Word problem

sams

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Quarters and Half Dollars equal value of $6.25. There are 16 coins. How many of each coin?

I know the answer: 7 quarters, 9 half dollars but getting confused on the equation.
 

stapel

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Pick a variable for the number of quarters, and another variable for the number of half-dollars; say, "q" and "h".

If you have one quarter, how many cents is that? If you have two, how many cents is that? What about three? Four? How about if you have "q" quarters?

Follow this reasoning to find an expression for the value represented by q quarters, and a similar expression for the value represented by h half-dollars.

If you have one quarter, how many of the sixteen coins are half-dollars? What if you have two quarters? Three? Four? How about if you have "q" quarters?

Follow this reasoning to express "h" in terms of "q". Then plug this expression in for "h" in your "value of the half-dollars" expression. This gives you expressions for the values of each coins, all in terms of the one variable.

Sum these two expressions, and set equal to "625", the total value (in cents). Solve.

Eliz.
 

sams

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I'm sorry, you lost me. What would the equation look like?
 

stapel

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Where are you lost? I provided the variables. Are you not from a country that uses these coins? One "quarter" is worth twenty-five cents, and a "half-dollar" is worth fifty cents. Can you make some progress now?

Eliz.
 

sams

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I'm trying to put it into equation format...

What I have tried:

.25x + .50x = 6.25 (doesn't work)
.50x + 2(.25x) = 16 (doesn't work)

or...if x is quarters, y is half dollars

x + y = 16
.25x + .50y = 6.25 ???
 

stapel

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Please review and try following the step-by-step instructions (provided above) for how to do this. Thank you.

Eliz.
 

tkhunny

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sams said:
.25x + .50x = 6.25 (doesn't work)
.50x + 2(.25x) = 16 (doesn't work)
You should not expect this to work.
1) You have provided no definition for 'x'. If you solve for it, what will you have?
2) You have used 'x' for both # of quarters and # of halves. The problem statement does not say the counts are the same.

if x is quarters, y is half dollars

x + y = 16
.25x + .50y = 6.25
This is better, but it remains an insufficient definition. Be very specific and explicit. If you had provided the exact problem statement, a better definition would be possible.

Can you solve this system of two variables in two equations?
 

sams

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To date, I still do not have a solution to this problem other than to do it trial and error. The exact problem states:

"There is a collection of coins of half dollars and quarters. The total number of coins is 16 and the value of the coins is $6.25."

My daughter has several problems like this on her algebra homework. Unfortunately, she was absent when the teacher went over the problems and has yet to ask the teacher for the proper solution....
 

stapel

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sams said:
To date, I still do not have a solution to this problem other than to do it trial and error.
So you haven't yet read the outline of how to solve this algebraically, in the first reply (above) to this thread...?

Eliz.
 

dagr8est

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sams said:
x + y = 16
.25x + .50y = 6.25 ???
You have the right system of equations there.

x+y = 16

0.25x+0.50y = 6.25
2(0.25x+0.50y) = 2(6.25)
0.5x+y = 12.5

Solve it by substitution or subtraction.
 

sams

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Thank you Dagr8est...your solution makes total sense. It's exactly what I was looking for. Aloha!
 
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