Paul, Jeremy and Perry had a total of 87 pencils. The ratio of Jeremy's pencils to Perry's pencils was 8 : 5 at first. Paul and Jeremy each gave away ¹₂ of their pencils. Given that the three boys had 56 pencils left, how many pencils did Paul have at first?

	Paul	Jeremy	Perry	Total
Comparing Jeremy and Perry at first		8 u	5 u
Before	2 p	2x4 = 8 u	5 u	87
Change	- 1 p	-1x4 = - 4 u		- 31
After	1 p	1x4 = 4 u	5 u	56

Total number of pencils that Paul and Jeremy gave away
= 87 - 56
= 31

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

1 p + 4 u = 87 - 56
1 p + 4 u = 31
1 p = 31 - 4 u --- (1)

1 p + 4 u + 5 u = 56
1 p + 9 u = 56
1 p = 56 - 9 u --- (2)

(1) = (2)
31 - 4 u = 56 - 9 u
9 u - 4 u = 56 - 31
9 u - 4 u = 25
5 u = 25

1 u = 25 ÷ 5 = 5

Substitute 1 u = 5 into (1).
1 p = 31 - 4 u
1 p = 31 - 4 x 5
1 p = 31 - 20

1 p = 11

Number of pencils that Paul had at first
= 2 p
= 2 x 11
= 22

Answer(s): 22