SOLUTION: VALUE OF COINS. Mary has $3.00 in nickels, dimes, and quarters. If she has twice as many dimes as quarters and five more nickels than dimes, how many coins of each type does she h
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-> SOLUTION: VALUE OF COINS. Mary has $3.00 in nickels, dimes, and quarters. If she has twice as many dimes as quarters and five more nickels than dimes, how many coins of each type does she h
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Question 634033: VALUE OF COINS. Mary has $3.00 in nickels, dimes, and quarters. If she has twice as many dimes as quarters and five more nickels than dimes, how many coins of each type does she have? (NOTE: nickel = 5 cents; dime = 10 cents; quarter = 25 cents) 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, we are going to get all the coins in
terms of d:
:
"Mary has $3.00 in nickels, dimes, and quarters."
.05n + .10d + .25q = 3.00
:
"If she has twice as many dimes as quarters"
d = 2q
or divide both sides by 2:
q = .5d
:
" and five more nickels than dimes,"
n = d + 5
:
In the Total Value equation, replace q with .5d and n with (d+5)
.05(d+5) + .10d + .25(.5d) = 3.00
.05d + .25 + .10d + .125d = 3.00
Combine like terms
.275d = 3.00 - .25
.275d = 2.75
d =
d = 10 dimes
then
n = 10 + 5
n = 15 nickels
and
q = .5(10)
q = 5 quarters
:
See if that checks out
.05(15) + .10(10) + .25(5) = 3.00
