Q.1  50 students were admitted to a Nursery Class. Some students can speak only Hindi. Ten students can speak both Hindi & English. If the no. of students who can speak English is 21, then how many students can speak Hindi & how many can speak only English?

A) 39, 29 & 11 respectively

B) 37, 27 & 13 respectively

C) 28, 18 & 22 respectively

D) 21, 11 & 29 respectively

Ans. A

Circle X represent English.

Circle Y represent Hindi

A = Students speaking English only

B = Students speaking both English & Hindi

C= Students speaking Hindi only

Total no. of students

=   A+B+C = 50……..(i)

B = 10………………(ii)

A+B=21……………….(iii)

Now, A=(A+B)-B

=21-10

=11

Thus the no. of students who can speak English only=11

No. of students who can speak Hindi=

(B+C)=(A+B+C)-A

=50-11=39……………..iv

No. of students who can speak Only Hindi = C

From iv (B+C)=39,Thus C=39-10

=29

Therefore, Number of students speaking Hindi=39

No. of students speaking Hindi only = 29

No. of students speaking English only = 11

Q.2 Out of 20 tickets sold in a lottery, 5 will have different prizes. In how many different ways, a person who purchased 4 tickets can win the prize(s)?

A) 24

B) 5

C) 31

D) 20

Ans. C

He can win zero, one, two, three or four prizes.

So, the answer is 5C0 + 5C1 + 5C2 + 5C3 + 5C4 = 31.

Q.3 In how many ways 4 persons out of 8 (A, B, C, D, E, F, G, H) can be selected such that if B is selected, D should not be selected?

A) 30

B) 64

C) 36

D) 55

Ans. D

If B is selected, D should not be. So, 3 more have to be selected from (8 – 2) = 6.

Or if B is not selected, 4 more have to be selected from (8 – 1) = 7

So, = 6C3 + 7C4 = 55

Q.4 In how many ways 8 persons can stand in a row, if one refuses to stand at the extreme end?

A) 6 × 7!

B) 6 × 8!

C) 7 × 8!

D) 7 × 7!

Ans. A

One person refuses to stand at the extreme. So, the remaining 7 persons can stand in 7! ways. The refused person can stand in 6 ways in between these 7 persons. So, the answer is 6 × 7!

Q.5 Five persons have to be selected out of ten (A to J). In how many ways can it be done such that if ‘I’ is selected, D and E cannot be selected and ‘H’ has to be selected?

A) 20

B) 120

C) 146

D) 126

Ans. C

If I is selected, three more (except I and H) have to be selected from the remaining (except I, D, E and H). The different ways are 6C3.

If I is not selected, then the five have to be selected from the remaining 9. This can be done in 9C5 ways.

So, the total number of ways are  6C3 + 9C5 = 146.