Quantitative aptitude Quiz exclusively for SSC CHSL. Attempt now and test your strength in Reasoning section!
SSC CGL  07
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Question 1 of 10
1. Question
Read the table given below and answer the question.
Train Total number of seats Percentage of seats booked X 1200 87 Y 1400 92 Z 1500 96 Find the total number of seats that is not booked in all the given trains together.
Correct
Required total number of seats that is not booked = 13% of 1200 + 8% of 1400 + 4% of 1500
= 156 + 112 + 60
= 328
Hence, option b.Incorrect
Required total number of seats that is not booked = 13% of 1200 + 8% of 1400 + 4% of 1500
= 156 + 112 + 60
= 328
Hence, option b. 
Question 2 of 10
2. Question

Question 3 of 10
3. Question
The rate at which pipe A and pipe B fill the same tank is in the ratio 2:3, respectively. If together they can fill 2/3^{rd} of the tank in 20 minutes, then find the time taken by pipe A alone to fill the tank.
Correct
Since 2/3 of the tank is filled in 20 minutes. So, the tank gets fully filled in 30 minutes.
Since, rate of filling a tank by a pipe is inversely proportional to the time taken.
Let, the time taken by A and B alone to fill the tank be 3x minutes and 2x minutes, respectively.
According to question,
1/2x + 1/3x = 1/30
=> x = 25 minutes
So, time taken by A alone to fill the tank = 3x = 75 minutes
Hence, option b.Incorrect
Since 2/3 of the tank is filled in 20 minutes. So, the tank gets fully filled in 30 minutes.
Since, rate of filling a tank by a pipe is inversely proportional to the time taken.
Let, the time taken by A and B alone to fill the tank be 3x minutes and 2x minutes, respectively.
According to question,
1/2x + 1/3x = 1/30
=> x = 25 minutes
So, time taken by A alone to fill the tank = 3x = 75 minutes
Hence, option b. 
Question 4 of 10
4. Question
Find the value of cos 40^{o} cos 20^{o} – sin 20^{o} sin 40^{o}.
Correct
Since, cosx cosy – sinx siny = cos(x + y)
So, cos 40^{o}cos 20^{o} – sin 20^{o}sin 40^{o}
= cos (40 + 20)^{o}
= cos 60^{o} = 1/2
Hence, option c.Incorrect
Since, cosx cosy – sinx siny = cos(x + y)
So, cos 40^{o}cos 20^{o} – sin 20^{o}sin 40^{o}
= cos (40 + 20)^{o}
= cos 60^{o} = 1/2
Hence, option c. 
Question 5 of 10
5. Question
If 2^{x+3} = 64 and 4^{y+1} = 256, then find the value of 1/x + 1/y.

Question 6 of 10
6. Question
Find the volume of a cone whose radius and slant height is 7 cm and 25 cm, respectively.

Question 7 of 10
7. Question
If (x – y) = 7, (y – z) = 4 and (z – x) = –11, then find the value of (x^{3} + y^{3} + z^{3} – 3xyz)/(x + y + z).
Correct
x^{3} + y^{3} + z^{3} – 3xyz = (x + y + z)(x^{2} + y^{2} + z^{2} – xy – yz – zx)
x^{3} + y^{3} + z^{3} – 3xyz = (1/2)(x + y + z){(x – y)^{2} + (y – z)^{2} + (z – x)^{2}}
x^{3} + y^{3} + z^{3} – 3xyz = (1/2)(x + y + z){(7)^{2} + (4)^{2} + (11)^{2}}
x^{3} + y^{3} + z^{3} – 3xyz = (1/2)(x + y + z){49 + 16 + 121)
x^{3} + y^{3} + z^{3} – 3xyz = (1/2)(x + y + z)(186)
x^{3} + y^{3} + z^{3} – 3xyz = 93(x + y + z)
So, (x^{3} + y^{3} + z^{3} – 3xyz)/(x + y + z) = 93(x + y + z) /(x + y + z) = 93
Hence, option b.Incorrect
x^{3} + y^{3} + z^{3} – 3xyz = (x + y + z)(x^{2} + y^{2} + z^{2} – xy – yz – zx)
x^{3} + y^{3} + z^{3} – 3xyz = (1/2)(x + y + z){(x – y)^{2} + (y – z)^{2} + (z – x)^{2}}
x^{3} + y^{3} + z^{3} – 3xyz = (1/2)(x + y + z){(7)^{2} + (4)^{2} + (11)^{2}}
x^{3} + y^{3} + z^{3} – 3xyz = (1/2)(x + y + z){49 + 16 + 121)
x^{3} + y^{3} + z^{3} – 3xyz = (1/2)(x + y + z)(186)
x^{3} + y^{3} + z^{3} – 3xyz = 93(x + y + z)
So, (x^{3} + y^{3} + z^{3} – 3xyz)/(x + y + z) = 93(x + y + z) /(x + y + z) = 93
Hence, option b. 
Question 8 of 10
8. Question
Find the interest rate at which, a person had deposited Rs. 40000 at compound interest to get a total amount of Rs. 46656 after 2 years.

Question 9 of 10
9. Question

Question 10 of 10
10. Question
Train A can cross train B,when running in the same direction, in 48 seconds, while train A cross train B when running in opposite direction in 36 seconds. Find the ratio of speed of train A to speed of train B.
Correct
Let the speed of train A be ‘a’ km/h and speed of train B be ‘b’ km/h.
Since the sum of length of both trains is same while crossing each other either from the same direction or opposite direction.
So, (a – b) x 48 = (a + b) x 36
=> 48a – 48b = 36a + 36b
12a = 84b
a:b = 7:1
Hence, option a.Incorrect
Let the speed of train A be ‘a’ km/h and speed of train B be ‘b’ km/h.
Since the sum of length of both trains is same while crossing each other either from the same direction or opposite direction.
So, (a – b) x 48 = (a + b) x 36
=> 48a – 48b = 36a + 36b
12a = 84b
a:b = 7:1
Hence, option a.