SOLUTION: Yuki Katanuma has saved $22.65 in nickels, dimes, and quarters. If Yuki has eight nickels fewer than four times the number of dimes and nine quarters more than two-sevenths the num

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Question 1199463: Yuki Katanuma has saved $22.65 in nickels, dimes, and quarters. If Yuki has eight nickels fewer than four times the number of dimes and nine quarters more than two-sevenths the number of dimes, how many coins of each type does Yuki have?
Found 2 solutions by greenestamps, josgarithmetic:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


I'll show two different ways to solve this problem using different setups, to demonstrate how taking the time to think about different possible setups can make the work of solving the problem easier.

A standard setup for the problem, using the information exactly as given, might look like this:

The numbers of nickels and quarters are both given in terms of the number of dimes, so...

let x = # of dimes
then 4x-8 = # of nickels
and (2/7)x+9 = # of quarters

The total value of the coins is $22.65, or 2265 cents:

10%28x%29%2B5%284x-8%29%2B25%28%282%2F7%29x%2B9%29=2265
10x%2B20x-40%2B%2850%2F7%29x%2B225=2265
30x%2B%2850%2F7%29x%2B185=2265
%28260%2F7%29x=2080
%281%2F7%29x=8
x=56

ANSWERS:
dimes: x = 56
nickels: 4x-8 = 216
quarters: (2/7)x+9 = 2*8+9 = 25

CHECK: 56(10)+216(5)+25(25) = 560+1080+625 = 2265

When I read the problem and am interested in solving the problem as easily as possible, I see the phrase "nine more than two-sevenths", and I know I don't want to start by writing an expression using that fraction 2/7.

So I would do this:

let 7x = # of dimes
then 4(7x)-8 = 28x-8 = # of nickels
and 2x+9 = # of quarters

Then

10%287x%29%2B5%2828x-8%29%2B25%282x%2B9%29=2265
70x%2B140x-40%2B50x%2B225=2265
260x%2B185=2265
260x=2080
x=8

ANSWERS:
dimes = 7x = 56
nickels = 28x-8 = 224-8 = 216
quarters = 2x+9 = 16+9 = 25

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Nickels, Dimes, Quarters
$22.65
n,d,q variables

system%285n%2B10d%2B25q=2265%2C4d-n=8%2C%282%2F7%29d%2B9=q%29

system%28n%2B2d%2B5q=453%2C-n%2B4d=8%2C2d%2B63=7q%29

system%28n%2B2d%2B5q=453%2C-n%2B4d=8%2C2d-7q=-63%29

system%28n%2B2d%2B5q=453%2C-n%2B4d=8%2C-2d%2B7q=63%29
If wish to use a system of equations, this is the one to start eliminating something.
n seems like a good choice.
Add E1 and E2.

system%286d%2B5q=461%2C-2d%2B7q=63%29----------two equations, two unknowns

system%286d%2B5q=461%2C-6d%2B21q=189%29
ADD,...
26q=650
q=650%2F26=25
highlight%28q=25%29
-
2d%2B63=7q
2d=7q-63
2d=7%2A25-63
d=%287%2A25-63%29%2F2
highlight%28d=56%29

Use the total money equation to find n.