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This is a very simple maths problem, which doesn’t seems simple to me when I just put some extra thoughts on it. So need to borrow some extra brains for this.

Question: If cost of coffee in my office was Dollar 0 till yesterday and today it is Dollar 2.5 then what is the percentage increase in the cost of coffee?

Initially it seemed like 2.5% increase, simply by distributing number from 0 to 100, but this is not the case.

Some suggested it is infinitely more than the previous value, because we are calculating percentage by dividing by zero. But what about this if the cost of coffee was $0.25? Isn’t still infinitely more! Something is wrong here.

PS: Yes, coffee were free in our office but not anymore, and we are still recovering from this shocker.

kishoredbn
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4 Answers4

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The relative increase from $a$ to $b$, which is given wrt $a$, is computed in percent as

$$100\,\frac{b-a}a.$$

Hence,

  • $0$ to $2.5\to100\,\dfrac{2.5-0}{0}=\infty\%$,

  • $0.25$ to $2.5\to100\,\dfrac{2.5-0.25}{0.25}=900\%$.

3

Let's say the initial cost of coffee in your office was $\$0$ per cup. Now a cup of coffee costs $\$2.5$. We could use the following formula to try to come up with the answer for the increase in the price of coffee in terms of percents:

$$ \text{cost_now}=\text{initial_cost}+\frac{\text{initial_cost}}{100}\cdot\text{percentage_change}. $$

$$$2.5=$0+\frac{$0}{100}\cdot p\implies 0\cdot p=2.5.$$

There is no real number such that when you multiply it by zero, it gives you a nonzero quantity. I guess, one possible interpretation of this result would be that with the initial cost of $\$0$, the concept of increase/decrease in percentage does not really apply. In other words, when you're talking about the increase/decrease in the cost of something percentage-wise, you necessarily need an initial cost that's greater than zero as the basis for your calculations. If there is no initial cost, there can be no calculations.

Michael Rybkin
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Short answer: percent change is undefined when the starting quantity is $0$. Calling it a $2.5\%$ increase makes no more sense than calling it a $\frac{\pi^2}{\sqrt{17}}\%$ increase (why would you distribute anything on a $(0,100)$ interval?)

If you want a bit of an intuition, imagine it raised from $\$1$ to $\$1.10$. Take the starting prize, increase it by $5\%$. You are still below the current prize. Go back and increse again, this time by $6\%$ instead. You are still below. When you get to $10\%$, you stop being below the current prize.

Now start at $0$. Increase $0$ by 10%. You are below the current prize. Now try $20\%$, $50\%$, $1,000,000\%$. You still haven't reached the current prize, and you never will. Therefore if makes sense to talk about an "infinitely big percentage increase"

David
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$$\text{margin}=1-({1\over {1+\text{markup}}}) $$ Or equivalently,$$\text{markup}= ({1\over {1-\text{margin}}})-1$$ where markup and margin are written in decimal form (0.25=25% etc.). If we have 100% margin (1 in decimal), then we hit a division by 0 error. It's undefined if the margin comes from simply setting a price after having it at 0.