I have some products that I want to increase in value such that a 20% discount gives their current value. It's been ~25 years since college algebra and so I'm a bit rusty on setting up the equation.

I've been trying to figure out how to solve for X being the percentage increase needed in order that 20% off would give the current value.

For example a product worth 100. I know a 20% increase would make it 120, but 20% off of that would be 96 which isn't 100.

I'd give a bounty for explaining the algebra and steps to figure it out, but I'm new to this exchange and am unable to award one - thanks if you spend the time to explain this to me!

And if someone wouldn't mind tagging this appropriately for this exchange I'd appreciate it.

  • $\begingroup$ You wouldn't be trying to mislead your customers, would you? Is that allowed where you come from? $\endgroup$ – TonyK May 31 '11 at 19:36
  • $\begingroup$ Actually, I'm thinking of setting up my website so independent sales people can sell my products, and their customers enter a code that gives them a 20% discount but then I also know who's code it is and they get their commission. So, their customers get the actual price and someone that doesn't use the sales code gets jacked (the 25% increase we figured out below), aka I'm not undercutting the sales people. $\endgroup$ – Chuck Savage May 31 '11 at 20:00
  • $\begingroup$ OK, sorry I misinterpreted you. $\endgroup$ – TonyK May 31 '11 at 20:22

Increasing an amount $A$ by $X$ percent means adding $A\cdot \frac{X}{100}$ to $A$, resulting in $A+A\cdot\frac{X}{100}=A\left(1+\frac{X}{100}\right)$. Decreasing an amount $B$ by $Y$ percent means subtracting $B\cdot\frac{Y}{100}$, resulting in $B-B\cdot\frac{Y}{100}=B\left(1-\frac{Y}{100}\right)$. To have an $X$ percent increase followed by a $20$ percent decrease with an initial amount $A$, you will first multiply by $1+\frac{X}{100}$ to obtain a new amount. If we call that amount $B$, then the next step is to decrease $B$ by $20$ percent by multiplying by $1-\frac{20}{100}$. At this point you will have $A\cdot \left(1+\frac{X}{100}\right)\cdot\left(1-\frac{20}{100}\right)$. For this to leave you where you started, you need to solve the equation $$A\cdot \left(1+\frac{X}{100}\right)\cdot\left(1-\frac{20}{100}\right) =A.$$ You can cancel $A$ from both sides, leaving an equation $$\frac{4}{5}\left(1+\frac{X}{100}\right)=1$$ with $X$ as the only unknown, which can then be solved by division, subtraction, and multiplication. Does that get you where you want to be?

  • $\begingroup$ I'm reading over it trying to figure it out. How come it isn't A^2 in the first equation, since you had an A for both the plus and the minus? $\endgroup$ – Chuck Savage May 31 '11 at 19:21
  • $\begingroup$ @ChuckSavage: When decreasing by 20%, the amount that had been called "$A$" when decreasing by $Y$ percent won't be the same as the original $A$, but instead will be the amount that results from the $X$ percent increase, namely $\displaystyle{A\cdot\left(1+\frac{X}{100}\right)}$. For example, if you were to start with $A$, then increase by 7%, then decrease by 23%, then increase by 12%, you would go through the sequence $A$, $A\cdot 1.07$, $A\cdot 1.07\cdot 0.77$, and finally $A\cdot 1.07\cdot 0.77\cdot 1.12$. I will edit to clarify. $\endgroup$ – Jonas Meyer May 31 '11 at 19:23
  • $\begingroup$ Yes - thank you. How do you write those equations... I came up for X = ((1/(1-20/100)-1)*100) and my math is so rusty, I don't know what to do with that right side to reduce it. $\endgroup$ – Chuck Savage May 31 '11 at 19:28
  • $\begingroup$ @Chuck: In general, simplifying $\displaystyle{\frac{1}{\left(1-\frac{Y}{100}\right)}}$ by hand won't be that neat, depending on what $Y$ is. In this case, it is a little easier if you simplify as follows: $\displaystyle{1-\frac{20}{100}=1-\frac{1}{5}=\frac{5}{5}-\frac{1}{5}=\frac{4}{5}}$, and $\displaystyle{\frac{1}{\left(\frac{4}{5}\right)}=\frac{5}{4}}$. Then $\frac{5}{4}-1=\frac{5}{4}-\frac{4}{4}=\frac{1}{4}$, and finally $100\cdot\frac{1}{4}=25$. Working this out in decimal form also wouldn't be too bad. $\endgroup$ – Jonas Meyer May 31 '11 at 19:36
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    $\begingroup$ @Jonas ty again, I worked through it on my own and came up with the same. These exchanges are great! If you ever have C#/asp.net coding questions head over to stackoverflow.com and I'd be happy to answer them! $\endgroup$ – Chuck Savage May 31 '11 at 19:52

Let's work first with your particular numbers. Suppose that an item is originally priced at 100 (dollars). Imagine that you inrease the price by $x$ percent. Then the new price is $100+x$.

You want to make sure that if you apply a 20 percent discount to this new price $100+x$, you end up with a price of exactly $100$ dollars.

A $20$ percent discount on a price of $100+x$ means that the price will be $80$ percent of $100+x$.

So the new price is $$(100+x)\frac{80}{100}, \qquad\text{or equivalently,} \qquad \frac{(100+x)(80)}{100}$$

It so happens that you want this new price to be $100$ dollars.

This gives you the equation $$\frac{(100+x)(80)}{100}=100.$$

You would like to "extract" $x$ from this equation.

First multiply both sides by $100$. On the left, you get simply $(100+x)(80)$. On the right, you get $(100)(100)$, which is $10000$.

So our new equation is $$(100+x)(80)=10000.$$

So something, namely $100+x$, multiplied by $80$, is $10000$. What is the something?

The idea is to divide both sides by $80$. On the left, you get $100+x$. On the right, you get $10000/80$. By calculator or by hand division, you get that $10000/80=125$.

So our new equation is $$100+x=125.$$

Now subtract $100$ from both sides. We get


This tells you that you must apply a $25$ percent markup so that a $20$ percent discount will leave the price unchanged.

This may look long, but that is only because I have done the calculations in great detail.

Note Since we are dealing with percentages, the answer is independent of the actual initial price, which for simplicity we took to be $100$.

Let's use the same reasoning to solve a different and harder problem. You want the final discount to be say $17$ percent. Let us ask what percent markup there should be so that at the end, after the discount, you end up selling the originally $100$ dollar item for $105$ dollars. The process will be almost exactly the same, except that the numbers will be a lot uglier, so you will have to use a calculator.

Let the desired markup be $x$ percent. Then the price is $100+x$. You want to apply a $17$ percent discount to that. So the new price would be $83$ percent of $100+x$. You want this new price to be $105$. So $83$ percent of $100+x$ is $105$.

That gives you the equation $$\frac{(100+x)(83)}{100}=105.$$

You want to extract $x$ from this equation. The procedure is in outline much the same as before. First multiply both sides by $100$. So our new equation is $$(100+x)(83)=10500.$$ Now divide both sides by $83$. We will not get a simple integer on the right, so I will round off, and sloppily still write "$=$" when I mean almost equal. If you are following this with a calculator, you should get something like $$100+x=126.506.$$ Now, like before, subtract $100$ from both sides. We get $$x=26.506.$$ Of course this is absurd precision. For all practical purposes, the markup should be $26.5$ percent.

I hope there is enough detail in the above calculations to enable you to solve problems of the same general kind with not much difficulty.

  • $\begingroup$ thanks for writing this all out, it was easy to follow. Also completely different from Jonas approach, but with the same answer. Reviewing his equation, his is the same as yours except = 1 and not 100. $\endgroup$ – Chuck Savage May 31 '11 at 20:52

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