Ahmad, Peter and Howard had a total of 159 pencils. The ratio of Peter's pencils to Howard's pencils was 6 : 7 at first. Ahmad and Peter each gave away ¹₂ of their pencils. Given that the three boys had 118 pencils left, how many pencils did Ahmad have in the end?

	Ahmad	Peter	Howard	Total
Comparing Peter and Howard at first		6 u	7 u
Before	2 p	2x3 = 6 u	7 u	159
Change	- 1 p	-1x3 = - 3 u		- 41
After	1 p	1x3 = 3 u	7 u	118

Total number of pencils that Ahmad and Peter gave away
= 159 - 118
= 41

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

1 p + 3 u = 159 - 118
1 p + 3 u = 41
1 p = 41 - 3 u --- (1)

1 p + 3 u + 7 u = 118
1 p + 10 u = 118
1 p = 118 - 10 u --- (2)

(1) = (2)
41 - 3 u = 118 - 10 u
10 u - 3 u = 118 - 41
10 u - 3 u = 77
7 u = 77

1 u = 77 ÷ 7 = 11

Substitute 1 u = 11 into (1).
1 p = 41 - 3 u
1 p = 41 - 3 x 11
1 p = 41 - 33

1 p = 8

Number of pencils that Ahmad had in the end
= 1 p
= 8

Answer(s): 8