Difference between revisions of "1998 AJHSME Problems/Problem 25"
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==Problem== | ==Problem== | ||
| − | Three generous friends, each with some money, redistribute the money as | + | Three generous friends, each with some money, redistribute the money as followed: |
Amy gives enough money to Jan and Toy to double each amount has. | Amy gives enough money to Jan and Toy to double each amount has. | ||
Jan then gives enough to Amy and Toy to double their amounts. | Jan then gives enough to Amy and Toy to double their amounts. | ||
| Line 19: | Line 19: | ||
==See Also== | ==See Also== | ||
| − | {{ | + | {{AJHSME box|year=1998|num-b=24|after=Last Question}} |
| + | * [[AJHSME]] | ||
| + | * [[AJHSME Problems and Solutions]] | ||
| + | * [[Mathematics competition resources]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 12:56, 18 August 2019
Problem
Three generous friends, each with some money, redistribute the money as followed: Amy gives enough money to Jan and Toy to double each amount has. Jan then gives enough to Amy and Toy to double their amounts. Finally, Toy gives enough to Amy and Jan to double their amounts. If Toy had 36 dollars at the beginning and 36 dollars at the end, what is the total amount that all three friends have?
Solution
If Toy had 
dollars at the beginning, then after Amy doubles his money, he has 
dollars after the first step.
Then Jan doubles his money, and Toy has 
dollars after the second step.
Then Toy doubles whatever Amy and Jan have. Since Toy ended up with , he spent 
to double their money. Therefore, just before this third step, Amy and Jan must have had 
dollars in total. And, just before this step, Toy had 
dollars. Altogether, the three had 
dollars, and the correct answer is 
See Also
| 1998 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Question | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
