Question

Kenny has nickels and dimes. He has $3.80 made from 44 coins. How many dimes are there?

Answer 1

There are `32` dimes and `12` nickels

Explanation:

We can set up a system of equations to solve this problem

Begin by assigning variables

nickels`=n`
dimes `=d`

So `n+d=44`

Nickels are worth 5 cents `=5n`
Dimes are worth 10 cents `=10d`
`$3.80 = 380 cents`

`5n + 10d = 380`

The system becomes

`n+d = 44`
`5n+10d=380`

Rearrange the first equation to isolate a variable
d = 44 -n

Now plug the first equation value into the second equation for `d`

`5n+10(44-n) = 380`

Use the distributive property
`5n+440-10n =380`

Combine like terms
`440-5n =380`

Use additive inverse to isolate the variable term
`cancel440 -5n cancel(-440) = 380-440`
`-5n = -60`

Use multiplicative inverse to isolate the variable
`cancel((-5)n)/cancel(-5) = (-60)/-5`

`n=12`

`d=44-n`
`d=44-12`
`d=32`

Answer 2

Just for fun! A different approach using ratio. Once you get used to how this works it is very fast.

Count of dimes `=32`
Count of nickels `=44-32=12`

Explanation:

10 dimes = $1
20 nickels = $1

Total coin count =44

Target value $3.80
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using the straight line graph approach and counting only dimes. Choose dimes as they give the highest value for 44 coins.

`color(blue)("If all 44 dimes the value is "$44/10 =$4.40)larr" 44 dimes"`

`color(red)("If all 44 nickels the value is "$44/20=$2.20) larr" 0 dimes"`

The slope (gradient) for part is the same as the slope for all of it

Let the count if dimes be `d`

Then the count of nickels is `44-d`

`($4.40-$2.20)/44=($3.80-$2.20)/d`

`d=(44(3.8-2.2))/(4.4-2.2) =32`