Question 1097741
<br>First an informal solution, using logical analysis and a little arithmetic....<br>
He has some number of dimes, that same number plus 10 more of nickels, and twice that number of quarters.<br>
So "take away" those 10 "extra" nickels, with a value fo 50 cents.  The remaining coins then have a value of $19.50.<br>
The remaining coins are equal numbers of dimes and nickels, and twice that many quarters.  So make "units" each consisting of 2 quarters, 1 dime, and 1 nickel. The value of each of those units is 65 cents.<br>
The number of those units required to make $19.50 is 19.50/.65 = 30.<br>
Those 30 "units" give you 60 quarters, 30 dimes, and 30 nickels.<br>
Now add back in those "extra" 10 nickels.<br>
You end up with 60 quarters, 30 dimes, and 40 nickels.<br><br>
Now a formal solution using algebra....<br>
let x = number of dimes
then 2x = number of quarters
and x+10 = number of nickels<br>
Since the total value is $20, or 2000 cents,<br>
{{{10(x)+25(2x)+5(10+x) = 2000}}}
{{{10x + 50x + 50 + 5x = 2000}}}
{{{65x + 50 = 2000}}}
{{{65x = 1950}}}
{{{x = 30}}}<br>
# of dimes: x = 30
# of quarters: 2x = 60
# of nickels: x+10 = 40