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  1. Home
  2. 12th grade
  3. Algebra I
  4. Exercise : Find a percent of change

Find a percent of change Algebra I

Before a sale, Victoria went to the store and saw a dress that cost $122. She decided to go back during the sale and sees that it is now $85.

What percent of the price has been cut, rounded to 10^{-2} ?

In order to find a percent change between two values v_{1} and v_{2}, calculate the ratio :

\dfrac{\left| v_{1}-v_{2} \right|}{v_{1}}

Here:

  • v_1=122
  • v_2=85

Therefore:

\dfrac{\left| v_{1}-v_{2} \right|}{v_{1}}=\dfrac{\left| 122-85 \right|}{122}\approx0.3\ 033

The price of the dress was reduced by approximately 30.33% of the original price.

Ben planted a tree in his yard that was 48 inches tall. After one year, the height of the tree was 69 inches.

What percent did the tree increase, rounded to 10^{-2} ?

In order to find a percent change between two values v_{1} and v_{2}, calculate the ratio :

\dfrac{\left| v_{1}-v_{2} \right|}{v_{1}}

Here:

  • v_1=48
  • v_2=60

Therefore:

\dfrac{\left| v_{1}-v_{2} \right|}{v_{1}}=\dfrac{\left| 48-60 \right|}{48}=43.75

The height of the tree increased 43.75%.

John's weight last year was 160 pounds. After one year of diet and exercise, John's weight is now 140 pounds.

What percent of weight did John lose, rounded to 10^{-2} ?

In order to find a percent change between two values v_{1} and v_{2}, calculate the ratio :

\dfrac{\left| v_{1}-v_{2} \right|}{v_{1}}

Here:

  • v_1=160
  • v_2=140

Therefore:

\dfrac{\left| v_{1}-v_{2} \right|}{v_{1}}=\dfrac{\left| 160-140 \right|}{160}=0.125

John's weight decreased 12.5%.

Last year, the average temperature in September in New York was 60 degrees Fahrenheit. This year, the average temperature in September in New York is 66 degrees Fahrenheit.

What percent did the average temperature in September increase this year compared to last year, rounded to 10^{-2} ?

In order to find a percent change between two values v_{1} and v_{2}, calculate the ratio :

\dfrac{\left| v_{1}-v_{2} \right|}{v_{1}}

Here:

  • v_1=60
  • v_2=66

Therefore:

\dfrac{\left| v_{1}-v_{2} \right|}{v_{1}}=\dfrac{\left| 60-66 \right|}{60}=0.1

The average temperature in September increased 10%.

Last year, the number of people that got the flu in United States was approximately 24 million. This year, the number of people that got the flu in United States is approximately 20 million.

What percent did the number of people with flu in United States decrease this year as compared to last year, rounded to 10^{-2} ?

In order to find a percent change between two values v_{1} and v_{2}, calculate the ratio :

\dfrac{\left| v_{1}-v_{2} \right|}{v_{1}}

Here:

  • v_1=24
  • v_2=20

Therefore:

\dfrac{\left| v_{1}-v_{2} \right|}{v_{1}}=\dfrac{\left| 24-20 \right|}{24}=0.1\ 667

The number of people with flu decreased 16.67%.

The number of gallons of gas used in United States increased in one year from approximately 120 million to 150 million.

What percent did the number of gallons of gas used increase, rounded to 10^{-2} ?

In order to find a percent change between two values v_{1} and v_{2}, calculate the ratio :

\dfrac{\left| v_{1}-v_{2} \right|}{v_{1}}

Here:

  • v_1=120
  • v_2=150

Therefore:

\dfrac{\left| v_{1}-v_{2} \right|}{v_{1}}=\dfrac{\left| 120-150 \right|}{120}=0.25

The number of gallons used increased 25%.

According to a research company survey, the number of people with diabetes in the world decreased in the last year from approximately 400 million to 380 million.

What percent did the number of people with diabetes decrease, rounded to 10^{-2} ?

In order to find a percent change between two values v_{1} and v_{2}, calculate the ratio :

\dfrac{\left| v_{1}-v_{2} \right|}{v_{1}}

Here:

  • v_1=400
  • v_2=380

Therefore:

\dfrac{\left| v_{1}-v_{2} \right|}{v_{1}}=\dfrac{\left| 400-380 \right|}{400}=0.05

The number of people with diabetes decreased 5%.

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See also
  • Course : Manipulating quantities
  • Exercise : Determine if two ratios are equivalent
  • Exercise : Find a unit rate
  • Exercise : Find a unit price
  • Exercise : Find a value after a certain percent of change
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