SOLUTION: Ally has a coin jar that contains nickels and dimes. The number of dimes is four more than twice the number of nickels. If she has $13.40, how many of each coin does she have?

Question 1165655: Ally has a coin jar that contains nickels and dimes. The number of dimes is four more than twice the number of nickels. If she has $13.40, how many of each coin does she have?
Found 2 solutions by Alan3354, greenestamps:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Ally has a coin jar that contains nickels and dimes. The number of dimes is four more than twice the number of nickels. If she has $13.40, how many of each coin does she have?
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d = 2n + 4
10d + 5n = 1340
etc
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

If a formal algebraic solution is not required, you can get some good mental exercise by solving this using logical reasoning and some mental arithmetic.

(1) Take the 4 "extra" dimes and set them aside; then arrange the remaining coins into groups of 2 dimes and 1 nickel.

(2) The value of the 4 dimes is $0.40; the value of all the coins is $13.40. So the value of the groups of dimes and nickels is $13.00.

(3) The value of 2 dimes and 1 nickel is $0.25, or 25 cents. The number of those groups required to make $13.00, or 1300 cents, is 1300/25 = 52.

(4) So there are 52 groups each containing 2 dimes and 1 nickel, plus the other 4 dimes. That makes the number of dimes 2(52)+4 = 104+4 = 108 and the number of nickels 52.

ANSWER: 108 dimes, 52 nickels.

CHECK: 108(10)+52(5) = 1080+260 = 1340

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