Dear Aspirants,
Quantitative Aptitude Quiz For IBPS PO/Clerk Prelims
Numerical Ability or Quantitative Aptitude Section has given heebie-jeebies to the aspirants when they appear for a banking examination. As the level of every other section is only getting complex and convoluted, there is no doubt that this section, too, makes your blood run cold. The questions asked in this section are calculative and very time-consuming. But once dealt with proper strategy, speed, and accuracy, this section can get you the maximum marks in the examination. Following is the Quantitative Aptitude quiz to help you practice with the best of latest pattern questions.
Q1. In how many different ways can the letters of the word ‘DRASTIC’ be arranged in such a way that the vowels always come together ?
720
360
1440
540
None of these
Solution:
Total letters = D, R, A, S, T, I, C (7)
Total vowels = A, I (2)
∴ Required no. of ways = 6! × 2!
= 1440
Total vowels = A, I (2)
∴ Required no. of ways = 6! × 2!
= 1440
Q2. In how many ways can Prabhat arrange the letters of the word ALLAHABAD ?
7650
7560
6750
5760
7660
Solution:
Q3. How many numbers between 2000 and 3000 can be formed with the digits 0, 1, 2, 3, 4, 5, 6, 7 (repetition of digits not allowed) ? 42
210
336
440
120
Solution:
Total required numbers between 2000 and 3000
= 1 × 7 × 6 × 5 (For eg. 2035, 2345)
= 210
= 1 × 7 × 6 × 5 (For eg. 2035, 2345)
= 210
Q4. In how many ways can a person sent invitation cards to 6 of his friends if he has four servants to distribute the cards ?
6⁴
4⁶
24
120
36
Solution:
Q5. A captain and a vice captain are to be chosen out of a team having eleven players. How many ways are there to achieve this ? 10.9
¹¹C₂
110
10.9!
11.10!
Solution:
Total ways = 11 × 10
= 110
= 110
Q6. In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
159
194
205
209
224
Solution:
Q7. 3 men and 3 women are to sit at a round table. In how many different ways can they sit so that no 2 women sit together? 16
12
18
21
10
Solution:
3 men can sit at a round table is (3 – 1)! Ways i.e., 2! = 2 ways
Now, 3 women are to sit so that no 2 women sit together.
They have to sit three places each between two men.
It can be done in
ways i.e., 3! = 3 × 2 × 1 = 6 ways.
Thus, required number of ways = 2 × 6 = 12
Now, 3 women are to sit so that no 2 women sit together.
They have to sit three places each between two men.
It can be done in
ways i.e., 3! = 3 × 2 × 1 = 6 ways.Thus, required number of ways = 2 × 6 = 12
Q8. If (n + 2)! = 2550 (n)!, find n.
49
-47
54
53
63
Solution:
Q9. How many four digits number can be formed by using the digits 0, 2, 4, 6, 7 if repetition of digits is allowed. 625
96
500
36
72
Solution:
Total digits = 5
First place can be filled up by using only one of 4 digits (except 0, since 0 at the first place is meaning less).
Second place can be filled up by using all the five digits (as repetition is allowed).
Similarly, third and fourth place can be filled up by using all the five digits.
Thus,
Places: 0 0 0 0
Digit: 4 5 5 5
Total numbers = 4 × 5 × 5 × 5 = 500
First place can be filled up by using only one of 4 digits (except 0, since 0 at the first place is meaning less).
Second place can be filled up by using all the five digits (as repetition is allowed).
Similarly, third and fourth place can be filled up by using all the five digits.
Thus,
Places: 0 0 0 0
Digit: 4 5 5 5
Total numbers = 4 × 5 × 5 × 5 = 500
Q10. Find the numbers between 100 and 1000 in which all digits are distinct.
548
648
748
448
684
Solution:
There are three digits numbers between 100 and 1000.
Total digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 which are 10.
First place can be filled up by using any one of 9 digits (except 0, since 0 at the first place is meaningless).
Second place can be filled up by using any one of 9 digits (as one digit has been used at first place)
Third place can be filled up by using only one of 8 digits.
Thus,
Places : 0 0 0
Digits : 9 9 8
Total number = 9 × 9 × 8 = 648
Total digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 which are 10.
First place can be filled up by using any one of 9 digits (except 0, since 0 at the first place is meaningless).
Second place can be filled up by using any one of 9 digits (as one digit has been used at first place)
Third place can be filled up by using only one of 8 digits.
Thus,
Places : 0 0 0
Digits : 9 9 8
Total number = 9 × 9 × 8 = 648
Q11. The letters of the word ‘PARADISE’ are to be arranged so that all vowels should not come together. Find the number of arrangements.
20160
18720
38880
16720
37440
Solution:
Q12. If P(5, 2) = P(n, 2), find n. 5
2
1
3
4
Solution:
Q13. Group of 6 students sitting around a circular table, find probability of 2 specified students sits together. 6!/2!
3
2!/5!
2/5
none of these.
Solution:
Q14. In how many ways the word ‘SCOOTER’ can be arranged such that ‘S’ and ‘R’ are always at two ends? 720
120
2520
5040
None of these
Solution:
Q15. Find total number of the 3 digits odd numbers by using the digits 2, 3, 4, 5 when repetitions of digits are not allowed. 12
22
15
18
24
Solution:
- https://www.bankersadda.com/2019/02/ibps-po-quantitative-aptitude-permutation-combination-16.html
No comments:
Post a Comment