Q: In a statistical investigation of 1003 families of Calcutta, it was found that 63 families has neither a radio nor a TV, 794 families has a radio and 187 has TV. The number of families in that group h

aligned Given, n(R) & =794 \\ n(T) & =187 \\ n(R T)^ & =63 \\ n( Total ) & =n(R T)+n(R T)^ \\ 1003 & =n(R T)+63 \\ n(R T) & =940 aligned By set theory array lc & n(R T)=n(R)+n(T)-n(R T) \\ & 940=794+187-n(R T) \\ & n(R T)=981-940=41 array

Question 1

In a statistical investigation of 1003 families of Calcutta, it was found that 63 families has neither a radio nor a TV, 794 families has a radio and 187 has TV. The number of families in that group h

Accepted Answer

aligned Given, n(R) & =794 \ n(T) & =187 \ n(R T)^ & =63 \ n( Total ) & =n(R T)+n(R T)^ \ 1003 & =n(R T)+63 \ n(R T) & =940 aligned By set theory array lc & n(R T)=n(R)+n(T)-n(R T) \ & 940=794+187-n(R T) \ & n(R T)=981-940=41 array

Question 2

Let R be the relation "is congruent to" on the set of all triangles in a plane is

Accepted Answer

Let S denote the set of all triangles in a plane. Let R be the relation on S defined by ( _1, _2 ) R triangle _1 _2 : (i) Let any triangle S, we have _1 _2 ( , ) R S R is reflexive on (ii) Let _1, _2 ~S , such that ( _1, _2 ) R, then _1 _2 _2 _1 ( _2, _1 ) R R is symmÄ™tric (iii) Again, let _1, _2, _3 ~S such that ( _1, _2 ) R and ( _2, _3 ) R _1 _2 _3 ( _1, _2 ) R R is transitive. R is an equivalence relation.

Question 3

Number of subsets of set of letter of word 'MONOTONE'.

Accepted Answer

Set created out of alphabet of word 'MONO- TONE' is \ M , N , O , T , E \ Total number of subsets =2^5=32

Question 4

In an examination, 62 \% of the candidates failed in English, 42% in Mathematics and 20% in both. The number of those who passed in both the subjects is

Accepted Answer

From the given condition, a + c =62 k ; b + c =42 k and c =20 k Hence a =42 k and b =22 k Number of students who failed in none means passed in both = n =100 k -( a + b + c ) =100 k -(42 k +22 k +20 k )=16 k . or 16 \%

Question 5

If n( ~A )=1000, n( ~B )=500 and if n( ~A B ) 1 and n( ~A B )=p, then

Accepted Answer

n ( A )=1000, n ( B )=500, n ( A B ) 1, n(A B)=p ; n(A B)=n(A)+n(B)-n(A B) p =1000+500- n ( A B ) 1 n ( A B ) 500 Hence p 1499 and p 1000 1000 p 1499

Question 6

A set A has 3 elements and another set B has 6 elements. Then

Accepted Answer

We have n(A B)= \ n(A), n(B)\ = \ 3,6\ =6 Max n(A B)=n(A)+n(B)=9 6 n(A B) 9

Question 7

If A=\ x: x is a multiple of 4\ and, B=\ x: x is a multiple of 6\ , then A B consists of multiples of

Accepted Answer

Let x A B. Then, x A and x B x is a multiple of 4 and x is a multiple of 6 x is a multiple of 4 and 6 both x is a multiple of 12

Sets and Relations — Previous Year Questions & Solutions