# Volume question on cuboid and sphere

algebra-precalculusvolume

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?

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.