The figure is not drawn to scale. Container M and Container N have base areas of 4000 cm
2 and 3500 cm
2 respectively. Water was poured into an empty rectangular Container M until it reached a height of 26 cm. Some of the water was then poured from Container M into Container N which contained 2.8 ℓ of water until the height of the water in both containers were the same.
- Find the new height of the water in Container M.
- How many litres of water were poured into Container N?
(a)
Volume of water in Container M
= 4000 x 26
= 104000 cm
3 (Base area x height)
1 ℓ = 1000 cm
3 2.8 ℓ = 2800 cm
3 Total volume of Container M and Container N
= 104000 + 2800
= 106800 cm
3 Total base area of Container M and Container N
= 4000 + 3500
= 7500 cm²
Height of Container M after
= 106800 ÷ 7500
= 14.24 cm
(b)
Volume of water in Container N after pouring
= 3500 x 14.24
= 49840 cm
3 Volume of water poured into Container N
= 49840 - 3500
= 46340 cm
3 46340 mℓ = 46.34 ℓ
Answer(s): (a) 14.24 cm; (b) 46.34 ℓ