0

A and B worked together to complete a certain task in three hours. If A works alone, he can complete the task in five hours. If B works alone, how much longer will it take him to complete the task?

I have the answer to this question, but I just don't get how that works. Please help me out.

Soha Farhin Pine
  • 663
MLIAL k
  • 27
  • You want to add the progress of both people together for a certain period of time (say one hour). What is done in one hour? – Element118 Oct 08 '15 at 15:45

2 Answers2

2

Assuming that their working speed adds if they work together:

Working Speed of person A = 1/5 task per hour

Working Speed of A + B = 1/3 task per hour

They finish the task in 3 hours, so A has done 3/5 of the task in that time, which means Person B must have done 2/5 of the task.

If we divide the work done by Person B in 3 hours by 3, we get the work Person B does in 1 hour.

(2/5)/3 = 2/15

The working speed of Person B is 2/15 of the task per hour, which means it takes Person B 7.5 hours of work to get the task done.

LionIsLoose
  • 94
  • Thanks! I wasn't sure of the 1/5 and 1/3 part from the answer, but as you put it 1/5 (task per hour) and so on, I understood it clearly. – MLIAL k Oct 09 '15 at 10:21
1

We use the fact that speed is inversely proportional to time. Let us denote:

$v_A$ - speed of $A$ working alone,

$v_B$ - speed of $B$ working alone,

$t_A$ - time needed for $A$ to finish the job alone,

$t_B$ - time needed for $B$ to finish the job alone,

$t$ - time needed for $A$ and $B$ to finish the job together.

Then by inverse proportionality we have $$ v_At_A = v_Bt_B = (v_A+v_B)t$$ and when we substitute $t_A = 5$, $t = 3$ we get $$5 v_A = 3v_A + 3v_B\implies v_A = \frac 32 v_B$$ and $$v_Bt_B = \left(\frac 32v_B\right)t_A \implies t_B = \frac 32 t_A\implies t_B = \frac{15}2$$ so $B$ needs to work $7$ and half hours to get the job done.

Ennar
  • 23,082