Question 1108361
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If all 79 coins were dimes, the total value would be $7.90; but it is only $6.00.  With all dimes, the total value is $1.90 too much.<br>
Each time we replace a dime with a nickel, the number of coins stays the same but the total value goes down by 5 cents.<br>
To get the total down to the required $6.00, the number of times we need to replace a dime with a nickel is {{{1.90/.05 = 190/5 = 38}}}.<br>
That means we end up with 38 nickels; the number of dimes we have left is 79-38 = 41.<br>
Answer: 41 dimes, 38 nickels.<br>
Algebraically....<br>
Let d = number of dimes
Let n = number of nickels
Then
{{{d+n = 79}}}  the total number of coins is 79
{{{10d+5n = 600}}}  the total value of the coins (in cents) is 600<br>
Multiply the first equation by 5 and subtract from the second equation to eliminate n:
{{{10d+5n = 600}}}
{{{5d+5n = 395}}}
{{{5d = 205}}}
{{{d = 205/5 = 41}}}<br>
The number of dimes is 41; then the number of nickels is 79-41 = 38.