Question 1175513
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He started with the same number x of each coin.<br>
He used 59 20c coins, so the number of 20c coins he had left was x-59.<br>
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.<br>
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.<br>
The total number of coins he had left was 88:<br>
{{{(x-59)+(x-28)+(x-28)/3 = 88}}}<br>
{{{2x-87+(x-28)/3 = 88}}}
{{{2x+(x-28)/3 = 175}}}
{{{6x+x-28 = 525}}}
{{{7x = 553}}}
{{{x = 79}}}<br>
The number of $1 coins he had left was (x-28)/3 = (79-28)/3 = 51/3 = 17.<br>
So the number of $1 coins he used was 79-17 = 62.<br>
ANSWER: Josh used 62 $1 coins.<br>