2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ If you substitute $n=30-5q$ into $5n+65q=260$ you get $150-25q+65q=260$. That means $40q=110$ or $q=11/4$. Therefore there is no solution to the problem, it is impossible. $\endgroup$ – Gregory Grant Aug 13 '16 at 0:05
  • $\begingroup$ Perhaps it's a typo and they meant $\$2.30$, which works out to $20$ nickels, $2$ quarters and $8$ dimes. $\endgroup$ – Gregory Grant Aug 13 '16 at 0:12
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.