SSC CHSL Quantitative Aptitude Questions with Solutions

SSC CHSL Quantitative Aptitude (Basic Arithmetic Skill) Questions with answers and Solutions for free online practice.

Mock Test : Maths (Quantitative Aptitude) MCQs
Exam : SSC CHSL
Medium : English
Questions : 25 ( New practice set in every attempt)
All questions are solved with short tricks
You can check the answer and solution of every question.

Results

#1. The quotient when $10^{100}$ is divided by $5^{75}$ is

$2^{25}\times 10^{75}$

$10^{25}$

$2^{75}$

$2^{75}\times 10^{25}$

#2. 25% of annual salary of A is equal to eighty percent of annual salary of B. The monthly salary of B is 40% of the monthly salary of C. The annual salary of C is Rs. 6 lac. What is the monthly salary of A?

Rs. 60.000

Rs. 62000

Rs.64000

Rs. 56,000

#3. The ratio of the length of a school ground to its width is 5 : 2. If the width is 40 m, then the length is :

200 m

100 m

50 m

80 m

#4. The mean of 20 items is 55. If two items such as 45 and 30 are removed, the new mean of the remaining items is :

65.1

56.9

65.3

#5. A sum was invested on simple interest at a certain rate for 2 years. Had it been put at 3% higher rate, it would have fetched Rs 72 more. The sum is :

Rs 1200

Rs 1500

Rs 1600

Rs 1800

#6. The LCM of two multiples of 12 is 1056. If one of the number is 132, the other number is

132

#7. The number, whose square is equal to the difference between the squares of 975 and 585 is :

780

390

1560

130

#8. The greatest number, which when divide 989 and 1327 leave remainders 5 and 7 respectively, is :

The largest number which when divide the numbers a, b and c give remainders as p, q, r respectively is given by H.C.F. of (a – p), (b – q) and (c – r)
Required number
= HCF of (989 – 5) and (1327 – 7)
= HCF of 984 and 1320 = 24
HCF = 24

#9. Of the three numbers, the first number is twice the second and the second is thrice the third number. If the average of these 3 numbers is 20, then the sum of the largest and the smallest numbers is :

#10. If one-ninth of a certain number exceeds its one-tenth by 4, the number is :

320

360

400

440

#11. The total number of prime factors in $4^{10}\times{7}^3\times{16}^2\times11\times{10}^2$ is

#12. Find the least multiple of 23, which when divided by 18, 21 and 24 leaves the remainder 7, 10 and 13 respectively.

3013

3024

3002

3036

LCM of 18, 21 and 24
LCM = 2 × 3 × 3 × 7 × 4 = 504
Now compare the divisors with their respective remainders. We observe that in all the cases the remainder is just 11 less than their respective divisor. So the number can be given by 504 K – 11 Where K is a positive integer
Since 23 × 21 = 483
We can write 504 K – 11
= (483 21) K – 11, = 483 K (21K – 11)
483 K is multiple of 23, since 483 is divisible by 23.
So, for (504K – 11) to be multiple of 23, the remainder (21K – 11) must be divisible by 23.
Put the value of K = 1, 2, 3, 4, 5,6, ….. and so on successively.
We find that the minimum value of K for which (21K – 11) is divisible by 23. is 6, (21 × 6 – 11)
= 115 which is divisible by 23.
Therefore, the required least number
= 504 × 6 – 11 = 3013

#13. Two numbers are in the ratio 3 : 5. If 9 is subtracted from each, then they are in the ratio 12 : 23. Find the smaller number.

#14. 4⁶¹+ 4⁶² + 4⁶³ + 4⁶⁴ is divisible by

#15. 5 – [4 – (3 – (3 – 3 – 6))] is equal to

#16. The product of two fractions is $\frac{14}{15}$ and their quotient is $\frac{35}{24}$ The greater of the fractions is :

$\frac74$

$\frac76$

$\frac73$

$\frac45$

#17. A shopkeeper fixes the price of an article at 30% higher than its actual cost. If he sells it at 10% discount on marked price then, the profit is :

18 %

19 %

17 %

20 %

#18. The length of a road is one kilometre. The number of plants required for plantation at a gap of 20 metres in both sides of the road is

102

100

#19. If A and B together can finish a piece of work in 20 days, B and C in 10 days and C and A in 12 days, then A, B and C jointly can finish the same work in :

$4\frac27$ days

30 day

$8\frac47$ days

$\frac{7}{60}$ days

#20. The simple interest on a sum after 4 years is $\frac15$ of the sum. The rate of interest per annum is :

4 %

5 %

6 %

8 %

#21. The ratio of the age of a father to that of his son is 5 : 2. If the product of their ages in years is 1000, then the father’s age (in years) after 10 years will be:

100

#22. The ratio of the incomes of A and B as well as B and C is 3 : 2. If one third of A’s income exceeds one fourth of C’s income by Rs.1000, what is B’s income in Rs. ?

3000

2500

3500

4000

#23. The compound interest on a certain sum of money at a certain rate for 2 years is Rs 40.80 and the simple interest on the same sum is Rs 40 at the same rate and for the same time. The rate of interest is :

2% per annum

3% per annum

4% per annum

5% per annum

#24. If the ratio of principal and the simple interest for 5 years is 10 : 3, then the rate of interest is :

5 %

6 %

8 %

3 %

#25. A sum of Rs 1600 gives a simple interest of Rs 252 in 2 years and 3 months. The rate of interest per annum is :

$5\frac12$ %

8 %

7 %

6 %

FINISH