Question 1161056: A vending machine contains nickels, quarters, and dimes with a total value of $83.75. There are twice as many dimes as nickels, and 50 more quarters than dimes. How many of each coin are there?
Found 4 solutions by ankor@dixie-net.com, josgarithmetic, ikleyn, MathTherapy:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A vending machine contains nickels, quarters, and dimes with a total value of $83.75.
.05n + .1d + .25q = 83.75
There are twice as many dimes as nickels,
d = 2n
or divide by2
n = .5d
and 50 more quarters than dimes.
q = d+50
:
How many of each coin are there?
replace n and q in the first equation
.05(.5d) + .1d + .25(d+50) = 83.75
.025d + .1d + .25d + 12.5 = 83.75
.375d = 83.75 - 12.5
..375d = 71.25
d = 71.25/.375
d = 190 dimes
then
n = .5(190)
n = 95 nickels
and
q = 190 + 50
q = 240 quarters
:
:
Check this with you calc
.05(95) + .1(190) + .25(240) = 83.75
Answer by josgarithmetic(39630)  (Show Source):
Answer by ikleyn(52903)  (Show Source):
You can put this solution on YOUR website! .
It can be solved M E N T A L L Y.
Take off mentally 50 quarters, for a minute.
Then you will have 0.25*50 = 12.50 dollars less than $83.75, i.e. 71.25 dollars in total, comprising of
x nickels, 2x dimes and 2x quarters.
It means that you can group these coins in x sets, containing one nickel, two dimes and two quarters, in each set.
Each set is worth 5 + 2*10 + 2*25 = 75 cents.
The number of these sets is, obviously, the quotient of dividing 7125 cents by 75 cents.
It is the only place to use your calculator  = 95.
So you have 95 such sets.
ANSWER. 95 nickels; 2*95 = 190 dimes; and 190 + 50 = 240 quarters.
CHECK. 5*95 + 10*190 + 240*25 = 8375 cents = $83.75, in total. ! Precisely correct !
Solved.
Answer by MathTherapy(10557)  (Show Source):
You can put this solution on YOUR website!
A vending machine contains nickels, quarters, and dimes with a total value of $83.75. There are twice as many dimes as nickels, and 50 more quarters than dimes. How many of each coin are there?
Let number of nickels be N
Then number of dimes and quarters are, 2N, and 2N + 50, respectively
Then we get: .05N + .1(2N) + .25(2N + 50) = 83.75
.05N + .2N + .5N + 12.5 = 83.75
.75N = 71.25
Number of nickels, or 
Do you think you can now find the number of dimes and quarters?
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