The hands of a clock are 10 cm and 7 cm respectively. The difference between the distance traversed by their extremities in 3 days 5 hours is
(a)4552.67 cm
(b)4555.67 cm
(c)4557.67 cm
(d)4559.67 cm
Answer
Answer (as printed): C
Explanation
Number of rounds completed by the minute hand in 3 days 5 hrs $=(3 \times 24+5)=77$. Number of rounds completed by the hour hand in 3 days $$5 \mathrm{hrs}=\left(3 \times 2+\frac{5}{12}\right)=6 \frac{5}{12} .$$ ∴ Difference between the distance traversed $$\begin{aligned} & =\left[77 \times\left(2 \times \frac{22}{7} \times 10\right)-6 \frac{5}{12} \times\left(2 \times \frac{22}{7} \times 7\right)\right] \mathrm{cm} \\ & =(4840-282.33) \mathrm{cm}=4557.67 \mathrm{cm} . \end{aligned}$$