If Marion gives Albert $15, the amount of money she has will be 80% of his. If Albert gives Marion $6, he will have the same amount of money as her. How much money did each of them have?
(a) Albert
(b) Marion

	Case 1		Case 2
	Albert	Marion	Albert	Marion
Before	5 u - 15	4 u + 15	4.5 u + 6	4.5 u - 6
Change	+ 15	- 15	- 6	+ 6
After	5 u	4 u	4.5 u	4.5 u

(a)
80% = ⁸⁰₁₀₀ = ⁴₅

In Case 1, when Marion gives to Albert, there is an internal transfer of money from Marion to Albert.  The total amount that both have remains the same.

In Case 2, when Albert gives to Marion, there is an internal transfer of money from Albert to Marion. The total amount that both have remains the same.

The total amount that both have in Case 1 and Case 2 are the same.

Total amount that Marion and Albert have
= 5 u + 4 u
= 9 u

Amount that Marion and Albert each has in the end for Case 2
= 9 u ÷ 2
= 4.5 u

Amount that Albert has at first is the same in Case 1 and Case 2.

5 u - 15 = 4.5 u + 6
5 u - 4.5 u = 6 + 15
0.5 u = 21 

1 u = 21 ÷ 0.5 = 42

Amount that Albert has
= 5 u - 15
= 5 x 42 - 15
= 210 - 15
= $195 

(b)
Amount that Marion has
= 4 u + 15
= 4 x 42 + 15
= 168 + 15
= $183

Answer(s): (a) $195; (b) $183