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SSC Sample Paper SSC CHSL (10+2) Sample Test Paper-5

  • question_answer
    In a race of one kilometre, A gives B a start of 100 metres and still wins by 20 seconds. But if A gives B a start of 25 seconds, B wins by 50 metres. The time taken by A to run one kilometre is

    A) \[17\,\,\operatorname{seconds}\]             

    B) \[\frac{500}{29}\operatorname{seconds}\]  

    C) \[\frac{1200}{29}\operatorname{seconds}\]

    D)        \[\frac{700}{29}\operatorname{seconds}\]

    Correct Answer: B

    Solution :

     Let A take \[x\] second in covering \[1000\,\,m\] and \[b\] takes \[y\] seconds According to the question,             \[x+20=\frac{900}{1000}y\] \[\Rightarrow \]   \[x+20=\frac{9y}{10}\]                         ? (i) and,             \[\frac{950}{1000}x+25=y\]                              ? (ii) From equation (i),             \[\frac{10x}{9}+\frac{200}{9}=y\] \[\Rightarrow \]   \[\frac{10x}{9}+\frac{200}{9}=\frac{950x}{1000}+25\] \[\Rightarrow \]   \[\frac{10x}{9}+\frac{200}{9}=\frac{19x}{20}+25\] \[\Rightarrow \]   \[\frac{10x}{9}-\frac{19x}{20}=25-\frac{200}{9}\] \[\Rightarrow \]   \[\frac{200x-171x}{180}=\frac{225-200}{9}\] \[\Rightarrow \]   \[\frac{29x}{180}=\frac{25}{9}\] \[\Rightarrow \]   \[x=\frac{25}{9}\times \frac{180}{29}=\frac{500}{29}\]seconds


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