Dana, Kimberly and Jean each had some balls and decided to play a game with their balls. In round 1, Dana lost ¹₂ of her balls to Kimberly. In round 2, Kimberly lost ¹₂ of her total number of balls to Jean. In round 3, Jean lost ¹₂ of her total number of balls to Dana. In the end, Dana, Kimberly and Jean had 132, 94 and 63 balls respectively. How many balls did Kimberly have at first?

	Dana	Kimberly	Jean
Before  1	2 b	119
Change 1	- 1 b	+ 1 b
After 1	1 b	188
Before 2	69	2 u
Change 2		- 1 u	+1
After 2		1 u	126
Before 3			2 p
Change 3	+ 1 p		- 1 p
After 3			1 p
Comparing Dana, Kimberly and Jean in the end	132	94	63

Working backwards.

1 p = 63

At the start of Round 3:
Number of balls that Jean had
= 2 p
= 2 x 63 
= 126

At the start of Round 2:
Number of balls that Dana had
= 132 - 1 p
= 132 - 63
= 69

At the end of Round 3:
Number of balls that Kimberly had = 94

1 u = 94

At the end of Round 1:
Number of balls that Kimberly had
= 2 u
= 2 x 94 
= 188

1 b = 69 

At the end of Round 1:
Number of balls that Dana had = 69

At the start of Round 1:
Number of balls that Kimberly had
= 188 - 1 b
= 188 - 69
= 119

Answer(s): 119