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Lecture Comments (3)

2 answers

Last reply by: Professor Fung
Mon May 20, 2013 2:33 AM

Post by Jeff Mitchell on January 7, 2011

I question the results on extra example II. The problem states that there was a decrease of 7 to 2. So, it appears that the original value would have been 7 + 2 or 9. the change was a decrease of 7 which would require the following solution 7/9 * p/100 or 77.7%
~Jeff

Percent of Change

  • A percent of change may be an increase or a decrease.
  • You can use a proportion to find a percent of change. Here is the formula:
  • Amount of Change/Original Amount = Percent of Change/100
  • Use these steps to find the percent of change.
    1. Find the amount of change
    2. Write a proportion.
    3. Find cross products
    4. Solve for the variable
    5. Change decimal into percent

Percent of Change

A class had 25 students in student council last year. This year, the class had 34 students in student council. Find the percent increase in the number of students in student council.
  • 34 − 25 = 9
  • [9/25] = [p/100]
  • 25p = 900
p = 36%
Find the percent increase from 75 to 165
  • 165 − 75 = 90
  • [90/75] = [p/100]
  • 75p = 9,000
p = 120%
Find the percent decrease from 180 to 49 to the nearest hundredth.
  • 180 − 49 = 131
  • [131/180] = [p/100]
  • 180p = 13,100
p = 72.78%
Drew grew 7 inches in one year and is now 5 ft 8 in. What is the percent increase in Drew's height to the nearest hundredth?
  • Drew used to be 5 ft 8 in - 7 in = 5 ft 1 in = 61 in
    Drew is now 5 ft 8 in = 68 in
  • 68 − 61 = 7
  • [7/61] = [p/100]
  • 61p = 700
p = 11.48%
The water level of the pool used to be at 65 in. Now, the water level is 9 in lower. Find the percent decrease in the level of water to the nearest hundredth.
  • [9/65] = [p/100]
  • 65p = 900
p = 13.85%
The record length for the long jump used to be 10 ft 5 in. The newest long jump record is now 11 ft 1 in. Find the percent increase of the long jump to the nearest hundredth.
  • 10 ft 5 in = ? in
  • 10 ft ×[12 in/1 ft] + 5 in = 125 in
  • 11 ft 1 in = ? in
  • 11 ft ×[12 in/1 ft] + 1 in = 133 in
  • 133 − 125 = 8
  • [8/125] = [p/100]
  • 125p = 800
p = 6.4%
Find the percent decrease from 87 to 30 to the nearest hundredth.
  • 87 − 30 = 57
  • [57/87] = [p/100]
  • 87p = 5,700
p = 65.52%
Find the percent increase of 46 to 109 to the nearest hundredth.
  • 109 − 46 = 63
  • [63/109] = [p/100]
  • 109p = 6,300
p = 57.80%
A company's annual earnings was $ 4.25 million in 2010 and $ 4.13 million in 2011. Find the percent decrease in the company's annual earnings to the nearest hundredth.
  • 4,250,000 − 4,130,000 = 120,000
  • [120,000/4,250,000] = [p/100]
  • 4,250,000p = 12,000,000
p = 2.82%
Find the percent increase from 4 to 132.
  • 132 − 4 = 128
  • [128/4] = [p/100]
  • 4p = 12,800
p = 3,200%

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Percent of Change

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • What You'll Learn and Why 0:05
    • Topics Overview
  • Vocabulary 0:17
    • Percent of Change
    • Amount of Change/ Original Amount
  • Finding a Percent of Increase 1:04
    • Example: Find the Percent of Increase
  • Converting Units 2:56
    • Example: Converting Units and Percent Increase
  • Finding a Percent of Decrease 5:08
    • Example: Find the Percent of Decrease
  • Extra Example 1: Find the Percent of Increase 6:32
  • Extra Example 2: Find the Percent of Decrease 7:30
  • Extra Example 3: Find the Percent of Increase 8:28
  • Extra Example 4: Find the Percent of Decrease 10:23