Maths Quiz for RBI & SBI Clerk Exams

1).Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random.What is the
probability that the ticket drawn has a number which is a multiple of 3 or 5?
A) 1/2
B) 3/5
C) 9/20
D) 8/15
E) None of these

2). If a box contains 10 bulbs,of which just three are defective. If a random sample of five bulbs is drawn, find the probability that the sample contains no defective bulbs.
A) 5/12
B) 7/12
C) 3/14
D) 1/12
E) None of these

3). Two dices are tossed. The probability that the total score is a prime number is:
A) 5/12
B) 1/6
C) 1/2
D) 7/9
E) None of these

4). A brother and a sister appear for an interview against two vacant posts in an office. The probability of the  brother’s selection is 1/5 and that of the sister’s selection is 1/3. What is the probability that only one of them is selected?

A) 1/5
B) 3/4
C) 2/5
D) 3/5
E) None of these

5). A card is drawn from a pack of 52 cards. The card is drawn at random. What is the probability that it is neither a spade nor a Jack?
A) 9/13
B) 4/13
C) 10/13
D) 8/13
E) None of these

6). 67.39 -11.78 + 19.63 =? + 22.41 
A) 52.73
B) 52.83
C) 65.78
D) 64.78
E) None of these

7). 22% of 250 + 35% of 460 = ? 
A) 216
B) 226
C) 232
D) 242
E) None of these

8).  0.6 x 1.8÷0.5 x 12 = ? 
A) 24.92
B) 25.92
C) 18.46
D) 17.46
E) None of these

9). 55 x (25)3  ÷
625 = (5)?
 

A) 6
B) 9
C) 8
D) 5
E) None of these

10).  ? % of 400 – (7)2 = 159 
A)55
B) 53
C) 51
D) 52
E) None of these

Answer & Explanations


1. C) 9/20
Here, S = {1, 2, 3, 4, …., 19, 20}.
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
P(E) =n(E)/n(S)=9/20.

2.  D) 1/12
Total number of elementary events =10C5
Number of ways of selecting no defective bulbs i.e., 5 non-defective bulbs out of 7 is7C5.
So,required probability =7C/10C5 = 1/12.

3.  A) 5/12
Clearly, n(S) = (6 x 6) = 36.
Let E = Event that the sum is a prime number.
Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5) }
n(E) = 15.
P(E) =n(E)/n(S)=15/36=5/12

4.  C) 2/5
Probability that only one of them is selected = (prob. that brother is selected) × (prob. that sister is not selected)+ (Prob. that brother is not selected) × (Prob. that sister is selected
=(1/5 x 2/3) + (4/3 x 1/3) = 2/5

5. A) 9/13
There are 13 spade and 3 more jack
Probability of getting spade or a jack:
=13+352=1652=413
So probability of getting neither spade nor a jack:
=1-413 = 9/13

6.   B
7.   A
8    B
9.   E
10. D