A fish tank is in the shape of a cuboid with base $80$cm by $40$cm. The height of the tank is $35$cm.

i) Find the volume of the water in the tank when it is filled up to the depth of $30$cm

$$V= 80 \times 40 \times 30 = 96000\ \text{cm}^3$$

ii) $2400$ glass marbles with a radius of $0.5$cm, are placed in the tank. How much will the water level rise?

The answer to this question is $0.393$cm ($3$sf). However, I just don’t know how to approach the working of this part of the question. This question is from a past paper of an exam I’m studying for. It provides the answers to the questions but not the actual working which is how I know the answer. Could anyone do and explain the working for this question to justify the answer?

## Best Answer

Firstly: find the Scale Factor($SF$):

$$30\cdot (SF)^{3} = 96000$$

Evaluating this we have that:

$$(SF)^{3} = \frac{96000}{30} = 14.736$$

Now that we have a $SF$ we need to find the volume of the marbles. Using the formula: $$\frac{4}{3}\pi r^{3}$$

Subbing in we find that $$\frac{4}{3}\cdot\pi\cdot0.5^{3} = 0.5236$$ Since there are $2400$ marbles we multiply by that quantity yielding $1256.64$.

Next we add this to our original volume of $96000$ which gives us $$96000 + 1256.64= 97,256.64 $$

Using our $SF$ to find the height again and letting Height $= h$:

$$h\cdot (SF)^{3} = 97,256.64$$

Therefore $h = 30.393$ after some simplification.

To get the final answer we have to subtract the original height from the final height. This gives us: $30.393 - 30 = 0.393$. Which according to your paper is the answer.