At first, Containers S, T and U contained water as shown. Some water from Containers T and U was poured into Container S without any spilling over. The volume of water left in Containers T and U is shown. What would be the volume of water in Container S?
1 ℓ = 1000 mℓ

Volume of water at each division on Container S
= 1000 ÷ 10
= 100 mℓ

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

Volume of water at each division on Container T
= 800 ÷ 10
= 80 mℓ

Volume of water in Container T
= 8 x 80
= 640 mℓ

Volume of water at each division in Container U
= 500 ÷ 10
= 50 mℓ

Amount of water contained in Container U
= 9 x 50
= 450 mℓ

Total volume of water in Container T and Container U
= 640 + 450
= 1090 mℓ

Volume of water in Container T after pouring
= 6 x 80
= 480 mℓ

Volume of water in Container U after pouring
= 6 x 50
= 300 mℓ

Volume of water in Container T and Container U
= 480 + 300
= 780 mℓ

Volume of water poured in Container S
= 1090 - 780
= 310 mℓ

Total volume of water in Container S
= 600 + 310
= 910 mℓ

Answer(s): 910 mℓ