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 tap can fill a cistern in 40 minutes and a second tap can empty the filled cistern in 60 minutes. By mistake without closing the second tap, the first tap was opened. In how many minutes will the empty cistern be filled?
#2. In two blends of mixed tea, the ratios of Darjeeling and Assam tea are 4 : 7and 2 : 5. The ratio in which these two blends should be mixed to get the ratio of Darjeeling and Assam tea in the new mixture as 6 : 13 is :
#3. If the cost price of 10 articles equals selling price of 9 articles, the gain or loss percent will be :
#4. The first odd number is 1, the second odd number is 3, the third odd number is 5 and so on. The 200th odd number is :
#5. A candidate secured 30% marks in an examination and failed by 6 marks. Another secured 40% marks and got 6 marks more than the bare minimum to pass. The maximum marks are :
#6. The average of six numbers is 32. If each of the first three numbers is increased by 2 and each of the remaining three numbers is decreased by 4, then the new average is :
#7. 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 :
#8. Mohan gets 3 marks for each correct sum and loses 2 marks for each wrong sum. He attempts 30 sums and obtains 40 marks. The number of sums solved correctly is
#9. Two trains of length 140 m and 160 m run at the speed of 60 km/hour and 40 km/hour respectively in opposite directions on parallel tracks. The time (in seconds) which they take to cross each other is :
#10. 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 :
#11. A sum of Rs 3200 invested at 10% p.a. compounded quarterly amounts to Rs 3362. Compute the time period
#12. A person can row 
km an hour in still water and he finds that it takes him twice as long to row up as to row down the river. The speed of the stream is :
#13. The average weight of a group of 20 boys was calculated to be 89.4 kg and it was later discovered that one weight was misread as 78 kg instead of 87 kg. The correct average weight is :
#14. 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 :
#15. What sum of money must be given as simple interest for six months at 4% per annum in order to earn Rs 150 interest?
#16. The sum of two numbers is 37 and the difference of their squares is 185, then the difference between the two numbers is :
#17. If 12 men working 8 hours a day complete the work in 10 days, how long would 16 men working hours a day take to complete the same work?
#18. Which of the following number is the greatest of all? 
#19. if
=80, then P is equal to
#20. 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
#21. 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 :
#22. A bag contains Rs. 90 coins in the denominations of 50 paise, 25 paise and 10 paise. If coins of 50 paise, 25 paise and 10 paise are in the ratio of 2 : 3 : 5, then the number of 25 paise coins in the bag is
#23. 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:
#24. Simplify: 
#25. On multiplying a number by 7, all the digits in the product appear as 3’s. The smallest such number is :
Thank you so much