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Speed of Lisa= 15 kmph
Speed of Linda= 20 kmph
Distance between them= 200 m= 0.2 km
Let them meet at x kms from where Linda is:
Therefore, Distance for Lisa= x, =>Time for Lisa = x/15
 Distance for Linda= 0.2 + x, => Time for Linda= (0.2 + x) / 10
Since both start moving at the same time until they meet-
Time taken by Lisa = Time taken by Linda
 x/15= (0.2 + x)/20
 4*x = 3* (0.2 + x)
x = 0.6 km = 600 m . Option E
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Bunuel
Linda and Lisa, who are 200 meters apart, start moving at the same direction. If Linda walks at the rate of 20 kmph and Lisa walks at the rate of 15 kmph. What is the distance covered by Lisa by the time Linda catches up?

A. 200 m
B. 300 m
C. 400 m
D. 500 m
E. 600 m

Since this is a catch-up problem, we have:

time = ∆ distance/∆ rate

∆ distance = 200 m = 0.2 km
∆ rate = 20 – 15 = 5 kmph
time = 0.2/5 hours

So, the distance covered by Lisa, whose rate is 15 kmph, is:

distance = rate × time = 15 × 0.2/5 = 0.6 km = 600 m

Answer: E
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