2(5c + 2d)(5c + 2d)
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$2, $1, 50c, 20c, 20c.
Well, isn't that a happy little math problem we have here! If Ann has more 50c coins than 10c coins and the total value is $5, we can figure out that she must have 4 10c coins. Let's give Ann a little encouragement as she counts her coins and solves this puzzle.
Tentatively I will say $4.35, with for instance: 19 x 20c 1 x 50c 1 x 5c It's possible that a rigorous method exists, but this is just a fudged guess. Any advances on it?
Indeed. 74 cents can be made with the following set of 7 coins: 50c 10c 10c 1c 1c 1c 1c
2 variables with 2 equations. A simultaneous equation problem. d+ n = 14 10d + 5n = 95 Multiply the first equation by -5 -5d + -5n = -70 10d + 5n = 95 add together 5d = 25 d = 5 ◄ plug that in the first equation to find n 5 + n = 14 n = 9 ◄ check 5 dimes plus 9 nickles is 50c+45c=95c ■