Subject

    Quantitative Aptitude

    Topic

    Speed & Distance

    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?

    ক)
    20 m
    খ)
    24.75 m
    গ)
    20.25 m
    ঘ)
    23.75 m
    ঙ)
    40 m

    Explanation

    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. ⇒100a=80b\Rightarrow \frac{{100}}{a} = \frac{{80}}{b} and 100a=61c\frac{{100}}{a} = \frac{{61}}{c} then , ab=10080  andac=10061\frac{a}{b} = \frac{{100}}{{80}}\;and\frac{a}{c} = \frac{{100}}{{61}} ⇒cb=6180×100100\Rightarrow \frac{c}{b} = \frac{{61}}{{80}} \times \frac{{100}}{{100}} ⇒cb=6180\Rightarrow \frac{c}{b} = \frac{{61}}{{80}} Multiply both denomination and numerator by 1.25 ⇒cb=6180×1.251.25\Rightarrow \frac{c}{b} = \frac{{61}}{{80}} \times \frac{{1.25}}{{1.25}} ⇒cb=76.25100\Rightarrow \frac{c}{b} = \frac{{76.25}}{{100}} ⇒100b=76.25c\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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