3.
A manufacturing company sold a commodity to its distributor for ‘22,000 including VAT. The distributor sold the commodity to a retailer for ‘ 22,000 excluding tax and the retailer sold it to the customer for ‘ 25,000 plan tax (under VAT). If the rate of tax is 10%. What was the:
(i) Sale price of the commodity for the manufacture?
(ii) An amount of tax received by the state government on the sale of the commodity?
(i) Let the price of the commodity for the manufacture be ‘x, tax is 10%.
∴ tax charged by the manufactures = 10% of C.P

∴ The sale price of the commodity for the manufacture = 20,000

(ii) VAT collected by the manufacture = tax charged by the manufacture

since the distributor has sold the commodity tot he retailer for ‘22,000
∴ tax collected by the distributor = 10% of22,000
= $\frac{10}{100}$ × 22000 = ‘2200 = paid by retailer
VAT to be deposited by the distributor = ‘2200 – ‘2000 = ‘200
Tax collected by retailer = 10% of 25000
= $\frac{10}{100}$ × 25,000 = ‘2500
∴ VAT to be paid by the retailer = ‘2500 – ‘2200 = ‘300
∴ Amount of the (under VAT) received by the state government
= ‘2000 + ‘200 + ‘300 = ‘2500

## 1st PUC Basic Maths Question Bank Chapter 10 Averages

Students can Download Basic Maths Chapter 10 Averages Questions and Answers, Notes Pdf, 1st PUC Basic 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 Basic Maths Question Bank Chapter 10 Averages

Question 1.
A student asked to find the arithmetic mean of the numbers 4, 19, 17, 12, 34,16, 7, – 23,18, 21, 25, and x. Find the mean table 18. What should be the number in place of x.

18 = $\frac{196+x}{2}$ = 216 = 196 + x
∴ x = 216 – 196 = 20

Question 2.
The average of 30 results is 20 and the average of other 20 results is 30. Find the average of results taken to gather.
Given n1 = 30, x̄1 = 20, n2 = 20 , x̄2 = 30

Question 3.
If the average of daily wages of workers of two factories is Rs. 53 and average wages to factory ‘A’ with 250 employees is Rs. 50. Find the average wage of factoring ‘B’ with 200 employees.
Given X̄ = 53, X̄A = 50, nA = 250, X̄B = 7, nB = 200

i.e., 200 X̄B = 23850 – 12500 = 11,350 ∴ X̄B = $\frac{11,350}{200}$ = 56/75
∴ Average wage of factory B = Rs. 56.75

Question 4.
Ten years ago the average age of the family of 4 members was 24 years. Two children have been born. The average age of the family is same to day. Find the present age of two children assuming that the children’s age differ by 2 years.