Question

A takes 8 more days than B to finish a work. A and B start this work and B leaves the work 9 days before the work is finished if A complete 75%of the work. how long would. A have taken to finish the work alone.​

Answer 1

A & B together can complete work in (25x20)/(25+20) =500/45 = 100/9 days.

They complete 5 x 9/100= 9/20 work in 5 days.

Remaining 11/20 work done by B in 11/20 x 20 = 11 days.

Answer 2

A will take 14 days to finish the work alone.​

Step-by-step explanation:

Let the time taken by B working alone = x days,

So, time taken by A working alone = (x + 8) days

∵ A complete 75% of the work,

So, the day he will work = 75% of (x+8) = $\frac{75(x+8)}{100}$ = $\frac{3(x+8)}{4}$,

⇒ B has to complete 25% of work,

That is, days he will work = 25% of x = $\frac{25x}{100}$ =  $\frac{x}{4}$,

Since, B leaves the work 9 days before the work is finished,

Days for which A worked - Days for which B worked = 9 days.

$\frac{3(x+8)}{4}-\frac{x}{4}=9$

$\frac{3x+24-x}{4}=9$

$\frac{2x+24}{4}=9$

$2x+24=36$

$2x=12$

$\implies x = 6$

Now, x + 8 = 6 + 8 = 14

Hence, time taken by A to finish the work alone would be 14 days.

#Learn more :

A take 10 days less than time taken by b to finish a piece of work. if a and b together can finish the work in 12 days find the time taken by b to finish the work

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