Question : To do a certain work, the ratio of the efficiencies of A and B is 7 : 5. Working together, they can complete the same work in $17 \frac{1}{2}$ days. A alone will complete 60% of the same work in:
Option 1: 18 days
Option 2: 15 days
Option 3: 16 days
Option 4: 21 days
Correct Answer: 18 days
Solution :
Let the efficiency of A be $E_A$ and the efficiency of B as $E_B$.
Given that the ratio of the efficiencies of A and B is 7 : 5, for some constant $x$.
$E_A = 7x$
$E_B = 5x$
The total amount of work done by A and B together in one day.
$E_A + E_B = 7x + 5x = 12x$
Given that A and B together can complete the work in $17\frac{1}{2}$ days.
The total work $W$,
$W = (E_A + E_B) \times \text{time} = 12x \times 17\frac{1}{2} = 210x$
The time it will take for A to complete 60% of the work alone.
The amount of work A needs to do = $0.6W = 0.6 \times 210x = 126x$
The time it takes for A to complete this amount of work $=\frac{126x}{7x} = 18$ days
Hence, the correct answer is 18 days.
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