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

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

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

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

Decrease in the volume of water in Container L
= 2 intervals of Container L
= 2 x 20 
= 40 mℓ

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

Volume of water in Container M at first
= 6 intervals of Container M
= 6 x 100
= 600 mℓ

Total decrease in volume of water in Containers K and L
= 100 + 40
= 140 mℓ

Volume of water in Container M in the end
= 600 + 140
= 740 mℓ

Answer(s): 740 mℓ