Tom, Jerry and Bill start from point A at the same time in their cars to go to B. Tom reaches point B first and turns back and meets Jerry at a distance of 9 miles from B. When Jerry reaches B, he too turns back and meets Bill at a distance of 7 miles from B. If 3 times the speed with which Tom drives his car is equal to 5 times Bill’s speed, what could be the distance between the points A and B?
A. 24 miles
B. 31 miles
C. 40 miles
D. 45 miles
E. 63 miles
Let distance between point A and B = D; Speed of Tom = \(S_t\), Speed of Jerry = \(S_j\) and Speed of Bill = \(S_b\) (considering all drive at constant speed)
Also given \(3S_t = 5S_b\) OR \(\frac{S_b}{S_t} = \frac{3}{5}\)
Here time taken by Tom and Jerry to cover their respective distances when they meet is equal. \(T_t\) = \(T_j\)
So, when they meet:
Distance covered by Tom = D + 9
Distance covered by Jerry = D - 9
Hence \(\frac{D + 9 }{S_t} = \frac{D - 9}{S_j}\) OR \(\frac{D + 9 }{D - 9} = \frac{S_t}{S_j}\) Eqn. 1
Similarly, when Jerry and Bill meet time taken by them is equal. \(T_j\)' = \(T_b\)
When Jerry and Bill meet:
Distance covered by Jerry = D + 7
Distance covered by Bill = D - 7
Hence \(\frac{D + 7 }{S_j} = \frac{D - 7}{S_b}\) OR Hence \(\frac{D + 7 }{D - 7} = \frac{S_j}{S_b}\) Eqn. 2
Multiplying Eqn. 1 and 2
Hence \(\frac{(D + 9)(D + 7)}{(D - 9)(D - 7)} = \frac{S_t}{S_b} = \frac{5}{3}\)
\(D^2 - 64D + 63 = 0\)
(D - 63)(D - 1) = 0
D ≠ 1 So
D = 63
Answer E.
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