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 and
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Question 127309This question is from textbook Introductory and Intermediate Algebra
: 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? This question is from textbook Introductory and Intermediate Algebra
You can put this solution on YOUR website! 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?
:
Let x = original number of dimes
Let y = original number of quarters
:
Let t = total value of the coins:
:
Original equation .10x + .25y = t
Switched equation .25x + .10y = t+.90
:
From this write an equation (subtracted t from both sides):
.10x + .25y = .25x + .10y - .90
:
.25y - 10y = .25x - .10x - .90
:
.15y = .15x - .90
:
y = x - 6; divide equation .15
:
or in the standard form:
x - y = 6
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Add to the total coin equation:
x + y = 20
x - y = 6
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2x = 26
x =
x = 13 dimes, the we have 7 quarters (originally)
:
To check our solution:
Take the switched value equation, subtract the original value equation:
.25(13) + .10(7) = 3.95
.10(13) + .25(7) = 3.05
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The difference is= .90 as given
