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 a school 
of the boys are same in number as 
of the girls and 
of the girls are same in number as of the boys. The ratio of the boys to girls in that school is :
#2. The smallest among![\sqrt[6]{12},\sqrt[3]4,\sqrt[4]5,\sqrt3](https://s0.wp.com/latex.php?latex=%5Csqrt%5B6%5D%7B12%7D%2C%5Csqrt%5B3%5D4%2C%5Csqrt%5B4%5D5%2C%5Csqrt3+&bg=ffffff&fg=000&s=0&c=20201002)
is
#3. 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 :
#4. A 200 metre long train is running at a speed of 72 km/hr. How long will it take to cross 800 metre long bridge?
#5. The cost of manufacturing an article was Rs 900. The trader wants to gain 25% after giving a discount of 10%. The marked price must be :
#6. 12 pumps working 6 hours a day can empty a completely filled reservoir in 15 days. How many such pumps working 9 hours a day will empty the same reservoir in 12 days?
#7. When
) is divided by 
), the result is
#8. 553 + 173 – 723 + 201960 is equal to
#9. The speed of a boat is 5 km per hour in still water and the speed of the stream is 3 km per hour. If the boat takes 3 hours to go to a place and come back, then the distance of the place is :
#10. 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
#11. If the product of two positive numbers is 1575 and their ratio is 7 : 9, then the greatest number is :
#12. The HCF (GCD) of a, b is 12, a, b are positive integers and a > b > 12. The smallest values of (a, b) are respectively
HCF of a and b = 12
Numbers = 12x and 12y where x and y are prime to each other.
a > b > 12
a = 36; b = 24
#13. A man loses Rs 55.50 yearly when the annual rate of interest falls from 11.5% to 10%. His capital (in rupees) is :
#14. 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
#15. The greatest number, that divides 122 and 243 leaving respectively 2 and 3 as remainders, is
Clearly, 122 – 2 = 120 and 243 – 3 = 240 are exactly divisible by the required number.
Required number
= HCF of 120 and 240 = 120
#16. If 13 + 23 + … + 103 = 3025, then 4 + 32 + 108 + … + 4000 is equal to
#17. The middle term(s) of the following series 2 + 4 + 6 + … + 198 is
#18. if
=80, then P is equal to
#19. Nita blends two varieties of tea one costing Rs 180 per kg and another costing Rs 200 per kg in the ratio 5 : 3. If she sells the blended variety at Rs 210 per kg, then her gain percent is :
#20. 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 :
#21. The average marks of 50 students in a class is 72. The average marks of boys and girls in that subject are 70 and 75 respectively. The number of boys in the class is :
#22. The product of two 2–digit numbers is 2160 and their HCF is 12. The numbers are
#23. 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
#24. Two types of tea costing Rs 180 per kg and Rs 280 per kg should be mixed in the ratio so that the mixture obtained was sold at Rs 320 per kg to earn a profit of 20% is
#25. In a class there are 30 boys and their average age is 17 years. When on one boy aged 18 years leaving the class and another joining, the average age becomes 16.9 years. The age of new boy is :
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