SOLUTION: I have 27 coins that total $3.30. I have 3 times as many quarters as dimes; how many nickels do I have? This is my question and I have tried several ways to set the problem up but
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Question 1105533: I have 27 coins that total $3.30. I have 3 times as many quarters as dimes; how many nickels do I have? This is my question and I have tried several ways to set the problem up but none give me the correct answer. How do I begin to set this up?
Thank You,
Iron Heart Found 3 solutions by Boreal, greenestamps, ikleyn:Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! dimes=x
quarters=3x
nickels=27-4x, in other words, everything else
.10*x+.25(3x)+.05(27-4x)=3.30
.10x+.75x+1.35-0.20x=3.30
0.65x=1.95
x=3 dimes (0.30)
3x=9 quarters (2.25)
27-4x=15 nickels (0.75) ANSWER
adds to $3.30
With the information given in this way, I can't tell whether substitution or elimination would be easier. Since the other tutor showed a solution using substitution, let me try elimination.
That means using two variables:
let x = number of dimes
then 3x = number of quarters
let y = number of nickels
the total number of coins is 27
(1) the total value of the coins (in cents) is 330
(2)
Subtracting (1) from (2):
So the number of dimes is x=3; the number of quarters is 3x=9; and since the total number of coins is 27, the number of nickels is 15.
HMMM.... Looks very different than the solution using substitution... and both methods seem to require about the same a mount of work....
You will find there the lessons for all levels - from introductory to advanced,
and for all methods used - from one equation to two equations and even without equations.
Read them and become an expert in solution of coin problems.
Regarding your specific problem, look into the lesson marked (*) in the list, which contains similar problems to yours.
