Question 366049: Joe has a collection of nickels and dimes that is worth $9.60. If the number of dimes was tripled and the number of nickels was decreased by 31, the value of the coins would be $6.70. How many nickels and dimes does he have?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Response to student comment, I can solve it algebraically, but the solution
will not make sense as you will see here
:
Let n = no. of nickels
Let d = no. of dimes
:
Write an equation for each statement:
:
"Joe has a collection of nickels and dimes that is worth $9.60."
.05n + .10d = 9.60
:
If the number of dimes was tripled and the number of nickels was decreased by 31, the value of the coins would be $6.70."
.05(n-31) + .10(3d) = 6.70
.05n - 1.55 + .30d = 6.70
.05n + .30d = 6.70 + 1.55
.05n + .30d = 8.25
:
Use elimination here, subtract the 1st equation from the above equation
.05n + .30d = 8.25
.05n + .10d = 9.60
----------------------subtraction eliminates n, find d
0 + .20d = -1.35
d = 
d = -6.75 dimes, which is, of course, ridiculous
:
Let's just think about it.
Triple the no. of dimes and get $2.90 less even if you subtract 31 nickels
(which is $1.55) does not seem right does it?
:
I did a quick calc using the statement:
"If the number of nickels was tripled and the number of dime was decreased
by 31, the value of the coins would be $6.70."
and got a valid solution of 2 nickels and 95 dimes
:
What do you think about it? Are there others in your class doing this. Call
them up and see what they think.
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