Bobby, Xavier and Ken had a total of 67 pencils. The ratio of Xavier's pencils to Ken's pencils was 2 : 7 at first. Bobby and Xavier each gave away ¹₂ of their pencils. Given that the three boys had 58 pencils left, how many pencils did Bobby have in the end?

	Bobby	Xavier	Ken	Total
Comparing Xavier and Ken at first		2 u	7 u
Before	2 p	2x1 = 2 u	7 u	67
Change	- 1 p	-1x1 = - 1 u		- 9
After	1 p	1x1 = 1 u	7 u	58

Total number of pencils that Bobby and Xavier gave away
= 67 - 58
= 9

The number of pencils that Xavier had at first is repeated.  Make the number of pencils that Xavier had at first the same.  LCM of 2 and 2 is 2.

1 p + 1 u = 67 - 58
1 p + 1 u = 9
1 p = 9 - 1 u --- (1)

1 p + 1 u + 7 u = 58
1 p + 8 u = 58
1 p = 58 - 8 u --- (2)

(1) = (2)
9 - 1 u = 58 - 8 u
8 u - 1 u = 58 - 9
8 u - 1 u = 49
7 u = 49

1 u = 49 ÷ 7 = 7

Substitute 1 u = 7 into (1).
1 p = 9 - 1 u
1 p = 9 - 1 x 7
1 p = 9 - 7

1 p = 2

Number of pencils that Bobby had in the end
= 1 p
= 2

Answer(s): 2