SOLUTION: A man has 35 coins consisting of dimes and quarters. The total value is $5.15. How many of each coin does he have? I know that there are 24 dimes and 11 quarters (working it bac
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Question 234633: A man has 35 coins consisting of dimes and quarters. The total value is $5.15. How many of each coin does he have? I know that there are 24 dimes and 11 quarters (working it backwards) but I can't figure out what the equation should be to solve for this problem. Can ANYONE please help me?
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A man has 35 coins consisting of dimes and quarters. The total value is $5.15. How many of each coin does he have?
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Quantity Equation: d + q = 35
Value Equation:::10d+ 25q = 5l5 cents
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Multiply thru the Quantity Equation by 10 to get:
10d + 10q = 350
---
Subtract that from the Value Equation and solve for "q":
15q = 165
q = 11 (# of quarters)
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Since d+q = 35, d+11 = 35 ; d = 24 (# of dimes)
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Cheers,
Stan H.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Let
represent the number of dimes. Let
represent the number of quarters. Each dime is worth 10 cents, so the value of the dimes, in cents is
. Likewise, the value of the quarters is
. And the total value of all the coins, in cents, is 515.
Now we can write an equation that describes the number of coins:
And another equation that describes the value of the coins:
Giving you a system of two equations in two variables. Solve for
and
. Since you have unity coefficients on the first equation, I would use the substitution method to solve this system.
Which leads me to the description of a shortcut method for solving coin and different value ticket problems. Instead of defining a second variable as above, i.e. "Let
represent the number of quarters" you can realize up front that there are 35 coins, so if
represents the number of dimes,
must represent the number of quarters. Then you can just write the single equation for the value thus:
Which is exactly the same place you would be if you solved the first equation for
and then substituted into the second equation.
John

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