SOLUTION: a man has 20 coins consisting of dimes and quarters. if the dimes were quarters and the quarters were dimes, he would have ninety cents more than he has now. how many dimes and qua

Click here to see ALL problems on Linear Algebra

Question 288473: a man has 20 coins consisting of dimes and quarters. if the dimes were quarters and the quarters were dimes, he would have ninety cents more than he has now. how many dimes and quarters does he have?
Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
a man has 20 coins consisting of dimes and quarters.
if the dimes were quarters and the quarters were dimes,
he would have ninety cents more than he has now.
how many dimes and quarters does he have?
:
let x = original number of dimes
let y = original number of quarters
:
total coin equation
x + y = 20
y = (20-x); use for substitution
:
"if the dimes were quarters and the quarters were dimes,
he would have ninety cents more than he has now."
:
Amt now = amt when reversed - .90
.10x + .25y = .10y + .25x - .90
.25y - 10y = .25x - .10x - 90
.15y = .15x - .90
Replace y with (20-x)
.15(20-x) = .15x - .90
3.00 - .15x = .15x - .90
3.00 + .90 = .15x + .15x
3.90 = .30x
x =
x = 13 dimes
then
y = 20-13 = 7 quarters
:
:
Check solution by finding the values
reversed: 7(.10) + 13(.25) = 3.95
original: 13(.10) + 7(.25) = 3.05
----------------------------------
difference: .90