The total number of balls in Basket S, Basket T and Basket U was 131. ³₈ of the balls from Basket S and 43 balls from Basket T were removed. More balls were then added into Basket U until the number of balls in it was quadrupled. The ratio of the number of balls in Basket S to Basket T to Basket U became 5 : 2 : 4.

1. How many more balls were there in Basket S than Basket T at first?
2. Find the total number of balls in Basket S and Basket U in the end.

	Basket S	Basket T	Basket U	Total
Before	8 u	2 u + 43	1 u	131
Change	- 3 u	- 43	+ 3 u
After	5 u		4 u
Comparing the balls  in the end	5 u	2 u	4 u

(a)
Total number of balls at first
= 8 u + 2 u + 43 + 1 u
= 11 u + 43

11 u + 43 = 131
11 u = 131 - 43
11 u = 88

1 u = 88 ÷ 11 = 8

Number of more balls in Basket S than Basket T at first 
= 8 u - (2 u + 43)
= 8 u - 2 u - 43
= 6 u - 43
= 6 x 8 - 43
= 48 - 43
= 5

(b)
Total number of balls in Basket S and Basket U in the end
= 5 u + 4 u
= 9 u
= 9 x 8
= 72 

Answer(s): (a) 5; (b) 72