This has to deal with money.

Question

A pile of coins contained nickels, dimes, and quarters. There were 4 times as many dimes as nickels and twice as many quarters as dimes. If the value of the pile is $4.90, how many quarters are in the pile? (Please simplify your answer if you respond.)

Answer 1

Since this is an algebra course, I imagine the teacher wants an equation. But in real life there are often faster ways to figure things out. My specialty is SAT/ACT prep, and any way you can get ot the correct answer fast is the best way.

Notice the problem specifies that there is at least one nickel. Let's try 1 nickel (05) for our answer. Then there will be 4 dimes (.40) and 8 quarters (2.00). That adds up to 2.45. We notice that this is exactly half of 4.90, so all we have to do is start with 2 nickels and double everything. There will be 16 quarters.

Answer 2

Hi Arlazza,

This problem requires solving systems of equations.

Let's break down this problem into several parts.

First let's represent nickels, dimes, and quarters into their own variables.

Number of nickels as n, Number of dimes as d, and Number of quarters as q

Since there are 4 times as many dimes as nickels, for every nickel there are 4 dimes: 4n= d

Since there are twice as many quarters as dimes, for ever dime there are 2 quarters: 2d = q

To keep it simple, let's make everything in terms of nickels, n.

Representing dimes in nickels, d=4n.

Representing nickels, it's just n

Representing quarters it's a bit trickier. Since we know 4n=d and 2d=q, we can solve for n in terms of q.

So since we know 4n=d (for every nickel there are 4dimes), it is also true that 8n=2d (for every 2 nickels

there are 4 dimes).

So since we now know 8n=2d and 2d=q, we can say q=8n (for every nickel there are 8 quarters), since

they are both equal to the same thing which is (2d)

So representing quarters in nickels, q=8n.

Since the total value of the pile is $4.90, we now have 2 things, the value of each coin (nickels are 5 cents, dimes are 10 cents and quarters are 25 cents) AND we have the number of coins in terms of the number of nickels.

So in the beginning we would have .05n+.1d+.25q=4.90.

But now after making all the coins in terms of number of nickels we have .05n+.1(4n)+.25(8n)=4.90

Simplifying this down we have

.05n+.1(4n)+.25(8n)=4.90

.05n+.4n+2n=4.90

2.45n=4.90

n=2

So now we have figured out how many nickels there are, 2 nickels!

Using this we can solve for the number of dimes, and quarters.

Since we know there are 4 times as many dimes are nickels (d=4n), we now know there are 8 dimes (d=8)

Since we know there are twice as many quarters as dimes (q=2d), we now know there are 16 quarters (q=16)

So overall, there are 2 nickels, 8 dimes, and 16 quarters.

Hope this helped!

Devin