Question 90005: Jennifer has nickels and quarters. If Jennifer had seven more nickels and four times the number of quarters she would have $22.85 more. How much money she have?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Jennifer has nickels and quarters. If Jennifer had seven more nickels and four times the number of quarters she would have $22.85 more. How much money she have?
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Let x = amt of money she has; n = no. of nickels; q = no. of quarters
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The basic equation:
.05n + .25q = x
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"seven more nickels and four times the number of quarters she would have $22.85 more."
.05(n+7) + .25(4q) = (x+22.85)
.05n + .35 + 1.00q = x + 22.85
.05n + 1.0q = x + 22.85 - .35
.05n + 1.0q = x + 22.50
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Subtract the basic equation from the the above equation, find q:
.05n + 1.0q = x + 22.50
.05n + .25q = x
-------------------------subtract, eliminates n and x
0n + .75q = 0x + 22.50
.75q = 22.50
q = 22.50/.75
q = 30 quarters originally
:
At this point it is apparent that there is no unique solutions for n and x.
although they are dependent on each other:
using equation .05n + 1.0q = x + 22.50
.05n + 1(30) = x + 22.50
.05n + 30 = x + 22.50
.05n = x + 22.50 - 30
.05n = x - 7.50
:
We know that x has to be greater than 7.50 and multiple of .05
For example, let x = 8
.05n = 8 - 7.50
n = .50/.05
n = 10 nickels
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Let x = 20
.05n = 20 - 7.50
n = 12.50/.05
n = 250 nickels
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Either one will satisfy the equation, try: x = 20, n = 250:
.05(250+7) + .25(4*30) = 20 + 22.85
.05(257) + .25(120) = 42.85
12.85 + 30 = 42.85
:
Does this help?
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