Question 1162779: A collection of nickels, dimes, and quarters consist of 8 coins with a total of $1.30. If the number of dimes is equal to the number of nickels, find the number of each type of coins.
Found 2 solutions by ankor@dixie-net.com, MathTherapy:
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
let q = no. of quarters
:
Write an equation for each statement
:
A collection of nickels, dimes, and quarters consist of 8 coins
n + d + q = 8
with a total of $1.30.
.05 + .10d + .25 = 1.30
If the number of dimes is equal to the number of nickels,
d = n
:
multiply the 2nd equation by 4 and subtract from the 1st equation
1n + 1d + 1q = 8
.2n + .4d + 1q = 5.20
----------------------subtraction eliminates q
.8n + .6d = 2.80
d = n, therefore
.8n + .6n = 2.80
1.4n = 2.80
n = 2.8/1.4
n = 2 nickels
then since d = n
d = 2 dimes
and
q = 8 - 2 - 2
q = 4 quarters
:
:
See if that checks out
.05(2) + .10(2) + .25(4) =
.10 + .20 + 1.00 = 1.30
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website! A collection of nickels, dimes, and quarters consist of 8 coins with a total of $1.30. If the number of dimes is equal to the number of nickels, find the number of each type of coins.
Quite a few tutors have been PREACHING on here, the need to solve problems like these using ONE VARIABLE. I guess some who continue to use multiple
variables - the likes of the other person who responded - don't feel that they need to learn the most efficient and least complex way to solutions.
I would suggest that you don't solve this problem his way!
Let the number of dimes be D
Then the number of nickels is also D
Therefore, the number of quarters = 8 - (D + D), or 8 - 2D
We then get the following VALUE equation: .1D + .05D + .25(8 - 2D) = 1.3
.1D + .05D + 2 - .5D = 1.3
- .35D = - .7
Number of dimes, or 
Number of nickles: 2 ALSO
Do you think you can now find the number of quarters?
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