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Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

If sinɸ + cosecɸ = 2, then the value of sin^{6}ɸ + cosec^{6}ɸ

Correct

Explanation:
sinɸ + 1/sinɸ = 2
(sin^{2}ɸ + 1)/sinɸ = 2
sin^{2}ɸ – 2 sinɸ + 1 = 0
(sinɸ – 1) ^{2} = 0
So sinɸ – 1 = 0
So sinɸ = 1 and then also 1/sinɸ = cosecɸ = 1
So sin^{6}ɸ + cosec^{6}ɸ = 1 + 1 = 2

Incorrect

Explanation:
sinɸ + 1/sinɸ = 2
(sin^{2}ɸ + 1)/sinɸ = 2
sin^{2}ɸ – 2 sinɸ + 1 = 0
(sinɸ – 1) ^{2} = 0
So sinɸ – 1 = 0
So sinɸ = 1 and then also 1/sinɸ = cosecɸ = 1
So sin^{6}ɸ + cosec^{6}ɸ = 1 + 1 = 2

Question 2 of 10

2. Question

1 points

Category: Quantitative Aptitude

A positive integer when divided by 26 gives a remainder 15. When the same number is divided by 4, the remainder will be

Correct

Explanation:
The remainder will be the same as is obtained by dividing 15 by 4 which is 3

Incorrect

Explanation:
The remainder will be the same as is obtained by dividing 15 by 4 which is 3

Question 3 of 10

3. Question

1 points

Category: Quantitative Aptitude

If the total surface area of a hemisphere is 75ᴨ sq. cm, then the volume of hemisphere is

Correct

Explanation:
2ᴨr^{2} + ᴨr^{2} = 75ᴨ
3ᴨr^{2} = 75ᴨ
So r^{2} = 25
So r = 5
So volume = (2/3)*ᴨr^{3} = (2/3)*(22/7)*5*5*5

Incorrect

Explanation:
2ᴨr^{2} + ᴨr^{2} = 75ᴨ
3ᴨr^{2} = 75ᴨ
So r^{2} = 25
So r = 5
So volume = (2/3)*ᴨr^{3} = (2/3)*(22/7)*5*5*5

Question 4 of 10

4. Question

1 points

Category: Quantitative Aptitude

If x + 1/x = 20, then the value of (625x)/(2x^{2} + 85x + 2) is

Correct

Explanation:
(625x)/(2x^{2} + 85x + 2)
Divide both numerator n den. by x gives
(625)/(2x + 85 + 2/x)
Which equals 625/[2x + 2/x + 85] = 625/[2(x + 1/x) + 85]
= 625/[2*20 + 85] = 5

Incorrect

Explanation:
(625x)/(2x^{2} + 85x + 2)
Divide both numerator n den. by x gives
(625)/(2x + 85 + 2/x)
Which equals 625/[2x + 2/x + 85] = 625/[2(x + 1/x) + 85]
= 625/[2*20 + 85] = 5

Question 5 of 10

5. Question

1 points

Category: Quantitative Aptitude

A person can row a distance of 10 km upstream in 5 minutes and downstream in 2 minutes. What is the speed of the stream?

Correct

Explanation:
Upstream speed = 10/(5/60) = 120 km/hr
Downstream speed = 10/(2/60) = 300 km/hr
So speed of the stream = (1/2)*(300-120)

Incorrect

Explanation:
Upstream speed = 10/(5/60) = 120 km/hr
Downstream speed = 10/(2/60) = 300 km/hr
So speed of the stream = (1/2)*(300-120)

Question 6 of 10

6. Question

1 points

Category: Quantitative Aptitude

A dishonest seller professes to sell his goods at the cost price but uses a false weight of 800 gm instead of 1 kg. If he then also gave a 10% discount, what is his profit%?

Correct

Explanation:
uses a false weight of 800 gm instead of 1000 g
so Gain% = (1000-800)/800 * 100 = 25%
now a discount of 10%, so use successive formula:
25 + (-10) + (25)(-10)/100 = 12.5

Incorrect

Explanation:
uses a false weight of 800 gm instead of 1000 g
so Gain% = (1000-800)/800 * 100 = 25%
now a discount of 10%, so use successive formula:
25 + (-10) + (25)(-10)/100 = 12.5

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

If O is the circumcentre of ∆ABC and OD is perpendicular to BC, then angle BOD must be equal to

Correct

Explanation:

Angle BOC = 2 * angle BAC
Now beacuse angle BOD = (1/2) * angle BOC
So angle BOD = angle BAC

Incorrect

Explanation:

Angle BOC = 2 * angle BAC
Now beacuse angle BOD = (1/2) * angle BOC
So angle BOD = angle BAC

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

The numerical of 7/(1+tan^{2}ɸ) + 1/(1+cot^{2}ɸ) + 6 sin^{2}ɸ

A certain sum of money will be tripled in 20 years at the rate of simple interest percent per annum of

Correct

Explanation:
Sum tripled means became 3P, so SI = 3P – P = 2P
So 2P = P*20*r/100
Solve, r = 10%

Incorrect

Explanation:
Sum tripled means became 3P, so SI = 3P – P = 2P
So 2P = P*20*r/100
Solve, r = 10%

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

A, B, and C can do a piece of work in 15, 20 and 30 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?

Correct

Explanation:
Work done on first day by A = (1/15)
Work done on second day by A = (1/15)
Work done on third day by A+B+C = (1/15 + 1/20 + 1/30) = 3/20
Now the cycle repeats
So in 3 days work done is 1/15 + 1/15 + 3/20 = 17/60
Multiply by 3 both sides
In 9 days work done is 51/60
Work left = 1 – 51/60 = 9/60 = 3/20
Now A’s turn on 10th day, he do 1/15 of work
Now work left = 3/20 – 1/15 = 5/60 = 1/12
Now A’s turn on 11th day again, he do 1/15 of work
Now work left = 1/12 – 1/15 = 5/60 = 1/60
Now (A+B+C)’s turn on 12th day, they do 3/20 of work
They do 3/20 work in 1 day, so 1/60 in (1/60)*(20/3) = 1/9 day
So total 11 1/9 days

Incorrect

Explanation:
Work done on first day by A = (1/15)
Work done on second day by A = (1/15)
Work done on third day by A+B+C = (1/15 + 1/20 + 1/30) = 3/20
Now the cycle repeats
So in 3 days work done is 1/15 + 1/15 + 3/20 = 17/60
Multiply by 3 both sides
In 9 days work done is 51/60
Work left = 1 – 51/60 = 9/60 = 3/20
Now A’s turn on 10th day, he do 1/15 of work
Now work left = 3/20 – 1/15 = 5/60 = 1/12
Now A’s turn on 11th day again, he do 1/15 of work
Now work left = 1/12 – 1/15 = 5/60 = 1/60
Now (A+B+C)’s turn on 12th day, they do 3/20 of work
They do 3/20 work in 1 day, so 1/60 in (1/60)*(20/3) = 1/9 day
So total 11 1/9 days