Question

David has 83 coins in nickels and dimes. He has a total 6.95. How many of each coin does he have?

Answer 1

David had `27` nickels and `56` dimes

Explanation:

Solving problems about "mixture of coins" depends on knowing two different facts:

1) The NUMBER of each kind of coin
2) The monetary VALUE of each kind of coin

Find a way to express the NUMBER of each kind of coin

There are `83` coins in all
Let `x` represent the number of nickels
Then all the rest of the coins are dimes

Nickels . . . . . . . . . . . .. . `x` `larr` number of nickels
All the rest . . . .`(83 - x)` `larr` number of dimes

Find a way to express the monetary VALUE of the coins

`x` nickels @ `5ȼ` ea . . . . . . . . . .. . `5x` `larr` value of nickels
`(83-x`) dimes @ `10ȼ` ea . . . . `10(83-x)` `larr` value of dimes

Write the equation
[value of nickels] + [value of dimes] = `695ȼ`
[ . . . . . . `5x` . . . . .] + [. `10(83 - x)`. . ] = `695`

`5x + 10(83 - x) = 695`
Solve for `x`, already defined as "the number of nickels"

1) Clear the parentheses by distributing the `10`
`5x + 830 - 10x = 695`

2) Combine like terms
`-5x + 830 = 695`

3) Subtract `830` from both sides to isolate the `-5x` term
`- 5x = - 135`

4) Divide both sides by `-5` to isolate `x`, already defined as "the number of nickels"
`x = 27` `larr` answer for "the number of nickels"

Out of `83` coins, `27` are nickels.

`83 - 27` ("all the rest") are dimes
`color(white)(,,)` `56` are dimes `larr` answer for "the number of dimes"

Answer:
David had `27` nickels and `56` dimes

Check
`27` nickels @ `5ȼ` ea . . . . . . ,`$1.35`
`56` dimes @ `10ȼ` ea . . . . . . `$5.60`
`color(white)`――――――――――――――――
`83` coins . . . . . . . . . . . . . . . , `$6.95`

`Check`