Question

A and b together can do a piece of work in 12 days.b and c can do the same work in 15 days.if a is twice as fast as c, how much time would b alone complete the work

Answer 1

If A is twice as fast as C and A& B together does a work in 12 days and B & C together does a work in 15 days then B alone would complete the work in 20 days.

Step-by-step explanation:

Step 1:

A and B together do work in 12 days.

So, (A+B)’s 1-day work = $\frac{1}{12}$

B and C together do the same work in 15 days

So, (B+C)’s 1-day work = $\frac{1}{15}$

It is also given that A is twice as fast as C i.e., we can conclude from this that

A’s work in 1 day = C’s work in 2 days ….. (i)

Step 2:

Now, we can also write

(A+B)’s 1 -day work = A’s 1-day work + B’s 1- day work

⇒ (A+B)’s 1 -day work = C’s work in 2 days + B’s 1- day work …… [from (i)]

⇒ (A+B)’s 1 -day work = C’s work in 1 day + (B+C)’s 1- day work

⇒ C’s work in 1 day = (A+B)’s 1 -day work - (B+C)’s 1- day work

⇒ C’s work in 1 day = [$\frac{1}{12}$ - $\frac{1}{15}$] ….. [substituting given values]

⇒ C’s work in 1 day = $\frac{5-4}{60}$

⇒ C’s work in 1 day = $\frac{1}{60}$ ….. (ii)

Step 3:

Therefore,

B’s 1-day work = [B+C]’s 1-day work – C’s 1-day work

⇒ B’s 1-day work = [$\frac{1}{15}$ – $\frac{1}{60}$]

⇒ B’s 1-day work = $\frac{4-1}{60}$

⇒ B’s 1-day work = $\frac{3}{60}$

⇒ B’s 1-day work = $\frac{1}{20}$

Thus, B alone can complete the work in 20 days.

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