Free Objective Test 02 Practice Test - 11th and 12th

Question 1

If A = [ x:x is a multiple of 3] and B = [x:x is a multiple of 5] , then A - B is ( $¯ A$ means complement of A)

$¯ A$ $\cap$ B

A $\cap$ $¯ B$ .

$¯ A$ $\cap$ $¯ B$ .

$\frac{}{A \cap B}$

SOLUTION

Solution : B

Try taking some values of A and B
We'll see that A - B = A $\cap$ $¯ B$ .

Question 2

If A = [(x,y): $x^{2} + y^{2} = 25$ ] And B = [(x,y): $x^{2} + 9 y^{2} = 144], t h e n A \cap B$ contains

One point

Three points

Two points

Four points

SOLUTION

Solution : D

A = Set of all values (x,y) : $x^{2} + y^{2} = 25 = 5^{2}$

B = $\frac{x^{2}}{144} + \frac{y^{2}}{16} = 1$ i.e., $\frac{x^{2}}{(12)^{2}} + \frac{y^{2}}{(4)^{2}} = 1$

Clearly , A $\cap$ B consists of four points.

Question 3

Sets A and B have 3 and 6 elements respectively. What can be the minimum number of elements in (A U B )

SOLUTION

Solution : B

n(A $\cup$ B) = n(A) + n(B) - n(A $\cap B$ ) = 3 + 6 - 1 $(A \cap B)$

Since maximum number of elements in A $\cap$ B = 3

$∴$ Minimum number of elements in A $\cup$ B = 9 - 3 = 6 .

Question 4

If $N_{a} = {a n : n \in N}$ , then $N_{3} \cap N_{4}$ =

$N_{7}$

$N_{12}$

$N_{3}$

$N_{4}$

SOLUTION

Solution : B

$N_{a} = {a n : n \in N}$
$N_{3} \cap N_{4} = {3, 6, 9, 12, 15 \dots \dots} \cap {4, 8, 12, 16, 20, \dots \dots}$

$= {12, 24, 36 \dots \dots} = N_{12}$

[ $∵$ 3,4 are relatively prime numbers ]
$∴$ $N_{3} \cap N_{4}$ = $N_{12}$

Question 5

If X = ${8^{n} - 7 n - 1 : n \in N} a n d Y = {49 (n - 1) : n \in N}$ , then

X $\subseteq$ Y

Y $\subseteq$ X

X = Y

None of these

SOLUTION

Solution : A

Since $8^{n} - 7 n - 1 = (7 + 1)^{n} - 7 n - 1$

= $7^{n} +^{n} C_{1} 7^{n - 1} +^{n} C_{2} 7^{n - 2} + . . . . . +^{n} C_{n - 1} 7 +^{n} C_{n} - 7 n - 1$

= $^{n} C_{2} 7^{2} +^{n} C_{3} 7^{3} + . . . +^{n} C_{n} 7^{n}, (^{n} C_{0} =^{n} C_{n}^{n} C_{1} =^{n} C_{n - 1} e t c,)$

= 49[ $^{n} C_{2} +^{n} C_{3} (7) + . . . . . . . . +^{n} C_{n} 7^{n - 2}]$

$∴$ $8^{n} - 7 n - 1$ is a multiple of 49 for n $\geq$ 2

For n = 1 , $8^{n} - 7 n - 1 = 8 - 7 - 1 = 0;$

For n = 2, $8^{n} - 7 n - 1 = 64 - 14 - 1 = 49$

$∴$ $8^{n} - 7 n - 1$ is a multiple of 49 for n $\geq$ N

$∴$ X contains elements which are multiples of 49 and clearly $γ$

contains all multiplies of 49. $∴$ X $\subseteq$ Y.

Question 6

$If n (\cup) = 700, n (A) = 200, n (B) = 300$ and $n (A \cap B) = 100,$ then $n (A^{^{'}} \cap B^{^{'}}) =$ ___.

400

240

300

500

SOLUTION

Solution : C

From de-Morgan's law of complementation, we have $A^{'} \cap B^{'} = (A \cup B)^{'} .$
$\Rightarrow n (A^{'} \cap B^{'}) = n ((A \cup B)^{'})$
But, $n ((A \cup B)^{'}) = n (U) - n (A \cup B)$ by definition of complement of a set.
$∴ n (A^{'} \cap B^{'}) = n (U) - n (A \cup B)$
$= n (U) - [n (A) + n (B) - n (A \cap B)]$
$= 700 - (200 + 300 - 100)$
$= 300$

Question 7

In a committee 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. The number of persons speaking at least one of these two languages is

SOLUTION

Solution : A

$n (S \cup F)$ = $n (S) + n (F) - n (S \cap F)$

$\Rightarrow n (S \cup F)$ = 20 + 50 -10 = 60

Question 8

If A={1,2,3,4,5,6}, B={1,2}, then $\frac{A}{(\frac{A}{B})}$ is equal to

A $\cap$ B

A $\cup$ B

SOLUTION

Solution : C

$\frac{A}{B}$ = A - B = {3,4,5,6}

= $\frac{A}{(\frac{A}{B})}$ = {1,2}

Question 9

Let $n (A - B) = 25 + x, n (B - A) = 2 x$ and $n (A \cap B) = 2 x .$ If $n (A) = 2 (n (B)),$ then $x$ = ___.

SOLUTION

Solution : B

Given that $n (A - B) = 25 + x, n (B - A) = 2 x$ and $n (A \cap B) = 2 x .$

We know that $n (A) = n (A - B) + n (A \cap B) .$ $= 25 + x + 2 x = 3 x + 25$

Similarly, $n (B) = n (B - A) + n (A \cap B)$
$= 2 x + 2 x$ = $4 x .$

$n (A) = 2 (n (B)) \Rightarrow 3 x + 25$ = $2 \times 4 x$

$\Rightarrow 5 x = 25 \Rightarrow x = 5$

Question 10

In a group of 70 people, 37 like coffee, 52 like tea and each person like at least one of the two drinks. The number of persons liking both coffee and tea is:

SOLUTION

Solution : C

$n (A \cup B)$ = $n (A) + n (B) - n (A \cap B)$

We have, 70 = 37 + 52 - $n (A \cap B)$
$n (A \cap B)$ = 19.

Question 11

Let $F_{1}$ be the set of all parallelograms, $F_{2}$ the set of rectangles, $F_{3}$ the set of rhombuses, $F_{4}$ the set of squares and $F_{5}$ the set of trapeziums in a plane, then $F_{1}$ is equal to

$F_{2} \cap F_{3}$

$F_{2} \cup F_{3} \cup F_{4}$

$F_{3} \cap F_{4}$

$F_{2} \cap F_{1}$

SOLUTION

Solution : B

Since every rectangle, rhombus and square is parallelogram so

$F_{1}$ = $F_{2} \cup F_{3} \cup F_{4}$

Question 12

Of the 200 candidates who were interviewed for a position at a call center, 100 had a two-wheeler, 70 had a credit card and 140 had a mobile phone. 40 of them had both two-wheeler and credit card, 30 had both credit card and mobile phone and 60 had both two wheeler and mobile phone and 10 had all three. How many candidates had none of the three?

SOLUTION

Solution : B

Number of candidates who had none of the three = Total number of candidates - number of candidates who had at least one of three devices.

Total number of candidates = 200 =n(U), where U is the universal set

Number of candidates who had at least one of the three $= n (A \cup B \cup C)$ , where A is the set of those who have a two wheeler, B the set of those who have a credit card and C the set of those who have a mobile phone.

We know that $n (A \cup B \cup C) = n (A) + n (B) + n (C)$
$- [n (A \cap B) + n (B \cap C) + n (C \cap A)] + n (A \cap B \cap C)$
Therefore, $n (A \cup B \cup C) = 100 + 70 + 140 - {40 + 30 + 60} + 10$
Or $n (A \cup B \cup C) = 190$ .
As 190 candidates who attended the interview had at least one of the three gadgets, $n (U) - n (A \cup B \cup C)$ = 200 - 190 = 10 candidates had none of three.

Question 13

$A = {x : x \in R, x^{2} = 16 and 2 x = 6}$ 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 14

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 15

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.

Free Objective Test 02 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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