Dear Aspirants,
We have introduced "the 17 days study plan to crack IPPB Scale-1 Prelims 2017". This plan is specifically dedicated to Quantitative Aptitude and cover all types of questions asked in Bank PO exams. This will help you to prepare in a more efficient and systematic way. So, follow this post to keep updated with the plan.
1. Find the probability that a number from 1 to 300 is divisible by 3 or 7 ?
A. 37/75
B. 32/75
C. 36/75
D. 28/75
E. 26/75
Sol. Between 1 and 300 there are 100 multiples of 3
and 42 multiples of 7, while 14 are common multiples of 3 and 7
So total no. of favourable cases = 100 + 42 − 14 = 128.
Required probability = 128/300 = 32/75.
2. There are 3 red balls, 4 blue balls and 5 white balls. 2 balls are chosen randomly. Find probability that 1 is red and the other is white ?
A. 5/22
B. 5/23
C. 7/22
D. 4/9
E. None of these
Sol. Required Probability = (3c1 × 5c1)/12c2 = 5/22.
Directions (3-5): Study the following information carefully to answer the questions that follow :
A box contains 2 blue caps, 4 red caps, 5 green caps and 1 yellow cap.
3. If one cap is picked at random, what is the probability that it is either blue or yellow ?
A. 2/9
B. 1/4
C. 3/8
D. 6/11
E. None of these
Sol. Required Probability = (2c1 + 1c1)/12c1 = 3/12 = 1/4.
4. If two caps are picked at random, what is the probability that at least one is red ?
A. 1/3
B. 16/21
C. 19/33
D. 7/19
E. None of these
Sol. Required Probability = (4c1×8c1 + 4c2)/12c2 = (38×2)/(12×11) = 19/33.
5. If three caps are picked at random, what is the probability that two are red and one is green ?
A. 9/22
B. 6/19
C. 1/6
D. 3/22
E. None of these
Sol. Required Probability = (4c2×5c1)/12c3 = 3/22.
6. Find the number of ways of distributing 8 identical balls into 3 boxes so that no box is empty and each box being large enough to accommodate all balls ?
A. 24
B. 28
C. 36
D. 21
E. 18
Sol. (8-1)C(3-1) = 7C2 = 7×6/2×1 = 21.
7. A group consists of 4 couples in which each of the 4 men have one wife each. In how many ways could they arranged in a straight line so that the men and women occupy alternate position ?
A. 576
B. 982
C. 1152
D. 1024
E. None of these
Sol. 4!×4! + 4!×4! = 576+576 = 1152.
8. A five digit number is formed with the digits 0,1,2,3 and 4 without repetition.Find the chance that the number is divisible by 2 ?
A. 1/2
B. 2/3
C. 1/3
D. 1/4
E. None of these
Sol. 5 digit number = 5! = 120
Divisible by 2 then the last digit should be 0, 2, 4
Then no. of cases with 0 as last digit = 4! = 24
no. of cases with 2 as last digit = 18
no. of cases with 4 as last digit = 18
Total cases = 24 + 18 + 18 = 60
P = 60/120 = 1/2.
9. From a group of 4 men and 3 women , 2 persons are selected at random. Find the probability that at least one woman is selected ?
A. 2/3
B. 3/5
C. 5/7
D. 1/3
E. 1/2
Sol. Total cases = 7c2 = 7!/5!×2! = 21
Cases when no women is selectd = 4c2 = 4!/2!×2! = 6
Cases when atleast 1 woman is selected = (21-6) = 15
Probability = 15/21 = 5/7.
10. 16 persons are participated in a party. In how many different ways can they host the seat in a circular table, if the 2 particular persons are to be seated on either side of the host ?
A. 16! × 2
B. 14! × 2
C. 18! × 2
D. 14!
E. None of these
Sol. (16 – 2)! × 2 = 14! × 2.
11. A bag contains 8 white and 9 black balls. Balls are drawn one by one. Two balls are drawn without replacement. What is the probability that one is white and other is black?
A. 8/15
B. 1/17
C. 8/17
D. 1/15
E. 9/17
Sol. Required Probability = (8C1×9C1)/(17C2) = 8/17.
12. A bag contains 3 red, 6 white and 7 black balls. Two balls are drawn at random. What is the probability that both are black ?
A. 5/40
B. 6/40
C. 7/40
D. 8/40
E. None of these
Sol. The reqd. probability is 7C2 / 16C2 = 7/40.
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