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Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

A 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

One 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: (120-x)/120 * 100 = 20
Solve, x = 96
So gain = 120-96 = Rs 24
Difference of gains is 24-20 = 4
When difference is 4, then SP = 120
When difference is 85, then SP = 120/4 * 85 = 2550

Incorrect

Explanation:
Let CP = 100, then SP = 120 for both
For first trader: (120-x)/120 * 100 = 20
Solve, x = 96
So gain = 120-96 = Rs 24
Difference of gains is 24-20 = 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 points

Category: Quantitative Aptitude

If 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√3

Incorrect

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 points

Category: Quantitative Aptitude

If 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√10

Incorrect

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 points

Category: Quantitative Aptitude

Saloni’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 = 11-8.05 = 2.95
So % decrease in savings = (3-2.95)/3 * 100

Incorrect

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 = 11-8.05 = 2.95
So % decrease in savings = (3-2.95)/3 * 100

Question 6 of 10

6. Question

1 points

Category: Quantitative Aptitude

A 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. cm

Incorrect

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 points

Category: Quantitative Aptitude

If 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 = 0

Incorrect

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 points

Category: Quantitative Aptitude

A 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 points

Category: Quantitative Aptitude

Rs 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) * 2000

Incorrect

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 points

Category: Quantitative Aptitude

If sin17 = x/y, then the value of sec17 – sin73 is