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Speed Time and Distance
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A runs $$\frac{7}{4}$$ times as fast as B. If A gives B a start of 300 m, how far must the winning post be if both A and B have to end the race at same time?

Answer & Solution
Correct Answer: Option B
.ABReason (ST=D)
Speed74Given
Time 47Since, Speed ∝ $$\frac{1}{{{\text{Time}}}}$$
Distance4 7Distance ∝ time


Now,
7x - 4x = 300 (A runs 7x m and B runs 4x)
x = 100
7x = 7 × 100 = 700 m
Winning post is 700 m away, As A runs 700 m to complete the race.
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1 Comment
Sohel Rana
Sohel Rana 7 years ago
let,
total distance be x
B speed y
A speed 7y/4
ATQ
x/7y/4=x-300/y
=>y= 700