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. In selling an article, the single discount equivalent to two successive discounts of 25% and 5% is :
#2. A manufacturer fixes his selling price at 33% over the cost of production. If the cost of production goes up by 12% and the manufacturer raises his selling price by 10%, his percentage profit is :
#3. The value of \( \frac{\sqrt{80}-\sqrt{112}}{\sqrt{45}-\sqrt{63}} \) is
#4. A shopkeeper sells an article at 15% gain. Had he sold it for Rs 18 more, he would have gained 18%. The cost price (in Rs) of the article is
#5. Between two consecutive years my income are in the ratio of 2 : 3 and expenses are in the ratio of 5 : 9. If my income in the second year is `45,000 and my expenses in the first year is ` 25,000 my total savings for the two years is :
#6. The current ages of Sonali and Monali are in the ratio 5 : 3. Five years from now, their ages will be in the ratio 10 : 7. Then, Monali’ s current age is
#7. 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
#8. The number, whose square is equal to the difference between the squares of 975 and 585 is :
#9. If 120 is 20% of a number, then 120% of that number will be :
#10. 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 :
#11. A cricketer had a certain average of runs for his 64 innings. In his 65th innings, he is bowled out for no score on his part. This brings down his average by 2 runs. His new average of runs is :
#12. Three numbers are in the ratio 2 : 3 : 4 and their HCF is 12. The L.C.M of the numbers is
Let the numbers be 2x, 3x and 4x respectively.
HCF = x = 12
Numbers are : 2 ×12 = 24
3 ×12 = 36, 4 ×12 = 48
LCM of 24, 36, 48
= 2 × 2 × 2 × 3 × 3 × 2 = 144
#13. A tap can fill a cistern in 8 hours and another tap can empty it in 16 hours. If both the taps are open, the time (in hours) taken to fill the tank will be :
#14. If the difference between S.I. and C.I. for 2 years on a sum of money lent at 5% is Rs 6, then the sum is :
#15. $ \sqrt[3]{(333)^3+(333)^3+(334)^3 -3\times333\times333\times334} $ is equal to
#16. 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 ?
#17. In a school $ \frac{1}{10} $ of the boys are same in number as $ \frac14 $ of the girls and $ \frac58 $ of the girls are same in number as $ \frac14 $ of the boys. The ratio of the boys to girls in that school is :
#18. A boat takes half time in moving a certain distance downstream than upstream. The ratio of the speed of the boat in still water and that of the current is :
#19. A grocery dealer cheats to the extent of 10% while buying as well as selling by using false weight. What is his increase in the profit %?
#20. $ (4 \frac{11}{15}+\frac{15}{71})^2-(4\frac{11}{15}-\frac{15}{71})^2$ is equal to
#21. Two numbers are in ratio 5 : 8. If their difference is 48, then the smaller number is :
#22. If Rahim deposited the same amount of Rs x in a bank at the beginning of successive 3 years and the bank pays a simple interest of 5% per annum, then the amount at his credit at the end of 3rd year will be :
#23. The average age of 40 students of a class is 15 years. When 10 new students are admitted, the average is increased by 0.2 year. The average age of the new students is :
#24. The next number of the sequence 51, 52, 56, 65 is
#25. The greatest number, by which 1657 and 2037 are divided to give remainders 6 and 5 respectively, is
The largest number which when divide the numbers a, b and c give remainders as p, q, r respectively is given by HCF of (a – p), (b – q) and (c – r)
Required number
We have to find HCF of
(1657 – 6 = 1651) and (2037 – 5 = 2032)
1651 = 13 × 127, 2032 = 16 × 127
HCF = 127 So, required number will be 127
Thank you so much