Home / United States / Math Classes / 7th Grade Math / Percent of Increase and Decrease
Percent is a term that we often use to express score, rate of interest, and other statistics. A change in the original quantity will lead to the change in the percentage value. Here we will learn to calculate the percent of change, percent of increase, and percent of decrease with the help of some real-life examples....Read MoreRead Less
Percents are often used to express scores, discounts, and various statistics. The word “percent” means “one part in every hundred”. In math, a percentage is a ratio or a number expressed as a fraction of 100. The symbol used to denote percent is “%”.
A percent is calculated by dividing the number by the whole and then multiplying the result by 100. To find the percentage of a fraction \( \frac{a}{b} \), we need to divide a by b, and then multiply the result by 100.
Suppose a student’s math score is 43 out of 50. To express this as a percentage, we need to express the student’s score as a fraction.
Math score \( =\frac{43}{50} \)
Now, we need to divide \( 43\div 50 \) and then multiply the result by 100.
Percentage of marks \( =\frac{43}{50}\times 100=86\% \)
A percent of change is a measure of the change in quantity from the original amount.
\(\text{Percent of change}=\frac{\text{Amount of change}}{\text{Original amount}}\times 100\)
Percent of increase and percent of decrease are measures of the percentage of change. As the name suggests, the percent of change is calculated only when there is a change in quantity.
When the change in the original quantity is positive, that is, when the quantity increases, the percent of change is known as the percent of increase.
\( \text{Percent of increase }=\frac{\text{New amount – Original amount}}{\text{Original amount}}\times 100 \)
For example, if the student in the previous example gets 46 marks instead of 43 marks, the percent of the score will change as well.
In this case, \( \text{Percent of increase }=\frac{\text{New amount – Original amount}}{\text{Original amount}}\times 100 \)
\( =\frac{46-43}{50}\times 100\)
\( =\frac{3}{50}\times 100\)
\( =0.06\times 100=6\%\)
Therefore, the percentage of increase in the marks is \( 6\%\).
When the change in the original quantity is negative, that is, it decreases, the percent of change is known as the percent of decrease.
\( \text{Percent of decrease }=\frac{\text{Original amount – New amount}}{\text{Original amount}}\times 100 \)
The difference between the assumed value and the exact value is known as the error. This error can be expressed as a percent, known as the percent error.
Suppose a political party was expected to win 58 seats in an election and ended up winning only 52 seats. The absolute error in the prediction is 6 seats. This error can also be expressed using percent.
Percent error can be calculated using the following formula.
\( \text{Percent error }=\frac{\text{Amount of error}}{\text{Actual amount}}\times 100 \)
Example 1: Kaleb made an error while counting the number of sheep in his farm. He counted 48 sheep, even though there are 52 sheep. Find the percentage error in the number of sheep.
Solution:
\( \text{Percent error }=\frac{\text{Amount of error}}{\text{Actual amount}}\times 100 \)
Here, the amount of the error in the number of sheep is the difference between the actual number of sheep and the number of sheep according to Kaleb’s observation.
Actual number of sheep = 52
Number of sheep according to Kaleb = 48
So, Amount of error = 52 – 48 = 4
\( \text{Percent error }=\frac{\text{Amount of error}}{\text{Actual amount}}\times 100 \)
\( = \frac{4}{52}\times 100 \)
\( = 7.7\% \)
Therefore, Kaleb made a \( = 7.7\% \) error while counting the number of sheep.
Example 2: The table shows the number of hours Natalie studied in the last week. What is the percent of change in the time she spent studying from Wednesday to Thursday?
Solution:
From the table, it is clear that the number of hours Natalie spent studying dropped by half an hour from Wednesday to Thursday.
\( \text{Percent of decrease}=\frac{\text{Original amount – New amount}}{\text{Original amount}}\times 100 \)
\( =\frac{2-1.5}{2}\times 100 \) Substitute values in equation
\( =\frac{0.5}{2}\times 100 \) Simplify
\( =0.25\times 100=25\% \)
This shows us that Natalie spent 25% less time studying on Thursday when compared to the time she spent studying on Wednesday.
Example 3: Oliver was 5 feet 6 inches tall when he was 15 years old. When he reached the age of 22, his height increased to 6 feet 2 inches. What is the percent of change in his height?
Solution:
To make the comparison of heights easy, let’s convert the given heights into a single unit.
1 foot is equivalent to 12 inches.
Oliver’s initial height = 5 feet 6 inches
= 5 × 12 + 6 = 60 + 6 = 66 inches
Oliver’s current height = 6 feet 2 inches
= 6 × 12 + 2 = 72 + 2 = 74 inches
\( \text{Percent of increase}=\frac{\text{New amount – Original amount}}{\text{Original amount}}\times 100 \)
\( =\frac{74-66}{66}\times 100\) Substitute values in equation
\( =\frac{8}{66}\times 100\) Simplify
\( =12.12\% \)
Therefore, Oliver’s height increased by \( 12.12\% \).
Example 4: A swimming pool in the shape of a rectangle is 65 feet long and 30 feet wide. If the length is increased to 75 feet and the width is increased to 40 feet, find the percent of increase in the area and the perimeter of the swimming pool.
Solution:
Original length of the swimming pool = 65 feet
Original width of the swimming pool = 30 feet
Original perimeter of the swimming pool = 2 \( \times \) (length + width)
= 2 \( \times \) (65 + 30) Substitute values in equation
= 2 \( \times \) 95 Solving the parenthesis
= 190 feet Multiply
Original area of the swimming pool = length \( \times \) width
= 65 \( \times \) 30 Substitute values in equation
= 1950 sq. feet Multiply
New length of the swimming pool = 75 feet
New width of the swimming pool = 40 feet
New perimeter of the swimming pool = 2 \( \times \) (length + width)
= 2 \( \times \) (75 + 40) Substitute values in equation
= 2 \( \times \) 115 Solving the parenthesis
= 230 feet Multiply
New area of the swimming pool = length \( \times \) width
= 75 \( \times \) 40 Substitute values in equation
= 3000 sq. feet Multiply
\( \text{Percent of increase in perimeter}=\frac{\text{New amount – Original amount}}{\text{Original amount}}\times 100 \)
\( =\frac{230-190}{190}\times 100 \) Substitute values in equation
\( =\frac{40}{190}\times 100 \) Simplify
\( =21\% \)
\( \text{Percent of increase in area}=\frac{\text{New amount – Original amount}}{\text{Original amount}}\times 100 \)
\( =\frac{3000-1950}{1950}\times 100 \) Substitute values in equation
\( =\frac{1050}{1950}\times 100 \) Simplify
\( =53.8\% \)
With the new dimensions, the perimeter of the swimming pool increased by \( 21\% \) and its area increased by \( 53.8\% \).
Let’s take a look at the formulas.
\( \text{Percent of increase}=\frac{\text{New amount – Original amount}}{\text{Original amount}}\times 100 \)
\(\text{Percent of decrease}=\frac{\text{Original amount – New amount}}{\text{Original amount}}\times 100\)
As you can see, the formulas are almost the same. The RHS of the percent of increase equation is actually the negative of the RHS of the percent of decrease formula. This is because we are supposed to get a negative sign for the percent decrease as the original amount is always bigger than the new amount.
Percent of change is often used to represent changes in the national population, GDP, employment rate, and so on. Percent error is quite common in scientific estimations like the diameter of the sun, the distance between two stars, and so on.
Percent of error can be either positive or negative. It depends on whether the error caused by the assumption is positive or negative.