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In a game of skill A can give B 10 points in 100, and B can give C 15 in 75 and D 9 in 90; how many points could A and D give B and C in a game to 362
to make an even match?
Solution: Suggested Solution (Amy Hutt & Calista Shelkin, WWU students)
This is a situation (apparently) of handicap. It means: for every 100 points A earns, B earns 90; for every 75 points B earns, C earns 60; and, for every 90 points B earns, D earns 81.
That is, A/B = 100/90 and B/C = 75/60 and B/D = 90/81
Or, 90A = 100B and 60B = 75C and 81B = 90D
Or, (90A)/100 = B and (60B)/75 = C and (81B)/90 = D
D + A = [(90A)/100][81/90] + A = 362
Thus, A = 200 pts, B = 180 pts, C = 144 pts, D = 162 pts, and B + C = 324.
For every 362 points A and D earn, B and C earn 324 pts. Or, A and D can give B and C 38 points to make an even match.
Source: Charles Pendlebury's Arithmetic (London: G. Bell & Son, 1918)
