SOLUTION: Collin has $1.27 in quarters, nickels, and pennies. He has twice as many quarters as pennies. He has 11 coins total. How many of each coin does he have total?

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Question 998159: Collin has $1.27 in quarters, nickels, and pennies. He has twice as many quarters as pennies. He has 11 coins total. How many of each coin does he have total?
Found 2 solutions by CubeyThePenguin, greenestamps:
Answer by CubeyThePenguin(3113) (Show Source):
You can put this solution on YOUR website!
p = number of pennies
n = number of nickels
q = number of quarters

p + 5n + 25q = 127
q = 2p
p + n + q = 11

Substitute q = 2p into the first and third equations.

p + 5n + 25(2p) = 127 ---> 51p + 5n = 127 (eqn 1)
p + n + 2p = 11 ---> 3p + n = 11 (eqn 2)

Multiply the second equation by 5.

51p + 5n = 127
-(15p + 5n = 55)
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36p = 72 ----> p = 2

Collin has 2 pennies, 5 nickels, and 4 quarters.

Answer by greenestamps(13203) (Show Source):
You can put this solution on YOUR website!

Solve the problem using logical reasoning and simple mental arithmetic.

(1) A total of 127 cents with 11 coins means there must be 2 pennies.

(2) Then, since the number of quarters is twice the number of pennies, the number of quarters is 4.
(3) That makes 6 coins with a total value of $1.02.
(4) The remaining 25 cents is made up of 5 nickels.

ANSWER: 2 pennies, 4 quarters, 5 nickels.

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If you need a solution using formal algebra, set the problem up using a single variable -- not three different variables for the numbers of pennies, nickels, and quarters.

Let x = # of pennies
Then 2x = # of quarters
Then 11-3x = # of nickels.

The total value of the coins is 127 cents.

ANSWER:
# of pennies: x = 2
# of quarters: 2x = 4
# of nickels: 11-3x = 5