Question 1187432: My piggy bank has 28 coins in it that are only nickels, dimes, and quarters. I have $3.25 total.
The number of quarters is one less than the number of nickels.
How many of each coin do I have?
Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! My piggy bank has 28 coins in it that are only nickels, dimes, and quarters. I have $3.25 total.
The number of quarters is one less than the number of nickels.
How many of each coin do I have?
----------------------
n + d + q = 28
5n + 10d + 25q = 325 ---> n + 2d + 5q = 65
q = n-1
-----
Sub for q
n + d + n-1 = 28
2n + d = 29
-----
n + 2d + 5q = 65
n + 2d + 5(n-1) = 65
6n + 2d = 70 ---> 3n + d = 35
==============
3n + d = 35
2n + d = 29
---------------------- Subtract
n = 6
q = 5
d = 17
Answer by ikleyn(52787)  (Show Source):
You can put this solution on YOUR website! .
My piggy bank has 28 coins in it that are only nickels, dimes, and quarters. I have $3.25 total.
The number of quarters is one less than the number of nickels.
How many of each coin do I have?
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My loving/preferable method is different.
It uses only one unknown and only one single equation.
Let x be the number of nickels.
Then the number of quarters is (x-1) and the number of dimes is (28-x - (x-1)) = 29-2x.
Now write the total money equation
5x + 10*(29-2x) + 25*(x-1) = 325 cents.
Simplify and find x
5x - 20x + 25x = 325 - 10*29 + 25
10x = 60
x = 60/10 = 6.
ANSWER. 6 nickels, 6-1 = 5 quarters and the rest coins, 28 - 6 - 5 = 17, are dimes.
Solved (without using any systems of algebraic equations), at the level,
accessible to young beginner students.
If you want to see million other similar (and different) solved problems, look into my lessons
- Advanced word problems to solve using a single linear equation
- HOW TO algebreze and solve these problems using one equation in one unknown
in this site.
Consider these lessons as your textbook, handbook, guide, tutorials and (free of charge) home teacher.
Happy learning (!)
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