SOLUTION: Joe has a collection of nickels and dimes that is worth $6.10. If the number of dimes were tripled and the number of nickels were decreased by 49, the total value of the coins woul
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Question 408040: Joe has a collection of nickels and dimes that is worth $6.10. If the number of dimes were tripled and the number of nickels were decreased by 49, the total value of the coins would be $14.05. How many nickels and dimes does he have? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Let n = no. of nickels
Let d = no. of dimes
:
Write an equation for each statement:
:
"Joe has a collection of nickels and dimes that is worth $6.10."
.05n + .10d = 6.10
.05n = 6.10 - .10d
divide by .05
n = (122-2d); use this form for substitution
:
" If the number of dimes were tripled and the number of nickels were decreased
by 49, the total value of the coins would be $14.05."
.05(n-49) + .10(3d) = 14.05
.05n - 2.45 + .3d = 14.05
.05n + .3d = 14.05 + 2.45
.05n + .3d = 16.50
Replace n with (122-2d)
.05(122-2d) + .3d = 16.50
6.10 - .10d + .3d = 16.50
-.10d + .3d = 16.50 - 6.10
.20d = 10.40
d =
d = 52 dimes originally
and
n = 122 - 2d
n = 122 - 2(52)
n = 122 - 104
n = 18 nickels originally
:
:
This satisfies the equations but looking at the statement:
"If the number of dimes were tripled and the number of nickels were decreased
by 49, the total value of the coins would be $14.05."
:
How can you decrease the number of nickels by 49 if you only have 18!!!
:
Plunging ahead though:
.10(3*52) + .05(18-49) =
.10(156) + .05(-31) =
15.60 - 1.55 = 14.05; the math is happy with it, but how about real life?
