Question

Please help! Medal to best answer!!

A collection of nickles, dimes, and quarters are worth $3.40. There are twenty-five coins in all. There are four times as many nickles as there are dimes. How many of the coins are quarters?

a 3
b 5
c 10
d 12

Answer 1

I would do that with \(q,d,n\) as variables. The second equation needs to have one more variable in it, too.

Answer 2

or one more equation

Answer 3

or you can just substitute and solve for z.

Answer 4

4x+y = 25 is not a correct equation, though - there are quarters, nickels and dimes making up the 25 coins

Answer 5

I'm so confused. :/

Answer 6

I'll let @jhannybean fix what they've done. if you're still confused after that, I'll try...

Answer 7

Okay.

Answer 8

Okay, here's how I would do it:

Let the number of nickels, dimes, and quarters be \(n, d, q\) respectively. We know we have 25 coins, so that gives us:
\[n+d+q =25\]
We also know that the value of all the coins together is $3.40 or 340 cents. We'll do the problem in cents to avoid fractions.
\[5n + 10d + 25q = 340\]The number of each type of coin times the value of that type of coin gives us the value of the coins, and adding up the three types must give us the total value.

Now, we know that there are 4x as many nickels as dimes, so we can write that as n = 4d, right?

Answer 9

Yes

Answer 10

If we substitute that into our previous equations, we get:
\[4d + d + q = 25\]\[5(4d)+10d+25q=340\]and simplifying:
\[5d+q=25\]\[30d+25q=340\]

Can you solve those two equations?

Answer 11

Is it 10 quarters?

Answer 12

10 quarters, yes. 3 dimes, 12 nickels

Answer 13

Thank you.

Answer 14

Thank you for clarifying the nickels and quarters thing, it threw me off.