Question 123690: Janelle has $2.81 in coins, all in pennies, nickels, dimes, and quarters. The number of nickels equals the number of dimes. There are 24 coins in all. In how many ways is this possible? List all possibilities. Note: this is not guess and check. You are expected to set up a system of equations and solve.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Janelle has $2.81 in coins, all in pennies, nickels, dimes, and quarters. The number of nickels equals the number of dimes. There are 24 coins in all. In how many ways is this possible? List all possibilities. Note: this is not guess and check. You are expected to set up a system of equations and solve.
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Let x = no. of nickels and number of dimes
Let p = no. of pennies
Let q = no. of quarters
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Total coins equation:
x + x + p + q = 24
2x + p + q = 24
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Value equation (in cents)
5x + 10x + p + 25q = 281
15x + p + 25q = 281
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Subtract the total coin equation from the value equation:
15x + p + 25q = 281
2x + p + q = 24
------------------- subtracting, eliminates p
13x + 0 + 24q = 257
:
13x + 24q = 257
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We can approach it like this:
257 - 13x = 24q
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We know that the value subtracted from 257, is a multiple of 13.
It has to be odd value to give an even remainder, that is a multiply of 24
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x = 3 did not work
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If x = 5: and 5 * 13 = 65
257 - 65 = 192 = 24q
24q = 192
q = 192/24
q = 8 quarters
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Therefore: 5 nickels and 5 dimes
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We have:
5 + 5 + p + 8 = 24
p = 24 - 18
p = 6 pennies
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Answer is: 5 nickels, 5 dimes, 6 pennies, and 8 quarters
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check in the value equation:
5(5) + 10(5) + 6 + 25(8)
25 + 50 + 6 + 200 = 281 cents
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Are there any other combination of values that will work? I don't think so.
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Please let me know, if I am wrong.
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