The ratio of the number of biscuits in Container T to the number of biscuits in Container U was 7 : 3. 20% of the biscuits in Container T and 0.7 of those in Container U were peach. After transferring the biscuits between the 2 containers, the number of vanilla biscuits in both containers are the same. Likewise, the number of peach biscuits in both containers are the same. If a total of 216 biscuits were moved, how many more biscuits were there in Container T than Container U at first?

Container T		Container U
7 u		3 u
Peach	Vanilla	Peach	Vanilla
1.4 u	5.6 u	2.1 u	0.9 u
+ 0.35 u	- 2.35 u	- 0.35 u	+ 2.35 u
1.75 u	3.25 u	1.75 u	3.25 u

Number of peach biscuits in Container T
= 20% x 7 u
= ²⁰₁₀₀ x 7 u
= 1.4 u

Number of vanilla biscuits in Container T
= 7 u - 1.4 u
= 5.6 u

Number of peach biscuits in Container U
= 0.7 x 3 u
= 2.1 u

Number of vanilla biscuits in Container U
= 3 u - 2.1 u
= 0.9 u

Number of peach biscuits in each container in the end
= (1.4 u + 2.1 u) ÷ 2
= 3.5 u ÷ 2
= 1.75 u

Number of vanilla biscuits in each container in the end
= (5.6 u + 0.9 u) ÷ 2
= 6.5 u ÷ 2
= 3.25 u

Number of biscuits moved
= 0.35 u + 2.35 u
= 2.7 u

2.7 u = 216

1 u = 216 ÷ 2.7 = 80

Number of more biscuits in Container T than Container U at first
= 7 u - 3 u
= 4 u
= 4 x 80
= 320

Answer(s): 320