Question 136799: BETSY HAS A JAR CONTAINING 80 COINS, ALL OF WHICH ARE EITHER QUARTERS OR NICKLES. THE TOTAL VALUE OF THE COINS IS $ 14.60. HOW MANY OF EACH TYPE OF COIN DOES SHE HAVE?
CAN YOU PLEASE SHOW ME A SIMPLE FORMAT TO FOLLOW TO SOLVE MY PROBLEM AND HOW TO COME OUT WITH THE CORRECT ANSWER?
THANK,
lj23kfuller@yahoo.com
Answer by jojo14344(1513)   (Show Source):
You can put this solution on YOUR website!
Let "x" = # of coins for quarters
"y" = # of coins for nickels
then, x + y = 80 coins, right? This will be equation 1.
Next, when we multiply 0.25(quarter) by "x" (# of coins) +
nickel (.05) multiply by "y" (# of coins) equals $14.60 isn't it?
Equating that fact,
0.25x+0.05y=$14.60 and this is equation 2.
Going back to equation 1, we get "x = $80-y" . Then substitue this value to x in equation 2,
0.25($80-y)+0.05y=$14.60
$20-0.25y+0.05y=$14.60
$20-$14.60=0.20y
y=27 coins (nickels)
For x = 80-y= 80-27
x= 53 coins (quarter)
To check:
53quarters+27nickels = 80 coins
also,
0.25(53)+0.05(27)=$14.60
$13.25+$1.35=$14.60
$14.60=$14.60
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