Question 286376: A woman has $2.15 in change in her purse, comprised entirely of dimes and quarters. Given that there are more quarters than dimes in her purse, what is the total number of coins?
A. 9 B. 10 C. 11 D. 12 E. 13
Found 2 solutions by stanbon, ptaylor:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A woman has $2.15 in change in her purse, comprised entirely of dimes and quarters. Given that there are more quarters than dimes in her purse, what is the total number of coins?
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q > d
10d + 25q = 215 cents
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q = 7 ; d= 4
d+q = 11
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Cheers,
Stan H.
A. 9 B. 10 C. 11 D. 12 E. 13
Answer by ptaylor(2198)  (Show Source):
You can put this solution on YOUR website! Let x=number of quarters
And y=number of dimes
We are told that:
($ understood)
x>y-------------------eq1
and
0.25x+0.10y=2.15 multiply each term by 100 and divide by each term by 5 (get rid of decimals and reduce size) :
5x+2y=43 solve for y
2y=43-5x
y=(43-5x)/2----------------------eq2
Now we know that the number of dimes and quarters has to be positive numbers as well as whole numbers
Looking at eq2, we see that x has to be less than or equal to 8, otherwise the
number of dimes will be negative
But x cannot be 8--that would make the number of dimes a fraction(3/2)
How about 7 for x? Then y=(43-35)/2=4
7 quarters and 4 dimes fill the requirement of eq1, but does this add up to $2.15?
$1.75+$0.40=$2.15---------------This is a solution. Are there others??
How about 6 for x?----No good, y is a fraction
How about 5 for x? ---Then y=(43-25)/2=9 No good because the requirement of eq1 is not satisfied even though these add up to $2.15
And anything smaller than 5 will make y>x
Ans: 7 quarters; 4 dimes=D --11 coins
Hope this helps---ptaylor
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