Is it better value to decrease the price of a box by $20$% or increase its size by $20$%? 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?
 A: 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.
A: 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.
