SOLUTION: Joe has a collection of nickels and dimes that is worth $6.00. If the number of dimes was doubled and the number of nickels was increased by 6, the value of the coins would be $9.9
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Question 380481: Joe has a collection of nickels and dimes that is worth $6.00. If the number of dimes was doubled and the number of nickels was increased by 6, the value of the coins would be $9.90. How many dimes does he have?
Found 2 solutions by stanbon, mananth:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Joe has a collection of nickels and dimes that is worth $6.00. If the number of dimes was doubled and the number of nickels was increased by 6, the value of the coins would be $9.90. How many dimes does he have?
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Value Eq: 5n + 10d = 600 cents
Value Eq: 5(n+6)+10(2d) = 990 cents
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Rearrange and modify to get:
5n+ 20d = 990-30
5n+ 10d = 600
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Subtract bottom from top and solve for "d":
10d = 360
d = 36 (# of dimes)
---
Substitute and solve for "n":
5n+10*36 = 600
5n+360 = 600
5n = 240
n = 48 (# of nickels)
===========================
Cheers,
stan H.
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
let nickels be x
dimes = y
..
5x+10y=600
/5
x+2y=120...................1
..
dimes 2y
nickels increased by 6 = x+6
..
5(x+6)+10*2y=900
5x+30+20y=900
5x+20y=900-30
5x+20y=870
/5
x+4y=174......................2
...
multiply (1) by -1
-x-2y=-120
add this to (2)
x-x+4y-2y=174-120
2y=54
/2
y=27 the number of dimes
plug value of y in (1)
x+2y=120
x+54=120
x=66 -------- nickels
..
CHECK
5*66+10*27=330+270=600
m.ananth@hotmail.ca
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