SOLUTION: a man has twenty 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 an
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Question 558082: a man has twenty 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 now?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
let x = original number dimes
Let y = original number of quarters
:
man has twenty coins consisting of dimes and quarters.
x + y = 20
:
if the dimes were quarters and the quarters were dimes, he would have ninety cents more than he has now.
(10y + 25x) - (10x + 25y) = 90
10y + 25x - 10x - 25y = 90
25x - 10x + 10y - 25y = 90
15x - 15y = 90
simplify, divide by 15
x - y = 6
x + y = 20; added the 1st equation
------------eliminates y find
2x = 26
x = 13 dimes originally
then
20 - 13 = 7 quarters
;
:
See if that checks our
13(25) + 7(10) = 395
13(10) + 7(25) = 305
----------------------
difference is 90
..how many dimes and quarters does he have now? 13, 7
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