Question 1162226: Sue has $1.60 in dimes and nickels. If she has 7 more dimes than nickels, how many of each coin does she have? Found 3 solutions by ankor@dixie-net.com, greenestamps, Alan3354:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! d = no. of dimes
n = no. of nickels
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Write an equation for each statement
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Sue has $1.60 in dimes and nickels.
.10d + .05n = 1.60
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If she has 7 more dimes than nickels,
d = n + 7
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In the first equation replace d with (n+7)
.10(n+7) + .05n = 1.60
distribute .1
.10n + .70 + .05n = 1.60
.15n = 1.60 - .70
.15n = .90
n = .90/.15
n = 6 nickels
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You can find the no. of dimes. (d = 6 + 7)
Check your solutions in the 1st equation
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how many of each coin does she have?
Solving the problem using two variables is a good exercise in algebraic techniques.
But a solution using a single variable is faster and more efficient.
Let x = # of nickels
Then x+7 = # of dimes
The total value of the coins is $1.60, or 160 cents:
ANSWER: She has x=6 nickels and x+7 = 13 dimes
Note solving the problem informally with logical reasoning is also good mental exercise:
(1) set aside the "extra" 7 dimes, with a total value of 70 cents.
(2) that leaves equal numbers of dimes and nickels, with a total value of 90 cents.
(3) Since the value of one dime and one nickel is 15 cents, the number of each coin she still has is 90/15 = 6.
(4)Now bring back the 7 dimes you counted initially, to find that she has 6 nickels and 6+7=13 dimes.
You can put this solution on YOUR website! Sue has $1.60 in dimes and nickels. If she has 7 more dimes than nickels, how many of each coin does she have?
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Set the "extra" 7 dimes aside.
then she has pairs of nickels & dimes worth 15 cents each and a total of 90 cents.
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90/15 = 6 pairs
---> 6 nickels and 13 dimes
