Lee, Neave and David had a total of 52 pens. The ratio of Neave's pens to David's pens was 8 : 9 at first. Lee and Neave each gave away ¹₂ of their pens. Given that the three boys had 35 pens left, how many pens did Lee have in the end?

	Lee	Neave	David	Total
Comparing Neave and David at first		8 u	9 u
Before	2 p	2x4 = 8 u	9 u	52
Change	- 1 p	-1x4 = - 4 u		- 17
After	1 p	1x4 = 4 u	9 u	35

Total number of pens that Lee and Neave gave away
= 52 - 35
= 17

The number of pens that Neave had at first is repeated.  Make the number of pens that Neave had at first the same.  LCM of 8 and 2 is 8.

1 p + 4 u = 52 - 35
1 p + 4 u = 17
1 p = 17 - 4 u --- (1)

1 p + 4 u + 9 u = 35
1 p + 13 u = 35
1 p = 35 - 13 u --- (2)

(1) = (2)
17 - 4 u = 35 - 13 u
13 u - 4 u = 35 - 17
13 u - 4 u = 18
9 u = 18

1 u = 18 ÷ 9 = 2

Substitute 1 u = 2 into (1).
1 p = 17 - 4 u
1 p = 17 - 4 x 2
1 p = 17 - 8

1 p = 9

Number of pens that Lee had in the end
= 1 p
= 9

Answer(s): 9