Coin word problem

Question

A bowl contains $30$ coins consisting of nickels, dimes, and quarters. The number of dimes is $4$ times the number of quarters. If the total value of the coins is $2.60$ dollar how many coins of each type are there?

I'm tutoring someone and I'm not quite sure why I keep getting this problem wrong. I set it up like this

$n = $number of nickels

$5n = $value of nickels

$d = $number of dimes

$10d= $value of dimes

$q = $number of quarters

$25q = $value of quarters

$d = 4q$

Then I set up two equations and had

$5n+10d +25q= 260$

which I rewrote as: $5n+65q=260 \tag1$

And then I had $n+q+d=30$

which I rewrote as $n+5q=30\tag 2$

Then I solved the system of equations and got $q$ to be approximately $2.75$ . I know this is wrong because it has to represent the number of quarters. I then rounded it off to 3 quarters which would mean I have $15$ nickels and $12$ dimes but with that number the total comes out to $2.70$ dollars. What am I doing wrong?

Answer 1

It's easy to see this is impossible even if you don't trust your linear algebra. You can exhaust all posibilities:

If there's one quarter there must be four dimes and $25$ nickels. That adds to $\$2.90$.

If there's two quarters there must be eight dimes and $20$ nickels. That adds to $\$2.30$.

If there's three quarters there must be $12$ dimes and $15$ nickels. That adds to $\$2.70$.

If there's four quarters there must be $16$ dimes and $10$ nickels. That adds to $\$3.10$.

If there's five quarters there must be $20$ dimes and $5$ nickels. That adds to $\$3.50$.

If there's six quarters there must be $24$ dimes and $0$ nickels. That adds to $\$3.90$.

If there's seven or more quarters there must be $28$ or more dimes which goes over $30$ coins. So the above list is exhaustive.