1st PUC

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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

You can Download Chapter 7 Introduction to C++ 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 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++

You can Download Chapter 6 Object-Oriented Concepts 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 6 Object-Oriented Concepts

1st PUC Computer Science Object-Oriented Concepts One Mark Questions and Answers

Question 1.
What does OOP mean?
Answer:
OOP means Object-Oriented Programming.

Question 2.
What is object oriented programming?
Answer:
Object-oriented programming (OOP) is a programming language model, organized around objects” rather than “actions” and data rather than a logic of the program.

Question 3.
Define object.
Answer:

• An object is a combination of code and data that can be treated as a unit.
• An object can be considered as any real-time entity, (existing thing) that can perform a set of related activities.

Question 4.
Define class.
Answer:
A class is simply a representation of a type of object. It is the blueprint/ plan/ template that describes the details of an object.

Question 5.
Which programming approach does the OOP follow?
Answer:
OOP follows the Bottom-up approach of programming.

Question 6.
What is the meaning of abstraction in OOP?
Answer:
Abstraction in OOP is “A model of a complex system that includes only the details essential to the perspective of the viewer of the system.”

Question 7.
Define encapsulation.
Answer:
The method of combining data, attributes, and methods in the same entity is called encapsulation.

Question 8.
What is polymorphism?
Answer:
Polymorphism is a Greek word that means Many Shapes. In OOP Polymorphism means the ability to take on many forms. The term is applied both to objects and to operations.

Question 9.
What is dynamic binding?
Answer:
Dynamic binding means a link exists between procedure call and code to be executed at run time when that procedure is called for. At run time when a procedure is called for, calls the link between procedure call and code is dynamically chosen and executed.

Question 10.
What is message passing?
Answer:
Message passing is a method by which an object sends data to another object or requests other objects to invoke a method. This is also known as interfacing.

Question 11.
What do you mean by inheritance in OOP?
Answer:
Inheritance is the process by which objects can acquire the properties of objects of some other class.

Question 12.
Write one advantage of OOP.
Answer:
The object-oriented programming provides improved software development productivity over traditional procedure based programming techniques, because of the factors like modularity, extensibility, and reusability.

Question 13.
Write one disadvantage of OOP.
Answer:
Larger program size:
Object-oriented programs typically involve more lines of code than procedural programs.

Question 14.
Mention one application of OOP.
Answer:
Main application areas of OOP are

• User interface design such as windows, menu
• Object-oriented databases

1st PUC Computer Science Object-Oriented Concepts Two/Three Marks Questions and Answers

Question 1.
Explain object-oriented programming briefly.
Answer:
Object-oriented programming is the method of programming, where a system is considered as a collection of objects that interact together to accomplish certain tasks. Objects are entities that encapsulate data and procedures that operate on the data.

Question 2.
Compare Structured Programming and Object-Oriented Programming languages.
Answer:
Malar difference between Structured Programming Language and Object-Oriented
Programming Language are:

Structured Programming Language	Object-Oriented Programming Language
(1) Follows a top-down approach in program design.	(1) Follows a bottom-up approach in program design.
(2) Data and Functions are not coupled with each other.	(2) Functions and data are coupled together.

Question 3.
Write any two important characteristics of OOP.
Answer:
Data encapsulation and inheritance are two important characteristics of OOP.

Question 4.
What is the purpose of object-oriented programming?
Answer:
The main purpose of object-oriented programming is to simplify the design, programming and most importantly debugging of a program. So to modify a particular data, it is easy to identify which function to use.

Question 5.
Briefly explain the class.
Answer:
A class may be defined as a collection of similar objects. In other words, it is a general name for all similar objects. In the OOPs concept the variables declared inside a class are known as “Data Members” and the functions are known as “Member Functions”.

Question 6.
Compare objects and classes.
Answer:
A class is a definition of an object. It is a data type like int. A class is a type, and an object of this class is just like a variable. In other words, the class is a blueprint and an object can be considered as any real-time entity (existing thing) that can perform a set of related activities.

Question 7.
Explain about message passing in object-oriented programming?
Answer:
Message passing is a method by which an object sends data to another object or requests another object to invoke a method. It acts like a messenger from one object to another object to convey specific instructions.

Question 8.
State the advantages of object-oriented programming.
Answer:
Some of the advantages of object-oriented programming are as follows:

1. A clear modular structure can be obtained which can be used as a prototype and it will not reveal the details of the design.
2. Easy maintenance and modifications to the existing objects can be done with ease.
3. A good framework is provided which facilitates creating rich GUI applications.

Question 9.
Explain about inheritance in OOPS?
Answer:
Inheritance is the process of creating a new class, called derived class from an existing or base class. The derived class inherits all the capabilities of the base class. Using inheritance some qualities of the base class are added to the newly derived class, apart from its own features. Inheritance permits code reusability.

Question 10.
Briefly explain about polymorphism.
Answer:
Polymorphism is the ability of an object to take more than one form in different instances. For example, one function name can be used for different purposes. Similarly, the same operator can be used for different operations. There are two types of polymorphism namely “compile-time polymorphism” and “run-time polymorphism”.

Question 11.
Briefly explain about dynamic binding.
Answer:
In OOPs dynamic binding refers to linking a procedure call to the code that will be executed only at run time. The code associated with the procedure in not known until the program is executed, which is also known as late binding.

Question 12.
Explain the usage of encapsulation?
Answer:
Encapsulation is the method of combining the data and functions inside a class. This hides the data from being accessed from outside a class directly, and only through the functions inside the class, it is able to access the information. This is also known as “Data Abstraction”, as it gives a clear separation between properties of data type and the associated implementation details.

Question 13.
How is message passing done in OOP? Explain briefly.
Answer:
Message passing is nothing but sending and receiving of information by the objects, similar to people exchanging information. In OOPs, message passing involves specifying the name of objects, the name of the function, and the information to be sent.
The following are the basic steps in message passing.

• Creating classes that define objects and their behaviour.
• Creating objects from class definitions.
• Establishing communication among objects.

1st PUC Computer Science Object-Oriented Concepts Five Marks Questions and Answers

Question 1.
Write a short note on the techniques of programming.
Answer:
1. Procedural programming technique:
It is a list of instructions telling a computer, step-by-step, what to do, usually having a linear order of execution from the first statement to the second and so forth with occasional loops and branches. Procedural programming languages include c, C++, etc.,
Some of the benefits of the procedural programming methodology are:

• Easy to read program code.
• Easily maintainable program code, as various procedures can be debugged in isolation.
• The code is more flexible as one can change a specific procedure that gets implemented across the program.

2. Structured programming technique:
Structured programming deals only with logic and code, and suggests making use of programming structures such as sequence, selection, iteration and modularity in programs.
Features:

• It focuses on techniques for developing good computer programs and problem solution.
• It is most important to consider single-entry and single-exit control in a program and structures.
• Structured code is like a page, which can be read from the top to bottom without any backward references.
• Read from top to bottom makes the code easy to read, test, debug, and maintain.

3. Object-oriented technique:
Object-oriented programming is a method of programming where a system is considered as a collection of objects that interact together to accomplish certain tasks. Objects are entities that encapsulate data and procedures that operate on the data. The main purpose of object-oriented programming is to simplify the design, programming and most importantly debugging a program.

So to modify a particular data, it is easy to identify which function to use. The important elements of OOP are objects, classes, abstraction, encapsulation, polymorphism, and inheritance.

Question 2.
What are the features of object-oriented programming languages?
Answer:
Programs are divided into objects

• Data structures are designed such that, they characterize the objects.
• Functions that operate on the data of an object are tied together in the data structure.
• Data is hidden and cannot be accessed by external functions.
• Objects may communicate with each other through functions.
• New data and functions can be easily added whenever necessary.
• It follows a bottom-up approach in program design.
• Concentration is on data rather than on procedures.

Question 3.
What are the benefits of OOP?
Answer:
Through inheritance, we can eliminate redundant code and extend the use of existing classes.

• It saves the program development time and higher productivity.
• The use of data hiding helps the programmer to build secure programs that cannot be invaded by code, in other parts of the program.
• It is possible to have multiple instances of an object to co-exist without any interference.
• It is possible to map objects in the program domain to those objects in the program.
• It is easy to partition the work in a project based on objects.
• It can be upgraded from small to large systems.
• It makes the interface descriptions with external systems much simpler by using message passing techniques.
• Software complexity can be easily managed.

Question 4.
Write a short note on the characteristics of OOP.
Answer:
Basic Characteristics of Object-Oriented Programming:
1. Objects:

• An object represents an individual, identifiable item, unit, or entity, either real or abstract, with a well-defined role in the problem domain/An object may be tangible things like a car, printer, is the sentence incomplete.
• In object-oriented programming, a problem is analyzed in terms of objects. Each object contains data and code to manipulate the data. It can be defined as Object = Data + Methods or functions

2. Classes:
A class may be defined as a collection of similar objects. In other words, it is a general name for all similar objects. For example, mango, apple, banana all may be described under the common name fruits. A class serves as a blueprint or a plan or a template.

3. Inheritance:
Inheritance is the process by which objects of one class acquire the properties of objects of another class. Inheritance allows to create classes which are derived from other classes so that they automatically include their “parent’s” properties, plus their own. The concept of inheritance provides the idea of reusability, which means additional features can be added to an existing class without modifying it.

4. Reusability:
The reusability implies the reuse of existing code in another program without modification to it. The concept of inheritance provides the basis for reusability in OOP.

5. Creating new data types:
Creating a class in object-oriented programming can be considered as creating new data types.

6. Polymorphism and overloading:
The property of object-oriented programming polymorphism is the ability to take more than one form in different instances. For example, the same function name can be used for different purposes. Similarly, the same operator can be used for different operations.

Question 5.
Explain the advantages and disadvantages of OOP.
Answer:
The main advantages of OOP are:

• It is easy to model a real system as real objects are represented by programming objects in OOP.
• With the help of inheritance, we can reuse the existing class to derive a new class such that the repetition of code is eliminated and the use of the existing class is extended. This saves time and the cost of a program.
• In OOP, data can be made private to a class such that only member functions of the class can access the data. This principle of data hiding helps the programmer to build a secure program.
• With the help of polymorphism, the same function or same operator can be used for different purposes. This helps to manage software complexity easily.
• Large problems can be reduced to smaller and more manageable problems. It is easy to partition the work in a project based on objects.
• It is possible to have multiple instances of an object to co-exist without any interference i.e. each object has its own separate member data and function.

The few disadvantages of OOP are:
1. Size:
Object-Oriented programs are much larger than other programs. In the early days of computing, space on hard drives, floppy drives and in memory was at a premium. Today we do not have these restrictions.

2. Effort:
Object-Oriented programs require a lot of work to create. Specifically, a great deal of planning goes into an object-oriented program well before a single piece of code is ever written. Initially, this early effort was felt by many to be a waste of time. In addition, because the programs were larger (see above) coders had to spend more time while writing the program.

3. Speed:
Object-Oriented programs are slower than other programs, partially because of their size. Other aspects of Object-Oriented programs also demand more system resources, thus further slowing down the program.

Question 6.
Bring out the differences between structured programming and object-oriented programming languages
Answer:
The differences between Structured Programming language and Object-Oriented Programming language are:
1. Structured Programming Language:

• It follows a top-down approach in program design.
• Data and Functions are not coupled with each other.
• Large programs are divided into smaller self-contained program segments known as functions.
• Data moves openly around the system from function to function.
• Functions are dependent so reusability is not possible.

2. Object-Oriented Programming Language:

• It follows a bottom-up approach in program design.
• Functions and data are coupled together.
• Programs are divided into entities called Objects.
• Data is hidden and cannot be accessed from outside that class.
• Functions are not dependent so reusability is possible.

Question 7.
Explain in detail about data encapsulation and inheritance.
Answer:
1. Encapsulation:
It is the method of combining the data and functions within a class. This hides the data from being accessed from outside a class directly and only through the functions inside the class one is able to access the information.

This is also known as “Data Abstraction”, as it gives a clear separation between properties of data type and the associated implementation details. There are two types, known as “function abstraction” and “data abstraction”. A Function when used without knowing how implemented function abstraction. Data abstraction is using data without knowing how the data is stored.

2. Inheritance:
Inheritance is the process of creating a new class, called derived class, from an existing or base classes. The derived class inherits all the capabilities of the base class but in addition, can have its own capability. Inheritance permits code reusability. Once a base class is written and debugged, it need not be touched. Reusing existing code saves time and money and increases a program’s reliability.

Inheritance can also help in the original conceptualization of a programming problem, and in the overall design of the program. A programmer can use a class created by another person or company, and without modifying it, derive at other classes which are suited to particular situations.

Question 8.
Write a note on polymorphism and dynamic binding.
Answer:
Polymorphism:
In object-oriented programming, polymorphism is a generic term that means ‘many shapes’, (from the Greek meaning “having multiple forms”). Polymorphism is briefly described as “one interface, many implementations.” There are two types of polymorphism one is compile-time polymorphism and the other is run time polymorphism.
For example, consider the operation of addition.

• If the operands are strings, then the operation would produce a third-string by concatenation.
• If the operands are numbers, it will generate a sum by dynamic binding.
• It refers to the linking of a procedure call to the code to be executed in response to the call.
• Dynamic binding means, that the code associated with a given procedure call is not known until the time of the call at run time.
• It is associated with polymorphism and inheritance.

Question 9.
Explain briefly dynamic binding and message passing.
Answer:
Binding refers to the linking of a procedure call to the code to be executed in response to the call.

1. Dynamic binding:
Dynamic binding (also known as late binding) means the code associated with a given procedure is not known until the time of the call at runtime.

2. Message passing:
An object-oriented program consists of a set of objects that communicate with each other the process of programming in an object-oriented language, and so it involves the following Basic steps

• Creating classes that objects and their behaviour.
• Creating objects from class definitions, and
• Establishing communication among objects.

You can Download Chapter 9 Input Output Operators 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 9 Input Output Operators

1st PUC Computer Science Input Output Operators One Mark Questions and Answers

Question 1.
What is a stream?
Answer:
A stream is an object, where a program can either insert or extract characters to/from it.

Question 2.
Give the classification of streams.
Answer:

1. Input streams
2. Output streams

Question 3.
What is an input stream?
Answer:
Input stream is a sequence of characters from any input device like a keyboard which is inserted to the program of a computer.

Question 4.
What is an output stream?
Answer:
Output stream is a sequence of characters taken out from the program to output device like a monitor, or printer.

Question 5.
What is meant by stream extraction?
Answer:
cin is used in conjunction with >> operator, known as extraction or get from the operator.

Question 6.
What is meant by stream insertion?
Answer:
cout is used in conjunction with << operator, known as insertion or put to the operator.

Question 7.
Give the syntax of cin statement.
Answer:
Syntax: cin >> variable.

Question 8.
Give the syntax of cout statement.
Answer:
Syntax: cout<< expression or manipulator

Question 9.
What is cascading in I/O operations?
Answer:
The cascading is a way to extract/insert multiple values from/into more than one variable using one cin/cout statement.

Question 10.
How is cascading useful in output operations?
Answer:
The cascading allows the user to use cout once only but use << operator several times to output many values.

Question 11.
What are manipulators?
Answer:
A manipulator in C++ is used to control the formatting of output and/or input values.

Question 12.
Give an example for manipulator.
Answer:
end1 is a manipulator.

1st PUC Computer Science Input Output Operators Five Marks Questions and Answers

Question 1.
Describe I/O operator.
Answer:
1. Input Operator:
The statement cin>> num;
is an input statement and causes the program to wait for the user to type in a number. The operator >>is known as extraction or get from the operator. It takes the value from the keyboard and assigns it to the variable on its right.

2. Output Operator:
The statement cout<< “ the numbers”;
uses the cout identifier that represents the standard output stream ( screen) in C++. The operator < – Sounds incomplete -Author.

Question 2.
Explain manipulators.
Answer:
A manipulator is a C++ is used to control the formatting of output and/or input values. Manipulators can only be present in input/output statements. The end1 manipulator causes a newline character to be output. end1 is defined in the <iostream> header file and can be used as long as the header file has been included.

Question 3.
Explain input and output cascading with an example each.
Answer:
The multiple use of << or >>in a one statement is known for cascading.
Cascading of output operator ( >>):
cout<<“ Hello “<<“ ISC
cout<< “Value of B=”<< b;
Cascading of input operator (>>):
int n1,n2,n3;
cin >> n1 >> n2 >> n3 ;
cin>>n1>>n2;

You can Download Chapter 10 Control Statements 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 10 Control Statements

1st PUC Computer Science Control Statements One Mark Questions and Answers

Question 1.
What is a statement?
Answer:
Statements are the instructions given to the computer to perform specific of actions. Action may be in the form of data movement, decision making, etc.

Question 2.
What are control statements?
Answer:
Control statements are elements in the program that can control and alter the sequence of flow of constructions in any program execution.

Question 3.
Write the classification of control statements.
Answer:
The control statements are classified as

1. Selection statements
2. Iterative statements and
3. Jump statements.

Question 4.
What is meant by a compound statement?
Answer:
A compound statement is a grouping of statements enclosed within a pair of braces ({ }), in which each individual statement ends with a semi-colon.

Question 5.
What are selection statements?
Answer:
The selection statements are used to execute certain block of statements after evaluating the condition.

Question 6.
What is a block?
Answer:
A group of statements separated by semicolons and grouped together and enclosed within brackets marked with { } is called a block.

Question 7.
What is the purpose of ‘if’ statement?
Answer:
The ‘if’ statement is used to decide on whether a statement or a block should be executed or not.

Question 8.
What is the other name of ‘if’ statement?
Answer:
The other name of ‘if’ statement is one-way branching statement.

Question 9.
Give the possible output of condition evaluation.
Answer:
The possible output of condition evaluation is either TRUE or FALSE.

Question 10.
What is the numerical equivalent of TRUE or FALSE?
Answer:
The numerical equivalent of TRUE is 1 and FALSE is 0.

Question 11.
What is the other name of ‘if-else’ statement?
Answer:
The other name of ‘if-else’ statement is two-way branching.

Question 12.
When is ‘if-else’ statement used?
Answer:
‘If else’ statement is a conditional branching statement used to execute a set of codes when the condition is true, otherwise executes a different set of the codes in the “else” part.

Question 13.
What is an ‘if-else-if’ statement?
Answer:
The ‘if-else if’ statement is an extension of the “if-else” conditional branching statement. When the expression in the “if’ condition is “false” another “if-else” construct is used to execute a set statement based on an expression.

Question 14.
What is ‘nested if’?
Answer:
An if statement nestling inside another ‘if’ statement is termed as ‘a nested if’ statement.

Question 15.
Give the name of multi-branch selection statement.
Answer:
The multi-branch selection statement is called as ‘switch-case’ statement.

Question 16.
Give the purpose of the switch statement.
Answer:
The ‘switch’ statement is used in C++ for testing whether a variable is equal to any one of a set of values.

Question 17.
Define the term looping.
Answer:
Looping means execution of a statement or block of statements repeatedly until the condition is true.

Question 18.
Why is ‘while’ loop called a pre-tested looping statement?
Answer:
The ‘while’ loop evaluates the condition in the beginning itself, to select a suitable statement for execution repeatedly. Therefore it is called a pre-tested looping statement.

Question 19.
What is post-tested looping statement?
Answer:
The loop evaluates the condition at the end of the looping structure and selects a statement for execution repeatedly. Therefore it is called a post-tested looping statement.

Question 20.
Name the post-tested looping statement.
Answer:
The ‘do-while’ statement is a post-tested looping statement.

Question 21.
Which structure is called a fixed-execution looping statement?
Answer:
The ‘for’ looping statement is called a fixed-execution looping statement.

Question 22.
When is ‘for’ used?
Answer:
If the programmer knows the exact number of repetitions to be carried out among the set of statements then ‘for’ conditional statement is used in a program.

Question 23.
What is nesting of loop?
Answer:
If the loop appears inside the body of another loop it is called a nested loop.

Question 24.
What is the purpose of ‘break’ statement?
Answer:
The ‘break’ statement can be used to terminate a repeated structure (loops) such as ‘while’, do-while ‘and’ ‘for’ and multi-branching statements like ‘switch’.

Question 25.
What is the use of ‘goto’ statement?
Answer:
The ‘goto’ statement transfers/jump control from one part of the program to some other part.

Question 26.
What is the function of ‘continue’ statement?
Answer:
The ‘continue’ statement is used in loops and causes a program to skip the rest of the body of the loop.

Question 27.
Why is the ‘exit() library’ function used?
Answer:
The execution of a program can be stopped at any point with ‘exit ( )’ and a status code can be informed to the calling program.

1st PUC Computer Science Control Statements Two/Three Marks Questions and Answers

Question 1.
What are control statements? How are they classified?
Answer:
Control statements are elements in the program that control the flow of program execution. Control statements are classified as,

1. Selection statements
2. Iterative statements
3. Jump statements

Question 2.
What are selection statements? Give the different types of selection statements.
Answer:
Selection statements are used to execute certain block of statements by evaluating the conditions.
The different selection statements are

1. ‘if’ statement
2. ‘if-else’ statement
3. ‘if-else-if’ (nested) statement
4. ‘switch’ statement

Question 3.
Give the syntax of ‘if’ statement.
Answer:
The syntax of if statement is
if (condition)
{
statement 1;
statement 2;
statement n;
}

Question 4.
Write the flowchart representation of ‘if’ statement.
Answer:

Question 5.
Write the syntax of ‘if-else’ statement.
Answer:

Question 6.
Write the flowchart representation of ‘if-else’ statement.
Answer:

Question 7.
Write the syntax of if-else-if statement.
Answer:

Question 8.
Write the flowchart representation of ‘if-else-if’ conditional structure.
Answer:

Question 9.
Write the syntax of‘switch-case’ statement.
Answer:

Question 10.
Write the comparison between ‘if-else’ and ‘switch-case’ statement.
Answer:
The ‘if-else’ statements permit two-way branching, whereas ‘switch’ statement permits multiple branching.

Question 11.
Write the syntax of ‘while’ loop.
Answer:
Syntax of ‘while’ loop:
while(condition)
{
statement(s);
}

Question 12.
Write the syntax of ‘do-while’ loop.
Answer:
Syntax of ‘do-while’ loop
do
{
statements;
} while (condition);

Question 13.
Write the flowchart of representation of while loop.
Answer:

Question 14.
Write the flowchart representation of ‘do-while’ loop.
Answer:

Question 15.
Give the difference between ‘break’ and ‘exit()’ statements.
Answer:
The ‘break’ statement can be used to terminate a repeated structure (loops) such as ‘while’, do-while and for and multi-branching statements like ‘switch’.
The execution of a program can be stopped at any point with ‘exit ( )’.

Question 16.
Give the difference between ‘goto’ statement and ‘continue’ statement.
Answer:
The ‘goto’ statement transfers/jump control from one part of the program some other part. The ‘continue’ statement is used in loops and causes a program to skip the rest of the body of the loop.

1st PUC Computer Science Control Statements Five Marks Questions and Answers

Question 1.
Explain the working of ‘ if’ statement and ‘if-else’ statement with examples.
Answer:
Example for ‘if’ statement:

Result:
B is greater than A
In the above example the condition in the if statement returns a true value and so the text “B is greater than A” is displayed.
Example for ‘if else’ statement:

Result:
B is greater than A.
In the above example, the “if’ condition is false so the code in the “else” part is executed.

Question 2.
Explain the working of if-else if statement with a suitable example.
Answer:
The ‘if-else if’ statement is an extension of the “if-else” conditional branching statement. When the expression in the “if’ condition is “false” another “if-else” construct is used to execute a set of statements based on the expression.
Syntax:
if(expression)
{ statements }
else if(expression)
{ statements }
else if(expression)
{ statements }
Example: for ‘if else if’ statement


Result:
Enter your Percentage::60
First Class
In the above example, the ‘if else if’ statement is used to check multiple conditions based on the percentage of marks got, as input. The user entered percentage as 60, then first condition evaluated and returned with false. This resulted in checking for second if condition and resulted in true which give the print “first class”.

Question 3.
Write a short note on ‘switch-case’ statement.
Answer:
A ‘switch’ statement allows a variable to be tested for equality against a list of values. Each value is called a case, and the variable being switched on is checked for each case.
Syntax:
The syntax for a switch statement:
switch(expression)
{
case constant-expression : statement(s);
break; //optional
case constant-expression : statement(s);
break; //optional
default                              : statement(s);                 //Optional
}
The following rules apply to a switch statement:

1. The expression used in a ‘switch’ statement must have an integer or enumerated type.
2. We can have any number of case statements within a ‘switch’. Each case is followed by the value to be compared to, and a colon.
3. The constant-expression for a case must be the same data type as the variable in the ‘switch’,
   and it must be a constant or a literal.
4. When the variable being switched on is equal to a case, the statements following that case will execute until a ‘break’ statement is reached.
5. When a ‘break’ statement is reached, the ‘switch’ terminates, and the flow of control jumps to the next line following the ‘ switch ’ statement.
6. Not every case needs to contain a break. If no break appears, the flow of control will fall through to subsequent cases until a ‘break’ is reached.
7. If no conditions match, then the code under the default statement is executed.
8. The default case is used for performing a task when none of the cases is ‘true’.
9. No break is needed in the default case.

Question 4.
Explain ‘switch-case’ statement with a suitable program example.
Answer:
‘ Switch’ statement compares the value of an expression against a list of integers or character constants. The list of constants are listed using the “case” statement along with a “break” statement to end the execution.
Example:


Result:
Enter the day of the week between 1-7::7
Sunday
In the above Control Structure example, the “switch” statement is used to find the day of the week from the integer input got from the user. The value present in the day is compared for equality with constants written in the word case. Since no equality is achieved in the above example (from 1 to 6), when 7 is entered default is selected and gives “Sunday” as a result.

Question 5.
Explain ‘while’ loop structure with an example.
Answer:
A ‘while’ loop statement repeatedly executes a statement or sequence of statements written in the flower brackets as long as a given condition returns the value ‘true’.
Syntax:
The syntax of a ‘while’ loop in C++ is:
while(condition)
{
statement (s) ;
}
Here the condition may be any expression, and for true is any non zero value. The loop iterates while the condition is true. When the condition becomes false, program control passes to the line immediately following the loop. During the first attempt, when the condition is tested and the result is false, the loop body will be skipped and the first statement after the ‘while’ loop will be executed.
Example:

When the above code is executed, it produces the following result:
value of a: 10
value of a: 11
value of a: 12
value of a: 13
value of a: 14

Question 6.
Explain the working of ‘for’ loop structure.
Answer:
A ‘for’ loop is a repetition control structure, that allows you to efficiently write a loop that needs to execute a specific number of times.
Syntax:
The syntax of a ‘for’ loop in C++ is:
for (initialization; condition; increment/decrement)
{
statement(s);
}
The working of a ‘for’ loop:

1. The initialization step is executed first, and only once in the beginning. It is used to declare and initialize loop control variables.
2. Next, the condition is evaluated. If it is true, the body of the loop is executed. If it is false, the body of the loop does not execute and the flow of control jumps to the next statement just after the ‘for’ loop.
3. After the body of the ‘for’ loop executes, the flow of control jumps back up to the increment/ decrement statement and update any ‘loop’ control variables.
4. Then the condition is evaluated again. If it is true, the ‘ loop’ executes and the process repeats itself (body of ‘loop’, then increment step, and then again condition). After the condition becomes false, the ‘for’ loop terminates.

Example:

When the above code is compiled and executed, it produces following result:
value of a: 10
value of a: 11
value of a: 12
value of a: 13
value of a: 14
value of a: 15

You can Download Chapter 5 Problem Solving Methodology 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 5 Problem Solving Methodology

1st PUC Computer Science Problem Solving Methodology One Mark Questions and Answers

Question 1.
Define problem-solving.
Answer:
It is the process of expressing the solution of a specific problem, in terms of simple operations that can be understood by the computer.

Question 2.
What is the problem definition?
Answer:
The process of understanding the given problem and what the solution must do is known as problem definition.

Question 3.
What are the steps involved in problem analysis?
Answer:
The steps involved in problem analysis are:

1. Data we need to provide (input) and
2. Information we want the program to produce (the output).

Question 4.
What is the important aspect of the design of a solution?
Answer:
The most important aspect of developing a solution is developing the logic to solve the specific problem.

Question 5.
Write the tools used in the design of a solution.
Answer:
The algorithm and flowchart tools are used in the design of a solution.

Question 6.
Define an algorithm.
Answer:
An algorithm is a “step by step procedure to solve a given problem infinite number of steps”.

Question 7.
Define the flowchart.
Answer:
A flowchart is a pictorial or graphical representation of a solution to any problem.

Question 8.
How are flowcharts classified?
Answer:
Flowcharts are classified as system flowchart and program flowchart.

Question 9.
What is a pseudo code?
Answer:
Pseudo code is structured English that consists of short, English phrases used to explain specific tasks within a program’s algorithm.

Question 10.
Define coding.
Answer:
The process of translating the algorithmic solution or flowchart solution into a set of instructions in a programming language is called as coding.

Question 11.
What is testing?
Answer:
It is the process of checking the program logic, by providing selected sample data and observing the output for correctness.

Question 12.
What do you mean by debugging?
Answer:
The process of detecting the errors and correcting the errors in a program is called as debugging.

Question 13.
What is the function of compiler?
Answer:
It is a translator software which converts source program into its equivalent machine language object program.

Question 14.
Define source program.
Answer:
The program written using high level language is called source program.

Question 15.
Define object program.
Answer:
A machine language program generated by the compiler is called object program.

Question 16.
What is syntax error?
Answer:
It refers to an error in the syntax of a sequence of characters or tokens that is intended to be written in a particular programming language.

Question 17.
What are semantic errors?
Answer:
An error, which occurs due to the incorrect logic and a solution is called a semantic error.

Question 18.
What are run time errors?
Answer:
The errors that may occur during execution of the program are called run time errors.

Question 19.
Name the two types of program documentation.
Answer:
The two types of documentation are internal documentation and external documentation.

Question 20.
Define program maintenance.
Answer:
Program maintenance is the process of periodic review of the programs and modifications based on user requirements.

Question 21.
What is sequential construct?
Answer:
The ability of executing the program statement one after another in sequence is called sequential construct.

Question 22.
Define selection.
Answer:
It is the process of selecting a certain set of statements based on a requirement for execution.

Question 23.
Define iteration.
Answer:
It is the process of repeating the execution of a certain set of statements again and again until a requirement is satisfied.

Question 24.
What is the simple if also called as?
Answer:
The simple if is also called a one-way branch.

Question 25.
What is the if-else construct is also called as?
Answer:
The if-else construct is also called a two-way branch.

Question 26.
What is the if-else-if construct is also called as?
Answer:
The if-else-if construct is also called a multiple-way branch.

Question 27.
When is the multiple selection construct used?
Answer:
If there are more than two alternatives to be selected for execution then multiple selection construct is used.

Question 28.
What are the two types of iterative constructs?
Answer:
The two iterative constructs are conditional looping and unconditional looping.

Question 29.
What is top-down design?
Answer:
It is the process of dividing a problem into subproblems and further dividing the subproblems into smaller subproblems and finally to problems that can be implemented as program statements.

Question 30.
What is bottom-up design?
Answer:
It is the process of beginning design at the lowest level modules or subsystems and progressing upwards to the design of the main module.

Question 31.
What is structured programming?
Answer:
It is an easy and efficient method of representing a solution to a given problem using sequence, selection and iteration control.

Question 32.
What is a modular design technique?
Answer:
In this technique, a given problem is divided into a number of self-contained independent program segments. Each program segment is called a ‘module’ and a module can be called for in another program or in another module.

1st PUC Computer Science Problem Solving Methodology Two/Three Marks Questions and Answers

Question 1.
What does the programming task involves?
Answer:
The programming task involves defining and analyzing the problem and developing the solution logically, using an algorithm.

Question 2.
Which activity is represented by a rectangle and a rhombus symbol in the flowchart?
Answer:

1. The rectangle symbol represents the process or calculation activity.
2. Rhombus symbol represents decision making or branching activity.

Question 3.
What is the use of the assignment statement? Give an example.
Answer:
The assignment statement is used to store a value in a variable. For example, let A = 25

Question 4.
What are the input and output statements?
Answer:
The input statement is used to input value into the variable from the input device and the output statement is used to display the value of the variable on the output device.

Question 5.
Give the general form of a simple if statement.
Answer:
If (test condition) then
Statement 1

Question 6.
Give the general form of if-else statement.
Answer:
If (test condition) then
Statement;
Else
Statement;

Question 7.
What is unconditional looping? Give an example.
Answer:
If a set of statements are repeatedly executed for a specified number of times, is called unconditional looping. For example, for a conditional statements.

Question 8.
What is the difference between a program flowchart and system flowchart?
Answer:
A program flowchart details the flow through a single program. Each box in the flowchart will represent a single instruction or a process within the program.
A system flowchart will show the flow through a system. Each box will represent a program or a process made up of multiple programs.

Question 9.
Give the general form of for conditional structure.
Answer:
The general form of for conditional structure is
For (initialization; condition; increment/decrement)
statement 1
statement2
statement

Question 10.
Give the characteristics of a good program.
Answer:
Modification and portability are the two important characteristics of a good program.

Question 11.
Give the list of statements that can be used in structured programming.
Answer:

• Sequence of sequentially executed statements.
• Conditional execution of statements.
• Iteration execution statements.

Question 12.
Give the list of statements that cannot be used in structured programming.
Answer:

1. go to statement
2. break or continue statement
3. multiple exit points.

Question 13.
Mention the advantages of modular programming.
Answer:
Code reusability, localized errors, and team, work are the few advantages of modular programming.

1st PUC Computer Science Problem Solving Methodology Five Mark Questions and Answers

Question 1.
Explain the stages of problem-solving methodology.
Answer:
The stages of problem-solving methodology are
1. Problem definition:
The problem should be clearly understood by the solution provider. One has to analyze what must be done rather than how to do it and then is requires to developing the exact specification of the problem.

2. Problem Analysis:
In problem analysis, we try to understand what are the inputs to be specified and what are the required outputs.

3. Design of a solution using design tools:
The design of a solution includes a sequence of well-defined steps that will produce the desired result (output). Algorithms and flowcharts are used as design tools and represent the solution to a problem.

4. Coding:
The process of writing program instructions for i.e., it is the process of transforming algorithm/flowchart into a program code using programming language instructions.

5. Debugging:
It is the process of detecting and correcting the errors in the program. This stage is also referred to as verification.

6. Program Documentation:
It is a reference material that contains details about a program and functions of different programs of software. Documentation helps other users to understand the program and use it conveniently more effectively.

Question 2.
Explain the characteristics of the algorithm.
Answer:
Characteristics of the algorithm

1. It must be simple.
2. Every step should perform a single task.
3. There should not be any confusion at any stage.
4. It must involve a finite number of instructions.
5. It should produce at least one output.
6. It must give a unique solution to the problem.
7. The algorithm must terminate and must not enter into infinity.

Question 3. What are the advantages and disadvantages of an algorithm?
Answer:
Advantages:

1. Easy to understand since it is written in universally a spoken language like English.
2. It consists of a finite number of steps to produce the result.
3. Easy to first develop the algorithm.
4. It is independent of any programming language, (universal).
5. Easy program maintenance.

Disadvantages:

1. It is time-consuming and difficult to understand for larger and complex problems.
2. Understanding complex logic through algorithms would be difficult.

Question 4.
Write the steps involved in developing a flowchart.
Answer:
Steps involved in developing a flowcharts

1. Understand the problem statement clearly before developing the flowchart.
2. Study the outputs to be generated and the required inputs to solve the problem.
3. Design the process in such a way that it produces the desired result.
4. Test the flowchart by giving test data.
5. Verify the result for correctness. Make suitable changes, if required, and repeat the process.

Question 5.
What are the advantages and disadvantages of a flowchart?
Answer:
Advantages:

• It is a means of communication and easy to understand.
• Easy to convert into a program code.
• Independent of programming language, i.e., A flowchart can be used to write programs using different programming languages.
• Easy to test the program for errors and easy removal of such errors.

Disadvantages:

• It is time consuming process as it makes use of a number of symbols.
• It is difficult to show the complex logic using a flowchart.
• Any changes in the flowchart needs redrawing the flowchart again.

Question 6.
Write a note on program errors.
Answer:
The different program errors are as follows;
1. Syntax error:
An error occurs when there is a violation of the grammatical rules of a programming language’s instructions. It happens at the time of compilation. Such errors need to be rectified before proceeding further.

2. Semantic errors:
An error, which occurs due to the incorrect logic in a solution is called semantic error. It also occurs due to the wrong use of grammar in the program.

3. Runtime Errors:
occur at run-time. Such an error causes a program to end abruptly or even cause system shut-down. Such errors are hard to detect and are known as ‘Bugs’.

4. Logical Error:
It may happen that a program contains no syntax or run-time errors but still, it doesn’t produce the correct output. It is because the developer has not understood the problem statement properly. These errors are hard to detect as well. It may need the algorithm to be modified in the design phase and changing sources code.

Question 7.
Write a short note on flowchart symbols.
Answer:
Symbols used in flowcharts

Question 8.
Explain the top-down approach in brief.
Answer:
Top-Down Approach:
It is based on a concept called divide and conquer. A given problem is solved by breaking it down into smaller manageable parts called modules. Hence it is also called as stepwise refinement. The subprograms are further divided into still smaller subproblems. Finally, the subproblems are solved individually, and all these give the solution to the overall problem.

Properties of Top-Down Analysis:

1. Understandability: The individual modules are organized to execute in a particular sequence.
2. This helps to understand the program behaviour more easily.
3. Clear Identification of tasks.
4. Easy program maintenance.
5. Removes duplication or repetition of coding in a problem.
6. Enhances the feature of code reusability.

Question 9.
Write a short note on structured programming.
Answer:
Structured Programming:
The concept was contributed by Professor Dijkstra and other colleagues made it popular. Structured Programming deals only with logic and code and suggests making use of programming structures such as sequence, selection, iteration and modularity in programs.
Features:

1. It focuses on techniques for developing good computer programs and problem-solving.
2. The structures can be repeated one within another.
3. It is most important to consider single-entry and single-exit control in a program and structure.
4. Structured code is like a page, which can be read from the top to bottom without any backward references.
5. Reading from top to bottom makes the code easy to read, test, debug, and maintain.

Advantages:

1. Programs are easy to write because the programming logic is well organized.
2. Programs can be functionally divided into smaller logical working units (modularity).
3. Modularity leads to understanding the program, test and debug easily.
4. Easy to maintain because of single entry and single exit features.
5. Eliminates the use of undisciplined controls (GOTO, BREAK, etc.,) in the program.

You can Download Chapter 10 Random Variables and Mathematical Expectation 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 10 Random Variables and Mathematical Expectation

1st PUC Random Variables and Mathematical Expectation Two Marks Questions and Answers

Question 1.
Define Random variable.
Answer:
Random variable is a function which assigns a real number to every sample point in the sample space.

Question 2.
Define discrete random variable
Answer:
A random variable ‘ X ’ which takes the specified values x₁x₂,…,x_n with a respective probabilities p₁, p₂,….., p₁ is a discrete random variable.

Question 3.
Define continuous random variable.
Answer:
A random variable which assumes all the possible values in its range is called a continuous random variable.

Question 4.
Define Mathematical expectation of a random variable/Mean of random variable.
Answer:
Let ‘X’ be a discrete random variable which can takes the values x₁, x₂, x₃,… ,x_n with respective probabilities p₁, p₂,p₃,… , p_n then, the mathematical expectation of ‘ X ’ is:E(X) = x₁p₁ + x₂P₂ + x₃P₃ + – + x_nP_n
Then E(X) = Σ x p(x) ;it is also called as mean of discrete random variable (X)

Question 5.
Define probability distribution.
Answer:
A systematic presentation of the values taken by a random variable with respective probabilities is called the probability distribution of a random variable’.

Question 6.
Probability mass function (pmf).
Answer:
Probability mass function (pmf): Let ‘ X ’ be a discrete random variable . And let p(x) be a function : p(x) = p(X=x) . Then p(x) is called the probability mass function of ‘ X ’. If the following conditions are satisfied,

• p(x) ≥ 0 for all values X, and
• Σ p(x) =1.

Question 7.
Probability density function (pdf).
Answer:
Probability density function (pdf): Let ‘ X ’ be a continuous random variable taking values in the interval [a, b], then,
A function f(x) is said to be the probability density function of the continuous random variable ‘ X ’, if it satisfies the following conditions :

• f(x) ≥ 0 for all ‘ X’ in the interval [a, b] .
• For two distinct numbers c & d in the interval [a, b]:
  P(c ≤ X ≤ d) = (Area under the probability curve between ordinates at X =c and X=d).
• Total area under the curve is 1. i.e. , P(- ∞ < x < ∞) = 1 .

Question 8.
Writedown the formulae of all Mathematical Expectations:
Answer:

• E(X) = XΣX.P (X)
• E(a) = a,
• E(aX) = a E(X) and
• E(aX+b) = aE(X)+b.

Question 9.
Write down the Formulae of Variances:
Answer:

• Var(X)= {E(X²)-[E(X)]²} ,
• V (a)=0,
• V(aX)=a² V(X),
• V(aX+b)=a²V(X), and S.D.(X)= \(\sqrt{\mathrm{v}(\mathrm{x})}\).

Question 10.
Write down the Formula of Co-Variance:
Answer:
Cov(X, Y) =E(XY) – E(X).E(Y),

Question 11.
Write down the co-efficient of correlation:
Answer:

Question 12.
State Addition theorem expectation
Answer:
Statement: Let X and Y be two random variables with respective expectations E(X) and E(Y). Then, E(X + Y)=E(X) + E(Y)

Question 13.
State Multiplication of expectation.
Answer:
statement: Let X and Y be two independent random variables with respective expectations E(X) and E(Y). Then expectation of the product of these random variables is; E(XY) = E(X) E(Y)

Question 14.
If E(X)=2, then find the value of E(-x/4)
Answer:
Given E(-x/4) = E(-l/4. x) =-l/4.E(X) = -1/4. 2 =-1/2 since E(aX) = a E(X)

Question 15.
If E(X)=2/5, find E(5x/3)
Answer:
Given E(5x/3) = E(5/3.x) = 5/3.E(x) = 5/3.2/5 = 2/3

Question 16.
If X is a random variable and E(X) =1.5, then what is the value of E(3 + 4X)?
Answer:
E(3 + 4X) = E(4x + 3) = 4.E(x) + 3 = 4.(1.5) + 3 = 6 +3 = 9 since
E(ax+b) = aE(x)+b

Question 17.
If E(X) = 2, find the value of E(3X – 6).
Answer:
E(3X – 6) = 3E(x) – 6 = 3(2) – 6 =6 – 6 =0

Question 18.
If E(X+Y) =7 and E(X) =4, then find the value of E(Y).
Answer:
We know that E(X+Y) = E(X) + E(Y) ;
7 = 4 + E(y)
E(y) = 7 – 4 = 3

Question 19.
If E(X) = 3 and E(Y) = 5, then find E(3X+2Y), E(3X -Y)
Answer:
E(3X + 2Y) = 3 E(x) + 2 E(y) = 3(3) + 2(5) = 9 + 10=19
E(3X – Y) = 3 E(X) + (-1) E(Y) = 3(3)- 5 = 9 – 5 = 4

Question 20.
If X and Y are two independent random variable, E(XY)=10 and E(X)= 4, then what is the value of E(Y)?
Answer:
We know for any two independent random variables:
E(XY) = E(X)E(Y);
10 = 4 E(Y); E(Y)= 10/4 = 2.5

Question 21.
What is the value of co-efficient of correlation for two independent variables?
Answer:
Zero ie., r = 0

Question 22.
For a random variable X, E(X²) = 50 and V(X)=14, then what is E(X)?
Answer:
We know that V(X) = E(X²)-[E(x)]²
14 = 50-[E(x)]²
[E(x)]² = 50 – 14 = 36
Squaring on both sides, we get E(x) = \(\sqrt{36}\) = 6

Question 23.
If E(X)=3 and E(X²)=45, find the value of S.D.
Answer:
We know S.D(x)
\(\begin{array}{l}{=\sqrt{\mathrm{V}(\mathrm{x})}=\sqrt{\mathrm{E}(\mathrm{X} 2)-[\mathrm{E}(\mathrm{x})] 2}} \\ {=\sqrt{45-[3]^{2}=\sqrt{45-9}}=\sqrt{36}=6}\end{array}\)

Question 24.
If E(X) = 4, E(X²) = 25, find V(5X – 9).
Answer:
V(5x-9)this is of the type V(aX+b)=a²V(X)
V(5x-9) = 5². V(x)
Here V(X) = E(X²) – [E(x)]²;
= 25 – 4² = 25 – 16 = 9
∴ V(5×-9) = 5². V(x) = 25(9) = 225

Question 25.
If V(X)=4 then find V(2X)
Answer:
V(2x) = 2²V(x) = 4 (4) = 16; Since V(aX) = a² V(X)

Question 26.
If V(X) = 5 then find V(-5X).
Answer:
V(-5X) = (5)² V(x) = 25(5) = 125; Since V(aX)=a²V(X)

Question 27.
For a random variable V(X) = 4. Find V(3X +4) and V(-3X).
Answer:
V(3X+4) = 3² V(x) = 9 (4) = 36;
V(-3X) = (-3)² V(x) = 9 (4) = 36; Since V(aX+b)=a² V(X),

Question 28.
If V(x)=1.6, then find V(2X – 5).
Answer:
V(2X-5) = 2² V(x) = 4 (1.6) = 6.4; Since V(aX+b)=a² V(X),

Question 29.
If E(X) = 5, E(Y) = 2, E(XY) = 11, then find Cov(X,Y).
Answer:
We know that Cov(X, Y) = E(XY) – E(X).E(Y)
= 11 – (5) (2)=11 -10= 1.

Question 30.
When X and Y are two discrete random variables with E(X)=5, E(Y)=6, E(X²)=34, E(Y²)=52, and E(XY)=25, then find Cov(X,Y) andin a bi-variate data
Answer:
Cov (X, Y) = E(XY)-E(X).E(Y) = 25 – (5) (6) = 25 – 30 = -5;

Question 31.
If V(X)=V(Y)=Cov(X,Y), then what is the value of ?
Answer:
γ = 1

Question 32.
If P(X = 3) = 2/3 and P(X = -2) = 1/3, then find E(X) and V(3X + 4)
Answer:
Given probability distribution is:

E(x) =8/3;
V(x) = E(X²)-[E(X)]²
= – (8/3)²
= 22/3-7.0756 = 7.3333-7.0756 = 0.2577
V(3X + 4) = 3². V(X) = 9 (0.2577) = 2.3193

Question 33.
If P(X) is a probability mass function (p.m.f) then what is ΣP (x)?
Answer:
ΣP(X)=1

Question 34.
Find the mean and standard deviation for the following probability distribution.

Answer:
We Know that : Mean = E(x) : Σx.P(X)=1
\(=(-2) \times \frac{1}{5}+(-1) \times \frac{2}{5}+0 \times \frac{1}{5}+2 \times \frac{1}{10}+4 \times \frac{1}{10}\)
E(x) =-0.4 -0.4 + 0 +0.2+ 0.4 = -0.2
Var(x) = E(x²) – [E(x)]²
Here E(x²) = [x². P(x)]
\(=(-2)^{2} \times \frac{1}{5}+(-1)^{2} \times \frac{2}{5}+0^{2} \times \frac{1}{5}+2^{2} \times \frac{1}{10}+4^{2} \times \frac{1}{10}\)
= 0.8+ 0.4+ 0 + 0.4+ 1.5
E(x²) = 3.2
∴ Var(x) =3.2-(-0.2)² = 3.2 – 0.04 = 3.16

Question 35.
If two coins are tossed once. Find the mean and variance of number of Heads.
Answer:
S = {HH, HT, TH, TT}
The probability of number of Heads is :

E(x) = Σx. P(x) = \(0 \times \frac{1}{4}+1 \times \frac{2}{4}+2 \times \frac{1}{4}\)
Var (x) = E(X²) [E(x)]²
E(x²) = Ex². P(x) .
\(=0^{2} \times \frac{1}{4}+1^{2} \times \frac{2}{4}+2^{2} \times \frac{1}{4}=0+0.5+1=1.5\)
∴ Var (x) = 1.5 – 1² = 0.5

Question 36.
A fair die thrown once. A person will gets ₹ 5 if the faces of a die results in mulitple of 2. other wise he loses ₹ 2. find his expected amount.
Answer:
Let x denote the expected amount is random variable which can take: 5 (die results in multiple of 2 : 2.46) and -2( die results in 1, 3, 5), with respective probabilities.
P(x = 5) = P(getting multiple of 2) = \(\frac{3}{6}=\frac{1}{2}\)
P(x = -2) = P(Not getting multiple 2) = \(\frac{3}{6}=\frac{1}{2}\)
The probability distribution is :

Expected amount: E(x) = Σx. P(x) = \(5 \times \frac{1}{2}-2 \times \frac{1}{2}=2.5-1\)

Question 37.
A person tosses a fair coin, he gets ₹ 50 if head turns up and loses ₹ 10 if tail appears. Find the expected amount, also find the standard deviation.
Answer:
Let x denote the person to gets the amount is a random variable, which takes the values 50 and -10, with respective probabilities:
P(x = 50) = P(getting Head) = \(\frac { 1 }{ 2 }\) and P(x = -10) = P(getting tail) = \(\frac { 1 }{ 2 }\)
The probability distribution is :

The expected amount: E(x) = Σx. P(x) = 50 × \(\frac { 1 }{ 2 }\) -10 × \(\frac { 1 }{ 2 }\) = 25 – 5 = ₹20

Here E(x²) = Ex². P(x) = (50)² × \(\frac { 1 }{ 2 }\) + (-10)² × \(\frac { 1 }{ 2 }\)
E(x²) = 1250 + 50= 1300
∴ S.D(x) = \(\sqrt{1300-(20)^{2}}=\sqrt{900}\) =₹ 30.

Question 38.
A box has 5 blue and 3 yellow marbles. Two marbles are drawn at random. Find the expected number of blue marbles drawn.
Answer:
Let x denote the number of blue marbles drawn, is a random varible which takes the values: 0, 1,2. The respective probabilities are calcualted as below:
P(x = 0) = P(drawing no blue marbles)
= P(2 yellow marbles) = \(\frac{^{3} \mathrm{C}_{2}}{^{8} \mathrm{C}_{2}}=\frac{3}{28}\)
P(x = 1) = P(drawing one blue and one yellow marbles)
\(=\frac{^{5} \mathrm{C}_{1} \times^{3} \mathrm{C}_{1}}{^{8} \mathrm{C}_{2}}=\frac{15}{28}\)
P(x = 2) = P(drawing two blue marbles) = \(\frac{^{5} \mathrm{C}_{2}}{^{8} \mathrm{C}_{2}}=\frac{10}{28}\)
Then the probability distribution is :

Expected number of blue marbles: E(x) = Σx P(x)
\(=0 \times \frac{3}{28}+1 \times \frac{15}{28}+20 \times \frac{10}{28}=0+0.5357+0.3571\)
∴ E(x) =0.8928

Question 39.
The following is the probability distribution of x. Find k and E(x).

Answer:
For a probability mass funciton : EP(x) = 1
\(\text { i.e., } \frac{1}{4}+\frac{1}{2}+\frac{3}{10}+k+\frac{1}{8}=1\)
\(\mathrm{k}=1-\left[\frac{1}{4}+\frac{1}{2}+\frac{3}{10}+\mathrm{k}+\frac{1}{8}\right]\) ∴K = 1 – 0.8925 = 0.1075
∴ The probability distribution is

E(x) = Σx. P(x) = (1 × 0.25) + (2 × 0.33) + (3 × 0.1875) + (4 × 0.1075) + (5 × 0.125) = 2.5275

Question 40.
From the following probability distribution of x, Find k, E(x), E(x + 3) and Uar (x).

Answer:
By the definition of p.m.f: Σ P(x) = 1
\(\therefore \quad \frac{1}{5}+\frac{2}{5}+\frac{1}{5}+k=1\)  \(\left.\mathrm{k}=-1-\left[\frac{1}{5}+\frac{2}{5}+\frac{1}{5}\right]=0.2+0.4+0.2\right]\)
k = 1 – 0.8 = 0.2
To find E(x), E(2x + 3), (x) and (2x + 3) we need compute from the probability distribution.

∴ E(x) = Σx. P(x) = 1.4
Var (x) = E(x²) – [E(x)]² = Σx². P(x) – (1.4)²
= 3 – 1.96= 1.04
E(x + 3) = Σ(x + 3) P(x) = 7.4

Question 41.
If x is a discrete random variable and let a, b be two constants, show that
1. E(a) = a
2. E(ax) = aE(x)
3. E(ax + b) = aE(x) + b
4. Var (a) = 0
5. Var (ax) = a² Var (x)
6. Var (ax + b) = a² Var (x).
Answer:
Proof: From the definitions of probability mass function and mathematical expectation
1. E(a) = Σa .P(a) = a. ΣP(x) = a.l = a. ∵ΣP(x) = 1

2. E(ax) = Σax P(x) = aΣx P(x) = a. E(x) ∵Σx P(x) = E(x)

3. E(ax + b) = E(ax + b). P(x) Removing bracket
= Σax. P(x) + Σb.P(x) = aΣx . P(x) + bΣP(x)
= a. E(x) + b. 1 = aE(x) + b

4. Var (a) = E[a – E(a)]²; ∴ V(x) = E[x – E(x)]²
= E[a – a]² = E(0) = 0

5. Var (an) = E[ax – E(ax)]² = E[ax – aE(x)]²; ∴E(ax) = aE(x)
= a²E(x – E(x)]² = a² Var (x)

6. Var (ax + b) = E[(ax + b) – E(ax + b)]²

= E[ax – aE(x)]² = a². E[x – E(x)]² = a² Var (x)

Question 42.
Define joint/ Bivariate probability function.
Answer:
Let x and y be two discrete random variables, then P(x, y) is called bivariate or joint probability function of x and y, if P[x = x] and P(y = y)

Question 43.
Define margined probability functionof x.
Ans.
Margined probability function of x: P(x) is obtained from the joint probability faction of P(x,y).

Question 44.
Define marginal probability function of y.
Ans.
Marginal probability function of y: P,(y) 1’s obtained from the joint probability function of P(x, y).

Question 45.
State and prove the addition theorem of mathematical expectation. ’
Answer:
Statement: Let x and y be random variables with respective expectations E(x) and E(y). Then, the expectations of the sum of the random variables is
E(x + y) =E(x) + E(y)
Proof: Let x and y be two discrete random variables, with their joint probability function : P(x, y) as,
P(x,y) =P(x = x, y = y)
The marginal probability distributions of x and y are:

Question 46.
State tne prove multiplication theorem of mathematical expectation.
Answer:
Statement: Let x and y be two independent discrete random variables with respective expectations E(x) and E(y). then the expectation of the product of these random variables is: E(xy) = E(x) . E(y)
Proof: Let x and y be two independent discrete random variables, then joint probability function P(x, y) is
P(x, y) = P₁(x). P₂(y) ……(i)
Then, by the defintion of mathematical expectation.

From the definition mathematical expectation
E(x y) = E(x) . E(y), Hence the proof.

Question 47.
Find, Mean, Variance of the following probability distribution.

Answer:
Mean = E(x) = Σx . P(x)
= (-2) × 0.05 + (-1) 0.2 + 0 × 0.25 + 1 × 0.4 + 2 × 0.1
= -0.1 – 0.2 + 0 + 0.4 + 0.2 = 0.3
Var(x) =E(x²) – [E(x)]²
E(x²) = Σx². P(x) = (-2)² x 0.05 + (-1)² x 0.2 + 0 x 0.25 + 1 x 0.4 + 2 x 0.1
= 0.2 + 0.2 + 0 + 0.4 + 0.2 = 1
∴ Var (x) = 1 – (0.3)² = – 0.09 = 0.91

Question 48.
For the following probability distribution find E(x), E(2x + 5), E(x2 + 3), V(x), V(2x).

Answer:
E(x) = Σx . P(x)
= 1 × 0.3 + 2 × 0.4 + 3 × 0.3
= 0.3 + 0.08 + 0.9 = 20
and E(x²) = Σx². P(x)
= 1² × 0.3 + 2² × 0.4 + 3² × 0.3
= 0.3 +1.6+ 2.7 = 4.6
Var (x) = E(x²) – [E(x)]² = 4.6 – (2)² = 0.6
E(2x + 5); From the results : E(ax + b) = aE(x) + b
= 2E(x) + 5; Since E(x) = 2
E(2x + 5)=2 × 2 + 5 = 9
E(x² + 3) = E(x²) + 3 = 4.6 + 3 = 7.6
∴ Var (2x) = 2² Var (x) = 4 x 0.6 = 2.4

Question 49.
From the Following probability distribution. find k, E(x), E\(\left(\frac{x}{3}\right)\), E(-4x) Var (x) Var (5x) var \(\left(\frac{2 x}{3}\right)^{3}\), Var (-3x).

Answer:
We know from the probability mass function. ΣP(x) = 1
\(\therefore \quad \frac{\mathrm{k}}{4}+\frac{\mathrm{k}}{8}+\frac{\mathrm{k}}{2}+\frac{\mathrm{k}}{8}=1\) \(k\left[\frac{1}{4}+\frac{1}{8}+\frac{1}{2}+\frac{1}{8}\right]=1\)
K.1 = 1; ∴K = 1
The probability distribution reduce now,

\(\mathrm{E}(\mathrm{x})=\Sigma \mathrm{x} . \mathrm{P}(\mathrm{x})=(-2) \times \frac{1}{4}+0 \times \frac{1}{8}+1 \times \frac{1}{2}+2 \times \frac{1}{8}\)
= -0.5 + 0+ 0.5 + 0.25
E(x) = 0.25
\(\mathrm{E}\left(\mathrm{x}^{2}\right)=(-2)^{2} \times \frac{1}{4}+0^{2} \times \frac{1}{8}+1^{2} \times \frac{1}{2}+2^{2} \times \frac{1}{8}\)
= 1 + 0 + 0.5 + 0.5 = 2
Var (x) = E(x²) – [E(x)]²
= 2 – (0.25)² = 2 – 0.0625 = 1.9375
\(\mathrm{E}\left(\frac{\mathrm{x}}{3}\right)=\mathrm{E}\left(\frac{1}{3} \cdot \mathrm{x}\right)\); From E(ax) = aE(x)
\(=\frac{1}{3} \mathrm{E}(\mathrm{x})=\frac{1}{3} \times 0.25=0.083\)
E(-4x) = -4E()x = -4 × 0.25 = -1
Var (5x) = 5² var (x); from Var (ax) = a² Var (x)
= 25 × 1.9375 = 48.4375
Var\(\left(\frac{2 x}{3}\right)\) = Var\(\left(\frac{2}{3} \cdot x\right)\)
= \(\left(\frac{2}{3}\right)^{2}\)Var(x) = \(\frac { 4 }{ 9 }\) x 1.9375 = 0.861
Var (-3x) = (-3)² Var (x) = 9 × 1.9375 = 17.4375

Question 50.
A purse has 5 Ten rupee coins, 3 five rupee coins and 2 two rupee coins. A coin is drawn at random from the purse. Find expectation of amount draw.
Answer:
Let x denote the expected amount drawn, is a discrete random variable which takes the values : x = 10, 5, 2 with the prababilities
P(x = 10) = P(drawing ₹ 10 coin) = \(\frac { 5 }{ 10 }\)
P(x=5)= \(\frac { 3 }{ 10 }\). P(x = 2)=\(\frac { 2 }{ 10 }\)
Then, the probability distribution is

Expected amount draw : E(x) = Ex. P(x)
\(=10 \times \frac{5}{10}+5 \times \frac{3}{10}+2 \times \frac{2}{10}=\frac{50}{10}+\frac{15}{10}+\frac{4}{10}\) = ₹ 6.90

Question 51.
In a lottery there are 1000 tickets costing ₹5 each. There is one first prize worth ₹ 1000. Two second prizes worth ₹ 500 each and ten third prizes worth ₹ 100 each. Find expected gain in buying one lottery ticket.
Answer:
Let x denote the expected gain is a discrete random variable which takes the values x : 995, 495, 95 and -5. Since he buys a ticket of for ₹ 5 is deducted). The respective probability can be calculated as below:
p(x = 995) = \(\frac { 1 }{ 1000 }\): p(x = 495) = \(\frac { 2 }{ 1000 }\)
p(x = 95) = \(\frac { 10 }{ 1000 }\) p(x = -5) = \(\frac { 987 }{ 1000 }\)
The probability distribution is :

Expected gain : E(x) = Σx. P(x)
\(=995 \times \frac{1}{1000}+495 \times \frac{2}{1000}+95 \times \frac{10}{1000}-5 \times \frac{987}{1000}\)
= 0.995 + 0.99 + 0.95 – 4.985 = -2.47
i.e., expected loss in buying a lottery ticket is ₹ 247.

Question 52.
A person enters into a fair by paying ₹10. In the fair if choose to hit a target with airgun by paying ₹ 1 per shot if he hits gets ₹ 50. The probability of hitting a target is 1/2. Find the expect amount.
Answer:
Let x denote the expect amount is a discrete random variable, which takes the values : -11, 39.
The respective probabilities
P(x = -11) = P(not hitting a target) = \(\frac { 2 }{ 3 }\)
P(x = 39) = P(hitting a target) = \(\frac { 1 }{ 3 }\)
Therefore, the expected amount is :
E(x) = Σx. P(x)
= -11 × \(\frac { 2 }{ 3 }\) + 39 × \(\frac { 1 }{ 3 }\) = -7.33 + 13 = Rs. 5.67 profit.

Question 53.
Find expected sum of numbers obtained in a throw of two dice.
Answer:
Let x and y be the numbers appeared on the faces of two dice is a discrete random variable, then each die show with numbers 1, 2,3, 4, 5, 6 with respective probabilities \(\frac { 1 }{ 6 }\) each.
Expected number obtained on the first die be : E( x) = Σx . P(x)
\(=1 \times \frac{1}{6}+2 \times \frac{1}{6}+3 \times \frac{1}{6}+\ldots \ldots+6 \times \frac{1}{6}=3.5\)
Similarly, expected number on the faces of 2^nd die be E(y) = 3.5
The expectation of sum of numbers obtained is
E(x + y) = E(x) + E(y) = 3.5 + 3.5 = 7
Note: Similarly, sum of’ ‘n’ dice of numbers can be calcualted.

Question 54.
A bag has ten tiny of dice. Two dice are randomly drawn. Find the expectation of the product of the numbers obtained on the faces of dice.
Answer:
Let x and y denoted the numbers obtained on the faces of dice, is a discrete random variable with values 1, 2,…. 6 and with the probability \(\frac { 1 }{ 6 }\) each.
Therefore E(x) = \(1 \times \frac{1}{6}+2 \times \frac{1}{6}+\ldots \ldots+6 \times \frac{1}{6}=3.5\)
Similarly, E(y) =3.5
Since, drawing dice are independens, therefore
E(x, y) = E(x) . E(y) = 3.5 × 3.5 = 12.25
Note: Similarly product of’n’ dice of numbers can also be calculated in the square way.

Question 55.
If in a bivoriate data, E(x) = 4, E(x²) = 25, E(y) = 8 and E(y²) = 100 and E(x y ) = 24. Find Var (x), Var (y), Cov (x, y) and the coefficient of correlation (3xy).
Answer:
Var (x) = E(x²) – [E(x)]² = 25 – 4² = 9
Var (x) = E(y²) – [E(y)]² = 100 – 8² = 36
COV (x, y) = E(xy) – E(x) . E(y) = 24 – 4 x 8 = -8

Question 56.
In a bivariate data E(x) = 1.3. E(x²) = 3.4, E(y) = 0 and E(y²) = 4. E(xy) = 0.1 Find S.D(x), SD(y) and rxy.
Answer:

Question 57.
For the following bivariate probaiblity distribution, find E(x + y) and E(xy).

Answer:

Here 0.08, 0.17, 0.25, 0.08, 0.25 and 0.17 are joint probability function P(x, y) of discrete random variables x and y.
E(x.y) =(-2) × 0 × 0.08 + 0 × 0 × 0+1 × 0 × 0.25 + (-2) × 1 × 0.17 + 0 × 1 × 0.25 + 1 × 1 × 0.17 +(2) × 2 × 0 + 0 × 2 × 0.08+ 1 × 2 × 0.
= (0 + 0 + 0) +(0.34 + 0 + 0.17) + (0 + 0 + 0)
∴ E(x.y) =0.17

= (-2 + 0) 0.08 + (0 + 0) 0 + (1 + 0) 0.25 + (-2 + 1) 0.17 + (0 + 1)0.25 + (1 + 1)0.17+ (-2 +2) +0 + (0 + 2) 0.08+ (1 +2)0 = (-0.16 + 0 + 0.25) + (-0.17 + 0.25 + 0.34) + (0 + 0.16 + 0)
= 0.09 + 0.42 + 0.16.
E(x + y) =0.67

Question 58.
From the following bivariate probability distribution, find the coefficient of correlation.

Answer:
The coefficient of correlation is :

Here Var(x) = E(x²) – [E(x)]²
‘To calculate, obtain the marginal probility distribution of x:

Here : 0.05 + 0.25 + 0.1 = 0.4 and 2 : 0.4 + 0 + 0.2 = 0.6
E(x) = Ex . P(x) = 1 × 0.4 + 2 × 0.6 = 1.6
E(x²) = Ex². P(x) = 1² × 0.4 + 2² × 0.06 = 2.8
∴ Var (x)= 2.8 — (1.6)² = 0.24

The marginal probability distribution of y:

E(y) = Ey P(y) = (-1) × 0.45 + (-2) × 0.25 + 0 × 0.3 = – 0.95
E(y2) = (-1 )² × 0.45 + (-2)² × 0.25 + 0² × 0.3 = 0.45 + 1 + 0 = 1.45
∴ Var(y) = E(y²) – [E(y)]² = 1.45 – (-0.95)² = 1.45 – 0.9025 = 0.5475
cov (x, y) = E(xy) – E(x) . E(y)

= (1 × (-1) × 0.05) + (2 × (-1) × 0.4) + (1 × (-2) × 0.25) + (2 × (-2) × 0) + (1 × 0 × 0.1) +(2 × 0 + 0.2)
= -0.05 -0.8 -0.5 + 0 + 0 + 0
E(xy) =-1.35
∴ cov(x,y) =-1.35- 1.6 x (-0.95) = -1.35 + 1.52 = -0.17

Question 59.
If E(x²) = 65, and E(x) = 4. Find standard deviation of x.
Answer:
Here,

Var(x) = E(x²) = [E(x)]² = 65 – (4²) = 65 – 16 = 49

Question 60.
IfVar(x)=0.4327, and E(x2) = 0.75. Find E(x).
Answer:
Var(x) = E(x²) – [E(x)]²
0.4327 = 0.75 – [E(x)]²
[E(x)]² =0.75-0.4327 = 0.3153
∴ E(x) = \(\sqrt{0.3153}\) =0.5615

Question 61.
If x and y be two independent random variable with E(xy) = 8, and E(x) = 2, then find E(y).
Answer:
Since x and y are independent then E(xy) = E(x). E(y)
i.e., 8 = 2. E(y) ∴ E(y) = 2

Question 62.
For a bivariate data if E(x) = 5, E(y) = 3, E(x²) = 45, E(y²) = 85 and E(x) = 18. Find yxy.
Answer:

so, Var (x) = E(x²) = [E(x)]² = 45 – (5)² = 45 – 25 = 20
Var (y) = E(y²) – [(y)]² = 85 – (3)² = 85 – 9 = 76
COV (x, y) = E(xy) – E(x) . E(y) =18-5 x 3 = 18- 15 = 3
∴ \(\gamma=\frac{3}{\sqrt{20} \times \sqrt{76}}=\frac{3}{4.472 \times 8.718}=\frac{3}{38.987}=0.0769\)

Question 63.
If E(x) = 5 and E(y) = 12, find E(x + y), E(x – y), E(4x + 2y), and E(-3x + 2y).
Answer:
E(x + y) = E(x) + E(y) , by theorem = 5 + 12= 17
E(x-y) = E(x) – E(y) = 5 – 12 = -7
E(4x + 2y) = 4E(x) + 2E(y) = 4 × 5 + 2 × 12 = 20 + 24 = 44
E(-3x + 2y) = (-3) E(x) + 2E(y) = (-3) × 5 + 2 × 12 = 15 + 24 = 9

Question 64.
If Var (x) = 3, find Var (5x), Var (-8x), Var (-x), Var (4x – 5) and Var (8 – 3x)
Answer:
V(5x); By the results ; Var (ax) = a² Var (x)
∴ Var (5x) = 5² Var (x) = 25 x 3 = 75
Var (-8x) = (-8)² Var (x) = 64 x 3 = 192
Var (-x) = Var (-1 .x) = (-1)2 Var (x) = 1 x 3 = 3.
Var (4x – 5); using the result; Var (ax + b) = a² Var (x)
∴ Var(4x – 5) = 4² Var (x) = 16 x 3 = 48
Var (8 – 3x) = Var (-3x + 8) = (-3)² Var (x) = 9 x 3= 27