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. By walking at $ \frac34 $ of his usual speed, a man reaches his office 20 minutes later than his usual time. The usual time taken by him to reach his office is :
#2. 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 :
#3. 84 Maths books, 90 Physics books and 120 Chemistry books have to be stacked topic wise. How many books will be there in each stack so that each stack will have the same height too ?
As the height of each stack is same, the required number of books in each stack
HCF of 84, 90 and 120
84 = 2 × 2 × 3 × 7, 90 = 2 × 3 × 3 × 5, 120 = 2 × 2 × 2 × 3 × 5
HCF = 2 × 3 = 6
#4. A shopkeeper sold his goods at half the list price and thus lost 20%. If he had sold on the listed price, his gain percentage would be.
#5. A train of length 120 m long, takes 6 seconds to pass a telegraph post, the speed of train is :
#6. A store offe Rs a variety of discounts that range between 20% and 25% inclusive. If a book is discounted to a price of Rs 270, then its greatest possible original price was :
#7. The middle term(s) of the following series 2 + 4 + 6 + … + 198 is
#8. . If x : y = 3 : 4, then 4x + 5y : 5x− 2y = ?
#9. 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:
#10. The average weight of the first 11 persons among 12 persons is 95 kg. The weight of 12th person is 33 kg more than the average weight of all the 12 persons. The weight of the 12th person is :
#11. The simplification of $ \frac18+\frac{1}{8^2}+\frac{1}{8^3}+\frac{1}{8^4}+\frac{1}{8^5} $up to three places of decimals yields
#12. 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 :
#13. 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 :
#14. The HCF of two numbers is 8. Which one of the following can never be their LCM ?
HCF of two numbers is 8.This means 8 is a factor common to both the numbers. LCM is common multiple for the two numbers, it is divisible by the two numbers. So, the required answer = 60
#15. A man spends Rs. 1800 monthly on an average for the first four months and Rs. 2000 monthly for the next eight months and saves Rs. 5600 a year. His average monthly income is :
#16. A man rows down a river 15 km in 3 hours with the stream and returns in $ 7\frac1 2 $ hours. The rate at which he swims 26 km downstream is :
#17. The least number, that must be added to 1720 so as to obtain a perfect cube is :
#18. The population of a town increases by 5% every year. If the present population is 9261, the population 3 years ago was :
#19. 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 :
#20. The ratio of cost price and selling price is 5 : 4, the loss percent is :
#21. 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
#22. The greatest number, which when divide 989 and 1327 leave remainders 5 and 7 respectively, is :
The largest number which when divide the numbers a, b and c give remainders as p, q, r respectively is given by H.C.F. of (a – p), (b – q) and (c – r)
Required number
= HCF of (989 – 5) and (1327 – 7)
= HCF of 984 and 1320 = 24
HCF = 24
#23. The value of $ 3\div[\left(8-5)\div{\{(4-2)+(2+\frac{8}{13})}\}\right] $is
#24. 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 :
#25. A can do a piece of work in 6 days. B is 25% more efficient than A. How long would B alone take to finish this work?
#26. The cost price of 100 books is equal to the selling price of 60 books. The gain percentage/loss percentage is :
#27. If a number is as much greater than 31 as it is less than 75, then the number is :
#28. Divide 50 into two parts so that the sum of their reciprocals is $ \frac{1}{12} $
#29. To gain 8% after allowing a discount of 10%, by what percent cost price should be hiked in the list price?
#30. The difference between the value of the number increased by 20% and the value of the number decreased by 25% is 36. Find the number :
#31. The sum of the squares of three consecutive natural numbers is 2030. Then, what is the middle number?
#32. The average of 15 numbers is 7. If the average of the first 8 numbers be 6.5 and the average of last 8 numbers be 9.5, then the middle number is :
#33. Out of 4 numbers, whose average is 60, the first one is one-fourth of the sum of the last three. The first number is :
#34. The list price of a clock is Rs 160. A customer buys it for Rs 122.40 after two successive discounts. If first discount is 10%, the second is ?
#35. The sum lent at 5% per annum (i.e., 365 days) simple interest that produces interest, of Rs 2.00 per day is :
#36. 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
#37. A reduction of 25% in the price of rice enables a person to buy 10 kg more rice for Rs.600. The reduced per kg price of rice
#38. A builder borrows Rs 2550 which is to be paid back with compound interest at the rate of 4% per annum by the end of 2 years in two equal yearly instalments. How much will each instalment be?
#39. A sum of money placed at compound interest doubles itself in 4 years. In how many years will it amount to four times itself ?
#40. Two numbers are in the ratio 3 : 4. Their LCM is 84. The greater number is
#41. A sum of Rs 1550 was lent partly at 5% and partly at 8% simple interest. The total interest received after 3 years is Rs 300. The ratio of money lent at 5% to that at 8% is :
#42. The state electricity board gives 15% discount on electric bills if it is paid before due date. One person gets Rs 54 as discount. The amount of actual bill was :
#43. 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
#44. The ratio of two numbers is 3 : 8 and their difference is 115. The smaller of the two numbers is :
#45. The Government reduced the price of sugar by 10 per cent. By this a consumer can buy 6.2 kg more sugar for Rs 837. The reduced price per kg of sugar is :
#46. A pipe can fill a tank in ‘x’ hours and another pipe can empty it in ‘y’ (y > x) hours. If both the pipes are open, in how many hours will the tank be filled?
#47. If $ \sqrt7 = 2.646 $, then the value of $ \frac{1}{\sqrt{28}} $up to three places of decimals is :
#48. The digit in unit’s place of the number (1570)2 + (1571)2 + (1572)2 + (1573)2 is :
#49. Arrange $ \frac45,\frac78,\frac67,\frac56 $ in the ascending order.
#50. 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