**Volume Transfer**

This math revision note aims to help your child become equipped with the ability to visualise liquid transfer and solve related math problems. In particular, the following 2 types of liquid transfer questions will be covered:

- Liquid transfer until equal height in both containers (Question 1)
- Liquid transfer until one container’s height is twice that of the other container (Question 2)

**Before you read on, you might want to download this entire revision notes in PDF format to print it out for your child, or to read it later.**

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__Question 1__

Y and Z are rectangular containers with base area 160 cm^{2} and 480 cm^{2} respectively.

At first, Y contained water to a height of 80 cm and Z was empty.

Kayla poured some water from Y into Z until the height of the water level in Z is the same as that of Y.

What was the height of the water level of each container in the end?

Find the volume of water available

80 cm x 160 cm^{2} = 12 800 cm^{3}

This amount of water is to be shared between both containers such that the water level in both containers becomes the same.

Since the water level eventually becomes the same, we can imagine Y and Z placed side by side, effectively merging both Y and Z into one container.

This container now has a **combined base area** in which the water level comes to a certain height.

We can use this combined base area to find the eventual water level.

Combined based area = 160 cm^{2} + 480 cm^{2 }= 640 cm^{2}

Water level in the end = 12 800 cm^{3} ^{ }÷ 640 cm^{2 }= 20 cm (Ans)

__Question 2__

Y and Z are rectangular containers with base area 160 cm^{2} and 480 cm^{2} respectively.

At first, Y contained water to a height of 80 cm and Z was empty.

Kyle poured some water from Y into Z. The height of the water level in Y became twice that of Z.

What was the height of the water level in Z?

Find the volume of water available 80 cm x 160 cm^{2} = 12 800 cm^{3}

This amount of water is to be shared between both containers such that the water level in Y is twice that of Z.

So, when we compare the eventual heights of both containers, the height of Y will be 2 units while the height of Z will be 1 unit.

Effectively, it means that for every 1 layer of 480 cm^{2} (Z’s base), there are 2 layers of 160 cm^{2} (Y’s base).

Using this logic, we can form a volume ratio with each container’s final height multiplied by the respective base area.

Ratio of the volume of water

Y | : | Z |

2u x 160 | : | 1u x 480 |

320u | : | 480u |

2u | : | 3u |

With the most simplified volume ratio, we can see that the water will be shared across Y and Z in this proportion:

Y has 2 units of water while Z has 3 units of water.

2u + 3u = 5u

The total units (5u) represent the volume of water to be shared (12 800 cm^{3}).

Since the question is asking for Z’s final water level, we can first find the eventual volume of water in Z, which is 3u.

5u = 12 800 cm^{3 }

1u = 12 800 cm^{3} ÷ 5 = 2560 cm^{3 }

3u = 2560 cm^{3} x 3 = 7680 cm^{3}

Then, to find the final water level in Z, we just need to divide it by its base area.

7680 cm^{3} ÷ 480 cm^{2} = 16 cm (Ans)

**Before you go, you might want to download this entire revision notes in PDF format to print it out for your child, or to read it later.**

**This will be delivered to your email inbox.**

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