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Expected Quant Questions for SSC CHSL 2020 PDF

Download SSC CHSL Expected  Quant questions with answers  PDF based on previous year papers very useful for SSC CHSL Exams. Top-15 Very Important Quant  Questions for SSC Exam

Question 1: If $\frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = \frac{4}{\sqrt{3}}, 0^\circ < \theta < 90^\circ$, then the value of $(\tan \theta + \sec \theta)^{-1}$ is:

a) $2 – \sqrt{3}$

b) $3 – \sqrt{2}$

c) $2 + \sqrt{3}$

d) $3 + \sqrt{2}$

Question 2: P and Q are two points on the ground on either side of a pole. The angles of elevation of the top of the pole as observed from P and Q are $60^\circ$ and $30^\circ$, respectively and the distance between them is $84\sqrt3$ m. What is the height (in m) of the pole?

a) 63

b) 73.5

c) 52.5

d) 60

Question 3: A man can row a distance of 900 metres against the stream in 12 minutes and returns to the starting point in 9 minutes. What is the speed (in km/h) of the man in still water?

a) $4\frac{1}{2}$

b) 6

c) $5\frac{1}{4}$

d) 5

Question 4: Walking at 60% of his usual speed, a man reaches his destination 1 hour 40 minuteslate, His usual time (in hours) to reach the destination is:

a) $2\frac{1}{2}$

b) $2\frac{1}{4}$

c) $3\frac{1}{8}$

d) $3\frac{1}{4}$

Question 5: 25 persons can complete a work in 60 days. They started the work. 10 persons left the work after x days. If the whole work was completed in 80 days, then whatis the value of x ?

a) 9

b) 8

c) 12

d) 15

SSC CHSL Study Material (FREE Tests)

Question 6: A person covers 40% of the distance from A to B at 8 km/h, 40% of the remaining distance at 9 km/h and the rest at 12 km/h. His average speed (in km/h) for the journey is:

a) $9\frac{5}{8}$

b) $9\frac{2}{3}$

c) $9\frac{3}{8}$

d) $9\frac{1}{3}$

Question 7: A sum is divided among A, B, C and D such that the ratio of the shares of A and is 2 : 3, that of B and C is 1 : 2 and that of C and D is 3 : 4. If the difference between the shares of A and is ₹648,then the sum of their shares is:

a) ₹2,052

b) ₹2,160

c) ₹2,484

d) ₹1,944

Question 8: A, B and C started a business, Thrice the investment of A is equalto twice the investment of B and also equal to four times the investment of C, If C’s share out of the total profit is ₹4,863,then the share ofA in the profit is:

a) ₹7,272

b) ₹6,484

c) ₹9,726

d) ₹8,105

Question 9: In a 100m race 64 A beats B by 10 mand B beats C by 10 m. By what distance does A beat C (in m)?

a) 19

b) 18

c) 20

d) 21

Question 10: If $x^4 + 2x^3 + ax^2 + bx + 9$ is a perfect square, where a and b are positive real numbers,then the value of a and b are

a) a = 5, b = 6

b) a = 6, b = 7

c) a = 7, b = 6

d) a = 7, b = 8

Question 11: If the selling price of an article is 32% more than its cost price and the discount offered on its marked price is 12%, then what is the ratio of its cost price to the marked price?

a) 4 : 5

b) 3 : 8

c) 2 : 3

d) 1 : 2

Question 12: Sudha bought 80 articlesat the same price. She sold some of them at 8% profit and the remaining at 12% loss resulting in an overall profit of 6%. The mimber of items sold at 8% profit is:

a) 64

b) 60

c) 72

d) 70

Question 13: Two positive numbers differ by 2001, When the larger number is divided by the smaller number, the quotient is 9 and the remainder is 41. The sum of the digits ofthe larger number is:

a) 15

b) 11

c) 10

d) 14

Question 14: The average of 18 numbers is 37.5. If six numbers of average X are added to them, then the average of all the numbers increases by one, The value of x is:

a) 40

b) 41.5

c) 42

d) 38.5

Question 15: 5 years ago, the ratio of the age of A to that of B was 4 : 5. Five years hence, the ratio of the age of A to that of B will be 6 : 7. If, at present, C is 10 years younger than B, then what will be the ratio of the present age of A to that of C?

a) 3 : 2

b) 5 : 4

c) 4 : 3

d) 5 : 3

Sum of the 18 numbers = 37.5 $\times$ 18 = 675

($\because$ average = $\frac{sum of total terms}{number of terms}$)

Sum of the 6 numbers = 6 $\times$ X = 6X

Average of all the numbers = 37.5 + 1 = 38.5

$\frac{675 + 6X}{24}$ = 38.5$675 + 6X = 24$\times\$ 38.5

6X = 924 – 675 = 249

X = 41.5