Linear Races - Swetabh Kumar


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    Q1) A can give B 100 meters start and C 200 meters start in a kilometer race. How much start can B give C in a kilometer race?

    OA: 1000/9
    Solution:
    Va/Vb = 1000/900 (since 100m start means B has to go 1000-100=900m)
    Va/Vc = 1000/800 (similar logic)
    Dividing, Vc/Vb = 8/9
    so when B goes 1000, C goes 1000*8/9 = 8000/9
    so B can give a start of 1000-8000/9 = 1000/9 to C. Hence, 1000/9

    Q2) In a race of 1000 m, A can beat B by 100 m, in a race of 800m, B can beat C by 100m. By how many meters will A beat C in a race of 600 m?

    OA: 127.5
    Solution:
    Va/Vb = 1000/900=10/9
    Vb/Vc = 800/700=8/7
    Multiplying, Va/Vc = 80/63 = A's distance/C's distance
    so C's distance = A's distance * 63/80 = 600 * 63/80 = 472.5
    so Margin = 600-472.5 = 127.5 m

    Q3) A and B take part in 100 m race. A runs at 5 kmph. A gives B a start of 8 m and still beats him
    by 8 seconds. The speed of B is ?

    OA: 1.15 m/s or 4.14 kmph
    Solution:
    5 kmph = 25/18 m/s
    92/Vb - 100/Va = 8 (Difference in time taken)
    92/Vb - 100/(25/18) = 8
    92/Vb = 80
    Vb = 92/80 = 1.15 m/s or 4.14 kmph

    Q4) At a game of billiards, A can give B 15 points in 60 and A can give C to 20 points in 60. How
    many points can B give C in a game of 90?

    OA: 10
    Solution:
    Va/Vb = 60/45 = 4/3
    Va/Vc = 60/40 = 3/2
    Dividing, Vb/Vc = 9/8 = 90/C so C=80. hence B can give 90-80=10 points

    Q5) In a game of tennis, A can give B 15 points, B can give C 12 points and A can give C 20 points. How many points make up the game ? (Self made to illustrate concept, so answer can be fractional as well)

    OA: 180/7
    Solution:
    Total points= P
    A gives B x points is equivalent to A giving B a headstart of x metres to B. So,
    A/B = P/(P-15)
    B/C = P/(P-12)
    Multiplying above: A/C = P^2/ (P-12)(P-15)
    BUT, A/C = P/(P-20) given
    so P^2/ (P-12)(P-15) = P/(P-20)
    -20P =-27P+180
    so P = 180/7 points.

    Q6) A runs (5/3) times as fast as B. If A gives B a start of 80 m, how far must the winning post be
    so that A and B might reach it at the same time?

    OA : 200
    Let the length be L
    so time taken is same
    L/A = (L-80)/(3A/5)
    3L = 5L-400
    L= 200 m

    Q7) A, B and C are three participants in a kilometre race. If A can give B a start of 40 metres and B can give C a start of 25 metres, how many metres A can give C a start ?

    OA: 64 m
    Solution:
    A/B = 1000/960 = 25/24
    B/C = 1000/975 = 40/39
    A/C = 25/24*40/39 = 1000/x (where X is distance by C in race with A)
    x= 936. so start = 1000-936 = 64 m

    Q8) P runs a kilometre in 4 minutes and Q in 4 minutes 10 seconds. How many metres start can P give Q in a kilometre race so that the race may end in a dead heat ?
    (Dead heat means a draw)

    Q should end in 4 minutes.
    so 1000/250 =4/1= (1000-d)/240 so d=40. Hence, 40 metres.

    As is clear, basically only speed-distance proportionality applied in different ways can be used to attempt questions. Hope it was a useful fundamental session


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