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 Post subject: word problemPosted: Fri, 8 Oct 2010 21:19:59 UTC
 S.O.S. Newbie

Joined: Fri, 8 Oct 2010 02:58:05 UTC
Posts: 2
Location: NH
Can someone tell me where I'm going wrong with this equation? The Question is:

A coin bank contains 25 coins in nickels, dimes, and quarters. There are four times as many dimes as quarters. The value of the coins is \$2.05. How many dimes are in the bank

For quarters I use the variable x, and get 25x, for dimes I use the equation 10(4x-25). For nickels I don't substitute anything because it seems unneeded for the equation, am I suppose to add a third variable?

So I get: 25x+40x-250=205
65x=455
x=7

I substitute 7 for the dimes equation and get 3 dimes in total which I know is wrong. Thanks for any help!

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 Post subject: Posted: Fri, 8 Oct 2010 22:06:55 UTC
 S.O.S. Oldtimer

Joined: Fri, 10 Oct 2008 15:35:23 UTC
Posts: 179
Location: Clarksville, ARkansas
You chose "x" to be the number of quarters in the bank. 25x is the value of the quarters in cents.

The number of dimes is 4 times the number of quarters, so that means 4x is the number of dimes in the bank. 10 times 4x is the value of the dimes in cents.

The total number of coins is 25, so the number of nickels must be 25 minus the number of dimes & quarters, that is: 25 - (4x + x) = 25 - 5x. Each nickel is worth 5 cents, so the value of the nickels is: 5 times (25 - 5x).

The value of all the coins adds up to 205 cents. Solving for x gives the number of quarters. Find the number of dimes once you know the number of quarters.

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