Question 534625: Barbara has 20 coins consisting of nickels and dimes. If the nickels were dimes and the dimes were nickels, she would have 30 cents more than she has now. How many dimes did she have to begin with?
Answer by Edwin McCravy(20055)   (Show Source):
You can put this solution on YOUR website! Barbara has 20 coins consisting of nickels and dimes. If the nickels were dimes and the dimes were nickels, she would have 30 cents more than she has now. How many dimes did she have to begin with?
Let N = the number of nickels
Let D = the number of dimes
$.05N + $.10D = value of the coins.
$.10N + $.05D = value the coins would be if the nickels were dimes and the dimes were nickels.
>>...Barbara has 20 coins...<<
N + D = 20
>>...If the nickels were dimes and the dimes were nickels,
she would have 30 cents more than she has now...<<
Value of coins = Value coins would be + $.30
$.10N + $.05D = $.10D + $.05N + $.30
So you have this system:
N + D = 20
$.10N + $.05D = $.10D + $.05N + $.30
Multiply the 2nd equation through by 100 to remove all the decimals,
and drop the dollar signs. Then collect terms and solve the
system by substitution.
You'll get 13 nickels and 7 dimes, which is $1.35.
If she had 13 dimes and 7 nickels instead, that would be $1.65,
which is 30 cents more.
Edwin
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