Ahmad, Xavier and Jenson shared $495. During a shopping trip, Ahmad spent ³₈ of his share, Xavier spent ²₅ of his share and Jenson spent ¹₄ of his share. Given that all three boys each spent the same amount of money, how much more money do Ahmad and Xavier have than Jenson at first?

	Ahmad	Xavier	Jenson	Total
Before	8x2 = 16 u	5x3 = 15 u	4x6 = 24 u	$495
Change	- 3x2 = - 6 u	- 2x3 = - 6 u	- 1x6 = - 6 u
After	5x2 = 10 u	3x3 = 9 u	3x6 = 18 u

The amounts that Ahmad, Xavier and Jenson spent is the same.  Make the amounts that each boy spent the same. LCM of 3, 2 and 1 is 6.

Amounts that the 3 boys had at first
= 16 u + 15 u + 24 u
= 55 u

55 u = 495

1 u = 495 ÷ 55 = 9

Amount that Ahmad and Xavier have at first
= 16 u + 15 u
= 31 u

Amount that Ahmad and Xavier have more than Jenson at first
= 31 u - 24 u
= 7 u
= 7 x 9
= $63

Answer(s): $63