Home Quantitative Aptitude Sets,Functions and Relations-Test 1

Sets,Functions and Relations-Test 1

Subject :- QT

Chapter :- Sets,Functions and Relations-Test 1

Questions :- 25

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Question 1 of 24

1. Question
1 points
Out of 20 members in a family,11 like to take tea and 14 like coffee.Assume that each one like at least one of the two drinks.Find how many like the both drinks.
- 2
- 3
- 4
- 5
Correct

$l e t T : s e t o f p e o p l e l i k e t e a a n d C : s e t o f p e o p l e l i k e c o f f e e T h e n n (T) = 11, n (C) = 14 a n d n (T \cup C) = 20 n (T \cup C) = n (T) + n (C) - n (T \cap C) n (T \cap C) = 11 + 14 - 20 = 5$

Incorrect

$l e t T : s e t o f p e o p l e l i k e t e a a n d C : s e t o f p e o p l e l i k e c o f f e e T h e n n (T) = 11, n (C) = 14 a n d n (T \cup C) = 20 n (T \cup C) = n (T) + n (C) - n (T \cap C) n (T \cap C) = 11 + 14 - 20 = 5$
Question 2 of 24

2. Question
1 points
In a group there are 70 people ,45 speak hindi,33 speak english and 10 speak neither english nor hindi.Find how many people can speak both the languages?
- 13
- 19
- 18
- 28
Correct

$l e t H : s e t o f p e o p l e s p e a k h i n d i a n d E : s e t o f p e o p l e s p e a k e n g l i s h ‘ s o, n (H) = 45, n (E) = 33, n (E \cup H) = 70 - 10 = 60 n (E \cup H) = n (H) + n (E) - n (E \cap H) n (E \cap H) = 45 + 33 - 60 = 18$

Incorrect

$l e t H : s e t o f p e o p l e s p e a k h i n d i a n d E : s e t o f p e o p l e s p e a k e n g l i s h ‘ s o, n (H) = 45, n (E) = 33, n (E \cup H) = 70 - 10 = 60 n (E \cup H) = n (H) + n (E) - n (E \cap H) n (E \cap H) = 45 + 33 - 60 = 18$
Question 3 of 24

3. Question
1 points
In a survey of 300 companies,the number of companies using different media-Newspapers(N),Radio(R) and Television(T) are as follows:n(N)=200,n(R)=100,n(T)=40,n(N∩R)=50,n(R∩T)=20,n(N∩T)=25 and n(N∩R∩T)=5.Find the number of companies using none of these media:
- 20
- 30
- 40
- 50
Correct

n(NURUT)=n(N)+n(R)+n(T)-n(N∩R)-n(N∩T)-n(R∩T)+n(N∩R∩T)

=200+100+40-50-25-20+5

=250

no.of companies not using any media

=n(S)-n(NURUT)

=300-250

=50

Incorrect

n(NURUT)=n(N)+n(R)+n(T)-n(N∩R)-n(N∩T)-n(R∩T)+n(N∩R∩T)

=200+100+40-50-25-20+5

=250

no.of companies not using any media

=n(S)-n(NURUT)

=300-250

=50
Question 4 of 24

4. Question
1 points
In a town of 20000 families it was found that 40% families buy newspaper A,20% families buy newspaper B and 10% families buy newspaper C,5% buy newspaper A and B,3% buy B and C and 4% buy A and C.if 2% families buy all the three newspapers,then the number of families which buy A only is:
- 6600
- 6300
- 5600
- 5300
Correct

n(S)=20000

n(A)=40% of 20000=8000

n(B)=20% of 20000=4000

n(C)=10% of 20000=2000

n(A∩B)=5% of 20000=1000

n(B∩C)=3% of 20000=600

n(C∩A)=4% of 20000=800

n(A∩B∩C)=2% of 20000=400

now,we have to find n(A∩B’∩C’)

n(A∩B’∩C’)=n[A∩(BUC)]

=n[A∩(BUC)’]

=n(A)-n[A∩(B∩C)]

=n(A)-n[(A∩B)U(A∩C)]

=n(A)-[n(A∩B)+n(A∩C)-n(A∩B∩C)]

=8000-(1000+800-400)

=6600

Incorrect

n(S)=20000

n(A)=40% of 20000=8000

n(B)=20% of 20000=4000

n(C)=10% of 20000=2000

n(A∩B)=5% of 20000=1000

n(B∩C)=3% of 20000=600

n(C∩A)=4% of 20000=800

n(A∩B∩C)=2% of 20000=400

now,we have to find n(A∩B’∩C’)

n(A∩B’∩C’)=n[A∩(BUC)]

=n[A∩(BUC)’]

=n(A)-n[A∩(B∩C)]

=n(A)-n[(A∩B)U(A∩C)]

=n(A)-[n(A∩B)+n(A∩C)-n(A∩B∩C)]

=8000-(1000+800-400)

=6600
Question 5 of 24

5. Question
1 points
$i f f : R - R b e s u c h t h a t f (x) = 2^{x}, t h e n f (x + y) =$
- f(x)+f(y)
- f(x).f(y)
- f(x)/f(y)
- none of these
Correct

$f (x) = 2^{x} f (x + y) = 2^{x . y} = 2^{x} 2^{y} = f (x) . f (y)$

Incorrect

$f (x) = 2^{x} f (x + y) = 2^{x . y} = 2^{x} 2^{y} = f (x) . f (y)$
Question 6 of 24

6. Question
1 points
Out of 150 students,45 passed in Accounts,30 passed in Economics and 50 passed in Maths,30 both in Accounts and Maths , 32 both in Maths and Economics , 35 both in Accounts and Economics,25 passed in all the 3 subjects.Find the numbers who passed at least in any one of the subjects:
- 63
- 53
- 73
- none of these
Correct

Incorrect
Question 7 of 24

7. Question
1 points
$I f f (x) = \frac{2 + x}{2 - x}, t h e n f^{- 1} (x) :$
- 2(x-1)/x+1
- 2(x+1)/x-1
- x+1/x-1
- x-1/x+1
Correct

$l e t f (x) = y \frac{2 + x}{2 - x} = y 2 + x = 2 y - x y x + x y = 2 y - 2 x (1 + y) = 2 (y - 1) x = \frac{2 (y - 1)}{(y + 1)} = f^{- 1} (y) t h e r e f o r e f^{- 1} (x) = \frac{2 (x - 1)}{x + 1}$

Incorrect

$l e t f (x) = y \frac{2 + x}{2 - x} = y 2 + x = 2 y - x y x + x y = 2 y - 2 x (1 + y) = 2 (y - 1) x = \frac{2 (y - 1)}{(y + 1)} = f^{- 1} (y) t h e r e f o r e f^{- 1} (x) = \frac{2 (x - 1)}{x + 1}$
Question 8 of 24

8. Question
1 points
If A={1,2,3,4}

B={2,4,6,8}

f(1)=2,f(2)=4,f(3)=6 and f(4)=8
$f : A \to B t h e n f^{- 1} i s$
- {(2,1),(4,2),(6,3),(8,4)}
- {(1,2),(2,4),(3,6)(4,8)}
- {(1,4),(2,2),(3,6),(4,8)}
- none of these
Correct

If A={1,2,3,4}

B={2,4,6,8}
$w h e n f : A \to B, f = {(1, 2), (2, 4), (3, 6), (4, 8)} f^{- 1} i m p l i e s f : B \to A f^{- 1} = {(2, 1), (4, 2), (6, 3), (8, 4)}$

Incorrect

If A={1,2,3,4}

B={2,4,6,8}
$w h e n f : A \to B, f = {(1, 2), (2, 4), (3, 6), (4, 8)} f^{- 1} i m p l i e s f : B \to A f^{- 1} = {(2, 1), (4, 2), (6, 3), (8, 4)}$
Question 9 of 24

9. Question
1 points
$i f f (x) = x^{2} + x - 1 a n d 4 f (x) = f (2 x) t h e n f i n d x$
- 4/3
- 3/2
- 1/2
- -3/4
Correct

$f (x) = x^{2} + x - 1 4 f (x) = f (2 x) 4 [x^{2} + x - 1] = 4 x^{2} + 4 x - 4 2 x = 3 x = 3 / 2$

Incorrect

$f (x) = x^{2} + x - 1 4 f (x) = f (2 x) 4 [x^{2} + x - 1] = 4 x^{2} + 4 x - 4 2 x = 3 x = 3 / 2$
Question 10 of 24

10. Question
1 points
If A={p,q,r,s}

B={q,s,t}

C={m,q,n}

find C-(A∩B)
- {m,n}
- {p,q}
- {r,s}
- {p,r}
Correct

If A={p,q,r,s}

B={q,s,t}

C={m,q,n}

C-(A∩B)={m,n}

Incorrect

If A={p,q,r,s}

B={q,s,t}

C={m,q,n}

C-(A∩B)={m,n}
Question 11 of 24

11. Question
1 points
X={x,y,w,z},Y={1,2,3,4}

H={(x,1),(y,2),(y,3),(z,4),(x,4)}
- H is a function from X to Y
- H is not a function from X to Y
- H is a relation from Y to X
- none of these
Correct

Any relation from X to Y which no two different ordered pairs have the same first element is called a FUNCTION

therefore,in the given question,H is NOT a function from X to Y because the different ordered pairs of H have the same first element.

Incorrect

Any relation from X to Y which no two different ordered pairs have the same first element is called a FUNCTION

therefore,in the given question,H is NOT a function from X to Y because the different ordered pairs of H have the same first element.
Question 12 of 24

12. Question
1 points
Given the function f(x)=(2x+3) then the value of f(2x)-2f(x)+3 will be
- 3
- 2
- 1
- 0
Correct

f(x)=(2x+3)

f(2x)-2f(x)+3

=[2(2x)+3]=[2(2x+3)]+3

=4x+3-4x-6+3

=4x-4x+6-6

=0

Incorrect

f(x)=(2x+3)

f(2x)-2f(x)+3

=[2(2x)+3]=[2(2x+3)]+3

=4x+3-4x-6+3

=4x-4x+6-6

=0
Question 13 of 24

13. Question
1 points
if f(x)=2x+h then find f(x+h)-2f(x)
- h-2x
- 2x-h
- 2x+h
- none of these
Correct

f(x)=2x+h

f(x+h)-2f(x)

=[2(x+h)+h]-[2(2x+h)]

=2x+2h+h-4x-2h

=-2x+h

=h-2x

Incorrect

f(x)=2x+h

f(x+h)-2f(x)

=[2(x+h)+h]-[2(2x+h)]

=2x+2h+h-4x-2h

=-2x+h

=h-2x
Question 14 of 24

14. Question
1 points
$i f A = {x : x^{2} - 3 x + 2 = 0} B = {x : x^{2} + 4 x - 12 = 0}, t h e n B - A =$
- -6
- 1
- 1,2
- 2,-6
Correct

Incorrect
Question 15 of 24

15. Question
1 points
$i f F : A \to R i s a r e a l v a l u e d f u n c t i o n d e f i n e d b y f (x) = \frac{1}{x} =$
- R
- R-1
- R-0
- R-N
Correct

$f : A \to R f (x) = \frac{1}{x} i f x = 0 f (x) w i l l b e u n d e f i n e d A = R - 0$

Incorrect

$f : A \to R f (x) = \frac{1}{x} i f x = 0 f (x) w i l l b e u n d e f i n e d A = R - 0$
Question 16 of 24

16. Question
1 points
In the set N if all natural numbers the relation R defined by a R b if and only if,a divide b,then the relation R is:
- partial order relation
- equivalence relation
- symmentric relation
- none of these
Correct

For a function to be a partial order relation,it should be

(1)Reflexive

(2)Antisymmetric

(3)Transitive

a divides b satisfies the above 3 relations as follows:

(1)a/a Reflexive

(2)a/b and b/a ,a=b Antisymmentric

Incorrect

For a function to be a partial order relation,it should be

(1)Reflexive

(2)Antisymmetric

(3)Transitive

a divides b satisfies the above 3 relations as follows:

(1)a/a Reflexive

(2)a/b and b/a ,a=b Antisymmentric
Question 17 of 24

17. Question
1 points
For any 2 sets A and B,A∩(A’UB)=____,where A’ represents the compliment of the set A
- A∩B
- AUB
- A'UB
- none of these
Correct

A∩(A’UB)

A’UB=A∩B

Incorrect

A∩(A’UB)

A’UB=A∩B
Question 18 of 24

18. Question
1 points
$i f f : R \to R, f (x) = x + 1 g : R \to R, g (x) = x^{2} + 1 t h e n f o g (- 2) =$
- 6
- 5
- 4
- -2
Correct

$f (x) = x + 1 g (x) = x^{2} + 1 f o g (- 2) = f [g (- 2)] = f [(- 2)^{2} + 1] = f [5] = 5 + 1 = 6$

Incorrect

$f (x) = x + 1 g (x) = x^{2} + 1 f o g (- 2) = f [g (- 2)] = f [(- 2)^{2} + 1] = f [5] = 5 + 1 = 6$
Question 19 of 24

19. Question
1 points
$i f A \subset B t h e n w h i c h o n e o f t h e f o l l o w i n g i s t r u e$
- A∩B=B
- AUB=B
- A∩B=A'
- A∩B=0
Correct

$A \subset B A U B = B w h e r e A i s s u b s e t o f B$

Incorrect

$A \subset B A U B = B w h e r e A i s s u b s e t o f B$
Question 20 of 24

20. Question
1 points
There are 40 students 30 of them passed in english , 25 of them passed in maths and 15 of them passed in both.Assuming that every student has passed at least in one subject.how many students passed in english only but not in maths
- 15
- 20
- 10
- 30
Correct

Given,total no.of students n(E∩M)=40

no.of students passed in english n(E)=30

no.of students passed in maths n(M)=25

no.of students passed in both n(E∩M)=15

only n(only E)=n(E)-n(E∩M)=30-15=15

Incorrect

Given,total no.of students n(E∩M)=40

no.of students passed in english n(E)=30

no.of students passed in maths n(M)=25

no.of students passed in both n(E∩M)=15

only n(only E)=n(E)-n(E∩M)=30-15=15
Question 21 of 24

21. Question
1 points
If A={±2,±3},B={1,4,9} and F={(2,4),(-2,4),(3,9),(-3,9)} then F is defined as:
- one to one function from A into B
- one to one function from A onto B
- many to one function from A onto B
- many to one function from A into B
Correct

$F : A \to B 2 \to 4 - 2 \to 4 3 \to 9 - 3 \to 9 m a n y o n e f u n c t i o n f r o m A o n t o B$

Incorrect

$F : A \to B 2 \to 4 - 2 \to 4 3 \to 9 - 3 \to 9 m a n y o n e f u n c t i o n f r o m A o n t o B$
Question 22 of 24

22. Question
1 points
If A=(1,2,3,4,5),B=(2,4) and C=(1,3,5) then (A-C)xB is
- {(2,2),(2,4),(4,2),(4,4),(5,2),(5,4)}
- {(1,2),(1,4),(3,2),(3,4),(5,2),(5,4)}
- {(2,2),(4,2)(4,4),(4,5)}
- {(2,2),(2,4),(4,2),(4,4)}
Correct

(A-C)={1,2,3,4,5}-{1,3,5}={2,4}

(A-C)xB={2,4}x{2,4}={(2,2),(2,4),(4,2),(4,4)}

Incorrect

(A-C)={1,2,3,4,5}-{1,3,5}={2,4}

(A-C)xB={2,4}x{2,4}={(2,2),(2,4),(4,2),(4,4)}
Question 23 of 24

23. Question
1 points
For any two sets A and B the set (AUB’)’ is equal to (where denotes compliment of the set)
- B-A
- A-B
- A'-B'
- B'-A'
Correct

(AUB’)’=B-A

Incorrect

(AUB’)’=B-A
Question 24 of 24

24. Question
1 points
The number of proper sub set of the set {3,4,5,6,7}
- 32
- 31
- 30
- 25
Correct

Given set A={3,4,5,6,7}

Cardinal no n(A)=5
$n o . o f p r o p e r s u b s e t = 2^{n} - 1 = 2^{5} - 1 = 32 - 1 = 31$

Incorrect

Given set A={3,4,5,6,7}

Cardinal no n(A)=5
$n o . o f p r o p e r s u b s e t = 2^{n} - 1 = 2^{5} - 1 = 32 - 1 = 31$

Sets,Functions and Relations-Test 1

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