What are common mistakes that students will make when solving "What's the original price" percentage problems?

Question

Take this question for an example:

A smartphone is now $\$500$ after a $20\%$ discount. What was its original price?

Now, this is an example of a type of math problem that students face usually in 7th or 8th grade¹, and usually get the question incorrect due to not being careful.

The solution to this specific one would be as follows:

Since it is $500$ dollars after a $20$ percent discount, that means that $500$ dollars is going to be $80$ percent of the original price. Therefore, we can set up this equation$$0.8x=500$$which then we can multiply both sides by $10/8$ (or $5/4$ or $1.25$) to get that the original price of the smartphone would be$$x=\$625$$

While I am not a teacher, I do remember a few specific mistakes that me and my friends would make on these types of problems back in the day:

1. Using the example question as an example, $\$500$ after a $20\%$ discount implies $0.2x=500$
2. When tasked to find the discount, for example a price that was knocked down to $\$300$ when it was originally $\$700$, doing $300x=700\implies x=\dfrac73\approx233\%$ discount.

However that is all that I can think of, so my question is, what are other common mistakes that students will make when solving "What's the original price" percentage problems?

While I have found a Blackpenredpen video on this, any other references would be appreciated.

I did attempt a few Google searches, but nothing turned up.

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¹In America, this is students age 12-13, although it might be different in other places.

Answer 1

The most common one I think is going from

"The product costs $x$ after a $y$ percent discount"

to trying to calculate the original price by

"original.price $ = x \times (100+y)/100$"

It possibly originates from a certain failure in realising that talking about discounts commonly involves a certain mixture of addition/subtraction and multiplication/division, and this subsequent mistaken application of 'inverse rules/operations'

Answer 2

I agree with ac15's answer (and fedja's comment) describing two specific instances of mistake categories, and I would like to propose a more general framing that explains why students make those mistakes and also suggests other mistakes.

The idea is that in general, many students who pattern-match (as opposed to reasoning from first principles) will tend to prefer solutions requiring fewer and simpler operations, especially if those solutions are

• "directionally correct" in the sense that they make a quantity increase or decrease as expected, and

• "order-of-magnitude correct" in the sense that the final result is on a sensible order of magnitude compared to expectations.

It's essentially the result of applying Occam's razor in the case when common sense has granted you some significant but imperfect level of predictive ability. (As ac15 points out in the comments, this might be better characterized as the principle of least effort.)

Explaining Previously Mentioned Mistake Categories

If the price is $\$500$ after a $20\%$ discount, then what was the original price?

Well, the original price was obviously higher, and the simplest way to get a sensibly higher number using the quantities given in the problem is to just add $20\%$ of $\$500$ to the original price.

If a $\$500$ price is discounted by $10\%$ today and again by $20\%$ tomorrow, what is the price tomorrow?

Well, the discounted price is obviously lower, and a simple way to get a sensibly lower number using the quantities given in the problem is just to total up the discounts ($10\% + 20\% = 30\%$) and then apply that discount.

Suggesting Another Mistake Category

A third category of mistake suggested by this framing (that nobody has mentioned yet) is that when applying a discount, you might also see a student divide by the markup multiplier. For instance, if a $\$500$ price is discounted by $20\%,$ you might see a student divide $\$500 \div 1.2.$

I used to see this occasionally with students who were first learning about percents. If a student gets used to solving two-step problems about percent increase (write $1.\textrm{something}$ and then multiply or divide), then it's easy for them to mistakenly apply the same technique to get a directionally-correct, order-of-magnitude correct answer to a percent decrease problem. That said, this mistake tends to get hammered out quicker than the previous ones because the corresponding problems are more basic / foundational and therefore students get way more practice with them.

Answer 3

Whatever their mistake, they should learn that working with percentages is always about performing multiplications:

• Take $20\%$ of 500 => $\frac{20}{100} \cdot 500 = 0.2 \cdot 500 = 100$
• Add $20\%$ to 500 => $(1+\frac{20}{200}) \cdot 500 = 1.2 \cdot 500 = 600$
• Subtract $20\%$ from 500 => $(1 - \frac{20}{100}) \cdot 500 = 0.8 \cdot 500 = 400$

This will make it more or less "obvious" to not make the "$20\% + 30\%$" mistake, as: $500 + 20\% + 30\% = 500 \cdot 1.2 \cdot 1.3 = 500 \cdot 1.56 = 780$.