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. A and B can complete a piece of work in 8 days, B and C can do it in 12 days, C and A can do it in 8 days. A, B and C together can complete it in :
#2. 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 :
#3. A group of workers can complete a piece of work in 50 days, when they are working individually. On the first day one person works, on the second day another person joins him, on the third day one more person joins them and this process continues till the work is completed. How many approximate days are needed to complete the work?
#4. Which term of the series 72, 63, 54 … is zero?
#5. By selling a tape-recorder for Rs 1040 a man gains 4%. If he sells for Rs 950, then his loss will be :
#6. The value of $ \frac{(3.2)^3-0.008}{(3.2)^2 + 0.64 +0.04} $is
#7. 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?
#8. The difference between compound and simple interest on a certain sum for 3 years at 5% per annum is Rs 122. The sum is :
#9. The product of the LCM and the HCF of two numbers is 24. If the difference of the numbers is 2, then the greater of the number is
#10. A retailer buys a radio for Rs 225. His overhead expenses are Rs 15. He sells the radio for Rs 300. The profit per cent of the retailer is :
#11. 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 :
#12. Two numbers are in the ratio 3 : 4. Their LCM is 84. The greater number is
#13. The average of 5 consecutive natural numbers is m. If the next three natural numbers are also included, then how much more than m will the average of these 8 numbers be?
#14. A, B and C can do a piece of work in 24, 30 and 40 days respectively. They began the work together but C left 4 days before completion of the work. In how many days was the work done ?
#15. A number when divided by 221 leaves a remainder 64. What is the remainder if the same number is divided by 13?
#16. On selling an article for Rs 105 a trader loses 9%. To gain 30% he should sell the article at
#17. The HCF of two numbers is 96 and their LCM. is 1296. If one of the number is 864, the other is
#18. Two numbers are in the ratio 1 : 3 If their sum is 240, then their difference is :
#19. 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 :
#20. $ \frac{5\frac{9}{14}}{5+\frac3{3+\frac1{\frac35}}} $ is equal to
#21. 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 :
#22. 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 :
#23. A pipe can fill a tank in ‘x’ hours and another pipe can empty it in ‘y’ (y > x) hours. If both the pipes are open, in how many hours will the tank be filled?
#24. 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?
#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