Question

A takes 5 days more than b to do a certain job and 9 days more than c; a and b together can do the job in same time as

c. How many days a would take to do it?

Answer 1

Answer:

Let the time taken by C be x days, in which case time taken by A would be x + 9 days, and B would be x + 4 days (5 hours lesser than A)

Now, we're given that the rate of work of A and B (combined) = rate of work of C

1/(x+9) + 1/(x+4) = 1/x

2x+13/(x^2+13x+36) = 1/x

2x^2 + 13x = x^2 + 13x + 36

x^2 = 36

Since the value of days cannot be negative, we get x = 6, and we have to find A which is x + 9 days, i.e. 6 + 9 = 15 days,

Answer 2

Answer:

C takes 15 days to do the job.

Step-by-step explanation:

Let the day taken by C alone be n

so in one day work done by $C= \frac{1}{n}$

Time taken by A to do the same work $= n+9 \: \: days$

So in one day work done by $A= \frac{1}{n + 9}$

Also A and B can together do the work in same time as C

So work done in one day by A and B together = work done by C in one day

$= \frac{1}{n}$

let work done by B in one day be b

$\frac{1}{n} = b + \frac{1}{n + 9}$

$b = \frac{9}{n(n + 9)}$

Hence B takes $\frac{n(n+9)}{9}$

days to finish the work alone.

According to Question days taken by A alone = days taken by B alone +5

So, $n + 9 = \: \frac{n(n+9)}{9 + 5}$

solving this ${n}^{2} = 36$

and n=6 days

Hence A alone can do the job in

$n+9=6+9=15$ days time