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Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

An article is sold at a discount of 12% making a profit of 10%. What would have been the profit percent if the article was sold without discount?

Correct

Explanation:
Let MP = Rs 100
at 12% discount gives: SP = Rs 88, so Discount = 12
Now profit = 10%
so CP = (100/110) * 88 = 80
No discount given means SP = MP = 100
So Profit% when no discount = [(100-80)/80]*100

Incorrect

Explanation:
Let MP = Rs 100
at 12% discount gives: SP = Rs 88, so Discount = 12
Now profit = 10%
so CP = (100/110) * 88 = 80
No discount given means SP = MP = 100
So Profit% when no discount = [(100-80)/80]*100

Question 2 of 10

2. Question

1 points

Category: Quantitative Aptitude

A is one and a half times more efficient than B. A alone can do a work in 10 days. What portion of total work can both of them together complete in 5 days?

Correct

Explanation:
Let efficiency of B is x, then of A is 1 + 1(1/2) = 5x/2
Ratio of efficiencies of A : B = 5x/2 : x= 5 : 2
So ratio of number of days of A : B = 2 : 5
A completes work in 10 days, so 2x = 10, x = 5
So B alone completes work in 5x days = 25 days
(A+B)’s 1 day’s work = 1/10 + 1/25 = 7/50
So in 5 days, pat of work completed is 5 * (7/50)

Incorrect

Explanation:
Let efficiency of B is x, then of A is 1 + 1(1/2) = 5x/2
Ratio of efficiencies of A : B = 5x/2 : x= 5 : 2
So ratio of number of days of A : B = 2 : 5
A completes work in 10 days, so 2x = 10, x = 5
So B alone completes work in 5x days = 25 days
(A+B)’s 1 day’s work = 1/10 + 1/25 = 7/50
So in 5 days, pat of work completed is 5 * (7/50)

Question 3 of 10

3. Question

1 points

Category: Quantitative Aptitude

180 sweets are divided among friends A, B, C and D in which B and C are brothers also such that sweets divided between A and B are in the ratio 3 : 4, between B and C in the ratio 1 : 2 and between C and D in ratio 2 : 3. What is the number of sweets received by the brothers together?

Correct

Explanation:
A/B = N1/D1 B/C = N2/D2 C/D = N3/D3
A : B : C : D = N1*N2*N3 : D1*N2*N3 : D1*D2*N3 : D1*D2*D3
A/B = 3/4 B/C = 1/2 C/D = 2/3
A : B : C : D
3*1*2 : 4*1*2 : 4*2*2 : 4*2*3
6 : 8 : 16 : 24 = 3 : 4 : 8 : 12
B and C together = [(4+8)/(3+4+8+12)] * 180

Incorrect

Explanation:
A/B = N1/D1 B/C = N2/D2 C/D = N3/D3
A : B : C : D = N1*N2*N3 : D1*N2*N3 : D1*D2*N3 : D1*D2*D3
A/B = 3/4 B/C = 1/2 C/D = 2/3
A : B : C : D
3*1*2 : 4*1*2 : 4*2*2 : 4*2*3
6 : 8 : 16 : 24 = 3 : 4 : 8 : 12
B and C together = [(4+8)/(3+4+8+12)] * 180

Question 4 of 10

4. Question

1 points

Category: Quantitative Aptitude

10 liters are drawn from a vessel full of spirit and it is filled with water. Then 10 liters of the mixture are drawn and the vessel is again filled with water. The ratio of the spirit to water in the vessel is 9 : 16. Now what is the capacity of vessel in litres?

Correct

Explanation:
let x litres initially
after 2 steps , spirit left = x [1 – (10/x)]^2
and total is x we know
now spirit to water is 9 : 16, so spirit to total becomes 9 : (9+16) = 9 : 25
so [ x [1 – (10/x)]^2 ] / x = 9 / 25
x got cancelled
so [1 – (10/x)]^2 = 9/25
[1 – (10/x)] = 3/5
Solve, x = 25 litres

Incorrect

Explanation:
let x litres initially
after 2 steps , spirit left = x [1 – (10/x)]^2
and total is x we know
now spirit to water is 9 : 16, so spirit to total becomes 9 : (9+16) = 9 : 25
so [ x [1 – (10/x)]^2 ] / x = 9 / 25
x got cancelled
so [1 – (10/x)]^2 = 9/25
[1 – (10/x)] = 3/5
Solve, x = 25 litres

Question 5 of 10

5. Question

1 points

Category: Quantitative Aptitude

A, B, and C invested Rs 2500, Rs 3500 and Rs 4500 in a business. After 8 months they all added rs 500 to their investments. In what respective ratio they will get the profits after a year?

Directions (6-10): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –

2x^{2} – x – 6 = 0, 2y^{2} – 3y – 9 = 0

Correct

Explanation:
2x^{2} – x – 6 = 0
2x^{2} – 4x + 3x – 6 = 0
Gives x = -3/2, 2
2y^{2} – 3y – 9 = 0
2y^{2} – 6y + 3y – 9 = 0 3
Gives y = -3/2, 3
Put on number line
-3/2… 2 … 3
When x = -3/2, x < y (3) and x = y (-3/2) so x ≤ y
But when x = 2, x > y (-3/2) and x < y (3) – here relation cant be determined.

Incorrect

Explanation:
2x^{2} – x – 6 = 0
2x^{2} – 4x + 3x – 6 = 0
Gives x = -3/2, 2
2y^{2} – 3y – 9 = 0
2y^{2} – 6y + 3y – 9 = 0 3
Gives y = -3/2, 3
Put on number line
-3/2… 2 … 3
When x = -3/2, x < y (3) and x = y (-3/2) so x ≤ y
But when x = 2, x > y (-3/2) and x < y (3) – here relation cant be determined.

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

Directions (6-10): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –

3x^{2} – 5x + 2 = 0, 4y^{2} + 17y + 4 = 0

Correct

Explanation:
3x^{2} – 5x + 2 = 0
3x^{2} – 3x – 2x + 2 = 0
Gives x = 2/3, 1
4y^{2} + 17y + 4 = 0
4y^{2} + 16y + y + 4 = 0
Gives y = -4, -1/4
Put on number line
-4… -1/4… 2/3…. 1

Incorrect

Explanation:
3x^{2} – 5x + 2 = 0
3x^{2} – 3x – 2x + 2 = 0
Gives x = 2/3, 1
4y^{2} + 17y + 4 = 0
4y^{2} + 16y + y + 4 = 0
Gives y = -4, -1/4
Put on number line
-4… -1/4… 2/3…. 1

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

Directions (6-10): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –

2x^{2} + x – 15 = 0, y^{2} – 5y + 6 = 0

Correct

Explanation:
2x^{2} + x – 15 = 0
2x^{2} + 6x – 5x – 15 = 0
Gives x = -3, 5/2
y^{2} – 5y + 6 = 0
y^{2} – 2y – 3y + 6 = 0
Gives y = 2, 3
Put on number line
-3… 2… 5/2… 3

Incorrect

Explanation:
2x^{2} + x – 15 = 0
2x^{2} + 6x – 5x – 15 = 0
Gives x = -3, 5/2
y^{2} – 5y + 6 = 0
y^{2} – 2y – 3y + 6 = 0
Gives y = 2, 3
Put on number line
-3… 2… 5/2… 3

Question 9 of 10

9. Question

1 points

Category: Quantitative Aptitude

Directions (6-10): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –

3x^{2} + x – 2 = 0, 3y^{2} + 7y + 4 = 0

Correct

Explanation:
3x^{2} + x – 2 = 0
3x^{2} + 3x – 2x – 2 = 0
Gives x = -1, 2/3
3y^{2} + 7y + 4 = 0
3y^{2} + 3y + 4y + 4 = 0
Gives y = -1, -4/3
Put on number line
-4/3…. -1… 2/3

Incorrect

Explanation:
3x^{2} + x – 2 = 0
3x^{2} + 3x – 2x – 2 = 0
Gives x = -1, 2/3
3y^{2} + 7y + 4 = 0
3y^{2} + 3y + 4y + 4 = 0
Gives y = -1, -4/3
Put on number line
-4/3…. -1… 2/3

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

Directions (6-10): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –

4x^{2} + 15x + 9 = 0, 4y^{2} – 5y – 6 = 0

Correct

Explanation:
4x^{2} + 15x + 9 = 0
4x^{2} + 12x + 3x + 9 = 0
Gives x = -3, -3/4
4y^{2} – 5y – 6 = 0
4y^{2} – 8y + 3y – 6 = 0
Gives y= -3/4, 2
Put on number line
-3…. -3/4… 2

Incorrect

Explanation:
4x^{2} + 15x + 9 = 0
4x^{2} + 12x + 3x + 9 = 0
Gives x = -3, -3/4
4y^{2} – 5y – 6 = 0
4y^{2} – 8y + 3y – 6 = 0
Gives y= -3/4, 2
Put on number line
-3…. -3/4… 2