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If A, B,C,D can do 150 sums in 6 days. Let us find in how many days will A and D finish 250 sums if each of them does equal number of sums everyday.

this much is given in the question, i dont knw how to solve with this much data.

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    Does "each of them" refer to $A$ and $D$ or $A$, $B$, $C$, and $D$? – 2017-02-04
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    i have no idea i gave the question as it is ! – 2017-02-04
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    I agrree that there are not enough information if it is meant that each of them is A and D and not A,B,C and D. Just do another exercise which is better phrased. – 2017-02-04
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    It likely refers to $A$, $B$, $C$, and $D$ because, otherwise, there isn't enough information to answer the question. – 2017-02-04
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    Then how can they do 150 sums in 6 days or 150/6 = 25 sums every day because 25 not divisible by 4 – 2017-02-04

3 Answers 3

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First lets take A,B,C,D as four people and so lets find out how many sums four people can do in one day which is 150/6 as 150 sums in 6 days so that is equal to 25 sums in one day. To get the number of days four people take to do 250 sums you have to do 250/25 which is equal to 10 days. So four people take 10 days to do 250 sums therefore if you half the number of people the number of days will double hence taking them 20 days to do 250 sums.(We are taking the assumption that all four people do equal number of sums hence even two people do equal number of sums).

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Hint:

Start by calculating how much time it will take for $A,B,C,D$ to do $250$ sums.

Then calculate the amount of time it will take for only $A,D$ to do them (It will take $2$ times as much time for $2$ people to do them instead of $4$ people).

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    Well the answer matches the answer via your logic. but wheres the use of (each of them does equal number of sums everyday) this part. – 2017-02-04
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    This is a very slick answer +1. The equal number comes about in the twice as long calculation (since $2$ people take will take twice the time as $4$ people, all working at the same rate). – 2017-02-04
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Sketch (assuming that everyone works at the same rate):

Let $r$ be the rate at which $A$, $B$, $C$, or $D$ individually do sums. In other words, $$r=\frac{\#\text{ of sums}}{\#\text{ of days}}.$$ In the first case, you know that four of them working (so a total rate of $4r$) completes $150$ in $6$ days, so $$ (4r)(6)=150. $$ From here, you can solve for $r$. Then, with two of them working to get $250$ sums, you need to solve $$ (2r)(d)=250 $$ for $d$.