Question

A , B and C complete a work in 24 days. The day in which A and C complete the work together B and C take 80/11 days less. A and B complete that work in 36 days. A will alone do the work in how many days?

Answer 1

a+b+c=24
a+c=b+c-80/11
a-b=-80/11
11a-11b=-80
a+b=36
11a+11b=396
22a=316
a=14.6
a alone will take 14.6 days

Answer 2

Answer: $\frac{1440}{119}$ days.

Step-by-step explanation:

Since, A , B and C complete a work in 24 days.

Thus, the work done by A, B and C in one day when they work simultaneously  = $\frac{1}{24}$

According to the question,

The day in which A and C complete the work together B and C take 80/11 days less.

That is,

$(\frac{1}{A}+\frac{1}{C})-(\frac{1}{B}+\frac{1}{C})=\frac{11}{80}$

( Where A, B, C are the time taken by A, B and C alone respectively)

$\frac{1}{A}-\frac{1}{B}=\frac{11}{80}$ ------ (1)

Now, A and B complete that work in 36 days.

The work done by A and B in one day = $\frac{1}{36}$

That is,

$\frac{1}{A}+\frac{1}{B}=\frac{1}{36}$ -------(2)

By adding equation (1) and (2),

We get,

$\frac{2}{A}=\frac{1}{36}+\frac{11}{80}$

$\frac{2}{A}=\frac{20+99}{720}$

$\frac{2}{A}=\frac{119}{720}$

$A=\frac{1440}{119} \text{ days}$