The total number of balls in Basket W, Basket X and Basket Y was 101. ⁴₇ of the balls from Basket W and 11 balls from Basket X were removed. More balls were then added into Basket Y until the number of balls in it was quadrupled. The ratio of the number of balls in Basket W to Basket X to Basket Y became 3 : 2 : 4.

1. How many less balls were there in Basket X than Basket W at first?
2. Find the total number of balls in Basket X and Basket Y in the end.

	Basket W	Basket X	Basket Y	Total
Before	7 u	2 u + 11	1 u	101
Change	- 4 u	- 11	+ 3 u
After	3 u		4 u
Comparing the balls  in the end	3 u	2 u	4 u

(a)
Total number of balls at first
= 7 u + 2 u + 11 + 1 u
= 10 u + 11

10 u + 11 = 101
10 u = 101 - 11
10 u = 90

1 u = 90 ÷ 10 = 9

Number of less balls in Basket X than Basket W at first 
= 7 u - (2 u + 11)
= 7 u - 2 u - 11
= 5 u - 11
= 5 x 9 - 11
= 45 - 11
= 34

(b)
Total number of balls in Basket X and Basket Y in the end
= 2 u + 4 u
= 6 u
= 6 x 9
= 54 

Answer(s): (a) 34; (b) 54