Welcome to Online Quant Test in AffairsCloud.com. We are starting SSC CGL 2016 Course and we are creating sample questions in Quant section, type of which will be asked in SSC CGL 2016 !!!

Let 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 * a

Incorrect

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 points

Category: Quantitative Aptitude

If 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 degrees

Incorrect

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 points

Category: Quantitative Aptitude

Find 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 points

Category: Quantitative Aptitude

Directions (5-8): 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:
(1000-400)/400 * 100

Incorrect

Explanation:
(1000-400)/400 * 100

Question 6 of 10

6. Question

1 points

Category: Quantitative Aptitude

Directions (5-8): 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:
(900-800)/900 * 100

Incorrect

Explanation:
(900-800)/900 * 100

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

Directions (5-8): 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: (1200-1000)/1000 * 100 = 20% – the lowest

Incorrect

Explanation:
Find for all given years in the options.
For 2014: (1200-1000)/1000 * 100 = 20% – the lowest

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

Directions (5-8): 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 points

Category: Quantitative Aptitude

The 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 = 90-60 , so after 5 years 30+5

Incorrect

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 = 90-60 , so after 5 years 30+5

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

A 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.]