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 difference between the compound interest and simple interest on a certain sum for 2 years at 10% per annum is 300. Find the sum.
#2. A policeman goes after a thief who has 100 metres start if the policeman runs a kilometre in 8 minute and the thief a km in 10 minute, then the distance covered by the thief before he is over-powered is :
#3. The product of two fractions is
and their quotient is
The greater of the fractions is :
#4. Out of 4 numbers, whose average is 60, the first one is one-fourth of the sum of the last three. The first number is :
#5. 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 :
#6. 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 :
#7. 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)
#8. Ramesh deposited Rs 15600 in a fixed deposit at the rate of 10% per annum simple interest. After every second year, he adds his interest earnings to the principal. The interest at the end of fourth year is :
#9. A man rows 750 m in 600 seconds against the stream and returns in
minutes. Its rowing speed in still water is (in km/hr).
#10. At what per cent per annum will Rs 3000 amounts to Rs 3993 in 3 years if the interest is compounded annually?
#11. What least number must be subtracted from 1936 so that the resulting number when divided by 9, 10 and 15 will leave in each case the same remainder 7 ?
LCM of 9, 10 and 15 = 90
The multiple of 90 are also divisible by 9, 10 or 15.
21 × 90 = 1890 will be divisible by them.
Now, 1897 will be the number
that will give remainder 7.
Required number = 1936 – 1897 = 39
Thank you so much