SSC CGL 2018 Practice Test Papers | Quantitative Aptitude (Day-20)

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SSC CGL 2018 | Quantitative Aptitude (Day-20)

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1) If A person travels with a speed of 60 kmph in certain distance, he will reach his destination in the estimated time. But He covered 2/3 of the distance in in 4/5 of estimated time.  Find the new speed by which he covers the remaining distance on time.

a) 100

b) 40

c) 80

d) 60

2) A and B starts a business in the ratio 4:5. A starts the business with an initial investment of Rs. 2500.After 3 years A withdraws Rs. 500 from his investment and B increases his investment by 500. Find the profit ratio after 5 years.

a) 61: 100

b) 133: 79

c) 92:133

d) 78: 151

3) A median bisects ________ of the triangle into two halves.

a) Area

b) Angle

c) Both angle and area

d) Either a or b

4) The sum of four positive numbers is odd. What is the minimum number of number of odd numbers to have sum of four numbers to be odd?

a) 1

b) 2

c) 3

d) 4

5) In how many ways a number 3780 are written as product of two different factors?

a) 20

b) 35

c) 30

d) 24

6) Find the value of 5/ (1+tan2α) +2/ (1+cot2α) +3 sin2α

a) 1

b) 6

c) 5

d) 2

7) Find how many triangles are there in the figure?

a) 13

b) 11

c) 12

d) 10

8) If a/b= c/d then the value of (ab+cd)2 is equal to______

a) (a2+ c2) (b2+ d2)

b) a2b2+ c2d2

c) a2d2+ b2c2

d) (a2+ d2)(b2+c2)

9) A car normally covers a distance of 360 km at a speed of 45 km/hr. Due to some mechanical failure it halted in two places for different time periods in the ratio 3:5 and its average speed reduced to 30 km/hr. How long the car stopped at first place?

a) 1 hour

b) 3.5 hours

c) 2 hours

d) 1.5 hours

10)  The average of 15 consecutive natural numbers starting from 333 to 361 is x. Find the average if 4 more consecutive natural numbers are added before 333.

a) 347

b) 345

c) 343

d) 351

Answers :

1). Answer:  a)

Let the estimated time= t, distance = x

S=60 kmph

S1=?, S2=?

X/t=60—-(i)

Sub (i) in (ii)

(2x/3)/(4t/5)= (10/12) (x/t)——–(ii)

S1=10/12 x60=50kmph

Remaining distance= x(1-2/3)= x/3

S2=(x/3)/(t/5)——(iii)

Sub (i) in (iii)

S2=5x/3t = 5/3* 60 =100 kmph

The new speed by which he covers the remaining distance on time = 100kmph

2). Answer: c)

Investment of B= 2500/4×5=3125

Profit ratio of A: B = (2500×3) + (2000×2): (3125x 3)+ (3625×2)
= (7500+4000) : (9375+7250)
= 11500 : 16725

= 92:133

3). Answer: a)

The line segment joining midpoint of a side to the vertex opposite to the side is called median. A median bisects area of the triangle intwo equal parts.

4). Answer: a)

Let the four numbers be a, b, c and d respectively.

If all the 4 numbers are even, sum = even

If 1 number is odd and 3 numbers are even, then the sum = odd

5). Answer : d)

Express the given number as prime factors.

3780= 22x 33x5x7

Total number of factors = (2+1) (3+1) (1+1) (1+1) = 3x4x2x2 = 48

No. of ways= 48/2=24

6). Answer: c)

7). Answer: d)

Number of triangles= 1+2+3+4 = 10

8). Answer: a)

Let a=c=2 and b=d=3

Substitute the value in (ab+cd)2

=144

Substitute the value in given options and find which of the equation value is equal to 144.

Option a) satisfies the condition

9). Answer: d)

Original time = 360/45 = 8 hours

Time taken if the avg. speed is 30 km/hr = 360/30 = 12 hours

Car halted for 4 hrs in the ratio 3:5

Halted Time at the first place=   4/8*3=1.5 hrs

10). Answer: b)

Average of numbers from 333 to 361= (333+361)/2=347

Average of all natural numbers if 4 numbers are added before 333= 347-2=345

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