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(C) Let $d$ be the distance of the epicenter from the seismograph.The time taken by $P$ waves is $t_p = \frac{d}{V_p} = \frac{d}{8.0} \, s$.The time t

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$2500$

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(C) Let $d$ be the distance of the epicenter from the seismograph.The time taken by $P$ waves is $t_p = \frac{d}{V_p} = \frac{d}{8.0} \, s$.The time taken by $S$ waves is $t_s = \frac{d}{V_s} = \frac{d}{4.5} \, s$.Given that the $P$ wave arrives $4.0 \, min$ before the $S$ wave,we have $t_s - t_p = 4.0 \, min = 240 \, s$.Substituting the values: $\frac{d}{4.5} - \frac{d}{8.0} = 240$.Taking $d$ as common: $d \left( \frac{8.0 - 4.5}{4.5 	imes 8.0} ight) = 240$.$d \left( \frac{3.5}{36} ight) = 240$.$d = \frac{240 	imes 36}{3.5} \approx 2468.57 \, km$.Rounding to the nearest given option,$d \approx 2500 \, km$.

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