Dear Aspirants, Here we have given the Important SSC CGL Exam 2018 Practice Test Papers. Candidates those who are preparing for SSC CGL 2018 can practice these questions to get more confidence to Crack SSC CGL 2018 Examination.
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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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