Sets(11th Grade > Mathematics ) Questions and answers for exam Preparation

11th Grade > Mathematics

SETS MCQs

Total Questions : 30 | Page 1 of 3 pages

Question 1.

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

One point
Three points
Two points
Four points

Discuss Question

Answer: Option D. -> Four points
:
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 2.

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

$N_{7}$
$N_{12}$
$N_{3}$
$N_{4}$

Discuss Question

Answer: Option B. ->

N_{12}

:
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 3.

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

Discuss Question

Answer: Option A. -> X

\subseteq

Y
:
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 4.

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

X
Y
N
None of these

Discuss Question

Answer: Option B. -> Y
:
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 5.

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

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

Discuss Question

Answer: Option C. -> 300
:
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 6.

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

Discuss Question

Answer: Option A. -> 60
:
A

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

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

Question 7.

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

Discuss Question

Answer: Option C. -> A ∩ B = ∅
:
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 8.

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