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Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

Direction: Q (1-5) Below is the performance of two candidates Rohan and Radha in various subjects. Study the following chart and answer the questions

What is the ratio of total marks scored by Radha to total marks secured by Rohan in all the subjects ?

Correct

Explanation:
250/225=10/9
10:9

Incorrect

Explanation:
250/225=10/9
10:9

Question 2 of 10

2. Question

1 points

Category: Quantitative Aptitude

Direction: Q (1-5) Below is the performance of two candidates Rohan and Radha in various subjects. Study the following chart and answer the questions

Had Rohan scored 50 % more marks than he has scored in Quants, then the marks scored by Rohan in Quants would have been how much percentage more than that of marks scored by Radha in General Awareness?

Correct

Explanation:
New marks of Rohan=90
Percentage=90-80/80*100=12.5%

Incorrect

Explanation:
New marks of Rohan=90
Percentage=90-80/80*100=12.5%

Question 3 of 10

3. Question

1 points

Category: Quantitative Aptitude

Direction: Q (1-5) Below is the performance of two candidates Rohan and Radha in various subjects. Study the following chart and answer the questions

If during re-checking it was found that the marks of Radha and Rohan were interchanged in English by mistake, hence the ratio of total marks scored by Rohan to Radha after correction

Correct

Explanation:
New marks of Rohan-250 Previous marks of Rohan-225
New marks of Radha-225 Previous marks of Rohan-250
New Ratio-10/9 Previous Ratio-9/10

Incorrect

Explanation:
New marks of Rohan-250 Previous marks of Rohan-225
New marks of Radha-225 Previous marks of Rohan-250
New Ratio-10/9 Previous Ratio-9/10

Question 4 of 10

4. Question

1 points

Category: Quantitative Aptitude

Direction: Q (1-5) Below is the performance of two candidates Rohan and Radha in various subjects. Study the following chart and answer the questions

If suppose a new subject Computer is introduced, Radha scores 50 marks out of 100 in computer, how much percent more marks than the marks scored by Radha in computer should Rohan score so that their overall marks becomes same.

Correct

Explanation:
Total marks of radha in all 5 subject-300
Marks of rohan in 4 subjects (except computer)-225
Therefore marks required by Rohan-300-225=75
Percentage=75-50/50*100

Incorrect

Explanation:
Total marks of radha in all 5 subject-300
Marks of rohan in 4 subjects (except computer)-225
Therefore marks required by Rohan-300-225=75
Percentage=75-50/50*100

Question 5 of 10

5. Question

1 points

Category: Quantitative Aptitude

Direction: Q (1-5) Below is the performance of two candidates Rohan and Radha in various subjects. Study the following chart and answer the questions

What is the overall percentage of Radha in all 4 subjects(Max marks in all subject 100) ?

Correct

Explanation:
Overall percentage-250/400*100=62.5%

Incorrect

Explanation:
Overall percentage-250/400*100=62.5%

Question 6 of 10

6. Question

1 points

Category: Quantitative Aptitude

Two liquids A and B are in ratio 4:1 in container X and 3:5 in container Y. In what ratio should the content of both container be mixed so that the resulting mixture has A and B in ratio 2:3?

Correct

Explanation:
Let the ratios in which they are mixed is x and y
Therefore A=4/5x+3/8y
B=1/5x+5/8y
Now A/B=2/3
On solving we get x:y=1:16

Incorrect

Explanation:
Let the ratios in which they are mixed is x and y
Therefore A=4/5x+3/8y
B=1/5x+5/8y
Now A/B=2/3
On solving we get x:y=1:16

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

8 women can complete the work in 10 days and 5 men can complete the work in 8 days where as 25 children can complete in 4 days. 16 women,4 men and 20 children work together for 2 days.If only women were to complete the remaining work in 1 day,how many women would be required?

Correct

Explanation:
(M1)(H1)(D1)/(W1)=(M2)(H2)(D2)/(W2)
Hence (8*10)W=(5*8)M=(25*4)C
4W=2M=5C
Now 16W+4M+20C=16W+8W+16W=40W
8W one day work=1/10
40W one day work=(1/10)*(40/8)=1/2
40W 2 day work=1
Remaining Work=0
Work is already completed.

Incorrect

Explanation:
(M1)(H1)(D1)/(W1)=(M2)(H2)(D2)/(W2)
Hence (8*10)W=(5*8)M=(25*4)C
4W=2M=5C
Now 16W+4M+20C=16W+8W+16W=40W
8W one day work=1/10
40W one day work=(1/10)*(40/8)=1/2
40W 2 day work=1
Remaining Work=0
Work is already completed.

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

A tank has a leak which would empty it in 6 hrs. A tap pumps water @ 8 litres/ minute into the tank, and it is now emptied in 12 hrs. What is the capacity of tank?

Correct

Explanation:
In the absence of leak time taken by tap to fill tank=12*6/12-6=12hour
Water filled in 1 hour=8*60=480L
Therefore water filled in 12 hour=12*480=5760L

Incorrect

Explanation:
In the absence of leak time taken by tap to fill tank=12*6/12-6=12hour
Water filled in 1 hour=8*60=480L
Therefore water filled in 12 hour=12*480=5760L

Question 9 of 10

9. Question

1 points

Category: Quantitative Aptitude

A man borrows Rs. 25,000 at 20% compound interest. At the end of every year he pays Rs. 5000 as part of repayment. How much does he still owe after three such installments?

Correct

Explanation:
C.I of 20000 in 3 years =25000*(1+20/100)^3=43200Rs
But as we are paying 2000Rs at the end of every year hence that should should be subtracted at the end of every year and the CI on remaining amount must be calculated.
Therefore CI of 2000Rs that is paid at the end of 1^{st} year=5000*(1+10/100)^2=7200
CI of 2000Rs that is paid at the end of 2nd year=5000*(1+10/100)^1=6000
Hence due amount after 3^{rd} payment=43200-(7200+6000+5000)=25000

Incorrect

Explanation:
C.I of 20000 in 3 years =25000*(1+20/100)^3=43200Rs
But as we are paying 2000Rs at the end of every year hence that should should be subtracted at the end of every year and the CI on remaining amount must be calculated.
Therefore CI of 2000Rs that is paid at the end of 1^{st} year=5000*(1+10/100)^2=7200
CI of 2000Rs that is paid at the end of 2nd year=5000*(1+10/100)^1=6000
Hence due amount after 3^{rd} payment=43200-(7200+6000+5000)=25000

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

In a triangle, two sides of right angle triangle are 8 cm and 6 cm. If the triangle is revolved along the 8 cm side, the curved surface area of the cone so formed will be

Correct

Explanation:
Radius of cone =8cm
Slant height=10cm
Curved surface area=πrl=22/7*6*10=188.4 cubic cm

Incorrect

Explanation:
Radius of cone =8cm
Slant height=10cm
Curved surface area=πrl=22/7*6*10=188.4 cubic cm