Question 1120729: Isabela has $2.55 in quarters,nickels and dimes. She has twice as many nickels as dimes and one less quarter than nickels. How many of each coin does she have?
Found 4 solutions by Shin123, MathTherapy, josgarithmetic, greenestamps:
Answer by Shin123(626)   (Show Source):
Answer by MathTherapy(10555)  (Show Source):
You can put this solution on YOUR website! Isabela has $2.55 in quarters,nickels and dimes. She has twice as many nickels as dimes and one less quarter than nickels. How many of each coin does she have?
There's absolutely NOTHING wrong with your post.
She has: 
You can do your own check!
BTW: You just need 1 (ONE) variable to solve for number of each coin. It's a lot easier and you're less prone to making errors!
Answer by josgarithmetic(39625)  (Show Source):
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
Presumably this is a problem from an algebra class, where you are supposed to provide a solution using formal algebra.
But don't underestimate the value of the brain exercise you get by working the problem informally using logical reasoning.
(1) Add one more quarter, so that now the numbers of nickels and quarters are the same; that makes the new total value $2.80.
(2) Now the numbers of quarters and nickels are both twice the number of dimes. So separate the coins into groups of 2 quarters, 2 nickels and 1 dime. The value of each group is then 50+10+10 = 70 cents.
(3) The new total value of the coins is $2.80; the value of each group of coins is 70 cents. That means the number of groups of coins is 280/70 = 4.
(4) So the total number of coins in the 4 groups is 8 quarters, 8 nickels, and 4 dimes.
(5) And last take away that "extra" quarter you added at the beginning. You have the final answer of the $2.55 being made up of 7 quarters, 4 dimes, and 8 nickels.
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