Question 50238This question is from textbook Intermediate algebra
: I'm working on this coin problem and I can't seem to get the formula right for the equation. Here is what the problem gives as information:
A box currently contains 44 coins, consisting of pennies and dimes and quarters. the number of pennies is equal to the number of dimes, and the total value is $4.37. How may of each denominatin of coin does the box contain?
They have a table:
# of coins|deommination | value
x | .01 | .01x
_______________________________
x | |
_______________________________
| 0.25 |
|4.37 <-total
How do I formulate the equation from this world problem? (one i get the equation i can solve it but getting the equation is tricky).
thanks
This question is from textbook Intermediate algebra
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A box currently contains 44 coins, consisting of pennies and dimes and quarters. the number of pennies is equal to the number of dimes, and the total value is $4.37. How may of each denominatin of coin does the box contain?
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Let the number of pennies be "x".
The value of those pennies is x cents.
the number of dimes is also "x".
The value of those dimes is 10x cents
The number of quarters is "44-x-x"=44-2x
The value of those quarters is 25(44-2x)=1100-50x cents
EQUATION:
value + value + value = 437 cents
x + 10x + 1100-50x = 437
-39x = - 663
x=17 ( number of pennies and number of dimes)
44-2y=44=34=10 ( number of quarters)
Cheers,
Stan H.
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