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Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?

A. 8 kmph

B. 11 kmph

C. 12 kmph

D. 14 kmph

Answer: Option C

Solution(By Examveda Team)

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{distance}}\,{\text{travelled}}\,{\text{by}}\,x\,{\text{km}} \cr & {\text{Then}},\,\frac{x}{{10}} - \frac{x}{{15}} = 2 \cr & \Rightarrow 3x - 2x = 60 \cr & \Rightarrow x = 60\,km \cr & {\text{Time}}\,{\text{taken}}\,{\text{to}}\,{\text{travel}}\,60\,km\,{\text{at}}\,10\,{\text{km/hr}} \cr & = {\frac{{60}}{{10}}} \,hrs = 6\,hrs \cr & {\text{So,}}\,{\text{Robert}}\,{\text{started}}\,{\text{6}}\,{\text{hours}}\,{\text{before}}\,2\,P.M.\,i.e.,\,at\,A.M. \cr & \therefore {\text{Required}}\,{\text{speed}} \cr & = {\frac{{60}}{5}} \,kmph = 12\,kmph \cr} $$

This Question Belongs to Arithmetic Ability >> Speed Time And Distance

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Comments ( 1 )

  1. Maiks Blank
    Maiks Blank :
    3 years ago

    This is how i do.
    Let time taken to reach home at 2= y
    Case 1: Distance covered= 10y
    case 2: Distance covered= 15(y-2)

    Since distance is same,
    10y=15(y-2)
    y=6hrs

    and, Distance= speed X time= 10 X 6= 60 kms

    To reach @1, (y-1)= (6-1)= 5 hrs

    required speed= 60÷5= 12 kms/hr

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