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Percent increase and decrease

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5115666
Read the news report and evaluate its claim. “Bicycle theft has risen dramatically in our town! The number of reported thefts increased by \(100\%\) from last year.” One bicycle theft was reported last year, and two were reported this year. Explain why the headline could create a misleading impression.

Hints

- Find the actual increase in the number of reports. - Compare the effect of saying “one more case” with saying “a \(100\%\) increase.” - Consider how a small starting value affects percent change.

Solution

1. The number increased by \(2 - 1 = 1\) theft. 2. Relative to last year’s \(1\) theft, the percent increase is \(\frac{1}{1} \times 100\% = 100\%\), so the numerical claim is correct. 3. However, the absolute number only changed from \(1\) to \(2\). With a very small starting value, a small absolute change can produce a large percent increase. The word “dramatically” may therefore exaggerate the scale of the situation.

Answer

The \(100\%\) increase is mathematically correct, but the headline is potentially misleading because the reported count rose by only one case, from \(1\) to \(2\).
5117486
Decrease a duration of \(1\,\text{h}\ 40\,\text{min}\) by \(15\%\). Give the result in hours and minutes.

Hints

- Would it be easier to convert the entire duration to minutes first? - How many minutes are in one hour? - A decrease means the percent amount must be subtracted.

Solution

1. Convert the duration to minutes: \(1\,\text{h}\ 40\,\text{min} = 60\,\text{min} + 40\,\text{min} = 100\,\text{min}\). 2. Find the decrease: \(15\%\) of \(100\,\text{min}\) is \(100\,\text{min} \times 0.15 = 15\,\text{min}\). 3. Subtract the decrease: \(100\,\text{min} - 15\,\text{min} = 85\,\text{min}\). 4. Convert back to hours and minutes: \(85\,\text{min} = 1\,\text{h}\ 25\,\text{min}\).

Answer

\(1\,\text{h}\ 25\,\text{min}\)
5127786
A smartphone battery has a capacity of \(4500\,\text{mAh}\). a) After one hour of heavy use, \(3330\,\text{mAh}\) remains. What percent of the original charge remains? b) After one more hour in power-saving mode, \(2830.5\,\text{mAh}\) remains. By what percent did the charge decrease during the second hour, based on the charge after the first hour?

Hints

- Identify the whole used in each part. - For part a, compare the remaining charge with the original capacity. - For part b, first find the change in charge, then compare it with the charge at the start of the second hour.

Solution

1. For part a, the remaining portion is \(\frac{3330}{4500} = 0.74 = 74\%\). 2. During the second hour, the charge decreases by \(3330\,\text{mAh} - 2830.5\,\text{mAh} = 499.5\,\text{mAh}\). 3. Relative to the \(3330\,\text{mAh}\) starting charge for that hour, the percent decrease is \(\frac{499.5}{3330} \times 100\% = 15\%\).

Answer

a) \(74\%\) of the original charge remains. b) The charge decreased by \(15\%\) during the second hour.
5127796
A small city registered \(120\) new electric vehicles last year and \(168\) this year. a) Find the percent increase in new electric-vehicle registrations. b) The city has \(2400\) registered vehicles altogether. What percent of all registered vehicles is represented by this year’s \(168\) new electric vehicles? c) Last year, electric vehicles made up exactly \(25\%\) of all \(480\) newly registered vehicles. Verify that this is consistent with the stated \(120\) electric vehicles.

Hints

- Identify the base value in each part. - For percent increase, compare the increase with last year’s number. - To verify part c, find \(25\%\) of the stated total.

Solution

1. For part a, the increase is \(168 - 120 = 48\). The percent increase is \(\frac{48}{120} \times 100\% = 40\%\). 2. For part b, \(\frac{168}{2400} \times 100\% = 7\%\). 3. For part c, \(480 \times 0.25 = 120\), so the statement is consistent.

Answer

a) Registrations increased by \(40\%\). b) This year’s new electric vehicles represent \(7\%\) of all registered vehicles. c) Yes. \(25\%\) of \(480\) is \(120\).
5127806
A company has two locations with different numbers of apprentices. Location A: \(80\) employees, including \(12\) apprentices. Location B: \(120\) employees, including \(15\) apprentices. a) Find the percent of employees who are apprentices at each location. Which location has the greater percent? b) Location A plans to increase its number of apprentices by \(25\%\). How many apprentices will it then have?

Hints

- Compare the number of apprentices with the total number of employees at each location. - A percent increase is added to the original amount.

Solution

1. At Location A, the apprentice percent is \(\frac{12}{80} \times 100\% = 15\%\). 2. At Location B, the apprentice percent is \(\frac{15}{120} \times 100\% = 12.5\%\). Location A has the greater percent. 3. A \(25\%\) increase in \(12\) apprentices is \(12 \times 0.25 = 3\). The new number is \(12 + 3 = 15\).

Answer

a) Location A: \(15\%\); Location B: \(12.5\%\). Location A has the greater percent. b) Location A will have \(15\) apprentices.
5354956
A utility company analyzes the annual electricity use of an average household. The household uses \(3500\,\text{kWh}\) per year, divided among the categories shown in the pie chart. a) How many kilowatt-hours are used for cooking? b) Replacing the household’s lights with efficient LEDs reduces the lighting electricity use by \(60\%\). How many kilowatt-hours are saved per year? c) What is the household’s new total annual electricity use after this savings?
Figure for problem 535495

Hints

- Read the current category percentages from the pie chart. - The \(60\%\) reduction applies only to lighting use, not to the total electricity use. - Subtract the saved amount from the original total.

Solution

1. For part a, cooking uses \(15\%\) of the total: \(3500\,\text{kWh} \times 0.15 = 525\,\text{kWh}\). 2. Lighting currently uses \(10\%\) of the total: \(3500\,\text{kWh} \times 0.10 = 350\,\text{kWh}\). 3. For part b, the savings is \(60\%\) of the lighting use: \(350\,\text{kWh} \times 0.60 = 210\,\text{kWh}\). 4. For part c, the new total use is \(3500\,\text{kWh} - 210\,\text{kWh} = 3290\,\text{kWh}\).

Answer

a) \(525\,\text{kWh}\) b) \(210\,\text{kWh}\) c) \(3290\,\text{kWh}\)

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