Question

Josh has 12 coins, he has four times as many dimes as quarters the total value is 1.00. How many dimes nickles and quarters does he have?

Answer 1

You have to solve this using a system of equations, one that deals with the NUMBER of coins and another that deals with the value of the coins. He has 12 coins total, 4 times as many dimes as quarters. That will be the first equation we will have to write.

Answer 2

4x=d

Answer 3

and x=q

Answer 4

Not sure where to go from there

Answer 5

@IMStuck

Answer 6

Sorry, I had to tend to something...I'm here. give me a sec, ok?

Answer 7

I'm here, just give me a sec...

Answer 8

ok

Answer 9

anyone?

Answer 10

@IMStuck You back?

Answer 11

There are 12 coins, nickels, dimes and quarters:

n + d + q = 12

There are 4 times as many dimes as quarters:

4q = d

The total values is 100 (just for simplicity's sake, using cents instead of dollars):

5n + 10d +25q = 100

There are three equations in three unknowns. Solve simultaneously.

Answer 12

OMG!!!!! I totally overlooked the fact that there were nickels in this problem, too! No Wonder it wasn't working out! Good God AlmightY!!!!!!!

Answer 13

Did you figure this out, Toasterino? Can you solve those 3 equations simultaneously?

Answer 14

No, the 5n+10d+25q is what is confusing me, if i solve each they will all get 100

Answer 15

Here are your equations, like LarsEighner said, ok:
n+d+q=12 (that is the number of nickels + number of dimes + number of quarters)
4q=d (because there are 4 times as many dimes as quarters)
5n+10d+25q=100 (because a nickel is 5 cents, a dime is 10 cents and a quarter is 25)
Start with the fact that d=4q and fill "4q" in for "d" in our "money equation" like this:

Answer 16

5n+10(4d)+25q=100...5n+40q+25q=100...5n+65q=100. That is our first equation in 2 terms. Now we will do the same with the equation that relates the number of coins to each other, keeping in mind the fact that d=4q:

Answer 17

n+d+q=12...n+4q+q=12...n+5q=12. That's our second equation in terms of two variables. So let's look at those 2 together, now:

Answer 18

From above we have
5n + 65q = 100 and then
n+ 5q = 12.
We need to solve those for either n or q. I picked q, so let's do it like this:

Answer 19

5n+65q=100
-5(n+5q=12)...

5n+65q=100
-5n -25q=-60

The n's cancel each other out (5n - 5n = 0n, right?) so we are left with
40q=40 (65q-25q=40q, and 100-60=40)
40q=40...q=1. So there is 1 quarter.

Answer 20

Going back to one of our equations in two terms, we can sub in 1 for q and solve for n like this:

Answer 21

4q=d...4(1)=d...d=4. So there are 4 dimes.

Answer 22

NOW, going back to our original problem that says n+d+q=12, we can fill in the fact that q=1, and d=4 to solve for n. Like this:

Answer 23

n+q+d=12...n+1+4=12...n+5=12...12-5=n...n=7.

Answer 24

When you fill all that back into our money equation, we have 7(.05)+4(.10)+1(.25)=1.00...
(there are 7 nickels and nickels are 5 cents each; there are 4 dimes and dimes are 10 cents each; there is one quarter and they are 25 cents each)... .35+.40+.25=1.00 See that?

Answer 25

@IMStuck I understand it now, what was confusing was the equation 5n+10d+25q but now I understand that I had to substitute thanks for everything! As well as @LarsEighner for providing the fundamentals

Answer 26

With a simple problem like this you can go it intuitively.

There cannot be 3 quarters because then there would have to be 12 dimes, but that is more than 12 coins total. So there must be 1 or 2 quarters (of there were 0 quarters, you would have 12 nickels = 100, which cannot be). Now if there are 2 quarters, there must be 8 dimes, but that 50 + 80 which is > 100. So there must be 1 quarter. Then you know there are 4 dimes. That is 65 cents, leaving 35 cents in nickels which is 7 nickels. 7 nickels, 4 dimes, 1 quarter = 12 coins..

Answer 27

Yay for all of us, then! ; )