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The ans is C
Let the work be 30 units
So eff of A=2 & eff B=3 units
Since A left 3 days before so A 3 days work=6 units
Total work= 30+6=36 units
Time taken= 36/(2+3)=36/5

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gkumar08
The ans is C
Let the work be 30 units
So eff of A=2 & eff B=3 units
Since A left 3 days before so A 3 days work=6 units
Total work= 30+6=36 units
Time taken= 36/(2+3)=36/5

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Hello, Gkumar08.

I don't really understand why you add 6 units of work (6+30). IMO, if the total work = 30 units, A leaves 3 days before finishing it at the combined rate (thus, has already worked 3 days and done 2units*3days= 6 units of work) we should extract 6 units, not add them.

What I would following your approach is:

A+B = 5 units of work/day
30 units/5 = 6 days
A leaves after the third day -> Work done by A and B in the first 3 days: 2*3 + 3*3= 6+9= 15.
Remaining units of work: 15 -> B has to do them on his own -> 15/3= 5 extra days.

Total number of days: 3 (A+B) + 5 (only B) = 8 days.

Please correct me if I'm missing something.
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Those who are saying that the answer is D, can u guys plz consider the following fact?

Does this question anywhere say A left 3 days before the scheduled date of completion(6 in that case)?? If not, why are you calculating this way?
.
Instead, we can assume A works for t days and B works t+3 days.
Make an equation,
t/15+(t+3)/10=1 which yields t=21/5
So, total time=(21/5)+3=36/5

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Hi KarishmaB & 12921
Could you please help me to identify the mistake.

I am unable to understand what mistake I am doing in my approach.

Total unit of work = 30 units
A works 2 units/day and B works 3 units/day
Together A and B work 5 units/ day

Let total time taken is t

(t-3)5 + 3*3 = 30
Hence t = 36/5

===============================================

I though understand logically that 30 units of work if together A&B work, then both of them are supposed to finish it in 30/5 = 6 days.

As A & B work for 3 days so they finish 15 units. And the rest 15 units of work is finished by B in 15/3 = 5 days

So total is 8 days

But why the algebraic approach is giving wrong answer?
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rsrighosh
Hi KarishmaB
Could you please help me to identify the mistake.

I am unable to understand what mistake I am doing in my approach.

Total unit of work = 30 units
A works 2 units/day and B works 3 units/day
Together A and B work 5 units/ day

Let total time taken is t

(t-3)5 + 3*3 = 30
Hence t = 36/5

===============================================

I though understand logically that 30 units of work if together A&B work, then both of them are supposed to finish it in 30/5 = 6 days.

As A & B work for 3 days so they finish 15 units. And the rest 15 units of work is finished by B in 15/3 = 5 days

So total is 8 days

But why the algebraic approach is giving wrong answer?

Posted from my mobile device
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rsrighosh
Hi KarishmaB & 12921
Could you please help me to identify the mistake.

I am unable to understand what mistake I am doing in my approach.

Total unit of work = 30 units
A works 2 units/day and B works 3 units/day
Together A and B work 5 units/ day

Let total time taken is t

(t-3)5 + 3*3 = 30
Hence t = 36/5

===============================================

I though understand logically that 30 units of work if together A&B work, then both of them are supposed to finish it in 30/5 = 6 days.

As A & B work for 3 days so they finish 15 units. And the rest 15 units of work is finished by B in 15/3 = 5 days

So total is 8 days

But why the algebraic approach is giving wrong answer?

The answer here is 36/5 days, not 8 days.
The algebraic approach used by you is correct above.

Assuming 30 units of work, A does 2 units and B does 3 units every day.
Last 3 days, only B works so he does 9 units in that time. So we are left with 21 units if work which A and B do together first.
21 units of work will take 21/5 days (since A and B are working together)
Total time taken will be 36/5 days.
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rsrighosh
Hi KarishmaB & 12921
Could you please help me to identify the mistake.

I am unable to understand what mistake I am doing in my approach.

Total unit of work = 30 units
A works 2 units/day and B works 3 units/day
Together A and B work 5 units/ day

Let total time taken is t

(t-3)5 + 3*3 = 30
Hence t = 36/5

===============================================

I though understand logically that 30 units of work if together A&B work, then both of them are supposed to finish it in 30/5 = 6 days.

As A & B work for 3 days so they finish 15 units. And the rest 15 units of work is finished by B in 15/3 = 5 days

So total is 8 days

But why the algebraic approach is giving wrong answer?

The logical approach you mentioned is incorrect. Note that they will take 6 days to complete the work ONLY when both work together for the whole 6 days. Since they don't, it will take more than 6 days to complete the work. So we cannot say when A left. He left 3 days before the work was actually completed. If it had taken 8 days to complete the work, he would have left after 5 days, not after 3 days.

8 days would be the answer if the question were different - if it were given that A works only for 3 days and B works for all days.
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