In a race of senior citizens, Mr. A can give his fiend Mr. B a start of 20 m and Mr. C a start of 39 m in a race of 100 m. How much start can Mr. B give Mr. C?
Solution
Correct Answer: Option D
Speed = distance/time
Let the speed of Mr. A, Mr. B, and Mr. C are a, b and c respectively. Given that, Mr. A can give his friend Mr. B a start of 20 m and Mr. C a start of 39 m in a race of 100 m.
Since Mr. A can give his friend Mr. B a start of 20 m
Time taken by them must be equal.
\(\Rightarrow \frac{{100}}{a} = \frac{{80}}{b}\) and \(\frac{{100}}{a} = \frac{{61}}{c}\)
then , \(\frac{a}{b} = \frac{{100}}{{80}}\;and\frac{a}{c} = \frac{{100}}{{61}}\)
\(\Rightarrow \frac{c}{b} = \frac{{61}}{{80}} \times \frac{{100}}{{100}}\)
\(\Rightarrow \frac{c}{b} = \frac{{61}}{{80}}\)
Multiply both denomination and numerator by 1.25
\(\Rightarrow \frac{c}{b} = \frac{{61}}{{80}} \times \frac{{1.25}}{{1.25}}\)
\(\Rightarrow \frac{c}{b} = \frac{{76.25}}{{100}}\)
\(\Rightarrow \frac{{100}}{b} = \frac{{76.25}}{c}\)
Thus, Mr. B covers 100m in the same time Mr. C covers 76.25 m.
Thus, Mr. B can give Mr. C ahead start of (100 – 76.25) m = 23.75 m
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