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 number when divided by 221 leaves a remainder 64. What is the remainder if the same number is divided by 13?
#2. A shopkeeper allows a discount of 10% on the marked price of a camera. Marked price of the camera, which costs him Rs 600, to make a profit of 20% should be :
#3. A sum of money at simple interest triples itself in 15 years. It will become 5 times of itself in :
#4. Three numbers are in the ratio 1 : 2 : 3. By adding 5 to each of them, the new numbers are in the ratio 2 : 3 : 4. The numbers are:
#5. A manufacturer sells an item to a wholesale dealer at a profit of 18%. The wholesaler sells the same to a retailer at a profit of 20%. The retailer in turn sells it to a customer for Rs 15,045 thereby earning a profit of 25%. The cost price of the manufacturer is :
#6. A sum was invested on simple interest at a certain rate for 2 years. Had it been put at 3% higher rate, it would have fetched Rs 72 more. The sum is :
#7. The sum of a natural number and its square equals the product of the first three prime numbers. The number is
#8. The digit in unit’s place of the number (1570)2 + (1571)2 + (1572)2 + (1573)2 is :
#9. On selling 17 balls at Rs 720, there is a loss equal to the cost price of 5 balls. The cost price (in Rs) of a ball is
#10. The ratio of the fifth and sixth terms of the sequence 1, 3, 6, 10, … is
#11. A sum of money becomes $ \frac76 $ of itself in 3 years at a certain rate of simple interest. The rate per annum is :
#12. 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 :
#13. The ratio of two numbers is 3 : 8 and their difference is 115. The smaller of the two numbers is :
#14. In a test a student got 30% marks and failed by 25 marks. In the same test another student got 40% marks and secured 25 marks more than the essential minimum pass marks. The maximum marks for the test were
#15. A sum of money is paid back in two annual instalments of Rs 17640 each, allowing 5% compound interest compounded annually. The sum borrowed was :
#16. $ \frac{4.41\times0.16}{2.1\times 1.6 \times 0.21} $ is simplified to
#17. If out of 10 selected students for an examination, 3 were of 20 years age, 4 of 21 and 3 of 22 years, the average age of the group is :
#18. At what rate percent per annum will the simple interest on a sum of money be $ \frac25 $ of the amount in 10 years ?
#19. 2 men and 1 woman can complete a piece of work in 14 days while 4 women and 2 men can do the same work in 8 days. If a man gets Rs 180 per day, then what amount will a woman get per day?
#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. $ \sqrt{110\frac14} $ is equal to
#22. The next number of the sequence 51, 52, 56, 65 is
#23. If the product of first 50 positive consecutive integers be divisible by $ 7^n $ , where n is an integer, then the largest possible value of n is :
#24. The average of five numbers is 7. When three new numbers are included, the average of the eight numbers becomes 8.5. The average of three new numbers is :
#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