Discussion Forum

A, B and C together can complete a piece of work in 10 days. All the three started working at it together and after 4 days A left. Then B and C together completed the work in 10 more days. A alone could complete the work in :

Answer:

Explanation:

Set a = the rate of A in (piece of work)/day

b = the rate of B in (piece of work)/day

c = the rate of C in (piece of work)/day

Below is the equation for all three working together.

($10$ days )$\times$ $(a+b+c)$ = 1(piece of work)

Next we have the equation were all work together for 4 days,

then only B and C continue for $10$ more days.

($4$ days) $\times$ $(a+b+c )$ + ($10$ days)$( b + c)$ = 1(piece of work)

Since each left hand side is equal to 1 piece of work,

we have
($10$ days) $\times$ $( a + b + c )$ = ($4$ days) $\times$ $(a+b+c)$ + ($10$ days)$(b+c)$

and

$10(a+b+c )$ $= 4(a+b+c )$ + $10(b+c)$

$10a + 10b + 10c$ $= 4a + 4b + 4c + 10b + 10c$

combining terms

$10a+10b+10c$ $=4a+14b+14c$

Notice that we can rewrite the above as

$10a + 10(b + c)$ $=4a+4(b+c)$

if we subtract $10(b + c)$ from each side

$10a$ $=4a+4(b+c)$

subtract $4a$ from each side

$6a$ $= 4(b + c)$

divide each side by $4$

$\frac{6a}{4}$ $=(b+c)$

Recall

($10$ days) $\times$ $(a+b+c)$ = 1(piece of work) or

$10(a+b+c)$ $= 1$

So

$10a+10(b+c)$ $=1$

Now substitute $\frac{6a}{4}$ for $(b+c)$ and we have

$10a + 10\left(\frac{6a}{4}\right)$ $= 1$

$10a+15a$ $= 1$

$25a=1$

This means that A at rate a will take 25 days to complete the work.