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Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

Direction: (1-5) The line chart given below shows the number of students (boys+girls) selected by a school for International Science Competition in six different years. Study the data carefully and answer the following questions.

The total number of boys selected in 2015 and 2017 together in what percent of the total number of students (boys+girls) selected in 2013.

Direction: (1-5) The line chart given below shows the number of students (boys+girls) selected by a school for International Science Competition in six different years. Study the data carefully and answer the following questions.

Find the difference between the total number of boys selected in 2013, 2015 and 2017 to the total number of girls selected in 2014, 2016 and 2018.

Direction: (1-5) The line chart given below shows the number of students (boys+girls) selected by a school for International Science Competition in six different years. Study the data carefully and answer the following questions.

Find the ratio of total number of students selected in 2017 to the total number of girls selected in 2015 and 2018?

Direction: (1-5) The line chart given below shows the number of students (boys+girls) selected by a school for International Science Competition in six different years. Study the data carefully and answer the following questions.

Find the total number of students selected in 2019, if the number of boys selected in 2019 are 125% of the number of girls selected in 2014 and number of girls selected in 2019 are 75% of the number of girls selected in 2016?

Direction: (1-5) The line chart given below shows the number of students (boys+girls) selected by a school for International Science Competition in six different years. Study the data carefully and answer the following questions.

Find the total number of students selected in 2013 and 2017?

Kavi invested a certain amount of money in a scheme offering 15% compound interest per annum for two years. Ritika invested Rs.3000 less than Kavi in another scheme offering 10% simple interest per annum for three years. Find the amount invested by Ritika, if the difference in the interest earned by Kavi and Ritika is Rs.3150.

Correct

Answer: 3) Rs. 97000
Explanation:
Let the amount invested by Kavi = x
Amount invested by Ritika = x-3000
Interest received by kavi = 32.25x/100 (15+15+15*15/100)
Interest received by Ritika =(x-3000)*30/100
ATQ,
32.25x/100 – (x-3000)*30/100 = 3150
32.25x -30x+90000 = 315000
2.25x = 225000
X = 100000
Amount invested by Ritika = 100000-3000 = 97000

Incorrect

Answer: 3) Rs. 97000
Explanation:
Let the amount invested by Kavi = x
Amount invested by Ritika = x-3000
Interest received by kavi = 32.25x/100 (15+15+15*15/100)
Interest received by Ritika =(x-3000)*30/100
ATQ,
32.25x/100 – (x-3000)*30/100 = 3150
32.25x -30x+90000 = 315000
2.25x = 225000
X = 100000
Amount invested by Ritika = 100000-3000 = 97000

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

Atul sells a boat to Bhanu at 2/5th the rate of profit at which Bhanu sells it to Chanda. Further, Chanda sells it to Diya at half the rate of profit at which Atul sold it to Bhanu. If Chanda earns a profit of 4% by selling it to Diya for Rs. 20800. Find the approximate cost price of Boat for Bhanu?

Correct

Answer: 1. 16700
Explanation:
Let the profit at which Bhanu sells to Chanda = 10x%
Profit at which Atul sells to Bhanu = 10x*2/5 = 4x%
Profit at which Chanda sells to Diya = 4x*1/2 = 2x%
2x% = 4%
X = 2%
Cost price for Chanda = 2 = 20800*100/104 = 20000
Profit earned by Bhanu = 10*2 = 20x%
CP for Bhanu = 20000*100/120 = 16666(approx)

Incorrect

Answer: 1. 16700
Explanation:
Let the profit at which Bhanu sells to Chanda = 10x%
Profit at which Atul sells to Bhanu = 10x*2/5 = 4x%
Profit at which Chanda sells to Diya = 4x*1/2 = 2x%
2x% = 4%
X = 2%
Cost price for Chanda = 2 = 20800*100/104 = 20000
Profit earned by Bhanu = 10*2 = 20x%
CP for Bhanu = 20000*100/120 = 16666(approx)

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

Out of a total of 90 children playing badminton or table tannins or both, the total number of girl in the group is 80% of boys in the group. The number of boys playing only badminton is 30% of the number of boys. The number of children playing only table tennis is 30% of the total numbers of children and a total of 22 children play badminton and table tennis both. What is the number of girls playing only badminton?

Correct

Answer: 5) 26
Explanation:
Let number of boys = 10x
Number of girls = 80% of 10x = 8x
18x = 90
X =5
Therefore, total boys = 10*5 =50
Total girls = 8*5 = 40
Boys playing only badminton = 30% of 50 = 15
Children playing only table tennis = 30% of 90 =27
Children playing badminton and table tennis both = 22
Girls playing only badminton = 90 – (15+27+22) = 90-64 = 26

Incorrect

Answer: 5) 26
Explanation:
Let number of boys = 10x
Number of girls = 80% of 10x = 8x
18x = 90
X =5
Therefore, total boys = 10*5 =50
Total girls = 8*5 = 40
Boys playing only badminton = 30% of 50 = 15
Children playing only table tennis = 30% of 90 =27
Children playing badminton and table tennis both = 22
Girls playing only badminton = 90 – (15+27+22) = 90-64 = 26

Question 9 of 10

9. Question

1 points

Category: Quantitative Aptitude

Two pipes A and B are opened together to fill a tank. Both pipes can fill the tank in a certain amount of time. If A separately takes 2 min more than the time taken by ( A + B ) and B takes 8 min more than the time taken by ( A + B ). Find the time taken by A and B to fill the tank when both the pipes are opened together.

Correct

Answer: 2. 4 minutes Explanation:
Let the time taken by (A+B) to fill the tank = x minute
Tank filled by (A+B) per minute = 1/x
Time taken by A = x+2
Tank filled by A per minute = 1/(x+2)
Time taken by B = x+8
Tank filled by B per minute = 1/(x+8)
Both the pipes are opened
1/(x+2) + 1/(x+8) = 1/x
{(2x+10)/(x+2)(x+8)} = 1/x
2x^{2}+10x = x^{2}+10x+16
X^{2} = 16
X =4

Incorrect

Answer: 2. 4 minutes Explanation:
Let the time taken by (A+B) to fill the tank = x minute
Tank filled by (A+B) per minute = 1/x
Time taken by A = x+2
Tank filled by A per minute = 1/(x+2)
Time taken by B = x+8
Tank filled by B per minute = 1/(x+8)
Both the pipes are opened
1/(x+2) + 1/(x+8) = 1/x
{(2x+10)/(x+2)(x+8)} = 1/x
2x^{2}+10x = x^{2}+10x+16
X^{2} = 16
X =4

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

In a river, the ratio of the speed of stream and speed of boat A in still water is 3: 11. Again, the ratio of the speed of stream and speed of boat B in still water is 2: 9. What is the ratio of the speeds of boat A to boat B in still water?

Correct

Answer: 3. 22/27
Explanation:
Stream: Boat A = 3:11
Stream: Boat B = 2:9
Since, the speed of stream will be same
Multiplying first ratio by 2 and second by 3, to make the speed of stream look same to get the ratio of speed boat A to the speed of boat B in still water
2*(3:11) = 6:22
3*(2:9) = 6:27
Therefore, ratio of speeds of boat A to boat B = 22/27

Incorrect

Answer: 3. 22/27
Explanation:
Stream: Boat A = 3:11
Stream: Boat B = 2:9
Since, the speed of stream will be same
Multiplying first ratio by 2 and second by 3, to make the speed of stream look same to get the ratio of speed boat A to the speed of boat B in still water
2*(3:11) = 6:22
3*(2:9) = 6:27
Therefore, ratio of speeds of boat A to boat B = 22/27

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