Hopefully, you have finished the first blog post I wrote about percentages because this directly follows that one. If you haven’t, it describes how to write percentages in the form of multiplication which is very useful when working with a complicated problem. Here we will discuss the more complicated percentage questions and how to go about solving them. The test likely won’t ask you to solve a simple question like “Decrease Y by X percentage” but is more likely to ask something that involves multiple steps. The two most common are “successive percentages” and “reversing percentages.” Keep reading below to find strategies for solving both of them.
Part 1: Successive Percentages
These questions involve multiple percentages to be taken of the same value to find the answer.
Ex: A book has a retail price P, it is on sale with a 20% discount. Dave uses a super coupon to get an extra 40% off the sale price. What percent of the retail price does Dave pay?
The most common incorrect response I see is P – 20% – 40% = P – 60%, so Dave received a 60% discount. This is wrong because the 40% is a reduction of the sale price, not of the original retail price.Thinking of percentages using the technique in the first blog post will save you here and make it very simple.
P reduced by 20% is P(0.8) and reduced by a further 40% is P(0.8)(0.6) = P(0.48)
So Dave pays 48% of the original retail price.
If you see a problem with a value changed by multiple percentages do not add or subtract the percentages but rewrite them so they can be multiplied!
Part 2: Reversing Percentages
These problems involve receiving the result of a percentage being taken and you need to solve for the original value.
Ex: A population decreased by 30% from 2010 to 2015. The population in 2015 is 150,000, what was the population in 2010?
The most common mistake for this question is simply increasing the population in 2015 by 30%. 150,000*(1.3) = 195,000. This may get you close to the right answer, but it will never be the right answer. Percentages are not reversible in this way. An easy example to think about is below:
100 decreased by 90% is 10. 10 increased by 90% is NOT 100, it’s 19.
Why does it work like this? The simple answer is that 90% of 100 is not equivalent to 90% of 10. The same way 30% of 150,000 is not equal to 30% of the population in 2010.
To solve this properly you should write it as an algebra equation. Let’s call the population in 2010 P. That gives us:
P * (0.7) = 150,000 solving gives us P = 214,285
Rewriting the equations or at least thinking about the algebraically should help a lot when you need to reverse percentages and will save you from making simple mistakes!
Practice: Put the techniques to use and solve these more complicated percentage problems!
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