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SSC Quantitative Aptitude Speed, Time and Distance Question Bank Speed Time and Distance (I)

  • question_answer
    A student walks from his house at \[2\frac{1}{2}km/h\] and reaches his school late by 6 min. Next day, he increases his speed by 1 km/h and reaches 6 min before school time. How far is the school from his house?

    A) \[\frac{5}{4}km\]

    B) \[\frac{7}{4}km\]

    C) \[\frac{9}{4}km\]

    D) \[\frac{11}{4}km\]

    Correct Answer: B

    Solution :

    [b] Let the required distance be x km. Then,             \[\frac{x}{(5/2)}-\frac{x}{(7/2)}=\frac{12}{60}\] [\[\therefore \] Difference between two times is 12 min] \[\Rightarrow \]            \[\frac{2x}{5}-\frac{2x}{7}=\frac{1}{5}\] \[\Rightarrow \]            \[14x-10x=7\] \[\Rightarrow \]            \[4x=7\]           \[\Rightarrow \] \[x=\frac{7}{4}\] Required distance \[=\frac{7}{4}\,km\]


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