SOLUTION: There were 26 nickels, dimes and quarters in all, and their value was $2.25. How many coins of each type were there if there were 10 times as many nickels as quarters?

Question 69314This question is from textbook An Incremental Development
: There were 26 nickels, dimes and quarters in all, and their value was $2.25. How many coins of each type were there if there were 10 times as many nickels as quarters? This question is from textbook An Incremental Development

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Write an equation for each sentence:
:
"there were 26 nickels, dimes and quarters in all,"
n + d + q = 26
:
"their value was $2.25."
.05n + .10n + .25q = 2.25
:
"there were 10 times as many nickels as quarters?"
n = 10q
:
How many coins of each type were there?
:
Substitute 10q for n and simplify both equations:
10q + d + q = 26
d + 11q = 26
and
.05(10q) + .10d + .25q = 2.25
.5q + .10d + .25q = 2.25
.10d + .75q = 2.25
:
Multiply the above equation by 10 and subtract it from d + 11q = 26:
:
d + 11q = 26
d + 7.5q = 22.50
--------------------subtracting eliminates d
0 + 3.5q = 3.5
q = 3.5/3.5
q = 1 quarter
:
Find d using d + 11q = 26
d + 11(1) = 26
d = 26 - 11
d = 15 dimes
;
Remember n = 10q
n = 10 nickels
:
Check our solution in the $ equation
.05(10) + .10(15) + .25(1)
.50 + 1.50 + .25 = 2.25
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