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Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

A part from Rs 8000 is deposited in scheme A which offers 6% per annum rate of interest and the remaining part is deposited in scheme B which offers 10% per annum rate of interest. If at the end of 4 years, a total of Rs 2400 is received as interest from both parts, what is the amount deposited at 10% per annum rate?

Correct

Explanation:
Find the rate % of 4 yrs:
2400 = 8000*r*4 / 100, this gives r = 15/2%
By the method of allegation:
6……………………………..…… 10
………….….15/2
10 – 15/2 = 5/2 ……………….15/2 – 6 = 3/2
5/2 : 3/2 or 5 : 3
Amount at 10% = 3/(5+3) * 8000 = 3000

Incorrect

Explanation:
Find the rate % of 4 yrs:
2400 = 8000*r*4 / 100, this gives r = 15/2%
By the method of allegation:
6……………………………..…… 10
………….….15/2
10 – 15/2 = 5/2 ……………….15/2 – 6 = 3/2
5/2 : 3/2 or 5 : 3
Amount at 10% = 3/(5+3) * 8000 = 3000

Question 2 of 10

2. Question

1 points

Category: Quantitative Aptitude

A and B starts a business and invests Rs 5000 and Rs 2000 respectively. It was decided that A will get 20% of the profit as his salary and after deducting his salary, the profit will be divided among them. If at the end of year, a profit of Rs 1400 was received, how much will A get?

Correct

Explanation:
5000 : 2000
5 : 2
A gets salary = (20/100)*1400 = 280
So profit left = 1400 – 280 = 1120
Now A gets his share as = (5/7) * 1120 = 800
So total of A = 280+800

Incorrect

Explanation:
5000 : 2000
5 : 2
A gets salary = (20/100)*1400 = 280
So profit left = 1400 – 280 = 1120
Now A gets his share as = (5/7) * 1120 = 800
So total of A = 280+800

Question 3 of 10

3. Question

1 points

Category: Quantitative Aptitude

A man can row 8 km/hr in still water. When the river is running at 4 km/hr, it takes him 4 hours to row to a place and come back. What total distance did the man travel?

Correct

Explanation:
B is speed of boat in still water, R is speed of stream
Time is total time taken for upstream and downstream
Distance = time * [B^2 – R^2] / 2*B
= 4 * [8^2 – 4^2] / 2*8 = 12
So total to and fro distance = 12+12

Incorrect

Explanation:
B is speed of boat in still water, R is speed of stream
Time is total time taken for upstream and downstream
Distance = time * [B^2 – R^2] / 2*B
= 4 * [8^2 – 4^2] / 2*8 = 12
So total to and fro distance = 12+12

Question 4 of 10

4. Question

1 points

Category: Quantitative Aptitude

A bucket contains 2 red balls, 4 blue balls, and 6 white balls. Two balls are drawn at random. What is the probability that at least one ball is white?

Correct

Explanation:
Three cases
Case 1: both white
Prob = ^{6}C_{2} / ^{12}C_{2} = 5/22
Case 2: one white, 1 red or blue
Prob = ^{6}C_{1} × ^{6}C_{1} / ^{12}C_{2} = 2/11.
Add all cases

Incorrect

Explanation:
Three cases
Case 1: both white
Prob = ^{6}C_{2} / ^{12}C_{2} = 5/22
Case 2: one white, 1 red or blue
Prob = ^{6}C_{1} × ^{6}C_{1} / ^{12}C_{2} = 2/11.
Add all cases

Question 5 of 10

5. Question

1 points

Category: Quantitative Aptitude

The outer and the inner perimeters of a circular path is in the ratio 12 : 5. If the path is 7 m wide, what is the diameter of the outer circle?

Correct

Explanation:
Outer/inner = 2ᴨr/2ᴨR = 12/5
So r/R = 12/5, so r = (12/5)R
r-R = 7, so (12/5)R – R = 7
R = 5, so r = 12 and then diameter is 24

Incorrect

Explanation:
Outer/inner = 2ᴨr/2ᴨR = 12/5
So r/R = 12/5, so r = (12/5)R
r-R = 7, so (12/5)R – R = 7
R = 5, so r = 12 and then diameter is 24

Question 6 of 10

6. 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} + 16x + 5 = 0, 2y^{2} + 9y – 5 = 0

Correct

Explanation:
3x^{2} + 16x + 5 = 0
3x^{2} + 15x + x + 5 = 0
Gives x = -5, -1/3
2y^{2} + 9y – 5 = 0
2y^{2} + 10y – y – 5 = 0
Gives y = -5, 1/2
Put on number line
-5… -1/3 … 1/2
When x = -1/3, x < y (1/2) and x > y (-5) – here relation cant be determined.

Incorrect

Explanation:
3x^{2} + 16x + 5 = 0
3x^{2} + 15x + x + 5 = 0
Gives x = -5, -1/3
2y^{2} + 9y – 5 = 0
2y^{2} + 10y – y – 5 = 0
Gives y = -5, 1/2
Put on number line
-5… -1/3 … 1/2
When x = -1/3, x < y (1/2) and x > y (-5) – 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 –

4x^{2} + 17x + 15 = 0, 4y^{2} – 3y – 10 = 0

Correct

Explanation:
4x^{2} + 17x + 15 = 0
4x^{2} + 12x + 5x + 15 = 0
Gives x = -3, -5/4
4y^{2} – 3y – 10 = 0
4y^{2} – 3y – 10 = 0
Gives y = -5/4, 2
Put on number line
-3… -5/4… 2

Incorrect

Explanation:
4x^{2} + 17x + 15 = 0
4x^{2} + 12x + 5x + 15 = 0
Gives x = -3, -5/4
4y^{2} – 3y – 10 = 0
4y^{2} – 3y – 10 = 0
Gives y = -5/4, 2
Put on number line
-3… -5/4… 2

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 –

8x^{2} + 10x + 3 = 0, 3y^{2} – 10y + 3 = 0

Correct

Explanation:
8x^{2} + 10x + 3 = 0
8x^{2} + 4x + 6x + 3 = 0
Gives x = -3/4, -1/2
3y^{2} – 10y + 3 = 0
3y^{2} – 10y + 3 = 0
Gives y = 1/3, 3
Put on number line
-3/4… -1/2… 1/3… 3

Incorrect

Explanation:
8x^{2} + 10x + 3 = 0
8x^{2} + 4x + 6x + 3 = 0
Gives x = -3/4, -1/2
3y^{2} – 10y + 3 = 0
3y^{2} – 10y + 3 = 0
Gives y = 1/3, 3
Put on number line
-3/4… -1/2… 1/3… 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 –

4x^{2} + 13x – 12 = 0, 2y^{2} – 7y + 3 = 0

Correct

Explanation:
4x^{2} + 13x – 12 = 0
4x^{2} + 16x -3x – 12 = 0
Gives x = -4, 3/4
2y^{2} – 7y + 3 = 0
2y^{2} – 6y – y + 3 = 0
Gives y = 1/2, 3
Put on number line
-4…. 1/2… 3/4 … 3

Incorrect

Explanation:
4x^{2} + 13x – 12 = 0
4x^{2} + 16x -3x – 12 = 0
Gives x = -4, 3/4
2y^{2} – 7y + 3 = 0
2y^{2} – 6y – y + 3 = 0
Gives y = 1/2, 3
Put on number line
-4…. 1/2… 3/4 … 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 –

2x^{2} – 5x + 2 = 0, 8y^{2} + 14y + 3 = 0

Correct

Explanation:
2x^{2} – 5x + 2 = 0
2x^{2} – 5x + 2 = 0
Gives x = 1/2, 2
8y^{2} + 14y + 3 = 0
8y^{2} + 2y + 12y + 3 = 0
Gives y= -3/2, -1/4
Put on number line
-3/2…. -1/4… 1/2…. 2

Incorrect

Explanation:
2x^{2} – 5x + 2 = 0
2x^{2} – 5x + 2 = 0
Gives x = 1/2, 2
8y^{2} + 14y + 3 = 0
8y^{2} + 2y + 12y + 3 = 0
Gives y= -3/2, -1/4
Put on number line
-3/2…. -1/4… 1/2…. 2