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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeA completes a piece of work in 20 days, and B completes the same work in 15 days. Both A and B started the work, but B left after some days. Now A completed the remaining work in 8 1/3 days. What is the number of days after which B left?
Correct
Explanation:
Let B left after ‘x’ days. A completed remaining work in 8 1/3 days or 25/3 days. So,
(1/20 + 1/15) * x + 1/20 * 25/3 = 1
(7/60) * x = 1 – 5/12
7/60 * x = 7/12
Solve, x = 5Incorrect
Explanation:
Let B left after ‘x’ days. A completed remaining work in 8 1/3 days or 25/3 days. So,
(1/20 + 1/15) * x + 1/20 * 25/3 = 1
(7/60) * x = 1 – 5/12
7/60 * x = 7/12
Solve, x = 5 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeA man starts from point P at 7:00 AM and reaches point Q at 9:00 AM. Another man starts from point Q at 8:00 AM and reaches point P at 12:00 Noon. At what time both men will meet?
Correct
Explanation:
Let the distance from A to B is ‘d’ km.
First train reaches destination in 2 hrs, so speed of first train = d/2 km/hr.
Speed of second train = d/4 km/hr
Speed of first train = d/2 km/hr, so after 1 hr, i.e. at 8AM, the first train has covered d/2 km.
The distance left = (d – d/2) km = d/2 km. relative speed = d/2 + d/4.
So time they will meet at = 8:00 AM + (d/2) / (d/2 + d/4)
= 8:00 AM + (2/3)*60 = 8:40 AMIncorrect
Explanation:
Let the distance from A to B is ‘d’ km.
First train reaches destination in 2 hrs, so speed of first train = d/2 km/hr.
Speed of second train = d/4 km/hr
Speed of first train = d/2 km/hr, so after 1 hr, i.e. at 8AM, the first train has covered d/2 km.
The distance left = (d – d/2) km = d/2 km. relative speed = d/2 + d/4.
So time they will meet at = 8:00 AM + (d/2) / (d/2 + d/4)
= 8:00 AM + (2/3)*60 = 8:40 AM 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative Aptitude90 litres of a mixture contains 20% water and the rest milk. What amount of water must be added so that the resulting mixture contains only 60% milk?
Correct
Explanation:
Mixture contains 20% water.
The water which will be added will be 100% water
The new mixture contains 60% milk, this gives 40% water
So by allegation rule
Mixture………………..Water
20…………………….. 100
……………..40
60……………………… 20
This gives 60 : 20 = 3 : 1
Let x litres of water to be added to 80 litres mixture.
So 90/x = 3/1
x = 30Incorrect
Explanation:
Mixture contains 20% water.
The water which will be added will be 100% water
The new mixture contains 60% milk, this gives 40% water
So by allegation rule
Mixture………………..Water
20…………………….. 100
……………..40
60……………………… 20
This gives 60 : 20 = 3 : 1
Let x litres of water to be added to 80 litres mixture.
So 90/x = 3/1
x = 30 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeHow many words can be made out of the letters of the word ADDINS such that all vowels occur together?
Correct
Explanation:
We have D D N S and AI
So 5! For these 5 letters (take AI as single letter because they are to be taken together). And since 2 Ds are there, so 5!/2
Now arrangement of AI is 2!
So total 5!/2 * 2!Incorrect
Explanation:
We have D D N S and AI
So 5! For these 5 letters (take AI as single letter because they are to be taken together). And since 2 Ds are there, so 5!/2
Now arrangement of AI is 2!
So total 5!/2 * 2! 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeA dealer gains 10% when an article is sold at 2/5th of its actual selling price. What will be the gain% if the same article is sold at 40% less than the actual selling price?
Correct
Explanation:
Profit is 10%. So,
(2/5 S.P. – C.P.)/C.P. * 100 = 20
This gives S.P. = 11/4 C.P.
Now product sold at 40% less S.P. i.e. 60/100 or 6/10 of S.P.
Profit/loss% = (6/10 S.P. – C.P.)/C.P. * 100
Put S.P. = 11/4 C.P. and find the final answerIncorrect
Explanation:
Profit is 10%. So,
(2/5 S.P. – C.P.)/C.P. * 100 = 20
This gives S.P. = 11/4 C.P.
Now product sold at 40% less S.P. i.e. 60/100 or 6/10 of S.P.
Profit/loss% = (6/10 S.P. – C.P.)/C.P. * 100
Put S.P. = 11/4 C.P. and find the final answer 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative Aptitude25 men are employed for a work which needs to be completed in 40 days. But after 24 days, it was found that only one third of the work was completed. Now in order to complete the work 4 days earlier, how many extra men needed to be employed?
Correct
Explanation:
4 days earlier the work is to be completed, this means in (404) = 36 days
After 24 days, the meaning work is to be completed in (3624) = 12 days
Remaining work = 1 – 1/3 = 2/3
Let x extra men are to be employed. So,
25 * 24 * 2/3 = (x+25) * 12 * 1/3
Solve, x = 75Incorrect
Explanation:
4 days earlier the work is to be completed, this means in (404) = 36 days
After 24 days, the meaning work is to be completed in (3624) = 12 days
Remaining work = 1 – 1/3 = 2/3
Let x extra men are to be employed. So,
25 * 24 * 2/3 = (x+25) * 12 * 1/3
Solve, x = 75 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeThere are 2 pipes A and B such that A fills the cistern in 8 hrs and B empties the cistern in 2 hrs. First the pipe A was opened and after 2 hours B was also opened. In how much total time will the cistern get empty?
Correct
Explanation:
In 1 hr, cistern filled by A is 1/8 so in 2 hrs 2/8 = 1/4
After 2 hr, cistern filled by A + B is 1/4 – 1/2 = 1/4
This means after 1/4th of cistern filled, in 2 hr, 1/4 gets empty in next one hour
So 2 + 1Incorrect
Explanation:
In 1 hr, cistern filled by A is 1/8 so in 2 hrs 2/8 = 1/4
After 2 hr, cistern filled by A + B is 1/4 – 1/2 = 1/4
This means after 1/4th of cistern filled, in 2 hr, 1/4 gets empty in next one hour
So 2 + 1 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative Aptitude15 men can complete a work in 40 days. 6 men started the work and after working for 20 days it was realized that the work will not get completed on time so 10 more persons are employed. Now they will complete the remaining work in how many days?
Correct
Explanation:
Shortcut:
15*40 = 6*20 + (6+10)* x
600 = 120 + 16x
Solve, x = 30Incorrect
Explanation:
Shortcut:
15*40 = 6*20 + (6+10)* x
600 = 120 + 16x
Solve, x = 30 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeThe incomes of 2 persons A and B is in the ratio 4 : 5. Their expenditures are in the ratio 8 : 15 respectively. If both save Rs 40,000, then what is B’s income?
Correct
Explanation:
Incomes 4x and 5x
Expenditures 8y and 15y
So 4x – 8y = 40000
And 5x – 15y = 40000
Solve both equations, x = 14,000
So B’s income = 5*14000Incorrect
Explanation:
Incomes 4x and 5x
Expenditures 8y and 15y
So 4x – 8y = 40000
And 5x – 15y = 40000
Solve both equations, x = 14,000
So B’s income = 5*14000 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeThe HCF and LCM of two numbers is 6 and 462 respectively. If the first number is 42, find the second one.
Correct
Explanation:
Product of two numbers = HCF * LCM
So 2nd number = 6*462/42Incorrect
Explanation:
Product of two numbers = HCF * LCM
So 2nd number = 6*462/42
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