# Notebook Cost

• Mar 31st 2007, 04:57 AM
aznmartinjai
Notebook Cost
John and his twin brothers each bought three identical notebooks when school started in August. John paid for the notebooks with a \$10 bill. He received less than \$0.50 in change, all in nickels. No sales tax was charged. What was the price paid for one notebook?
• Mar 31st 2007, 07:26 AM
ecMathGeek
Quote:

Originally Posted by aznmartinjai
John and his twin brothers each bought three identical notebooks when school started in August. John paid for the notebooks with a \$10 bill. He received less than \$0.50 in change, all in nickels. No sales tax was charged. What was the price paid for one notebook?

The total cost of all three notebooks was some amount of money less than \$10 but greater than \$9.50. So we need some amount between those two numbers that is divisible by 0.05 (since all the change was in nickles) AND is divisible by 3 (since this is the price for three books).

If we divide \$10 by 3, we get \$3.333333... (repeats)
If we divide \$9.50 by 3, we get \$3.1666666... (repeats)

We could use trial and error to see what the amount is by testing numbers between 9.50 and 10.00 that are multiples of 0.05:

\$9.55/3 doesn't work
\$9.60/3 = \$3.20
\$9.65/3 doesn't work
\$9.70/3 doesn't work
\$9.75/3 = \$3.25
\$9.80/3 doesn't work
\$9.85/3 doesn't work
\$9.90/3 = \$3.30
\$9.95/3 doesn't work

There are three answers that work: \$3.20, \$3.25, \$3.30. I'm not sure if I missed something in your explanation of the problem.
• Mar 31st 2007, 04:53 PM
aznmartinjai
3 answers? huh..? are you sure?
• Mar 31st 2007, 05:17 PM
Jhevon
Quote:

Originally Posted by aznmartinjai
John and his twin brothers each bought three identical notebooks when school started in August. John paid for the notebooks with a \$10 bill. He received less than \$0.50 in change, all in nickels. No sales tax was charged. What was the price paid for one notebook?

ecMathGeek's logic is correct, however, the mistake was he was trying to account for three notebooks when he should have accounted for nine. Recall, John and his twin brothers each bought three identical notebooks. so that's John and two other boys, so that's 3 per boy which is 9 notebooks. so following ecMathGeek's logic of trial and error, we find:

\$9.55/9 doesn't work
\$9.60/9 = doesn't work
\$9.65/9 doesn't work
\$9.70/9 doesn't work
\$9.75/9 = doesn't work
\$9.80/9 doesn't work
\$9.85/9 doesn't work
\$9.90/9 = \$1.10 each notebook
\$9.95/9 doesn't work
• Mar 31st 2007, 06:46 PM
ecMathGeek
Quote:

Originally Posted by Jhevon
ecMathGeek's logic is correct, however, the mistake was he was trying to account for three notebooks when he should have accounted for nine. Recall, John and his twin brothers each bought three identical notebooks. so that's John and two other boys, so that's 3 per boy which is 9 notebooks. so following ecMathGeek's logic of trial and error, we find:

\$9.55/9 doesn't work
\$9.60/9 = doesn't work
\$9.65/9 doesn't work
\$9.70/9 doesn't work
\$9.75/9 = doesn't work
\$9.80/9 doesn't work
\$9.85/9 doesn't work
\$9.90/9 = \$1.10 each notebook
\$9.95/9 doesn't work

Ha! I didn't see that. :o

Good job, Jhevon. :D
• Mar 31st 2007, 06:52 PM
Jhevon
Quote:

Originally Posted by ecMathGeek
Ha! I didn't see that. :o

Good job, Jhevon. :D

Aw, shucks, let's just call it luck
• Apr 12th 2007, 08:10 PM
mykidsmom
Would that be "Lucky Number Jhevon"?
;)
Quote:

Originally Posted by Jhevon
Aw, shucks, let's just call it luck