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?
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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
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
