SSC CGL 2018 Practice Test Papers | Quantitative Aptitude (Day-14)

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SSC CGL 2018 | Quantitative Aptitude (Day-14)

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  1. The circumference of a circle is increased by 10 % due to increase in radius. The radius increases by
  1. 10
  2. 100
  3. 21
  4. d. 25
  1. The circumradius is always_______________ median.
  1. greater than
  2. less than
  3. equal to
  4. d. twice
  1. A builder decided to build a building in 50 days. He employed 200 men in the beginning and 150 more men after 30 days and finished the work in 50 days. If he had not employed the additional men, how many days are required more than the estimated duration?
  1. 30
  2. 15
  3. 20
  4. 14
  1. The average of four consecutive odd numbers A, B, C and D is 18. Find A (B+C).
  1. 550
  2. 660
  3. 540
  4. 780
  1. The H.C.F. and L.C.M. of two numbers are 3 and 2310 respectively. If one of the numbers is 21, the other is

(a) 360

(b) 330

(c) 240

(d) 960

  1. What is that number whose 125% is 1600?

(A) 5000

(B) 12000

(C) 10000

(D) 20000

  1. Two numbers are in the ratio 7: 12. If the larger number is 25 more than the smaller number, what would be the product of the two numbers?

(A) 2500

(B) 2100

(C) 1255

(D) 1140

  1. If x +y =7√( xy) , what is the value of x/y+ y/ x?

(A) 7/47

(B) 47/7

(C) 7

(D) 47

  1. Rakesh cycles at 4 km/h more than his usual speed and reached the destination 80 km away 1 hour earlier. What is his usual speed?

(1) 15 km/h

(2) 16 km/h

(3) 25 km/h

(4) 10 km/h

  1. What will be the minimum value of (x – 5) (x – 13)?

(A) – 10

(B) 3

(C) 8

(D) – 16

Answers :

1). Answer: c

Increase in circumference of circle= 10%

Increase of radius = 10%

Increase radius = [2×10+102/100] % = 21

2). Answer: d

The circumradius is always twice the median.

3). Answer : b

Or M 1 D 1 = M 2 D 2

200 x X = 350 x 20

X =35

Work will be completed 35 – 20 = 15 days behind the schedule

4). Answer: c

Let the four numbers A, B, C, D be x-2, x, x+2, x+4 respectively.

X – 2 + x + x +2 + x + 4 = 18 x 4 = 72

4x + 12 = 12

4x = 60

X = 15

A = 15;            B = 17;            C = 19;            d = 21

A (B+C) = 15 (17+19) = 540

5). Answer: B

First number × second number = HCF × LCM

21 × second number = 3 × 2310

Second number = 330

6). Answer: D

Suppose that number is = x.

∴ 125% of x = 125x/100 ⇒ 125x/100 = 1600

∴ x = 1600*100/125 = 20000

7). Answer: (B)

Let the numbers be 7x and 12x, respectively.

According to the question, 12x = 7x + 25

5x = 25

x = 25/5 = 5

Therefore, numbers are 35 and 60 Product of numbers = 35 × 60 = 2100

8). Answer: D

(x+y)2 = x2+y2+2xy = 49xy

x2+y2 = 49xy – 2xy = 47xy

(x2+y2)/xy = 47

x2/xy + y2/xy = 47

x/y + y/x = 47

9). Answer: (b)

Let the usual speed be x km/h.

∴ Usual time = 80/x h

New time = 80/(x+4) h

∴ 80/x − 80/(x + 4) = 1

x2 + 4x – 320 = 0

x = 16,-20

x=-20 is neglected since it is negative.

Usual speed of rakesh =16 km/h

10). Answer: (D)

(x – 5) (x – 13) = x2 – 18x + 65

For a quadratic equation, ax2 + bx + c = 0,

The minimum value of the equation is calculated as (4ac- b2)/4a.

Minimum value = (4x1x65) – (-18)2)/ 4×1

= (260-324)/ 4 =-64/4 = -16

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