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

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

Decrease in the volume of water in Container L
= 1 interval of Container L
= 50 mℓ

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

Decrease in the volume of water in Container M
= 4 intervals of Container M
= 4 x 20 
= 80 mℓ

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

Volume of water in Container N at first
= 4 intervals of Container N
= 4 x 100
= 400 mℓ

Total decrease in volume of water in Containers L and M
= 50 + 80
= 130 mℓ

Volume of water in Container N in the end
= 400 + 130
= 530 mℓ

Answer(s): 530 mℓ