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Suppose I am selling boxes of cereal.

Each box has $m$ grams of cereal and costs $p$ pounds.

Two possible promotions:

  • Decrease the price of the box by $20$%
  • Put $20$% more cereal in each box.

Which offer would be better value for money?

My attempt:

Value for money is usually $v = \frac{m}{p}$ grams per pound.

In scenario 1:

Value for money changes to $\frac{m}{0.8p} = 1.25v$

In scenario 2:

Value for money changes to $\frac{1.2m}{p} = 1.2v$

So scenario 1 is better value for money?

Is my answer correct?

The reason I am unsure is that it doesn't seem 'obvious' to me that why reducing the price is better than increasing the size of the box. Since I am applying a $20$% promotion in both cases, I would have thought that both promotions are the same value for money but the maths doesn't say that.

Is this outcome 'obvious' to anyone?

Khashayar Baghizadeh
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Ben Derby
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    Not really an answer, but maybe taking it to the extreme case will help intuition. Which is a better value for the buyer: Increase the size of the box by $100%$, or reduce the price by $100%$? Increasing the box size doubles the value, whereas if the box is free the value is infinite. – DreiCleaner Aug 09 '20 at 11:53

2 Answers2

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Suppose we have $x$ amount in a box initially, and to make things concrete let's say we have a price of 5 pounds, and a total budget of 20 pounds.

Initially I get $\frac{20}{5}=4$ boxes, so $4x$ amount.

After the 20% price reduction, we have a new price of 4 pounds, so I get 5 boxes, and so $5x$ amount of cereal.

After the volume increase (keeping the price) we get 4 boxes still of $\frac{6}{5}$x each so $\frac{24}{5}x = 4.8x$ amount.

So for the same budget I get more cereal after the price decrease, so I'd call that better "value for money", as the same money buys me more stuff.

Henno Brandsma
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In your two scenarios, you have

Case $1$ (decrease price): $v=\frac{m}{p}\cdot\frac{1}{0.8}$

Case $2$ (increase quantity): $v=\frac{m}{p}\cdot\frac{1.2}{1}$

So it comes down to which of $\frac{1}{0.8}$ or $\frac{1.2}{1}$ is bigger. Evidently $\frac{1}{0.8}$ is the winner.

In the general case, we take $0<x<1$ and compare $\frac{1}{1-x}$ and $\frac{1+x}{1}$.

We can cross multiply to compare. We can see $1>1-x^2$, which indicates that $\frac{1}{1-x}$ is the greater of the two fractions.

DreiCleaner
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