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 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

#2. In an examination there are three subjects of 100 marks each. A student scores 60% in the first subject and 80% in the second subject. He scored 70% in aggregate. His percentage of marks in the third subject is :

#3. A cistern has two pipes. One can fill it with water in 8 hours and other can empty it in 5 hours. In how many hours will the cistern be emptied if both the pipes are opened together when
of the cistern is already full of water?

#4. Water tax is increased by 20% but its consumption is decreased by 20%. Then the increase or decrease in the expenditure of the money is :

#5. In a school, the ratio of boys to girls is 4 : 3 and the ratio of girls to teachers is 8 : 1. The ratio of students to teachers is :

#6. 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 :

#7. On multiplying a number by 7, all the digits in the product appear as 3’s. The smallest such number is :

#8. 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 :

#9. If 120 is 20% of a number, then 120% of that number will be :

#10. Two successive discounts of 10% and 5% is given on a bill of Rs 110. Find the net amount of money payable to clear the bill. (Answer to the nearest rupee)

#11. At an election there were two candidates. A candidate got 38% of votes and lost by 7200 number of votes. The total number of valid votes were :

#12. The value of
is

#13. The least number that should be added to 2055, so that the sum is exactly divisible by 27 is :

#14. If in a sale, the discount given on a saree is equal to one-fourth the marked price and the loss due to this discount is 15%, then the ratio of the cost price to the selling price is :

#15. Krishnan bought a camera and paid 20% less than its original price. He sold it at 40% profit on the price he had paid. The percentage of profit earned by Krishnan on the original price was :

#16. If the product of two positive numbers is 1575 and their ratio is 7 : 9, then the greatest number is :

#17. A boat goes 12 km downstream and comes back to the starting point in 3 hours. If the speed of the current is 3 km/hr, then the speed (in km/hr) of the boat in still water is :

#18. If A and B are the HCF and LCM respectively of two algebraic expressions x and y, and A + B = x + y, then the value of
is

#19. In an examination, 52% students failed in Hindi and 42% in English. If 17% failed in both the subjects, what percentage of students passed in both the subjects?

#20. If the compound interest on a certain sum for two years at 12% per annum is Rs 2544, the simple interest on it at the same rate for 2 years will be :

#21. The square root of

#22. The value of
is

#23. If the selling price of 4 articles is equal to the cost price of 5 articles, the profit per cent is :

#24. 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
#25. If
, then the value of x is
