A total of j dollars is given to Abby, Bill, and Carla and divided equ

Abby ............ Bill ............... Carla ............. Total

\(\frac{j}{3}\) .............. \(\frac{j}{3}\) .................. \(\frac{j}{3}\) ................... j

Abby gives Bill k dollars

\(\frac{j}{3}\) - k .............. \(\frac{j}{3}\) + k .................. \(\frac{j}{3}\) ................... j

Bill gives Carla 2k dollars

\(\frac{j}{3}\) - k .............. \(\frac{j}{3}\) + k - 2k .................. \(\frac{j}{3}\) + 2k ................... j

Carla gives Abby 3k dollars

\(\frac{j}{3}\) - k + 3k .............. \(\frac{j}{3}\) - k .................. \(\frac{j}{3}\) + 2k - 3k ................... j

Abby has exactly half of all the dollars given to the three people

\(\frac{j}{2} = \frac{j}{3} + 2k\)

\(\frac{j}{6} = 2k\)

j = 12k

Answer = D

Official Solution:

A total of \(j\) dollars is given to Abby, Bill, and Carla and divided equally among them. Then Abby gives Bill \(k\) dollars, Bill gives Carla \(2k\) dollars, and Carla gives Abby \(3k\) dollars. Afterwards, Abby has exactly half of all the dollars given to the three people. In terms of \(k\), how much money was given originally to Abby, Bill, and Carla?

A. \(4k\)
B. \(6k\)
C. \(8k\)
D. \(12k\)
E. \(16k\)


Taking a "direct algebra" approach, we can see that each of the three people receives \(\frac{j}{3}\) dollars. During the swaps, Abby gives away \(k\) dollars but receives \(3k\) dollars, so her total increases by \(2k\) dollars. Her final total is half of the original amount of money, or \(\frac{j}{2}\). Now we can write an equation:
\(\frac{j}{3} + 2k = \frac{j}{2}\)

Now solve for \(j\) in terms of \(k\). First, multiply through by 6 to eliminate fractions:
\(2j + 12k = 3j\)
\(12k = j\)

This is our answer. We can also solve by picking a number for \(j\), but realize that we cannot separately pick numbers for \(j\) and \(k\) - after all, that would determine the very relationship the question is asking us for. Moreover, it's difficult to know ahead of time what would be a good test number for \(j\). Seeing 2 and 3 as coefficients within the problem, we might pick $6 as the total. Then each of the people receives $2. Abby has to wind up with $3 (half of $6) when all is said and done, so she has to increase her total by $1. Since she gives away \(k\) dollars but receives \(3k\) dollars, she increases her total by \(2k\) dollars. This tells us that \(k\) is $0.50, so \(j\) is 12 times bigger. However, this reasoning doesn't save us a whole lot of work. Direct algebra is just as fast.


Answer: D.

This question is all about keeping track of monetary movement. Below should be what you arrived at for the main information given in the problem.

Abby --> (j/3)-k+3k
Bill --> (j/3)+k-2k
Carla --> (j/3)+2k-3k

Lastly, we find out Abby has J/2, so we set her amount equal to this and solve.

(j/3)-k+3k =(j/2)
j/6 = 2k
j = 12k

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