Question

A can do a piece of work in 16 days, B in 10 days. A and B together work for 6 days and then C finishes it in 3 days, then in how many days could C have done it alone ?

a) 40 days
b) 80 days
c) 90 days
d) 120 days ​

Answer 1

Answer:

120 days

Step-by-step explanation:

Total work =80

days......efficiency

A----16--------5

B----10--------8

W.D. =13×6=78

W.L.=2

C.....2W......3Days

C.....80W.....3×80/2= 120 days

Answer 2

Concept

In this type of question find one's day work of workers and then made the equation according to the statement given .

Given

We have given that work done by the A is 16 days and work done by the B is 10 days while A and B finishes the work in 6 days together and C finishes it in 3 days.

Find

We are asked to determine the days in which C will finish the work alone.

Solution

A can do a piece of work in 16 days and B do the same work in 10 days.

Then A can do the work in one day $=\frac{1}{16}$

And B can do the work in one day $=\frac{1}{10}$

Then A & B's one day's work $=\frac{1}{16} +\frac{1}{10}=\frac{13}{80}$

According to the statement

$6(A+B)+3C=1\\\\6(\frac{13}{80} )+3C=1\\\\\frac{78}{80} +3C=1\\\\3C=\frac{2}{80} \\\\3C=\frac{1}{40}\\\\C=\frac{1}{120}$

Therefore, C finishes the work alone in 120 days.

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