1998 AMC 8 Problem 25
Attempt Problem 25 of the 1998 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1998 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
25.
Three generous friends, each with some cash, redistribute their money as follows: Ami gives enough money to Jan and Toy to double the amount that each has. Jan then gives enough to Ami and Toy to double their amounts. Finally, Toy gives Ami and Jan enough to double their amounts. If Toy has when they begin and when they end, what is the total amount that all three friends have?
Answer: D
Small hint:
Track Toy’s amount through the first two exchanges.
Big hint:
Before Toy’s final turn, the amount he gives equals Ami and Jan’s combined amount.
Solution:
Toy begins with After Ami doubles Toy’s amount, Toy has After Jan doubles Toy’s amount, Toy has
On Toy’s final turn, Toy ends with so Toy gives away That gift doubles the combined amount of Ami and Jan, so Ami and Jan together had just before Toy’s final turn.
The total amount of money is constant, so the total is
Thus, the correct answer is D .
Problem 25 in other years
- 1985 AMC 8
- 1986 AMC 8
- 1987 AMC 8
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