Linear Races  Swetabh Kumar

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 1000100=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 10008000/9 = 1000/9 to C. Hence, 1000/9Q2) 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 = 600472.5 = 127.5 mQ3) 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 kmphQ4) 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 9080=10 pointsQ5) 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/(P15)
B/C = P/(P12)
Multiplying above: A/C = P^2/ (P12)(P15)
BUT, A/C = P/(P20) given
so P^2/ (P12)(P15) = P/(P20)
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 = (L80)/(3A/5)
3L = 5L400
L= 200 mQ7) 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 = 1000936 = 64 mQ8) 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= (1000d)/240 so d=40. Hence, 40 metres.As is clear, basically only speeddistance proportionality applied in different ways can be used to attempt questions. Hope it was a useful fundamental session