Lee, Justin and Ken had a total of 81 pens. The ratio of Justin's pens to Ken's pens was 6 : 7 at first. Lee and Justin each gave away ¹₂ of their pens. Given that the three boys had 58 pens left, how many pens did Lee have at first?

	Lee	Justin	Ken	Total
Comparing Justin and Ken at first		6 u	7 u
Before	2 p	2x3 = 6 u	7 u	81
Change	- 1 p	-1x3 = - 3 u		- 23
After	1 p	1x3 = 3 u	7 u	58

Total number of pens that Lee and Justin gave away
= 81 - 58
= 23

The number of pens that Justin had at first is repeated.  Make the number of pens that Justin had at first the same.  LCM of 6 and 2 is 6.

1 p + 3 u = 81 - 58
1 p + 3 u = 23
1 p = 23 - 3 u --- (1)

1 p + 3 u + 7 u = 58
1 p + 10 u = 58
1 p = 58 - 10 u --- (2)

(1) = (2)
23 - 3 u = 58 - 10 u
10 u - 3 u = 58 - 23
10 u - 3 u = 35
7 u = 35

1 u = 35 ÷ 7 = 5

Substitute 1 u = 5 into (1).
1 p = 23 - 3 u
1 p = 23 - 3 x 5
1 p = 23 - 15

1 p = 8

Number of pens that Lee had at first
= 2 p
= 2 x 8
= 16

Answer(s): 16