Xavier, Ahmad and Oliver had a total of 121 erasers. The ratio of Ahmad's erasers to Oliver's erasers was 4 : 9 at first. Xavier and Ahmad each gave away ¹₂ of their erasers. Given that the three boys had 101 erasers left, how many erasers did Xavier have in the end?

	Xavier	Ahmad	Oliver	Total
Comparing Ahmad and Oliver at first		4 u	9 u
Before	2 p	2x2 = 4 u	9 u	121
Change	- 1 p	-1x2 = - 2 u		- 20
After	1 p	1x2 = 2 u	9 u	101

Total number of erasers that Xavier and Ahmad gave away
= 121 - 101
= 20

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

1 p + 2 u = 121 - 101
1 p + 2 u = 20
1 p = 20 - 2 u --- (1)

1 p + 2 u + 9 u = 101
1 p + 11 u = 101
1 p = 101 - 11 u --- (2)

(1) = (2)
20 - 2 u = 101 - 11 u
11 u - 2 u = 101 - 20
11 u - 2 u = 81
9 u = 81

1 u = 81 ÷ 9 = 9

Substitute 1 u = 9 into (1).
1 p = 20 - 2 u
1 p = 20 - 2 x 9
1 p = 20 - 18

1 p = 2

Number of erasers that Xavier had in the end
= 1 p
= 2

Answer(s): 2