Percentage Increase & Decrease

This article is dedicated to the precise mathematical meaning of common everyday phrases ”$p$ percent larger” and ”$p$ percent smaller.” Despite these being common ways to make comparisons, even many adults struggle to accurately understand their precise meaning. Standardized exams such as the SAT and ACT like to assess this understanding.

Percentage Increase

Percentage Increase

We say ”$A$ is $p$ percent larger than $B$” when

$$A=\left(1+\frac{p}{100}\right)B.$$

For example, if we say ”$A$ is $40\%$ larger than $B$,” we mean that $A=(1+\frac{40}{100})B.$ Put more simply, $A=1.4B$. To illustrate,

• $A=140$ is $40\%$ larger than $B=100$ because $140=1.4\cdot 100$ and
• $A=50.4$ is $40\%$ larger than $B=36$ because $50.4=1.4\cdot 36.$

Example

Question: $50$ is $31\%$ larger than what number?

Solution: Let the unknown number be $x$. So,

$$\begin{array}{c}50=\left(1+\frac{31}{100}\right)x\\ 50=1.31x\\ x=\frac{50}{1.31}=\boxed{38.1679\dots }\end{array}$$

Percentage Decrease

Percentage Increase

We say ”$A$ is $p$ percent smaller than $B$” when

$$A=\left(1-\frac{p}{100}\right)B.$$

For example, if we say ”$A$ is $40\%$ smaller than $B$,” we mean that $A=(1-\frac{40}{100})B.$ Put more simply, $A=0.6B$. To illustrate,

• $A=60$ is $40\%$ smaller than $B=100$ because $60=0.6\cdot 100$ and
• $A=21.6$ is $40\%$ smaller than $B=36$ because $21.6=0.6\cdot 36.$

Warning: "percent larger" and "percent smaller" are not reversible

Note that we saw earlier that $140$ is $40\%$ larger than $100$. However, the reverse is not true. $100$ is not $40\%$ smaller than $140.$ This is explored further in the examples below.

Example 1

Question: What number is $40\%$ smaller than $140$?

Solution: $(1-\frac{40}{100})\cdot 140=0.6\cdot 140=\boxed{84}$

Example 2

Question: What percentage smaller is $100$ than $140$?

Solution: Let the unknown percentage be $p$. Now, use the definition of percentage decrease to set up an equation to solve for $p$.

$$\begin{array}{c}100=\left(1-\frac{p}{100}\right)140\\ \frac{100}{140}=1-\frac{p}{100}\\ \frac{p}{100}=1-\frac{100}{140}\\ p=100\left(1-\frac{100}{140}\right)\approx 28.57\end{array}$$

So, $100$ is approximately $\boxed{28.57\%}$ smaller than $140.$

Quick Drills

Hover over Tap on any question to see the answer.

1. 390 is 95% larger than 200.
2. 52 is 4% larger than 50.
3. 81 is 80% larger than 45.
4. 39 is 56% larger than 25.
5. 66 is 32% larger than 50.
6. 108 is 20% larger than 90.
7. 125 is 25% larger than 100.
8. 63 is 75% larger than 36.
9. 84 is 50% larger than 56.
10. 99 is 50% larger than 66.
11. 80 is 20% smaller than 100.
12. 30 is 76% smaller than 125.
13. 144 is 10% smaller than 160.
14. 6 is 94% smaller than 100.
15. 72 is 20% smaller than 90.
16. 3 is 97% smaller than 100.
17. 51 is 32% smaller than 75.
18. 3 is 85% smaller than 20.
19. 90 is 40% smaller than 150.
20. 196 is 2% smaller than 200.

Practice Problems

Problem 1

For what value $p$ and for what values of $A$ and $B$ can we find that both of the following statements are simultaneously true?

• $A$ is $p\%$ smaller than $B$
• $B$ is $p\%$ larger than $A$

Solution 1

As equations, both of these statements can be written as follows.

$$\{\begin{array}{l}A=\left(1-\frac{p}{100}\right)B\\ B=\left(1+\frac{p}{100}\right)A\end{array}$$

Substitute the second equation into the first to eliminate $B$.

$$\begin{gathered} \cancel{A}=\left(1-\frac{p}{100}\right)\left(1+\frac{p}{100}\right)\cancel{A} \\ 1=\left(1-\frac{p}{100}\right)\left(1+\frac{p}{100}\right) \\ 1=1-\frac{p^2}{10000} \\ 0=-\frac{p^2}{10000} \\ \boxed{p=0} \end{gathered}$$

So, evidently, both of these statements can be true only when

• $A$ is $0\%$ smaller than $B$
• $B$ is $0\%$ larger than $A$

In such a case, any values of $A$ and $B$ such that $\boxed{A=B}$ will do.

Problem 2

A T-shirt is on sale at a 15% discount. The original price is therefore what percentage larger than the sale price? Hint: the answer is not 15%.

Solution 2

Let the original price of the T-shirt be $x.$ The discounted price is thus $x(1-\frac{15}{100})=.85x$. We seek to determine the value of $p$ such that $x$ is $p$ percent larger than $.85x.$ As an equation, this reads as follows.

$$\begin{array}{c}\cancel{x}=\left(1+\frac{p}{100}\right).85\cancel{x}\\ \frac{1}{.85}=1+\frac{p}{100}\\ p=100\left(\frac{1}{.85}-1\right)=17.64705882\dots \end{array}$$

Hence, the original price of the shirt is roughly $\boxed{17.65\%}$ larger than the discounted price.

Problem 3 (Challenging)

Consider three nonzero numbers, $x$, $y$, and $z$.

• $x$ is $10\%$ smaller than the sum of the other two numbers
• $y$ is $30\%$ smaller than the sum of the other two numbers

By what percent smaller is $z$ than the sum of the other two numbers?

Solution 3

The two bulleted statements can be summarized as equations in the following manner.

$$\{\begin{array}{l}x=0.9(y+z)\\ y=0.7(x+z)\end{array}\Rightarrow \{\begin{array}{l}x=\frac{9}{10}(y+z)\\ y=\frac{7}{10}(x+z)\end{array}$$

Let $s=x+y$, the sum of the $x$ and $y$. Substitute $y=s-x$ and then work to eliminate $x$ from the equation, leaving us with only $s$ and $z$.

$$\{\begin{array}{l}x=\frac{9}{10}(s-x+z)\\ s-x=\frac{7}{10}(x+z)\end{array}$$ $$\{\begin{array}{l}\frac{10}{9}x=s-x+z\\ s-x=\frac{7}{10}x+\frac{7}{10}z\end{array}$$ $$\{\begin{array}{l}\frac{19}{9}x=s+z\\ s-x=\frac{7}{10}x+\frac{7}{10}z\end{array}$$ $$\{\begin{array}{l}x=\frac{9}{19}(s+z)\\ s-x=\frac{7}{10}x+\frac{7}{10}z\end{array}$$

Substitute the first equation into the second and solve for $z$.

$$\begin{array}{c}s-\frac{9}{19}(s+z)=\frac{7}{10}\frac{9}{19}(s+z)+\frac{7}{10}z\\ \frac{10}{19}s-\frac{9}{19}z=\frac{63}{190}s+\frac{98}{95}z\\ z=\frac{37}{286}s\end{array}$$

When we say that $z$ is $p\%$ smaller than the sum of $x$ and $y$, we mean $z=\left(1-\frac{p}{100}\right)s.$ Comparing this with the result $z=\frac{37}{286}s,$ we can see that

$$\begin{array}{c}1-\frac{p}{100}=\frac{37}{286}\\ p=\frac{12450}{143}\approx 87.06\end{array}$$

Therefore, $z$ is approximately $\boxed{87.06\%}$ smaller than the sum of $x$ and $y$.