Ans: (a) 10\displaystyle \frac{5}{7}

Explanation:

\begin{aligned} \text{Total Work}&=\text{LCM }(7\frac{1}{2},25)\\[1em] &=\text{LCM }(\frac{15}{2},25)\\[1em] &=\frac{\text{LCM }(15,25)}{\text{HCF }(2,1)}\\[1em] &=\frac{75}{1}\\[1em] &=15 \text{ units} \end{aligned}

A and B efficiency = \displaystyle \frac{75}{15}\times 2 = 10 units/day

B’s efficiency = \displaystyle \frac{75}{25} = 3 units/day

Now, A’s efficiency = 10 – 3 = 7 units/day

Time taken by A to complete the work

= \displaystyle \frac{75}{7}=10\displaystyle \frac{5}{7} \text{ days}

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