SOLUTION: I need help with this problem.
I found a handful of change (consisting of nickels, dimes and quarters) totaling $4.30. There were 3 times as many nickels as quarters and 2 more th
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I found a handful of change (consisting of nickels, dimes and quarters) totaling $4.30. There were 3 times as many nickels as quarters and 2 more th
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Question 294101: I need help with this problem.
I found a handful of change (consisting of nickels, dimes and quarters) totaling $4.30. There were 3 times as many nickels as quarters and 2 more than twice as many nickels than dimes. How many of each type of coin were found? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Let n, d, q, = no. of each coin
:
Write an equation for each statement
:
a handful of change (consisting of nickels, dimes and quarters) totaling $4.30.
.05n + .10d + .25q = 4.30
multiply by 100 to get rid of the decimals
5n + 10d + 25q = 430
:
There were 3 times as many nickels as quarters
n = 3q
q = n
:
and 2 more than twice as many nickels than dimes.
n = 2d + 2
n - 2 = 2d
2d = (n-2)
divide both sides by 2
d = (.5n-1)
:
How many of each type of coin were found?
Substitute for d and q
5n + 10d + 25q = 430
5n + 10(.5n-1) + 25(n) = 430
multiply by 3 to get rid of the fraction, results
15n + 30(.5n-1) + 25n = 1290
15n + 15n - 30 + 25n = 1290
55n = 1290 + 30
55n = 1320
n =
n = 24 nickels
then
q = 24
q = 8 quarters
and
d = .5(24) - 1
d = 11 dimes
:
:
Check solution
.05(24) + .10(11) + .25(8) =
1.20 + 1.10 + 2.00 = 4.30
