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

2 Answers
May 15, 2017

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.

May 15, 2017

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.