APPLIED ALGEBRA!

Auckland Star, Volume LXIII, Issue 67, 19 March 1932, Page 2 (Supplement)

APPLIED ALGEBRA!

Dear Peter Pan—l thought you would bo interested in this old, but amusing, puzzle, because it makes a tricky, but genuine, problem: —

"There was a rope hanging over a pulley with u weight on one end and a monkey of equal weight 011 the other end. The rope weighed four ounces per foot; the age of the monkey and the age of its mother were together equal to four years; the weight of the monkey was as many pounds as its mother was years old; the mother was twice as old as the monkey was when the mother was half as old as the monkey will be when the monkey is three times as old as the mother was when the mother was three times as old as the monkey. The weight of the weight and the weight of the rope was half as much again -as the difference between the weight of the weight and the weight of the weight and the weight of the monkey. What was the length of the rope?" Answer: When the mother is three times as old as the monkey let their respective ages be 3x and x. '1 lilis the difference of their ages is 2x (constant). When the monkey is three times as old as the mother then was, he will be When the mother is half this age, she will be 4ijx, and the monkey, therefore, 2ix. Twice this, i.e., sx, is the mother's age at the time of the problem, and the monkey's ago is therefore 3x. Now, 5x plus 3x equals 4, therefore x equals and ox equals Ihciefore weight of monkey equals weight of weight, equals 2Alb, equals 40oz. And we are told (expressing all weights in ounces) 40 plus weight of Tope equals 1$ (40 plus 40 minus 40). Therefore weight of rope .equals 00 minus equals 20oz. Therefore length of rope equals sft. Kind regards. Yours sincerely, Jack Tunnicliffc. Hukapapa.