# Free Objective Test 01 Practice Test - 11th and 12th

In a town of 10,000 families it was found that 40% family buy newspaper A, 20% buy newspaper B and 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, then number of families which buy A only is

A.

3100

B.

3300

C.

2900

D.

1400

#### SOLUTION

Solution : B

n(A) = 40% of 10,000 = 4,000

n(B) = 20% of 10,000 = 2,000

n(C) = 10% of 10,000 = 1,000

n(A ∩ B) = 5% of 10,000 = 500

n(B ∩ C) = 3% of 10,000 = 300

n(C ∩ A) = 4% of 10,000 = 400

n(A ∩ B ∩ C) = 2% of 10,000 = 200

We want to find the number of families which buy only A = n(A) - [n(A ∩ B) + n(A ∩ C) - n(A ∩ B ∩ C)]

=4000 - [500 + 400 - 200] = 4000 - 700 = 3300

Let n(U) = 700, n(A) = 200, n(B) = 300 and n(A ∩ B) = 100,

Then n(AcBc) =

A.

400

B.

600

C.

300

D.

200

#### SOLUTION

Solution : C

n(Ac ∩ Bc) = n(U) - n(A ∪ B)

= n(U) - [n(A) + n(B) - n(A ∩ B)]

= 700 - [200 + 300 - 100] = 300.

If X = {4n - 3n - 1 : n ∈ N} and Y = { 9(n-1) : n ∈ N}, then X ∪ Y is equal to

A.

X

B.

Y

C.

N

D.

None of these

#### SOLUTION

Solution : B

Since
4n3n1=(3+1)n3n1=3n+nC13n1+nC23n2+.....+nCn13+nCn3n1=nC232+nC3.33+...+nCn3n,(nC0=nCn,nCn1=nC1.....so on.)=9[nC2+nC3(3)+......+nC43n1]
4n3n1 is a multiple of 9 for n2.
For n=1,4n3n1=431=0For n=2,4n3n1=1661=94n3n1 is a multiple of 9 for all nϵN
X contains elements, which are multiples of 9, and clearly Y contains all multiples of 9.
XY i.e.,XY=Y

Let A = { x : x ∈ R, |x| < 1};  B = {x : x ∈ R, |x-1| ≥ 1} and

A ∪ B = R - D, then the set D is

A.

[x : 1 < x ≤ 2]

B.

[x : 1  ≤ x < 2]

C.

[x : 1 ≤ x ≤ 2]

D.

None of these

#### SOLUTION

Solution : B

A = {x: x ∈ R, -1 < x < 1}

B = {x: x ∈ R:x-1 ≤ -1 or x-1 ≥ 1}

= {x : x ∈ R:x ≤ 0 or  x ≥ 2}

∴ A ∪ B = R - D, where D = {x : x ∈ R, 1 ≤ x < 2}.

If the sets A and B are defined as

A = {(x, y) : y =  1x, 0 ≠ x ∈ R}

B = {(x, y) : y = -x, x ∈ R}, then

A.

A ∩ B = A

B.

A ∩ B = B

C.

A ∩ B = ∅

D.

None of these

#### SOLUTION

Solution : C

Since y =   1x, y = -x meet when -x =   1x  ⇒ x2 = -1,

which does not give any real value of x.

Hence, A ∩ B = ∅.

If A and B  are two given sets, then A ∩  (AB)cis equal to

A.

A

B.

B

C.

D.

A ∩ (Bc).

#### SOLUTION

Solution : D

A ∩ (AB)c) = A ∩ (Ac ∪ Bc)

= (A ∩ (Ac ) ∪ (A  ∩ (Bc)

= ∅ ∪ (A ∩ (Bc) = A ∩ (Bc).

If a set A has n  elements, then the total number of subsets of A is

A.

n

B.

n2

C.

2n

D.

2n

#### SOLUTION

Solution : C

Number of subsets of A = nC0nC1  + .............+ nCn2n

A={x:xR, x2=16 and 2x=6} can be represented in the roster form as _________ .

A.

A = {}

B.

A = { 14, 3, 4 }

C.

A = { 3 }

D.

A = { 4 }

#### SOLUTION

Solution : A

x2=16x=±4

2x=6x=3

There is no value of x which satisfies both the given equations. The set A is an empty set or a null set.

Thus, A = {}.

Which of the following is an empty set

A.

{x:x is a real number and x21=0}

B.

{x:x is a real number and x2+1=0}

C.

{x:x is a real number and x29=0}

D.

{x:x is a real number and x2=x+2}

#### SOLUTION

Solution : B

Since x2 + 1 = 0, gives x2=1

x=±i

x is not real but x has to be is real (given)

∴ No real value of x is possible in this case.

In a city 20 percent of the population travels by car, 50 percent travels by bus and 10 percent travels by both car and bus. Then the percentage of population travelling by car or bus is

A.

80 percent

B.

40 percent

C.

60 percent

D.

70 percent

#### SOLUTION

Solution : C

n(C) = 20, n(B) = 50, n(C ∩ B) = 10

Now n(C ∪ B) = n(C) + n(B) - n(C ∩ B)

= 20 + 50 - 10 = 60.

The group of intelligent students in a class is __________.

A.

a null set

B.

a finite set

C.

a well defined collection

D.

not a well defined collection

#### SOLUTION

Solution : D

Intelligence cannot be defined for students in a class. Hence, the group of intelligent students is not a well defined collection.

The number of proper subsets of the set {1,2,3} is  ___.

A.

8

B.

7

C.

6

D.

5

#### SOLUTION

Solution : B

A set of n elements has 2n subsets.
Every set is a subset of itself.
Thus, the number of proper subsets of a set is 2n1 subsets.
The set {1,2,3}  has 3 elements and hence, 231=7 proper subsets.

If the sets A and B are defined as A = {(x, y) : y = ex, x ∈ R};

B = {(x, y) : y = x, x ∈ R}, then

A.

B ⊆ A

B.

A ⊆ B

C.

A ∩ B = ∅

D.

A ∪  B = A

#### SOLUTION

Solution : C

Since, y =  ex and y = x do not meet for any x ∈ R

A ∩ B = ∅ .

In a survey of 200 students from 7 different schools, 50 people do not play NFS, 40 people do not play Dota and 10 people play no online game. Then find the no. of people out of 200 people who do not play both the games provided these are the only two games on offer.

A.

80

B.

70

C.

60

D.

50

#### SOLUTION

Solution : A

Let the no. of people who do not play NFS be  n(NI)  = 50  (Given)
Similarly no. of people who do not play Dota be  n(DI)  = 40  (Given)
And the no. of people who do not play any game n(NI ∩ DI)  = 10  (Given)
We have to find the no. of people who do not play both the games =  n(N∩ D)I

We know from the Demorgan's law

(A∩ B)I  =  AI ∪ BI
So,
n(N∩ D)I  =  n(NI) ∪ n(DI)

n(NI) ∪ n(DI) =  n(NI)  + n(DI) - n(NI ∩ DI)

n(NI) ∪ n(DI)  =  50 + 40 - 10
=   80

If A and B are any two sets, then  A(AB)___.

A.

A

B.

B

C.

AC

D.

BC

#### SOLUTION

Solution : A

If A and B are any two sets, then ABA.
Also, AA(AB)
AA(AB)A
A(AB)=A