Question

A and B alone can complete work in 9 days and 18 days respectively. They worked together. However 3 days before the completion of the work B left. In how many days the work was completed

Answer 1

Hi user 

Here's the answer 
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Time taken by A = 9 days

Rate of work done in 1 day = 1 / 9 of whole work

Time taken by B = 18 days

Rate of work done in 1 day = 1 / 18 of whole work 

But A leaves before 3 days of completion of work

Let the total number of days taken be " t "

Work done by A = 1 / 9 * ( t - 3 )

Work done by B = 1 / 18 * ( t )

Work done by A + Work done by B = 1

= ( t - 3 ) / 9 + ( t ) / 18 = 1

Taking LCM, we get

 Multiplying by 2 we get,

= 2 ( t - 3 ) / 9 * 2 + ( t ) / 18 = 1

= 2t - 6 / 18 + t / 18 = 1

= 2t + t - 6 / 18 = 1

= 3t - 6 = 18 * 1

= 3t - 6 = 18

= 3t = 18 + 6

= 3t = 24

=> t = 24 / 3 = 8

Hence total time taken ( t ) = 8 days
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Hope my answer is correct

Have a good day :-)

Answer 2

If a and b worked together than they can finish it in=1/18+1/9
=1+2/18
=3/18
=1/6
So they can finish it in 6 days but before 3 days of completion b left means they worked together for {6-3}=3 days
Now,
A and b worked for 3 days so their work is=
{1/18+1/9}×3
=1/6×3
=1/2
Now the no. Of work is 1 so
1-1/2
=1/2
Now only b works so
1/18x=1/2
X=18/2
=9