View Full Version : COINS~word problem

emmaiskool242

05-17-2006, 07:23 PM

Marvella has 23 coins in nickels,dimes, and quarters. If she has twice as many nickels as dimes and has $2.45, how many of each does she have?

Well I tried like three times to set up this problem and I couldn't figure out exactly how to do it, so if you could show me how to set it up I could probaly figure it out~thanks

tkhunny

05-17-2006, 07:41 PM

What did you try? Let's see how close you get.

emmaiskool242

05-17-2006, 07:53 PM

So I tried setting it up like 2n=d, d+n+q=23 so then you would put....2n+n+q=23,3n+q=23,but how you would solve that?

So then I tried doing this~2(0.05n)+0.05n+0.25q=$2.45. But that doesn't make much sense either.

So then I figure to solve it, I think it has to be some combination of the both...

So thats what I did...neither make much sense but I tried

Could you help me?

\L

\begin{array}{r}

n + d + q = ? \\

.05n + .10d + .25q = ? \\

n = 2d \\

\end{array}

Denis

05-17-2006, 11:51 PM

"So I tried setting it up like 2n=d, d+n+q=23 so then you would put....2n+n+q=23,3n+q=23,but how you would solve that?"

Well start by being CAREFUL: n = 2d ; NOT 2n = d;

that'll result in 3d + q = 23, so q = 23 - 3d [1]

"So then I tried doing this~2(0.05n)+0.05n+0.25q=$2.45. But that doesn't make much sense either."

you have 2d nickels, so .05(2d)

you have d dimes, so .10(d)

you have q quartwes, so .25(q)

.05(2d) + .10(d) + .25(q) = 2.45

.20d + .25q = 2.45 ; make it easier by multiplying by 100:

20d + 25q = 245 ; simplify by dividing by 5:

4d + 5q = 49 [2]

Now substitute [1] in [2] and solve for d

emmaiskool242

05-18-2006, 08:02 PM

wow thanks for the help i'm going to try that =)

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