Question 928114: Please help me solve this: A man has 52 coins, all of which are dimes, quarters, and nickels. The total value of his change is $8.20. If the number of quarters is twice the number of nickels, how many of each does he have?
So far I have two equations set up: x+2y+z=52 and 0.10x +0.25y +0.05z=8.20
but I don't think those are right.
Found 2 solutions by rothauserc, Alan3354:
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! let x be the number of nickels and y be the number dimes, then we have
2x +x +y = 52
.25(2x) + .05x + .10y = 8.20
solve first equation for y
y = 52 -3x
substitute for y in second equation
.25(2x) + .05x + .10(52-3x) = 8.20
.50x + .05x + 5.20 - .30x = 8.20
.25x = 3.00
x = 12
therefore,
there are 24 quarters, 12 nickels, 16 dimes
now check the answers
24*.25 + 12*.05 + 16*.10 = 8.20
6.00 + .60 + 1.60 = 8.20
8.20 = 8.20
answer checks :-)
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! A man has 52 coins, all of which are dimes, quarters, and nickels. The total value of his change is $8.20. If the number of quarters is twice the number of nickels, how many of each does he have?
================
So far I have two equations set up:
x+2y+z=52 and
0.10x +0.25y +0.05z=8.20
but I don't think those are right.
----------
Use d, q & n, makes it clearer
d + q + n = 52
10d + 25q + 5n = 820
q = 2n
--------
Sub for q in the 1st 2 eqns
---
d + 3n = 52 times 2 --> 2d + 6n = 104
10d + 50n + 5n = 820
10d + 55n = 820 /5 --> 2d + 11n = 164
------------
2d + 6n = 104
2d + 11n = 164
------------------- Subtract
-5n = -60
n = 12 nickels
etc
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