A parcel's shipping cost is proportional to the square of its volume in cubic meters. Suppose we break the parcel into 3 packages in the ratio $2:3:5$ of its original volume, to qualify for lower shipping cost. If the savings are the equivalent of $\sqrt{6200}$ cubic meters, what was the original volume of the parcel?

Let the original volume be $V$ cubic meters.

Since shipping cost is proportional to the square of volume:

Original shipping cost $= kV^2$ (where $k$ is a constant)

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The parcel is broken into 3 packages with volumes in ratio $2:3:5$.

Total ratio parts $= 2 + 3 + 5 = 10$

Package volumes:

• Package 1: $\dfrac{2}{10}V = \dfrac{1}{5}V$
• Package 2: $\dfrac{3}{10}V$
• Package 3: $\dfrac{5}{10}V = \dfrac{1}{2}V$

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New shipping costs:

• Package 1 cost: $k\left(\dfrac{1}{5}V\right)^2 = k \cdot \dfrac{V^2}{25}$
• Package 2 cost: $k\left(\dfrac{3}{10}V\right)^2 = k \cdot \dfrac{9V^2}{100}$
• Package 3 cost: $k\left(\dfrac{1}{2}V\right)^2 = k \cdot \dfrac{V^2}{4}$

Total new cost $= kV^2\left(\dfrac{1}{25} + \dfrac{9}{100} + \dfrac{1}{4}\right)$

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Converting to common denominator of 100:

$\dfrac{1}{25} = \dfrac{4}{100}$, $\dfrac{9}{100} = \dfrac{9}{100}$, $\dfrac{1}{4} = \dfrac{25}{100}$

$\dfrac{1}{25} + \dfrac{9}{100} + \dfrac{1}{4} = \dfrac{4 + 9 + 25}{100} = \dfrac{38}{100} = \dfrac{19}{50}$

Total new cost $= kV^2 \cdot \dfrac{19}{50}$

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Savings $= kV^2 - kV^2 \cdot \dfrac{19}{50}$

$= kV^2\left(1 - \dfrac{19}{50}\right)$

$= kV^2 \cdot \dfrac{31}{50}$

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The savings are equivalent to $\sqrt{6200}$ cubic meters of shipping cost:

Savings $= k(\sqrt{6200})^2$

$= k \cdot 6200$

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$kV^2 \cdot \dfrac{31}{50} = k \cdot 6200$

$V^2 \cdot \dfrac{31}{50} = 6200$

$V^2 = 6200 \cdot \dfrac{50}{31}$

$V^2 = \dfrac{310000}{31}$

$V^2 = 10000$

$V = 100$

The original volume was $100$ cubic meters.

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The original response sheet shows this:

The question seems broken. The savings figure $\sqrt{6200}$ does not give a proper answer.

The answer key says 10, which works only if savings $= 62$ (i.e. $k=1, V=10 \Rightarrow \frac{31}{50} \times 100 = 62$), meaning the question likely intended to write 62 or $\sqrt{3844}$, not $\sqrt{6200}$.

However, option 4 was awarded marks.