Question

Suppose Tom has 24 coins totaling $4.35. If he has only dimes and quarters, how many of each type does he have?

Answer 1

Tom has `13` quarters and `11` dimes.

Explanation:

Convert this word problem into numbers. Let `d` be dimes and `q` be quarters. We know that he has `24` coins made up of only dimes and quarters. That means:

`d+q=24`

Also, we know that a dime is worth `10` cents and a quarter is worth `25` cents. We also know that their total value is `4.35`. That means:

`.1d+.25q=4.35`

So we have our two equations:

`d+q=24`
`.1d+.25q=4.35`

Now we have to solve these equations. You could do it either by substitution or elimination. I like elimination. Let's get rid of dimes. Multiply the bottom equation by `10`. Our two equations are now:

`d+q=24`
`d+2.5q=43.5`

Now subtract the two equations from each other. You're left with:

`-1.5q=-19.5`

Solve for `q`:

`q=13`

Now plug this back into one of the original equations of your choice. I think plugging into the first one will be easier as I can avoid decimals:

`d+13=24`

Solve for `d`

`d=11`

This means that Tom has `13` quarters and `11` dimes.

Answer 2

If you want to do substitution instead:

Explanation:

Recall our equations are:

`d+q=24`
`.1d+.25q=4.35`

Solve for either `d` or `q` in the first equation. I will choose to solve for `d`:

`d=24-q`

Plug this into the other equation:

`.1(24-q)+.25q=4.35`

Simplify:

`2.4-.1q+.25q=4.35`

Subtract `2.4` on both sides:

`-.1q+.25q=4.35-2.4`

Combine like terms on both sides:

`.15q=1.95`

Solve for `q`:

`q=13`

Plug into an original equation. I will plug into the first one:

`d+13=24`

Solve for `d`:

`d=11`

So Tom has `13` quarters and `11` dimes.