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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeA man goes downstream and comes back to original point in 5 hours. If the speed of the boat in still water and the stream are 15 km/hr and 12 km/hr respectively, the total distance travelled by the man is
Correct
Explanation:
Use shortcut:
Distance = time * [B^2 – R^2] / 2*B
= 5 * [15^2 – 12^2] / 2*15 = 1.5Incorrect
Explanation:
Use shortcut:
Distance = time * [B^2 – R^2] / 2*B
= 5 * [15^2 – 12^2] / 2*15 = 1.5 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeOne trader calculates the profit on selling price and another at cost price. If they sell the article at same price, then the difference in their profits is Rs 85 and both gain 20%. What is the selling price of each article?
Correct
Explanation:
Let CP = 100, then SP = 120 for both
For first trader: (120x)/120 * 100 = 20
Solve, x = 96
So gain = 12096 = Rs 24
Difference of gains is 2420 = 4
When difference is 4, then SP = 120
When difference is 85, then SP = 120/4 * 85 = 2550Incorrect
Explanation:
Let CP = 100, then SP = 120 for both
For first trader: (120x)/120 * 100 = 20
Solve, x = 96
So gain = 12096 = Rs 24
Difference of gains is 2420 = 4
When difference is 4, then SP = 120
When difference is 85, then SP = 120/4 * 85 = 2550 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeIf the difference between areas of the circumcircle and the incircle of an equilateral triangle is 66 sq. cm, then the area of the triangle in sq. cm is
Correct
Explanation:
Let the side of equilateral triangle be 2x, then radius of incircle = x/√3 and radius of incircle = 2x/√3
According to question:
ᴨ(2x/√3)^{2} – ᴨ(x/√3)^{2} = 66
ᴨx^{2} = 66
solve, x^{2} = 21
area of equi. Triangle = (√3/4)* side^2 = (√3/4)* (2x)^2 = 21√3Incorrect
Explanation:
Let the side of equilateral triangle be 2x, then radius of incircle = x/√3 and radius of incircle = 2x/√3
According to question:
ᴨ(2x/√3)^{2} – ᴨ(x/√3)^{2} = 66
ᴨx^{2} = 66
solve, x^{2} = 21
area of equi. Triangle = (√3/4)* side^2 = (√3/4)* (2x)^2 = 21√3 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeIf x – 1/x = 6, then (x + 1/x) equals
Correct
Explanation:
x – 1/x = 6
square both sides
x^{2} + (1/x)^{2} – 2 = 36
add 4 both sides
x^{2} + (1/x)^{2} + 2 = 40
(x + 1/x) ^{2} = 40
So (x + 1/x) = 2√10Incorrect
Explanation:
x – 1/x = 6
square both sides
x^{2} + (1/x)^{2} – 2 = 36
add 4 both sides
x^{2} + (1/x)^{2} + 2 = 40
(x + 1/x) ^{2} = 40
So (x + 1/x) = 2√10 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeSaloni’s expenditures and savings are in the ratio 7 : 3. If her income increases by 10% and expenditure by 15%, then by how much percent do her savings decrease?
Correct
Explanation:
E : S = 7 : 3, so income = 7+3 = 10
New income = (110/100)*10 = 11
New exp. = (115/100)*7 = 8.05
So new savings are = 118.05 = 2.95
So % decrease in savings = (32.95)/3 * 100Incorrect
Explanation:
E : S = 7 : 3, so income = 7+3 = 10
New income = (110/100)*10 = 11
New exp. = (115/100)*7 = 8.05
So new savings are = 118.05 = 2.95
So % decrease in savings = (32.95)/3 * 100 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeA straight line parallel to the base BC of the triangle ABC intersects AB and AC at the points D and E resp. If the area of triangle ABE is 45 sq. cm, then the area of the triangle ACD is
Correct
Explanation:
∆DBC is similar to ∆BEC because they lie on same base and between same parallel lines.
So ∆ABC – ∆DBC = ∆ABC – ∆BEC
∆ADC = ∆ABE = 45 sq. cmIncorrect
Explanation:
∆DBC is similar to ∆BEC because they lie on same base and between same parallel lines.
So ∆ABC – ∆DBC = ∆ABC – ∆BEC
∆ADC = ∆ABE = 45 sq. cm 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeIf cos^{2}a + cos^{2}b = 2, then the value of tan^{3}a + sin^{5}b is
Correct
Explanation:
cos^{2}a + cos^{2}b = 2
1 – sin^{2}a + 1 – sin^{2}b = 2
So sin^{2}a + sin^{2}b = 0
This means sin a = sin b = 0
So a = b = 0
So tan^{3}a + sin^{5}b = 0Incorrect
Explanation:
cos^{2}a + cos^{2}b = 2
1 – sin^{2}a + 1 – sin^{2}b = 2
So sin^{2}a + sin^{2}b = 0
This means sin a = sin b = 0
So a = b = 0
So tan^{3}a + sin^{5}b = 0 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeA job is completed by 5 men or 7 women in 40 days, then in how many days thrice the job as previous will be completed by 5 men and 3 women?
Correct
Explanation:
5 m or 7 w
5m + 3w
Cross multiply and put in denominator
Days = 40*5*7/ [5*3 +7*5]Incorrect
Explanation:
5 m or 7 w
5m + 3w
Cross multiply and put in denominator
Days = 40*5*7/ [5*3 +7*5] 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeRs 2000 is lent in 2 parts, 1 part at 8% per annum and 2nd part at 15% per annum. At the end of a year Rs 216 is received as simple interest. What is the part lent at 15% p.a.?
Correct
Explanation:
216 = 2000*r*1/100
r = 10.8%
By the method of allegation:
1st part …………………….. 2nd part
8………………………………………15
………………….10.8
4.2………………………………….2.8
4.2 : 2.8 = 3 : 2
At 15% = (2/5) * 2000Incorrect
Explanation:
216 = 2000*r*1/100
r = 10.8%
By the method of allegation:
1st part …………………….. 2nd part
8………………………………………15
………………….10.8
4.2………………………………….2.8
4.2 : 2.8 = 3 : 2
At 15% = (2/5) * 2000 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeIf sin17 = x/y, then the value of sec17 – sin73 is
Correct
Explanation:
sec17 – sin73
= sec17 – sin(9073)
= sec17 – cos17
= 1/cos17 – cos17
= (1 – cos^{2})17/cos17
= sin^{2}17/ cos17
= (x^2/y^2)/√(1 – x^2/y^2)Incorrect
Explanation:
sec17 – sin73
= sec17 – sin(9073)
= sec17 – cos17
= 1/cos17 – cos17
= (1 – cos^{2})17/cos17
= sin^{2}17/ cos17
= (x^2/y^2)/√(1 – x^2/y^2)
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