Caden, Howard and Michael had a total of 61 erasers. The ratio of Howard's erasers to Michael's erasers was 6 : 7 at first. Caden and Howard each gave away ¹₂ of their erasers. Given that the three boys had 41 erasers left, how many erasers did Caden have in the end?

	Caden	Howard	Michael	Total
Comparing Howard and Michael at first		6 u	7 u
Before	2 p	2x3 = 6 u	7 u	61
Change	- 1 p	-1x3 = - 3 u		- 20
After	1 p	1x3 = 3 u	7 u	41

Total number of erasers that Caden and Howard gave away
= 61 - 41
= 20

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

1 p + 3 u = 61 - 41
1 p + 3 u = 20
1 p = 20 - 3 u --- (1)

1 p + 3 u + 7 u = 41
1 p + 10 u = 41
1 p = 41 - 10 u --- (2)

(1) = (2)
20 - 3 u = 41 - 10 u
10 u - 3 u = 41 - 20
10 u - 3 u = 21
7 u = 21

1 u = 21 ÷ 7 = 3

Substitute 1 u = 3 into (1).
1 p = 20 - 3 u
1 p = 20 - 3 x 3
1 p = 20 - 9

1 p = 11

Number of erasers that Caden had in the end
= 1 p
= 11

Answer(s): 11