Quantitative Aptitude Questions with Answers and Solution for SSC CGL- Mock Test of Maths MCQs set for online practice of upcoming Competitive exams.
Subject : Quantitative Aptitude (Mathematics)
Medium : English
Level : SSC CGL
All type Questions with Solution
As per latest exam pattern and syllabus
Set of 25 Questions – New Questions practice Set in Every Attempt
Results
#1. The sum of a pair of positive integer is 336 and their HCF is 21.The number of such possible pairs is
#2. The difference between the selling prices of an article at a profit of 15% and at a profit of 10% is Rs 10. The cost price of the article is
#3. A merchant sold an article for Rs 75 at a profit per cent equal to its cost price. The cost price of the article was :
#4. The rate of simple interest per annum of bank being decreased from 5% to
% the annual income of a person from interest was less by Rs 105. The sum deposited at the bank was :
#5. 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:
#6. The quotient when 
is divided by 
is
#7. The Government reduced the price of sugar by 10 per cent. By this a consumer can buy 6.2 kg more sugar for Rs 837. The reduced price per kg of sugar is :
#8. A supply of juice lasts for 35 days. If its use is increased by 40%, then the number of days would the same amount of juice lasts is :
#9. By selling an article for Rs 102, there is a loss of 15%, when the article is sold for Rs 134.40. The net result in the transaction is
#10. If the square of the sum of two numbers is equal to 4 times of their product, then the ratio of these numbers is :
#11. By selling 144 hens Mahesh suffered a loss equal to the selling price of 6 hens. His loss percent is :
#12. The sum of a natural number and its square equals the product of the first three prime numbers. The number is
#13. 
is equal to
#14. A certain number of men can do a work in 40 days. If there were 8 men more, it could be finished in 10 days less. How many men were there initially?
#15. 553 + 173 – 723 + 201960 is equal to
#16. The sum 5 + 6 + 7 + 8 + … + 19 is equal to
#17. The first odd number is 1, the second odd number is 3, the third odd number is 5 and so on. The 200th odd number is :
#18. Find the value of 
#19. A and B can together finish a work in 30 days. They worked at it for 20 days and then B left. The remaining work was done by A alone in 20 more days A alone can finish the work in :
#20. A double bed is marked at Rs 7500. The shopkeeper allows successive discounts of 8%, 5% and 2% on it. What is the net selling price?
#21. The average weight of the first 11 persons among 12 persons is 95 kg. The weight of 12th person is 33 kg more than the average weight of all the 12 persons. The weight of the 12th person is :
#22. The next number of the sequence 2, 5, 10, 14, 18, 23, 26, 32 … is
#23. The product of two fractions is 
and their quotient is 
The greater of the fractions is :
#24. The greatest number, by which 1657 and 2037 are divided to give remainders 6 and 5 respectively, is
The largest number which when divide the numbers a, b and c give remainders as p, q, r respectively is given by HCF of (a – p), (b – q) and (c – r)
Required number
We have to find HCF of
(1657 – 6 = 1651) and (2037 – 5 = 2032)
1651 = 13 × 127, 2032 = 16 × 127
HCF = 127 So, required number will be 127
#25. Find the least multiple of 23, which when divided by 18, 21 and 24 leaves the remainder 7, 10 and 13 respectively.
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
Thank you so much