Bunuel wrote:
Two friends Phil and Andrew started running simultaneously from a point P in the same direction along a straight running track. The ratio of speeds of Phil and Andrew was 2:5. Two hours later, Andrew turned back and started running running backwards at one-fifth of his original speed. He met Phil at a distance of 10 km from the point P. what was Phil's running speed ?
(A) 1.25 km/hr
(B) 2.25 km/hr
(C) 2.5 km/hr
(D) 3.75 km/hr
(E) 6.25 km/hr
Are You Up For the Challenge: 700 Level QuestionsThe above solutions consider relative speeds of the Andrew and Phil. However, if you are someone like me who gets confused by relative speeds, check out this simple solution which employs the basic distance formula we all are well versed with:
distance = speed * time
OR
time = \(\frac{distance }{ speed}
\)
Let's talk about Andrew first. Let his speed be 5x (where x is a real constant).
When Andrew travels for the first 2 hours at 5x speed, his distance covered will be 2*5x = 10x
Now, he stops after covering 10x distance and starts running backwards till he meets Phil at 10 km away from P.
This means that the distance covered by Andrew during the backward run is 10x-10. Also, as mentioned in the question, speed is 1/5th the original speed - > which is \(\frac{5x}{5}\) = x.
Now, let's move to Phil.
He only covers a distance of 10km with speed 2x (where x is a real constant). Thus, the time he takes is \(\frac{10}{2x}\) = \(\frac{5}{x}\)
Remember that the total time taken by Phil to cover 10km is equal to the total time taken by Andrew to cover 10x (when moving forward)+ 10x - 10 (when moving backward)
Now, this helps me set up the following equation:
\(\frac{5}{x} = \frac{10x }{ 5x} + \frac{10x - 10 }{ x}
\)
Solving this equation, we get:
5 = 2x + 10x - 10
12x = 15
x = 5/4
Phil's speed = 2x = 2*(5/4) = 5/2 = 2.5 km/hr
Answer (C)
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