JEE MATHEMATICS MCQ: SET THEORY

1 Easy

1 of 1

Application +4 | -1

Suppose

A_1, A_2, A_3, \ldots \ldots . A_{30}

are thirty sets each with 5 elements and

B_1, B_2, B_3, \ldots \ldots . B_n

are n sets. Each with three elements such that

\mathop U\limits_{i = 1}^{30} A_i=\mathop U\limits_{j = 1}^{n} B_j=S

. If each element of

{S}

belongs to exactly 10 of the

A_i

's and exactly 9 of the

B_j

's, then the value of

{n}

View answer and explanation

Correct option: C

Option C (45) is correct because it is derived from equating the total number of elements in S calculated using the information from sets A_i and B_j.

More Questions from SET THEORY

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Let

S

be the set of all those 2007-place decimal integers

2 a_1 a_2 a_3 \ldots . . a_{2006}

which contain odd number of digit 9 in each sequence

a_1, a_2, a_3, \ldots ., a_{2006}

. The cardinal number of

S

is If

a N=\{a n: n \in N: b N=\{n b: n \in N\}

c N=\{c N: n \in N\}

and

b N \cap c N=d N

, where

a, b, c \in N

and

b, c

are coprime, then Which of the following cannot be the number of elements in the power set of any finite set? If

A=\{x: x

is a multiple of 3

\}

and

B=\{x: x

is a multiple of 5}, then

A-B

is equal to If

F(n)

denotes the set of all divisors of

n

except 1 , what is the least value of y satisfying

[\mathrm{F}(20) \cap \mathrm{F}(16)] \subseteq \mathrm{F}(\mathrm{y})

? In a town of 10000 families, it was found that

40 \%

families buy newspaper

A, 20 \%

families buy newspaper B and

10 \%

families buy newspaper

C

, 5% families buy newspapers

A

and B, 3% buy newspapers B and

C

and

4 \%

buy newspapers

A

and

C

. If

2 \%

families buy all the three newspapers, then number of families which buy

A

only is