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In a 400 m race, A gives B a start of 5 seconds and beats him by 15 m. In another race of 400 m, A beats B by $$7\frac{1}{7}$$ seconds. Their respective speed are -

A. 6 m/sec, 7 m/sec

B. 5 m/sec, 7 m/sec

C. 8 m/sec, 7 m/sec

D. 9 m/sec, 7 m/sec

Answer: Option C

Solution(By Examveda Team)

Suppose A covers 400 m in t seconds
Then, B covers 385 m in (t + 5) seconds
$$\eqalign{ & \therefore {\text{ B covers 400 m in}} \cr & = \left\{ {\frac{{\left( {t + 5} \right)}}{{385}} \times 400} \right\}{\text{ sec}} \cr & = \frac{{80\left( {t + 5} \right)}}{{77}}{\text{ sec}} \cr & {\text{Also, B covers 400 m in}} \cr & = \left( {t + 7\frac{1}{7}} \right){\text{ sec}} \cr & = \frac{{\left( {7t + 50} \right)}}{7}{\text{ sec}} \cr & \therefore \frac{{80\left( {t + 5} \right)}}{{77}} = \frac{{7t + 50}}{7} \cr & \Rightarrow 80\left( {t + 5} \right) = 11\left( {7t + 50} \right) \cr & \Rightarrow \left( {80t - 77t} \right) = \left( {550 - 400} \right) \cr & \Rightarrow 3t = 150 \cr & \Rightarrow t = 50 \cr & \therefore {\text{ A's speed}} \cr & = \frac{{400}}{{50}}\,m/\sec \cr & = 8\,m/\sec \cr & \therefore {\text{B's speed}} \cr & = \frac{{385}}{{55}}\,m/\sec \cr & = 7\,m/\sec \cr} $$

This Question Belongs to Arithmetic Ability >> Races And Games

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Comments ( 1 )

  1. Orko Abir
    Orko Abir :
    4 years ago

    SUPPOSE time taken by A = x and time taken by B is = x+5
    distance covered by A= 400 meter and distance covered by B =400-15 =385 meter
    Now A's speed = (400/x) and B's speed =( 385/(x+5))

    From second condition we found both A and B cover 400 meter thus A beat B by (50/7) sec ,

    developing equation we get ,
    (400/(365/(x+5)))- (400/(400/x))=(50/7) [ we know (distance/time ) = time taken ]
    solving it we will get x =50 sec

    thus speed of A = (400/50)=8 sec
    speed of B =(385/(50+5)=(385/55)=7 sec
    Answer: C
    [ Note for Examveda team , please rectify the question where given A beat B by (15/2) sec; actually it must be (50/7) sec ]

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