Quantitative Aptitude Questions with answers and Solutions for SSC CPO- SI in CPMF Exam. Maths MCQs Mock Test for free online practice of SI CAPF Exam.
Practice Set : Mathematics (Quantitative Aptitude)
Questions : 50
Medium : English
Level : SSC CPO SI Exam
All Type and topic MCQs – Solved by short tricks
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#1. 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 :
#2. There is a ratio of 5 : 4 between two numbers. If 40 per cent of the first is 12, then 50% of the second number is :
#3. Product of two co-prime numbers is 117. Then their LCM is
HCF of two-prime numbers = 1
Product of numbers = their
LCM = 117
117 = 13 × 9 where 13 & 9 are
co-prime. L.C.M (13,9) = 117
#4. If the discount is equal to one fifth of the marked price and the loss is half the discount, then the percentage of loss is :
#5. In an election there were only two candidates. One of the candidates secured 40% of votes and is defeated by the other candidate by 298 votes. The total number of votes polled is :
#6. A group of 75 men are employed to lay down a railway line in 3 months. Due to certain emergency conditions, the work was to be finished in 18 days. How many more men should be employed to complete the work in the desired time ?
#7. Unit digit in (264)102 + (264)103 is :
#8. 25% of annual salary of A is equal to eighty percent of annual salary of B. The monthly salary of B is 40% of the monthly salary of C. The annual salary of C is Rs. 6 lac. What is the monthly salary of A?
#9. If 90% of A = 30% of B and B = 2x% of A, then the value of x is
#10. The ratio of the length of a school ground to its width is 5 : 2. If the width is 40 m, then the length is :
#11. A cricket player after playing 10 tests scored 100 runs in the 11th test. As a result, the average of his runs is increased by 5. The present average of runs is :
#12. LCM of 12 and 16 Prime factorisation of 
× 3 = 22 × 3 Prime factorisation of 16 = 2 × 2 × 2 × 2 = 24
[latex]e^{i pi} 1 = 0[/latex]
[latex]e^{i pi} 1 = 0[/latex]
[latex]e^{i pi} 1 = 0[/latex]
#13. If the ratio of cost price to selling price is 10: 11, then the rate of percent of profit is
#14. 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 :
#15. A man, a woman and a boy can together complete a piece of work in 3 days. If a man alone can do it in 6 days and a boy alone in 18 days, how long will a woman alone take to complete the work?
#16. A man goes downstream with a boat to some destination and returns upstream to his original place in 5 hours. If the speed of the boat in still water and the stream are 10 km/hr and 4 km/hr respectively, then the distance of the destination from the starting place is :
#17. If A’s income is 50% less than that of B’s, then B’s income is what per cent more than that of A?
#18. 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 :
#19. The simple interest on a sum after 4 years is 
of the sum. The rate of interest per annum is :
#20. A man, a woman and a boy together finish a piece of work in 6 days. If a man and a woman can do the same work in 10 and 24 days respectively then the number of days taken by a boy to finish the work is :
#21. The distance between two cities A and B is 330 km. A train starts from A at 8 a.m. and travels towards B at 60 km/hr. Another train, starts from B at 9 a.m. and travels towards A at 75 km/hr. At what time will they meet?
#22. In a race of 200 metres, B can give a start of 10 metres to A and C can give a start of 20 metres to B. The start that C can give to A, in the same race is :
#23. The ratio of the number of boys to that of girls in a village is 3: 2. If 30% of boys and 70% of girls appeared in an examination, the ratio of the number of villagers, appeared in the examination to that not appeared in the same examination is :
#24. If a boat goes 100 km downstream in 10 hours and 75 km upstream in 15 hours, then the speed of the stream is :
#25. A table with marked price Rs 1200 was sold to a customer for Rs 1100. Find the rate of discount allowed on the table.
#26. The printed price of an article is Rs 900 but the retailer gets a discount of 40%. He sells the article for Rs 900. The retailer’s gain percent is :
#27. The greatest value among the fractions 
#28. The difference between compound and simple interest on a certain sum for 3 years at 5% per annum is Rs 122. The sum is :
#29. A bicycle, marked at Rs 2,000 is sold with two successive discounts of 20% and 10%. An additional discount of 5% is offered for cash payment. The selling price of the bicycle at cash payment is :
#30. The value of is 
#31. . A train is running at 36 km/hr. If it crosses a pole in 25 seconds, then its length is :
#32. A sum of Rs 10,000 lent partly at 8% and remaining at 10% per annum. If the yearly interest on the average is 9.2%, the two parts are :
#33. A train passes two bridges of lengths 500 m and 250 m in 100 seconds and 60 seconds respectively. The length of the train is :
#34. 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
#35. Which term of the series 72, 63, 54 … is zero?
#36. The total number of prime factors in 
is
#37. A boy and girl together fill a cistern with water. The boy pours 4 litres of water every 3 minutes and the girl pours 3 litres every 4 minutes. How much time will it take to fill 100 litres of water in the cistern?
#38. Rs 500 was invested at 12% per annum simple interest and a certain sum of money is invested at 10% per annum simple interest. If the sum of the interest on both the sum after 4 years is Rs 480, the latter sum of money is :
#39. 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 ?
#40. (461 + 462 + 463) is divisible by
#41. If a number x is 10% less than another number y and y is 10% more than 125, then x is equal to :
#42. If X = 
, Y = 
, Z =
, then
#43. In a school having roll strength 286, the ratio of boys and girls is 8 : 5. If 22 more girls get admitted into the school, the ratio of boys and girls become
#44. 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
#45. The product of the LCM and the HCF of two numbers is 24. If the difference of the numbers is 2, then the greater of the number is
#46. In an examination there are three subjects of 100 marks each. A student scores 60% in the first subject and 80% in the second subject. He scored 70% in aggregate. His percentage of marks in the third subject is :
#47. The next number of the sequence 2, 5, 10, 14, 18, 23, 26, 32 … is
#48. A sum of money is divided among A, B, C and D in the proportion of 7 : 6 : 3 : 5. If B gets `270 more than C, then the share of D is :
#49. 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
#50. 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 :
