Problem: A gives B as many cents as B has and C as many cents as C has. Similarly, B then gives A and C as many cents as each then has. C, similarly, then gives A and B as many cents as each then has. If each finally has 16 cents, with how many cents does A start?
Answer Choices:
A. 24
B. 26
C. 28
D. 30
E. 32
Solution:
If a,b,c, respectively, represent the initial amounts of A,B,C, then the given conditions lead to the following:
After TransactionIIIIIIA hasa−b−c2(a−b−c)4(a−b−c)B has2b2b−(a−b−c)−2c(=3b−a−c)2(3b−a−c)C has2c4c4c−2(a−b−c)−(3b−a−c)(=7c−a−b)
Consequently 4(a−b−c)=16,6 b−2a−2c=16,7c−a−b=16. Solving this set of equations, you obtain a=26, b=14,c=8
or
Working from the last condition to the first, we may set up the following table:
AmountsaccStep 2161616Step 38832Step 442816Step 126 (required)148