A alone can finish a piece of work in 42 days. B is 20% more efficient

1.	A alone can finish a piece of work in 42 days. B is 20% more efficient than A and C is 40% more efficient than B. In how many days B and C working together can finish the same piece of work? (in days)1). $$11\frac{5}{12}$$2). $$13\frac{5}{12}$$3). $$15\frac{1}{12}$$4). $$14\frac{7}{12}$$
Answer» Solution Let rate at which A finishes the work = $100x$ units/day => Rate at which B finishes the work = $\FRAC{120}{100} \times 100x = 120x$ units/day Rate at which C finishes the work = $\frac{140}{100} \times 120x = 168x$ units/day Work done by A in 42 DAYS = $42 \times 100x = 4200x$ units/day Rate at which B and C finishes the work = $120x + 168x = 288x$ units/day $\therefore$ Time taken by B and C TOGETHER to finish the same work = $\frac{4200x}{288x}$ = $\frac{175}{12} = 14\frac{7}{12}$ days

A alone can finish a piece of work in 42 days. B is 20% more efficient than A and C is 40% more efficient than B. In how many days B and C working together can finish the same piece of work? (in days)1). $$11\frac{5}{12}$$2). $$13\frac{5}{12}$$3). $$15\frac{1}{12}$$4). $$14\frac{7}{12}$$

Answer»

Solution

Let rate at which A finishes the work = $100x$ units/day

=> Rate at which B finishes the work = $\FRAC{120}{100} \times 100x = 120x$ units/day

Rate at which C finishes the work = $\frac{140}{100} \times 120x = 168x$ units/day

Work done by A in 42 DAYS = $42 \times 100x = 4200x$ units/day

Rate at which B and C finishes the work = $120x + 168x = 288x$ units/day

$\therefore$ Time taken by B and C TOGETHER to finish the same work = $\frac{4200x}{288x}$

= $\frac{175}{12} = 14\frac{7}{12}$ days

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