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 wrong number of the sequence 36, 81, 144, 225, 256, 441 is
#2. Every Sunday, Gin jogs 3 miles. For rest of the week, each day he jogs 1 mile more than the previous day. How many miles Gin jogs in 2 weeks?
#3. The compound interest on a certain sum for 2 year at 10% per annum is Rs 525. The simple interest on the same sum for double the time at half the rate percent per annum is :
#4. 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 :
#5. 
+
is equal to
#6. 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 :
#7. The next number of the sequence 2, 5, 10, 14, 18, 23, 26, 32 … is
#8. The square root of 
#9. At what rate percent per annum will the simple interest on a sum of money be 
of the amount in 10 years ?
#10. The marked price of a radio is Rs 4800. The shopkeeper allows a discount of 10% and gains 8%. If no discount is allowed, his gain per cent will be :
#11. If a = 4011 and b = 3989, then the value of ab = ?
#12. If a number is as much greater than 31 as it is less than 75, then the number is :
#13. The average per day income of A, B and C is Rs. 450. If the average per day income of A and B be Rs. 400 and that of B and C be Rs. 430, the per day income of B is :
#14. A and B can do a work in 12 days, B and C in 15 days and C and A in 20 days. If A, B and C work together, they will complete the work in
#15. 4 boys and 3 girls spent Rs.120 on the average, of which the boys spent Rs.150 on the average. Then the average amount spent by the girls is:
#16. 31% of employees pay tax in the year 2008. The number of non-tax paying employees are 20,700. The total number of employees is :
#17. A store offe Rs a variety of discounts that range between 20% and 25% inclusive. If a book is discounted to a price of Rs 270, then its greatest possible original price was :
#18. If a boy walks from his house to school at the rate of 4 km per hour, he reaches the school 10 minutes earlier than the scheduled time. However, if he walks at the rate of 3 km per hour, then he reaches 10 minutes late. Find the distance of his school from his house :
#19. A and B together can do a piece of work in 6 days. If A can alone do the work in 18 days, then the number of days required for B to finish the work is :
#20. A, B and C can do a piece of work in 30, 20 and 10 days respectively. A is assisted by B on one day and by C on the next day, alternately. How long would the work take to finish?
#21. The average of 20 numbers is calculated as 35. It is discovered later, that while calculating the average, one number, namely 85, was read as 45. The correct average is :
#22. A student multiplied a number by 
instead of 
. What is the percentage error in the calculation?
#23. If 20% of A = 50% of B, then what per cent of A is B?
#24. The total discount on Rs 1860 due after a certain time at 5% is Rs 60. Find the time after which it is due :
#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