SOLUTION: Hi Josh had the same number of 20c 50c and $1 coins at first. After using 59 20c coins some 50c coins and some $1 coins, he had 88 coins left. There were 3 times as many 50c coins

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: Hi Josh had the same number of 20c 50c and $1 coins at first. After using 59 20c coins some 50c coins and some $1 coins, he had 88 coins left. There were 3 times as many 50c coins      Log On

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Question 1175513: Hi
Josh had the same number of 20c 50c and $1 coins at first. After using 59 20c coins some 50c coins and some $1 coins, he had 88 coins left. There were 3 times as many 50c coins as $1 coins left. There were also 31 more 50c coins than 20c coins left.
How many $1 coins did Josh use.
Thanks

Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


He started with the same number x of each coin.

He used 59 20c coins, so the number of 20c coins he had left was x-59.

He was left with 31 more 50c coins than 20c coins, so the number of 50c coins he had left was (x-59)+31 = x-28.

He was left with 3 times as many 50c coins as $1 coins, so the number of $1 coins he had left was (x-28)/3.

The total number of coins he had left was 88:

%28x-59%29%2B%28x-28%29%2B%28x-28%29%2F3+=+88

2x-87%2B%28x-28%29%2F3+=+88
2x%2B%28x-28%29%2F3+=+175
6x%2Bx-28+=+525
7x+=+553
x+=+79

The number of $1 coins he had left was (x-28)/3 = (79-28)/3 = 51/3 = 17.

So the number of $1 coins he used was 79-17 = 62.

ANSWER: Josh used 62 $1 coins.