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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
| . | A | B | Reason (ST=D) |
| Speed | 7 | 4 | Given |
| Time | 4 | 7 | Since, Speed ∝ $$\frac{1}{{{\text{Time}}}}$$ |
| Distance | 4 | 7 | Distance ∝ time |
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Logintotal distance be x
B speed y
A speed 7y/4
ATQ
x/7y/4=x-300/y
=>y= 700