Numerical Ability Practice Questions for IBPS RRB (Set – 1)

Numerical Ability Practice Questions for IBPS RRB (Set – 1)
1. 45316 + 52131 – 65229 =? + 15151
a) 17063
b) 17073
c) 17076
d) 17067
e) None of these
2. (23 × 23 × 23 × 23 × 23 × 23)^5 × (23 × 23)^2 ÷ (23)^2 = (23)?
a) 32
b) 30
c) 9
d) 7
e) 11
3. (62.5 × 14 × 5) ÷ 25 + 41 = (?)^3
a) 4
b) 5
c) 9
d) 8
e) 6
4. (41+7×9)÷7+(25.153)=?
a) 40
b) 23
c) 25
d) 35
e) None of these
5. 4 × 566 ÷ 5 + 24.2 – 36 = (?)^2
a) 20
b) 21
c) 22
d) 23
e) 25
6. If the square of a number is subtracted from 4052 and the difference is multiplied by 15, the answer so obtained is 41340. What is the number?
a) 36
b) 1024
c) 32
d) 1296
e) None of these
7. The average of 20 numbers is zero. Of them, How many of them may be greater than zero, at the most?
A) 0
B) 1
C) 20
D) 19
E) 21
8. What is the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder?
A) 1108
B) 1683
C) 2007
D) 3363
E) 1997
9. A man is 24 years older than his son. In two years, his age will be twice the age of his son. What is the present age of his son?
A) 24 years
B) 23 years
C) 22 years
D) 21 years
E) 20 years
10. The population of a town increased from 1,75,000 to 2,62,500 in a decade. What is the average percent increase of population per year?
A) 4%
B) 5%
C) 6%
D) 7%
E) 50%
1. D)   2. A)   3. E)   4. A)   5. B)
6. A) (4052 – x^2) × 15 = 41340 => x^2 = 4052 – 2756 = 1296 => x = 36
7. D) Average of 20 numbers = 0
=> Sum of 20 numbers/20 = 0
=> Sum of 20 numbers = 0
Hence at the most, there can be 19 positive numbers.
(Such that if the sum of these 19 positive numbers is x, 20th number will be ‘-x’)
8. B) LCM of 5, 6, 7 and 8 = 840
Hence the number can be written in the form (840k + 3) which is divisible by 9.
If k = 1, number = (840 × 1) + 3 = 843 which is not divisible by 9.
If k = 2, number = (840 × 2) + 3 = 1683 which is divisible by 9.
Hence 1683 is the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder.
9. C) Let the present age of the son = x years
Then, present age of the man = (x + 24) years
Given than, in 2 years, man’s age will be twice the age of his son
=> (x + 24) + 2 = 2 (x + 2) => x = 22 years
10.  B) Increase in the population in 10 years = 262500 – 175000 = 87500
Percentage increase in the population in 10 years = 87500/175000 × 100 = 8750/175 = 50%
Average percent increase of population per year = 50%/10 = 5%