# Difference between revisions of "1998 AJHSME Problems/Problem 25"

## Problem

Three generous friends, each with some money, redistribute the money as follow: 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?

$\textbf{(A)}\ 109\qquad\textbf{(B)}\ 180\qquad\textbf{(C)}\ 216\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 288$

## Solution

If Toy had $36$ dollars at the beginning, then after Amy doubles his money, he has $36 \times 2 = 72$ dollars after the first step.

Then Jan doubles his money, and Toy has $72 \times 2 = 144$ dollars after the second step.

Then Toy doubles whatever Amy and Jan have. Since Toy ended up with $36$, he spent $144 - 36 = 108$ to double their money. Therefore, just before this third step, Amy and Jan must have had $108$ dollars in total. And, just before this step, Toy had $144$ dollars. Altogether, the three had $144 + 108 = 252$ dollars, and the correct answer is $\boxed{D}$

 1998 AJHSME (Problems • Answer Key • Resources) Preceded byProblem 24 Followed by1998 AMC 8 First 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