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 !!!

You have already completed the quiz before. Hence you can not start it again.

Quiz is loading...

You must sign in or sign up to start the quiz.

You have to finish following quiz, to start this quiz:

Results

0 of 10 questions answered correctly

Your time:

Time has elapsed

You have reached 0 of 0 points, (0)

Categories

Quantitative Aptitude0%

1

2

3

4

5

6

7

8

9

10

Answered

Review

Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

The percentage decrease in the surface area of a cube when each side if halved is

Correct

Explanation:
Surface area = 6a^{2}
So when side becomes a/2, Surface area = 6(a/2)^{2} = (3/2)a^{2}
So % decrease = (6a^{2} – (3/2)a^{2})/6a^{2} * 100
= (9/12) *100

Incorrect

Explanation:
Surface area = 6a^{2}
So when side becomes a/2, Surface area = 6(a/2)^{2} = (3/2)a^{2}
So % decrease = (6a^{2} – (3/2)a^{2})/6a^{2} * 100
= (9/12) *100

Question 2 of 10

2. Question

1 points

Category: Quantitative Aptitude

Pipe A and B can fill the tank in 6 hours and 8 hours respectively. In what time the tank will get filled if they are opened for alternate hours starting with B?

Correct

Explanation:
Total work = LCM(6,8) = 24
So efficiency of A = 24/6 = 4, efficiency of B = 24/8 = 3
Total work done in 2 hrs = 3+4 = 7
Multiply both sides by 3
So Total work done in 6 hrs = 21
Now work left = 24-41 = 3
Next hr, B’s turn, efficiency of B = 3, so he will do that left 3 work in 3/3 = 1 day
So total 6+1 = 7 hrs

Incorrect

Explanation:
Total work = LCM(6,8) = 24
So efficiency of A = 24/6 = 4, efficiency of B = 24/8 = 3
Total work done in 2 hrs = 3+4 = 7
Multiply both sides by 3
So Total work done in 6 hrs = 21
Now work left = 24-41 = 3
Next hr, B’s turn, efficiency of B = 3, so he will do that left 3 work in 3/3 = 1 day
So total 6+1 = 7 hrs

Question 3 of 10

3. Question

1 points

Category: Quantitative Aptitude

The list price of an article is 30% more than its cost price. Then the rate of discount offered so that a person gains 4% profit is

Correct

Explanation:
Use formula: MP = [(100+p%)/(100-d%)] * CP
So using this:
(130/100) * CP= [(100+4)/(100-d%)] * CP
13/10 = 104/(100-d%)
80 = 100-d%
d = 20

Incorrect

Explanation:
Use formula: MP = [(100+p%)/(100-d%)] * CP
So using this:
(130/100) * CP= [(100+4)/(100-d%)] * CP
13/10 = 104/(100-d%)
80 = 100-d%
d = 20

Question 4 of 10

4. Question

1 points

Category: Quantitative Aptitude

A and B invested Rs 16000 and Rs 12000 respectively in a business. After 3 months A withdrew Rs 5000 while B invested Rs 5000 more. After 3 more months C joined them with Rs 21000. After 1 year, out of a total profit of Rs 26,400, the share of B exceeded that of C by

Correct

Explanation:
A : B : C
16000*3 + 11000*9 : 12000*3 + 17000*9 : 21000*6
7 : 9 : 6
So, difference in shares of B and C = (9-6)/(7+9+6) * 26400 = 3600

Incorrect

Explanation:
A : B : C
16000*3 + 11000*9 : 12000*3 + 17000*9 : 21000*6
7 : 9 : 6
So, difference in shares of B and C = (9-6)/(7+9+6) * 26400 = 3600

Question 5 of 10

5. Question

1 points

Category: Quantitative Aptitude

In ∆ABC, the sides AB, BC and CA are produced to E, F and G respectively. If angle CBE = angle ACF = 125 degree, then the value of angle GAB in degrees is

Correct

Explanation:

CBE = 125, so ABC = 180-125 = 55
Similarly ACB = 55
So GAB = ABC+ACB = 55+55 = 110

Incorrect

Explanation:

CBE = 125, so ABC = 180-125 = 55
Similarly ACB = 55
So GAB = ABC+ACB = 55+55 = 110

Question 6 of 10

6. Question

1 points

Category: Quantitative Aptitude

In a rhombus, if angle A = 60 degree and AB = 10 cm, then the diagonal BD is

Correct

Explanation:

Diagonals of rhombus intersect at right angle. Also the diagonals bisect the angle and themselves also. Angle A = angle C = 60 degree
So angle B = angle D = 120 degree
In rt angled ∆AOB
Angle AOB = 90
Angle OAB = 60/2 = 30 and angle ABO = 120/2 = 60
Sin 30 = perpendicular/hypotenuse = OB/AB = OB/10
So 1/2 = OB/10 , so OB = 5cm
So BD = 5*2 = 10 cm

Incorrect

Explanation:

Diagonals of rhombus intersect at right angle. Also the diagonals bisect the angle and themselves also. Angle A = angle C = 60 degree
So angle B = angle D = 120 degree
In rt angled ∆AOB
Angle AOB = 90
Angle OAB = 60/2 = 30 and angle ABO = 120/2 = 60
Sin 30 = perpendicular/hypotenuse = OB/AB = OB/10
So 1/2 = OB/10 , so OB = 5cm
So BD = 5*2 = 10 cm

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

The minimum value of 2 tan^{2}ɸ + 8 cot^{2}ɸ is

Correct

Explanation:
We have formula, Minimum value = 2√(ab)
So Minimum value = 2√(2*8) = 2*4 = 8

Incorrect

Explanation:
We have formula, Minimum value = 2√(ab)
So Minimum value = 2√(2*8) = 2*4 = 8

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

If cosa/sinb = n and cosa/cosb = m, then the value of cos^{2}b equals

Correct

Explanation:
We want only cosb term
So first divide both equations, this gives
cosb/sinb = n/m
now observe the options, there are square terms
so we should square both sides, this gives
cos^{2}b/sin^{2}b = n^{2}/m^{2}
now see we have m^{2} + n^{2} term
so lets do componendo and dividend, this gives
[cos^{2}b + sin^{2}b]/[ cos^{2}b – sin^{2}b] = [n^{2} + m^{2}]/ [n^{2} – m^{2}]
This gives
1/[ cos^{2}b – sin^{2}b] = [n^{2} + m^{2}]/ [n^{2} – m^{2}]
Or [ cos^{2}b – sin^{2}b] = [n^{2} – m^{2}]/ [n^{2} + m^{2}]
Or [ cos^{2}b – (1 – cos^{2}b)] = [n^{2} – m^{2}]/ [n^{2} + m^{2}]
So 2cos^{2}b – 1= [n^{2} – m^{2}]/ [n^{2} + m^{2}]
2cos^{2}b = [(n^{2} – m^{2})/ (n^{2} + m^{2})] + 1
Or 2cos^{2}b = [n^{2} – m^{2} + n^{2} + m^{2}]/ [n^{2} + m^{2}]
Or 2cos^{2}b = 2n^{2} / [n^{2} + m^{2}]

Incorrect

Explanation:
We want only cosb term
So first divide both equations, this gives
cosb/sinb = n/m
now observe the options, there are square terms
so we should square both sides, this gives
cos^{2}b/sin^{2}b = n^{2}/m^{2}
now see we have m^{2} + n^{2} term
so lets do componendo and dividend, this gives
[cos^{2}b + sin^{2}b]/[ cos^{2}b – sin^{2}b] = [n^{2} + m^{2}]/ [n^{2} – m^{2}]
This gives
1/[ cos^{2}b – sin^{2}b] = [n^{2} + m^{2}]/ [n^{2} – m^{2}]
Or [ cos^{2}b – sin^{2}b] = [n^{2} – m^{2}]/ [n^{2} + m^{2}]
Or [ cos^{2}b – (1 – cos^{2}b)] = [n^{2} – m^{2}]/ [n^{2} + m^{2}]
So 2cos^{2}b – 1= [n^{2} – m^{2}]/ [n^{2} + m^{2}]
2cos^{2}b = [(n^{2} – m^{2})/ (n^{2} + m^{2})] + 1
Or 2cos^{2}b = [n^{2} – m^{2} + n^{2} + m^{2}]/ [n^{2} + m^{2}]
Or 2cos^{2}b = 2n^{2} / [n^{2} + m^{2}]

Question 9 of 10

9. Question

1 points

Category: Quantitative Aptitude

If x + (1/x) = 3, then [x^{2} + 4x + 1] / [x^{2} + 18x + 1] =

Correct

Explanation:
[x^{2} + 4x + 1] / [x^{2} + 18x + 1]
Take common x from both num and den and cancel
So = [x + 4 + 1/x] / [x + 18 + 1/x]
= [3+4]/[3+18] = 7/21

Incorrect

Explanation:
[x^{2} + 4x + 1] / [x^{2} + 18x + 1]
Take common x from both num and den and cancel
So = [x + 4 + 1/x] / [x + 18 + 1/x]
= [3+4]/[3+18] = 7/21

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

The sides of a triangle are in the ratio 1/2 : 1/3 : 1/4. If the perimeter of the triangle is 65 cm, then the largest side is

Correct

Explanation:
1/2 : 1/3 : 1/4 = 6 : 4 : 3,
So 6x+4x+3x = 65
13x = 65, so x = 5
So largest side is = 6x = 6*5

Incorrect

Explanation:
1/2 : 1/3 : 1/4 = 6 : 4 : 3,
So 6x+4x+3x = 65
13x = 65, so x = 5
So largest side is = 6x = 6*5