PSLE
At first, Containers T, U and V contained water as shown in Figure 1. Some water from T and U was poured into Container V without spilling. The amount of water left in Container T and U is shown in Figure 2. What would be the amount of water in Container V?
1 ℓ = 1000 mℓ

Container T 
10 intervals = 500
1 interval = 500 ÷ 10 = 50

Decrease in the volume of water in Container T
= 2 intervals of Container T
= 2 x 50
= 100 mℓ

Container U
10 intervals = 200
1 interval = 200 ÷ 10 = 20

Decrease in the volume of water in Container U
= 3 intervals of Container U
= 3 x 20 
= 60 mℓ

Container V
10 intervals = 1000
1 interval = 1000 ÷ 10 = 100

Volume of water in Container V at first
= 5 intervals of Container V
= 5 x 100
= 500 mℓ

Total decrease in volume of water in Containers T and U
= 100 + 60
= 160 mℓ

Volume of water in Container V in the end
= 500 + 160
= 660 mℓ

Answer(s): 660 mℓ