Tommy, Sean and Daniel had a total of 73 pencils. The ratio of Sean's pencils to Daniel's pencils was 4 : 3 at first. Tommy and Sean each gave away ¹₂ of their pencils. Given that the three boys had 50 pencils left, how many pencils did Tommy have at first?

	Tommy	Sean	Daniel	Total
Comparing Sean and Daniel at first		4 u	3 u
Before	2 p	2x2 = 4 u	3 u	73
Change	- 1 p	-1x2 = - 2 u		- 23
After	1 p	1x2 = 2 u	3 u	50

Total number of pencils that Tommy and Sean gave away
= 73 - 50
= 23

The number of pencils that Sean had at first is repeated.  Make the number of pencils that Sean had at first the same.  LCM of 4 and 2 is 4.

1 p + 2 u = 73 - 50
1 p + 2 u = 23
1 p = 23 - 2 u --- (1)

1 p + 2 u + 3 u = 50
1 p + 5 u = 50
1 p = 50 - 5 u --- (2)

(1) = (2)
23 - 2 u = 50 - 5 u
5 u - 2 u = 50 - 23
5 u - 2 u = 27
3 u = 27

1 u = 27 ÷ 3 = 9

Substitute 1 u = 9 into (1).
1 p = 23 - 2 u
1 p = 23 - 2 x 9
1 p = 23 - 18

1 p = 5

Number of pencils that Tommy had at first
= 2 p
= 2 x 5
= 10

Answer(s): 10