I have been stuck on this problem for 30+ minutes and I can't seem to get the correct answer; there must be something that I am missing/doing wrong!!
- You own $19.75 in dimes and quarters
- There are 100 coins in all
- How many dimes are there?
My variables:
$d$ = dimes, $q$ = quarters
My equations:
$$19.75 = 10d + 25q$$
$$100 = d + q$$
I have done problems like this before, system of equations, but I cannot, for some unknown reason (to me), find the correct answer.
It always comes out with like 3 decimal places, and numbers that are not divisible.
Any help, even hints, would be truly appreciated!
Thanks!
What I have tried:
$-10(100) = (d + q)-10 [=] -1000 = -10d - 10q$
$19.75 = 10d + 25q$
$+$
$-1000 = -10d - 10q$
If all 100 of your coins were quarters, you'd have \$25.00. That's too much; you're only supposed to have \$19.75, which is \$5.25 less. So you need to replace some of those 100 quarters with dimes. Every time you replace a quarter with a dime, the amount of money decreases by \$0.15. How many of these \$0.15 decreases are needed in order to get a total decrease of \$5.25? Well, 5.25 divided by 0.15 is 35, so you'll need 35 quarter-to-dime replacements. The result will be 65 quarters and 35 dimes. (Since I don't trust my ability to do arithmetic, I should check the answer: $(65\times0.25)+ (35\times0.10)=16.25+3.50=19.75$.)