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A pre-algebra coin word problem

A girl saved nickels, dimes and quarters in a jar. She had as many quarters as dimes, but twice as many nickles as dimes. If the jar has 844 coins, how much money had she saved?
I am not sure how to put this into an equation or how to explain it. I already know that the quarters = 25 cents, etc. 
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1 Answer

First, we know that she has 844 coins in all, we know that she has the same amount of quarters as she does dimes, and she has twice the amount of nickels as she has dimes. The value of the coins does not matter right now, that comes later. We can start with this general equation:
n + d + q = 844
where the letters stand for what they obviously stand for.
Now, we can start to equate some things. We know that she has the same amount of quarters and dimes. Thus, we can produce the equality:
and I have added x here just to make the math clearer. X, here, will represent the number of both quarters and dimes.
Moving on, we know that she has twice as many nickels as dimes, and since the number of dimes is equal to x, then:
n = 2x
Now we can substitute the original terms in our first equation:
  n + d + q = 844
would turn into:
2x + x + x = 844
We can then solve for x like this:
2x + x + x = 844
4x = 844
x = 211
Since x equals the number of dimes, then she has 211 dimes. And since she has as many dimes as quarters, then she has 211 quarters as well. And since she has twice as many nickels as she does dimes, the she has 221 * 2 = 422 nickels. If you want to check yourself, then you can add the number of coins up and see that 422 + 211 + 211 = 844 coins.
Now we can find the amount of money she has. We can multiply the value of the coins by the number of coins:
Dimes:    211 * $0.10 = $21.10
Quarters: 211 * $0.25 = $52.75
Nickels:   422 * $0.05 = $21.10
Adding all of those values up, we get $94.95


Do they solve equations in one variable in pre-algebra classes? I answered this question using a simple ratio, division, and multiplication.