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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeIf xy(x+y) = 1, then find the value of [1/x^{3} y^{3}] – x^{3} – y^{3} is
Correct
Explanation:
x+y = 1/xy
Cube both sides
x^{3} + y^{3} + 3xy(x+y) = 1/x^{3} y^{3}
so x^{3} + y^{3} + 3xy * 1/xy = 1/x^{3} y^{3}
so [1/x^{3} y^{3}] – x^{3} – y^{3} = 3Incorrect
Explanation:
x+y = 1/xy
Cube both sides
x^{3} + y^{3} + 3xy(x+y) = 1/x^{3} y^{3}
so x^{3} + y^{3} + 3xy * 1/xy = 1/x^{3} y^{3}
so [1/x^{3} y^{3}] – x^{3} – y^{3} = 3 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeLet two circles with centres O and K having same radius ‘a’. They pass through the centres of each other and cut each other at A and B. Find the area of the quadrilateral AOBK.
Correct
Explanation:
OK = a, OA = a, OP = a/2
So AP = √(a^{2} – a^{2}/4) = √3a/2
So AB = √3a
Area of AOBK = ½ * d1 * d2 = ½ * a * √3 * aIncorrect
Explanation:
OK = a, OA = a, OP = a/2
So AP = √(a^{2} – a^{2}/4) = √3a/2
So AB = √3a
Area of AOBK = ½ * d1 * d2 = ½ * a * √3 * a 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeIf cos^{2}ɸ/( cot^{2}ɸ – cos^{2}ɸ) = 3, then the value of ɸ is
Correct
Explanation:
cos^{2}ɸ/( cot^{2}ɸ – cos^{2}ɸ) = 3
cos^{2}ɸ = 3(cot^{2}ɸ – cos^{2}ɸ)
so 4 cos^{2}ɸ = 3 cot^{2}ɸ = 3 cos^{2}ɸ/sin^{2}ɸ
so cos^{2}ɸ (4 – 3/sin^{2}ɸ) = 0
so if cos^{2}ɸ = 0, then ɸ = 90 degrees
and if 4 – 3/sin^{2}ɸ = 0, solve ɸ = 60 degreesIncorrect
Explanation:
cos^{2}ɸ/( cot^{2}ɸ – cos^{2}ɸ) = 3
cos^{2}ɸ = 3(cot^{2}ɸ – cos^{2}ɸ)
so 4 cos^{2}ɸ = 3 cot^{2}ɸ = 3 cos^{2}ɸ/sin^{2}ɸ
so cos^{2}ɸ (4 – 3/sin^{2}ɸ) = 0
so if cos^{2}ɸ = 0, then ɸ = 90 degrees
and if 4 – 3/sin^{2}ɸ = 0, solve ɸ = 60 degrees 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeFind the minimum value of 4cos^{2}ɸ + 6sec^{2}ɸ
Correct
Explanation:
For this: a cos^{2}ɸ + b sec^{2}ɸ,
Minimum value = 2√(ab)
The maximum value can go up to infinity.
So for 4cos^{2}ɸ + 6sec^{2}ɸ, min value is 2√(4*6)Incorrect
Explanation:
For this: a cos^{2}ɸ + b sec^{2}ɸ,
Minimum value = 2√(ab)
The maximum value can go up to infinity.
So for 4cos^{2}ɸ + 6sec^{2}ɸ, min value is 2√(4*6) 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeDirections (58): Study the following bar graph and answer accordingly.
The chart shows the production of product S by company ABC during given years.
The production in 2013 in comparison to the production in 2009 increased by
Correct
Explanation:
(1000400)/400 * 100Incorrect
Explanation:
(1000400)/400 * 100 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeDirections (58): Study the following bar graph and answer accordingly.
The chart shows the production of product S by company ABC during given years.
The production decreased from 2011 to 2012 by
Correct
Explanation:
(900800)/900 * 100Incorrect
Explanation:
(900800)/900 * 100 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeDirections (58): Study the following bar graph and answer accordingly.
The chart shows the production of product S by company ABC during given years.
The year in which production increased the lowest as compared to the previous year
Correct
Explanation:
Find for all given years in the options.
For 2014: (12001000)/1000 * 100 = 20% – the lowestIncorrect
Explanation:
Find for all given years in the options.
For 2014: (12001000)/1000 * 100 = 20% – the lowest 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeDirections (58): Study the following bar graph and answer accordingly.
The chart shows the production of product S by company ABC during given years.
The production from 2010 to 2014 increased by
Correct
Incorrect

Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeThe average of the ages of 3 friends A, B, and C was 27 years before 3 years ago. If the average of the ages of A and C was 25 years 5 years ago, what will be the age of B after 5 years?
Correct
Explanation:
Sum of ages of A+B+C 3 yrs ago = 27*3 = 81
So after 3 yrs, i.e. at present their total = 81 + 3*3 = 90
Similarly sum of present age of A&C = 25*2 +5*2 = 60
So present age of B = 9060 , so after 5 years 30+5Incorrect
Explanation:
Sum of ages of A+B+C 3 yrs ago = 27*3 = 81
So after 3 yrs, i.e. at present their total = 81 + 3*3 = 90
Similarly sum of present age of A&C = 25*2 +5*2 = 60
So present age of B = 9060 , so after 5 years 30+5 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeA pipe ‘A’ can empty the tank in 30 minutes and pipe ‘B’ can fill the same tank in 15 minutes. The tank is filled with water and then pipe ‘A’ is opened. After 10 minutes, pipe B is also opened. Find in how much total time the tank will be full again?
Correct
Explanation:
Emptying pipe A is opened first for 10 minutes, so in 10 minutes the part of tank it has emptied is (1/30)*10 = 1/3
Now filling pipe is also opened, now since only 1/3 of the tank is empty so 1/3 is only to be filled by both pipes, let it take now x minutes, so
(1/15 – 1/30)*x = 1/3
Solve, x= 10
So total = 10+10 = 20 minutes [total time is asked – 10 minutes when emptying pipe was only opened, 10 minutes when both were operating.]Incorrect
Explanation:
Emptying pipe A is opened first for 10 minutes, so in 10 minutes the part of tank it has emptied is (1/30)*10 = 1/3
Now filling pipe is also opened, now since only 1/3 of the tank is empty so 1/3 is only to be filled by both pipes, let it take now x minutes, so
(1/15 – 1/30)*x = 1/3
Solve, x= 10
So total = 10+10 = 20 minutes [total time is asked – 10 minutes when emptying pipe was only opened, 10 minutes when both were operating.]
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