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

Question 1

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

3100

3300

2900

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

Question 2

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

Then n $(A^{c} \cap B^{c})$ =

400

600

300

200

SOLUTION

Solution : C

n( $A^{c}$ ∩ $B^{c}$ ) = n(U) - n(A ∪ B)

= n(U) - [n(A) + n(B) - n(A ∩ B)]

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

Question 3

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

None of these

SOLUTION

Solution : B

Since
$4^{n} - 3 n - 1 = (3 + 1)^{n} - 3 n - 1 = 3^{n} +^{n} C_{1} 3^{n - 1} +^{n} C_{2} 3^{n - 2} + . . . . . +^{n} C_{n - 1} 3 +^{n} C_{n} - 3 n - 1 =^{n} C_{2} 3^{2} +^{n} C_{3} {.3}^{3} + . . . +^{n} C_{n} 3^{n}, (^{n} C_{0} =^{n} C_{n},^{n} C_{n - 1} =^{n} C_{1} . . . . . s o o n .) = 9 [^{n} C_{2} +^{n} C_{3} (3) + . . . . . . +^{n} C_{4} 3^{n - 1}]$
$∴ 4^{n} - 3 n - 1$ is a multiple of 9 for $n \geq 2$ .
$F o r n = 1, 4^{n} - 3 n - 1 = 4 - 3 - 1 = 0 F o r n = 2, 4^{n} - 3 n - 1 = 16 - 6 - 1 = 9 ∴ 4^{n} - 3 n - 1$ is a multiple of 9 for all $n ϵ N$
$∴ X$ contains elements, which are multiples of 9, and clearly $Y$ contains all multiples of 9.
$∴ X \subset Y i . e ., X \cup Y = Y$

Question 4

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

A ∪ B = R - D, then the set D is

[x : 1 < x ≤ 2]

[x : 1 ≤ x < 2]

[x : 1 ≤ x ≤ 2]

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}.

Question 5

If the sets A and B are defined as

A = {(x, y) : y = $\frac{1}{x}$ , 0 ≠ x ∈ R}

B = {(x, y) : y = -x, x ∈ R}, then

A ∩ B = A

A ∩ B = B

A ∩ B = ∅

None of these

SOLUTION

Solution : C

Since y = $\frac{1}{x}$ , y = -x meet when -x = $\frac{1}{x}$ ⇒ $x^{2}$ = -1,

which does not give any real value of x.

Hence, A ∩ B = ∅.

Question 6

If A and B are two given sets, then A ∩ $(A \cap B)^{c}$ is equal to

∅

A ∩ ( $B^{c}$ ).

SOLUTION

Solution : D

A ∩ $(A \cap B)^{c}$ ) = A ∩ ( $A^{c}$ ∪ $B^{c}$ )

= (A ∩ ( $A^{c}$ ) ∪ (A ∩ ( $B^{c}$ )

= ∅ ∪ (A ∩ ( $B^{c}$ ) = A ∩ ( $B^{c}$ ).

Question 7

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

$n^{2}$

$2^{n}$

SOLUTION

Solution : C

Number of subsets of A = $^{n} C_{0}$ + $^{n} C_{1}$ + .............+ $^{n} C_{n}$ = $2^{n}$

Question 8

$A = {x : x \in R, x^{2} = 16 and 2 x = 6}$ can be represented in the roster form as _________ .

A = {}

A = { 14, 3, 4 }

A = { 3 }

A = { 4 }

SOLUTION

Solution : A

$x^{2} = 16 \Rightarrow x = \pm 4$

$2 x = 6 \Rightarrow x = 3$

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 = {}.

Question 9

Which of the following is an empty set

{ $x : x$ is a real number and $x^{2} - 1 = 0$ }

{ $x : x$ is a real number and $x^{2} + 1 = 0$ }

{ $x : x$ is a real number and $x^{2} - 9 = 0$ }

{ $x : x$ is a real number and $x^{2} = x + 2$ }

SOLUTION

Solution : B

Since $x^{2}$ + 1 = 0, gives $x^{2} = - 1$

⇒ $x = \pm i$

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

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

Question 10

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

80 percent

40 percent

60 percent

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.

Question 11

The group of intelligent students in a class is __________.

a null set

a finite set

a well defined collection

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.

Question 12

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

SOLUTION

Solution : B

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

Question 13

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

B = {(x, y) : y = x, x ∈ R}, then

B ⊆ A

A ⊆ B

A ∩ B = ∅

A ∪ B = A

SOLUTION

Solution : C

Since, y = $e^{x}$ and y = x do not meet for any x ∈ R

$∴$ A ∩ B = ∅ .

Question 14

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.

SOLUTION

Solution : A

Let the no. of people who do not play NFS be n(N^I) = 50 (Given)
Similarly no. of people who do not play Dota be n(D^I) = 40 (Given)
And the no. of people who do not play any game n(N^I ∩ D^I) = 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 =}A^I ∪ B^I

So, n(N∩ D)^{I =}n(N^I) ∪ n(D^I)

n(N^I) ∪ n(D^I) = n(N^I) + n(D^I) - n(N^I ∩ D^I)

n(N^I) ∪ n(D^I) = 50 + 40 - 10
= 80

Question 15

If A and B are any two sets, then $A \cup (A \cap B)$ = ___.

$A^{C}$

$B^{C}$

SOLUTION

Solution : A

If A and B are any two sets, then $A \cap B \subseteq A .$
Also, $A \subseteq A \cup (A \cap B)$
$\Rightarrow A \subseteq A \cup (A \cap B) \subseteq A$
$\Rightarrow A \cup (A \cap B) = A$

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

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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