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Question

P, Q, R, S, and T run a 100 metre race. P beats Q by 20 m, R by 25 m, S by 40 m, and T by 50 m. Q, R, S, and T now run another 100 metre race. Assume that they run at exactly the same speeds as before. What is the approximate total of the distances by which Q beats R, S, and T? (Write your answer to the nearest possible natural number) Metre
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Solution

From the first race, we know that, when P has run 100 m, Q, R, S, and T have run 80 m, 75 m, 60 m, and 50 m respectively.

In the second race, Q completes 100 m. Hence R will complete 75×(10080) = 93.75 m, and will be beaten by 6.25 m.

Similarly, S will run  60×(10080) = 75 m, and will be beaten by 25 m.

T will run  50×(10080) = 62.5 m, and will be beaten by 37.5 m.

The total of the three distances is (6.25 + 25 + 37.5) = 68.75 m, or approximately 69 m.
Alternatively,
In the first race, when Q has run 80 m, R, S and T have run 75 m, 60 m and 50 m respectively. So, the sum of the distances by which Q is ahead of R, S, and T is (5 + 20 + 30) = 55 m

In the second race, Q runs 100 m. So the sum of distances by which Q beats R, S, and T is  55×(10080) = 68.75 m, or approximately 69 m.

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