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Grade 6 math worksheets

Build your own math worksheets from 35,000+ problems for grades 3 to 12, from fractions to AP Calculus. Every problem comes with step-by-step solutions.

Find part given percent and whole

Learning goals

  • Interprets percent as a rate per 100.
  • Connects percent, fraction, and decimal representations.
  • Finds the part when the percent and whole are known.
  • Checks the part against the size of the whole and percent.

Click problems to add them to your worksheet.

Find part given percent and whole 1
Find each percent amount. a) \(10\%\) of \(350\,\text{m}\) b) \(25\%\) of \(\$48\)

Hints

- Write the percent as a decimal or fraction. - Multiply the whole by the percent. - Is there an easy division for \(10\%\) or \(25\%\)?

Solution

1. For a), \(350\,\text{m} \times 0.10 = 35\,\text{m}\). 2. For b), \(\$48 \times 0.25 = \$12\).

Answer

a) \(35\,\text{m}\) b) \(\$12\)
Find part given percent and whole 2
A box holds \(18\) oranges. A label says \(50\%\) are ripe. How many oranges are ripe? Explain how the percent helps you find the number.

Hints

- How is the given percentage related to a familiar fraction? - What equal share of the box does that fraction describe?

Solution

1. \(50\%\) means half of the oranges. 2. Split \(18\) into two equal groups: \(18\div2=9\) oranges.

Answer

\(9\) oranges; \(50\%\) is one half of the box.
Find part given percent and whole 3
Ten percent of the \(90\) seedlings in a tray have sprouted. How many seedlings have sprouted?

Hints

- How could you describe \(10\%\) as a fraction of the whole? - Can equal groups make this easier to work out?

Solution

1. Ten percent is one tenth of the whole. 2. \(90\div10=9\) seedlings have sprouted.

Answer

\(9\) seedlings
Find part given percent and whole 4
A board game has \(20\) cards. Exactly \(25\%\) are action cards. How many action cards are there?

Hints

- What simple fraction is the same as \(25\%\)? - Use that fraction to split the cards into equal groups.

Solution

1. \(25\%=\frac{1}{4}\). 2. A quarter of \(20\) cards is \(20\div4=5\).

Answer

\(5\) action cards
Find part given percent and whole 5
Two fruit drinks are available. Drink A has a total volume of \(400\,\text{mL}\) and is \(20\%\) fruit juice. Drink B has a total volume of \(500\,\text{mL}\) and is \(15\%\) fruit juice. Which bottle contains more fruit juice? Justify your answer with calculations.

Hints

- Find the amount of fruit juice in each bottle separately. - Write each percent as a decimal or fraction. - Compare the two amounts.

Solution

1. Find the fruit juice in Drink A: \(20\%\) of \(400\,\text{mL}\) is \(0.20 \times 400\,\text{mL} = 80\,\text{mL}\). 2. Find the fruit juice in Drink B: \(15\%\) of \(500\,\text{mL}\) is \(0.15 \times 500\,\text{mL} = 75\,\text{mL}\). 3. Since \(80\,\text{mL} > 75\,\text{mL}\), Drink A contains more fruit juice.

Answer

Drink A contains more fruit juice: \(80\,\text{mL}\), compared with \(75\,\text{mL}\) in Drink B.
Find part given percent and whole 6
A smartphone battery has a total capacity of \(3200\,\text{mAh}\). The phone shows that \(14\%\) of the charge remains. How many milliamp-hours of charge remain?

Hints

- What are the whole amount and the percent in this situation? - How do you find a part when the whole and percent are known? - You could first find \(1\%\) or \(10\%\).

Solution

1. Write the percent as a decimal: \(14\% = 0.14\). 2. Multiply the whole by the percent: \(3200 \times 0.14 = 448\). 3. The battery has \(448\,\text{mAh}\) of charge remaining.

Answer

\(448\,\text{mAh}\) of charge remains.
Find part given percent and whole 7
For a whole amount of \(\$480\), find the amounts represented by \(1\%\), \(10\%\), \(25\%\), \(50\%\), and \(75\%\). Briefly explain how the \(75\%\) amount can be found from the \(25\%\) and \(50\%\) amounts.

Hints

- Think about the fraction of the whole represented by each percent. - Can one percent amount be built by adding other percent amounts? - How does finding \(1\%\) help?

Solution

1. \(1\%\) of \(\$480\): \(\$480 \times 0.01 = \$4.80\). 2. \(10\%\) of \(\$480\): \(\$480 \times 0.10 = \$48\). 3. \(25\%\) of \(\$480\): \(\$480 \times 0.25 = \$120\). 4. \(50\%\) of \(\$480\): \(\$480 \times 0.50 = \$240\). 5. \(75\%\) of \(\$480\): \(\$480 \times 0.75 = \$360\). 6. Since \(25\% + 50\% = 75\%\), add the two amounts: \(\$120 + \$240 = \$360\).

Answer

\(1\%: \$4.80\); \(10\%: \$48\); \(25\%: \$120\); \(50\%: \$240\); \(75\%: \$360\). The \(75\%\) amount is the sum of the \(25\%\) and \(50\%\) amounts.
Find part given percent and whole 8
A school has \(600\) students. Find the number of students in each club. Use results from earlier parts when helpful. a) Sports club: \(10\%\) of the students b) Drama club: \(5\%\) of the students c) Choir: \(20\%\) of the students d) Chess club: \(15\%\) of the students

Hints

- Start with an easy percent such as \(10\%\). - What are half and twice \(10\%\)? - Can you write one percent as a sum of percents you already found?

Solution

1. For a), \(10\%\) of \(600\) is \(600 \times 0.10 = 60\). 2. For b), \(5\%\) is half of \(10\%\), so \(60 \div 2 = 30\). 3. For c), \(20\%\) is twice \(10\%\), so \(60 \times 2 = 120\). 4. For d), \(15\% = 10\% + 5\%\), so \(60 + 30 = 90\).

Answer

a) \(60\) students b) \(30\) students c) \(120\) students d) \(90\) students
Find part given percent and whole 9
Find each part. a) \(3\%\) of \(1200\,\text{kg}\) b) \(22\%\) of \(\$450\) c) \(150\%\) of \(60\,\text{m}\) d) \(4.5\%\) of \(200\,\text{L}\) e) \(0.8\%\) of \(5000\,\text{m}^2\)

Hints

- Write each percent as a decimal first. - What does a percent greater than \(100\%\) mean for the result? - Be careful when converting \(0.8\%\) to a decimal. - Include the correct unit in each answer.

Solution

1. Write the percents as decimals: \(0.03\), \(0.22\), \(1.5\), \(0.045\), and \(0.008\). 2. Multiply each whole by its decimal: a) \(1200\,\text{kg} \times 0.03 = 36\,\text{kg}\) b) \(\$450 \times 0.22 = \$99\) c) \(60\,\text{m} \times 1.5 = 90\,\text{m}\) d) \(200\,\text{L} \times 0.045 = 9\,\text{L}\) e) \(5000\,\text{m}^2 \times 0.008 = 40\,\text{m}^2\)

Answer

a) \(36\,\text{kg}\) b) \(\$99\) c) \(90\,\text{m}\) d) \(9\,\text{L}\) e) \(40\,\text{m}^2\)
Find part given percent and whole 10
Find each percent amount. 1) \(19\%\) of \(300\) 2) \(51\%\) of \(80\) 3) \(124\%\) of \(50\) 4) \(5\%\) of \(140\)

Hints

- Convert each percent to a decimal or fraction. - A part can be more than the whole when the percent is more than \(100\%\). - Compare the sizes of the percent and the whole to check each answer.

Solution

1. \(0.19\times300=57\). 2. \(0.51\times80=40.8\). 3. \(1.24\times50=62\). 4. \(0.05\times140=7\).

Answer

1) \(57\) 2) \(40.8\) 3) \(62\) 4) \(7\)
Find part given percent and whole 11
Investigate these percent problems. a) Which is greater: \(15\%\) of \(60\) or \(60\%\) of \(15\)? Justify your answer with calculations. b) Find \(120\%\) of \(75\). Explain why the result is greater than \(75\).

Hints

- Calculate both expressions in part a). - What does a percent greater than \(100\%\) mean? - You can also solve by setting up a proportion.

Solution

1. For a), \(15\%\) of \(60\) is \(0.15 \times 60 = 9\). Also, \(60\%\) of \(15\) is \(0.60 \times 15 = 9\). The amounts are equal. 2. For b), \(120\%\) of \(75\) is \(1.2 \times 75 = 90\). 3. Since \(120\%\) is greater than \(100\%\), the part is greater than the whole amount, \(75\).

Answer

a) The amounts are equal; both are \(9\). b) The amount is \(90\). It is greater than \(75\) because \(120\%\) is greater than \(100\%\).
Find part given percent and whole 12
A juice pitcher holds \(2\,\text{L}\). It is currently \(75\%\) full of apple juice. a) How many milliliters of apple juice are in the pitcher? b) What portion of the pitcher is empty? Write the answer as a fraction in simplest form and as a percent.

Hints

- How many milliliters are in one liter? - What simple fraction is equivalent to \(75\%\)? - Subtract the filled percent from \(100\%\) to find the empty portion.

Solution

1. Convert the capacity: \(2\,\text{L} = 2000\,\text{mL}\). 2. Find \(75\%\) of the capacity: \(2000\,\text{mL} \times 0.75 = 1500\,\text{mL}\). 3. The empty portion is \(100\% - 75\% = 25\%\). 4. Convert the empty portion to a fraction: \(25\% = \frac{25}{100} = \frac{1}{4}\).

Answer

a) \(1500\,\text{mL}\) b) \(\frac{1}{4}\), or \(25\%\)
Find part given percent and whole 13
A survey asked students to choose their favorite recess snack. The results were \(50\%\) fruit, \(25\%\) granola bars, and \(25\%\) sandwiches. a) Find the central angle for each category in a circle graph. b) Describe the shape of the fruit sector and explain why it has that shape.

Hints

- A full circle is \(360^\circ\). - Multiply \(360^\circ\) by each percent written as a decimal. - Think about the part of a circle represented by \(50\%\).

Solution

1. The fruit angle is \(360^\circ\times0.50=180^\circ\). 2. The granola-bar angle is \(360^\circ\times0.25=90^\circ\). 3. The sandwich angle is \(360^\circ\times0.25=90^\circ\). 4. The fruit sector is a semicircle because \(180^\circ\) is half of \(360^\circ\).

Answer

a) Fruit: \(180^\circ\); granola bars: \(90^\circ\); sandwiches: \(90^\circ\) b) The fruit sector is a semicircle because its angle is half of a full circle.
Find part given percent and whole 14
A class surveyed how students travel to school: \(40\%\) walk, \(20\%\) ride a bicycle, \(30\%\) take the bus, and \(10\%\) ride in a car. a) Find the central angle for each category in a circle graph. b) Andrés says, “The bicycle sector must be exactly twice as large as the car sector.” Use the angles to determine whether Andrés is correct.

Hints

- A full circle corresponds to \(100\%\) and \(360^\circ\). - Relate each category's percent to the same fraction of a full turn. - For part b), compare the bicycle and car angles multiplicatively.

Solution

1. Walking: \(360^\circ\times0.40=144^\circ\). 2. Bicycle: \(360^\circ\times0.20=72^\circ\). 3. Bus: \(360^\circ\times0.30=108^\circ\). 4. Car: \(360^\circ\times0.10=36^\circ\). 5. Since \(72^\circ=2\times36^\circ\), the bicycle sector is exactly twice as large as the car sector.

Answer

a) Walking: \(144^\circ\); bicycle: \(72^\circ\); bus: \(108^\circ\); car: \(36^\circ\) b) Andrés is correct because \(72^\circ=2\times36^\circ\).
Find part given percent and whole 15
A school has \(240\) seventh-grade students. A survey asked about their hobbies so the school can plan after-school clubs. a) \(25\%\) of the students said they play soccer. How many students is that? b) One-third of the students play a musical instrument. How many students is that? c) \(15\%\) of the students are interested in drama club. How many students is that? d) Which of the three groups is largest? Briefly explain.

Hints

- To find a part of a whole, multiply the whole by the fraction or percent. - How can you use one-third in a calculation with the total number of students? - Write each percent as a decimal before multiplying. - Compare the three numbers you find.

Solution

1. The number of soccer players is \(240 \times 0.25 = 60\). 2. The number of students who play an instrument is \(240 \div 3 = 80\). 3. The number interested in drama club is \(240 \times 0.15 = 36\). 4. Since \(80 > 60 > 36\), the group that plays a musical instrument is largest.

Answer

a) \(60\) students b) \(80\) students c) \(36\) students d) The musical-instrument group is largest, with \(80\) students.
Find part given percent and whole 16
A metal spoon has a mass of \(45\,\text{g}\). It is \(80\%\) silver, and the rest is copper. How many grams of pure copper are in the spoon?

Hints

- The silver and copper portions make up the whole spoon. - Find the percent that is copper. - Multiply the total mass by the copper portion.

Solution

1. The copper portion is \(100\% - 80\% = 20\%\). 2. The copper mass is \(45\,\text{g} \times 0.20 = 9\,\text{g}\).

Answer

The spoon contains \(9\,\text{g}\) of copper.
Find part given percent and whole 17
An electric scooter can travel \(25\) miles under ideal conditions. In very cold weather, its range decreases to \(72\%\) of the ideal range. Find the scooter's range in cold weather.

Hints

- Identify the whole distance. - Write \(72\%\) as a decimal and multiply by the ideal range.

Solution

The cold-weather range is \(72\%\) of \(25\) miles: \(25\,\text{mi} \times 0.72 = 18\,\text{mi}\).

Answer

The scooter's cold-weather range is \(18\) miles.
Find part given percent and whole 18
Jordan receives \(\$40\) each month as an allowance. The circle graph shows how Jordan plans to use the money. a) How many dollars does Jordan save each month? b) How many dollars does Jordan spend on entertainment? c) What fraction of the allowance is planned for the phone? Write the fraction in simplest form.
Figure for problem 531882

Hints

- Describe what the circle graph represents. - How do you find a percent of a total amount? - Write the phone percent as a fraction over \(100\), then simplify.

Solution

1. Jordan saves \(30\%\) of \(\$40\): \(\$40 \times 0.30 = \$12\). 2. Jordan spends \(35\%\) of \(\$40\) on entertainment: \(\$40 \times 0.35 = \$14\). 3. The phone portion is \(25\%\). As a fraction, \(25\% = \frac{25}{100} = \frac{1}{4}\).

Answer

a) \(\$12\) b) \(\$14\) c) \(\frac{1}{4}\)
Find part given percent and whole 19
A school surveyed \(120\) students about their favorite free-time activity. The circle graph shows the percent distribution of the responses. a) How many students chose sports? b) How many students chose making music?
Figure for problem 531884

Hints

- Read the needed percents from the circle graph. - What number represents the whole group? - Multiply the whole by each percent written as a decimal. - You can also set up a proportion.

Solution

1. From the graph, sports represents \(40\%\) and music represents \(15\%\). The whole group is \(120\) students. 2. Sports: \(120 \times 0.40 = 48\). 3. Music: \(120 \times 0.15 = 18\).

Answer

a) \(48\) students chose sports. b) \(18\) students chose making music.
Find part given percent and whole 20
A total of \(800\) students took a survey about their favorite free-time activity. The circle graph shows the percent distribution of the responses. Find the exact number of students in each of the five categories.
Figure for problem 531891

Hints

- What number represents the whole group? - How many students represent \(1\%\) of the group? - Use simple fractions for percents such as \(25\%\) when helpful. - Check that your category counts add to \(800\).

Solution

1. The whole group is \(800\) students. 2. Sports: \(800 \times 0.35 = 280\). 3. Video games: \(800 \times 0.25 = 200\). 4. Listening to music: \(800 \times 0.20 = 160\). 5. Reading: \(800 \times 0.15 = 120\). 6. Other: \(800 \times 0.05 = 40\).

Answer

Sports: \(280\) students Video games: \(200\) students Listening to music: \(160\) students Reading: \(120\) students Other: \(40\) students
Find part given percent and whole 21
All Grade 6 students at a school were surveyed about their favorite sport. A total of \(200\) students participated. The circle graph shows the percent distribution of the responses. Find the exact number of students in each category.
Figure for problem 531898

Hints

- Multiply the total number of students by each percent. - What simple fractions are equivalent to \(25\%\) and \(20\%\)? - Finding \(10\%\) first may make some calculations easier. - Check that the counts add to \(200\).

Solution

1. Use the whole group of \(200\) students. 2. Soccer: \(200 \times \frac{40}{100} = 80\) students. 3. Gymnastics: \(200 \times \frac{25}{100} = 50\) students. 4. Swimming: \(200 \times \frac{20}{100} = 40\) students. 5. Other: \(200 \times \frac{15}{100} = 30\) students.

Answer

Soccer: \(80\) students Gymnastics: \(50\) students Swimming: \(40\) students Other: \(30\) students
Find part given percent and whole 22
A class is planning a field trip with a total cost of \(\$4000\). The circle graph shows how the total cost is divided among four categories. Find the actual cost of each category.
Figure for problem 531905

Hints

- Read each percent from the circle graph. - What amount represents \(100\%\)? - Multiply the total cost by each percent. - Finding \(10\%\) first may help.

Solution

1. The whole amount is \(\$4000\). 2. Lodging and meals: \(\$4000 \times 0.45 = \$1800\). 3. Transportation: \(\$4000 \times 0.30 = \$1200\). 4. Admission fees: \(\$4000 \times 0.15 = \$600\). 5. Other: \(\$4000 \times 0.10 = \$400\).

Answer

Lodging and meals: \(\$1800\) Transportation: \(\$1200\) Admission fees: \(\$600\) Other: \(\$400\)
Find part given percent and whole 23
A package of trail mix weighs \(350\,\text{g}\). The circle graph shows the percent of each ingredient. Find the number of grams of each ingredient in the package.
Figure for problem 531909

Hints

- Read the percent for each ingredient. - What is the total weight of the package? - Multiply the total weight by each percent. - Finding \(10\%\) first may help. - Check that the ingredient weights add to \(350\,\text{g}\).

Solution

1. The whole weight is \(350\,\text{g}\). 2. Peanuts: \(350\,\text{g} \times 0.40 = 140\,\text{g}\). 3. Raisins: \(350\,\text{g} \times 0.30 = 105\,\text{g}\). 4. Almonds: \(350\,\text{g} \times 0.20 = 70\,\text{g}\). 5. Hazelnuts: \(350\,\text{g} \times 0.10 = 35\,\text{g}\).

Answer

Peanuts: \(140\,\text{g}\) Raisins: \(105\,\text{g}\) Almonds: \(70\,\text{g}\) Hazelnuts: \(35\,\text{g}\)
Find part given percent and whole 24
A class of \(40\) students was surveyed about their favorite pets. The circle graph shows the percent of students who chose each pet. a) How many students chose dogs? b) How many students chose cats? c) How many students chose rodents, birds, or reptiles altogether?
Figure for problem 531911

Hints

- Identify the total number of students. - Read the percent for each requested group from the circle graph. - To find a part when you know the whole and the percent, multiply the whole by the percent written as a decimal. - For part c), first add the percentages for the three pets.

Solution

1. Dogs represent \(45\%\) of the class: \(40 \times 0.45 = 18\). 2. Cats represent \(30\%\) of the class: \(40 \times 0.30 = 12\). 3. Rodents, birds, and reptiles represent \(15\% + 5\% + 5\% = 25\%\). Then \(40 \times 0.25 = 10\).

Answer

a) \(18\) students b) \(12\) students c) \(10\) students
Find part given percent and whole 25
A school surveyed how its \(240\) students travel to school in the morning. The circle graph shows the results. a) How many students take the bus? b) How many students either ride a bicycle or walk?
Figure for problem 531936

Hints

- Read the percent for “Bus” from the graph. - To find a part of a total, multiply the total by the percent written as a decimal. - For part b), you may add the two percentages first or find both student counts separately and then add them.

Solution

1. The bus represents \(40\%\) of the students: \(240 \times 0.40 = 96\). 2. Riding a bicycle or walking represents \(20\% + 25\% = 45\%\). Then \(240 \times 0.45 = 108\).

Answer

a) \(96\) students b) \(108\) students
Find part given percent and whole 26
In a survey, \(250\) people were asked which country they would most like to visit on vacation. The circle graph shows the results. a) How many people chose Germany? b) How many people chose Italy, and how many chose Spain?
Figure for problem 535486

Hints

- Identify the whole in this problem. - To convert a percent to a number of people, multiply the total number of people by the percent written as a decimal. - You may use a proportion or multiply directly.

Solution

1. Germany represents \(40\%\) of the responses: \(250 \times 0.40 = 100\). 2. Italy represents \(30\%\) of the responses: \(250 \times 0.30 = 75\). 3. Spain represents \(20\%\) of the responses: \(250 \times 0.20 = 50\).

Answer

a) \(100\) people b) Italy: \(75\) people; Spain: \(50\) people
Find part given percent and whole 27
A farm sold \(4500\,\text{kg}\) of produce last month. The circle graph shows the percent of the total mass represented by each crop. a) How many kilograms of potatoes were sold? b) How many kilograms of onions were sold? c) How many kilograms of apples and carrots were sold altogether?
Figure for problem 535494

Hints

- The whole is the total mass of produce sold. - Write each percent as a decimal before multiplying. - When two categories refer to the same whole, you can add their percentages before finding the combined amount.

Solution

1. Potatoes represent \(40\%\) of the total: \(4500\,\text{kg} \times 0.40 = 1800\,\text{kg}\). 2. Onions represent \(10\%\) of the total: \(4500\,\text{kg} \times 0.10 = 450\,\text{kg}\). 3. Apples and carrots together represent \(30\% + 20\% = 50\%\). Then \(4500\,\text{kg} \times 0.50 = 2250\,\text{kg}\).

Answer

a) \(1800\,\text{kg}\) b) \(450\,\text{kg}\) c) \(2250\,\text{kg}\)
Find part given percent and whole 28
A household produced \(600\,\text{kg}\) of waste last year. The circle graph shows how the waste was divided among five categories. a) How many kilograms of food and yard waste were collected? b) How many kilograms of trash were collected? c) Which categories each weighed more than \(100\,\text{kg}\)?
Figure for problem 535496

Hints

- Multiply the total mass by each percent written as a decimal. - For part c), compare each calculated mass with \(100\,\text{kg}\).

Solution

1. Food and yard waste: \(600\,\text{kg} \times 0.30 = 180\,\text{kg}\). 2. Trash: \(600\,\text{kg} \times 0.10 = 60\,\text{kg}\). 3. Find the other category amounts: paper is \(600\,\text{kg} \times 0.25 = 150\,\text{kg}\), plastic is \(600\,\text{kg} \times 0.20 = 120\,\text{kg}\), and glass is \(600\,\text{kg} \times 0.15 = 90\,\text{kg}\). 4. The categories above \(100\,\text{kg}\) are food and yard waste, paper, and plastic.

Answer

a) \(180\,\text{kg}\) b) \(60\,\text{kg}\) c) Food and yard waste, paper, and plastic
Find part given percent and whole 29
Jordan has \(\$40\) to spend each month. The circle graph shows how Jordan used the money last month. Find the amount spent or saved in each category.
Figure for problem 535540

Hints

- The whole circle represents \(\$40\). - Multiply \(\$40\) by each category’s percent written as a decimal.

Solution

1. Snacks: \(\$40 \times 0.15 = \$6\). 2. Movies: \(\$40 \times 0.25 = \$10\). 3. Games: \(\$40 \times 0.40 = \$16\). 4. Savings: \(\$40 \times 0.20 = \$8\). 5. Check: \(\$6 + \$10 + \$16 + \$8 = \$40\).

Answer

Snacks: \(\$6\); movies: \(\$10\); games: \(\$16\); savings: \(\$8\)
Find part given percent and whole 30
Three roommates share a monthly household budget of \(\$1200\). The circle graph shows how they plan to spend it. How much do they budget for groceries, and how much do they budget for entertainment?
Figure for problem 535560

Hints

- Read the two percentages from the circle graph. - Multiply the total budget by each percent written as a decimal. - The entire circle represents \(100\%\) of the budget.

Solution

1. Groceries are \(30\%\) of the budget: \(\$1200 \times 0.30 = \$360\). 2. Entertainment is \(20\%\) of the budget: \(\$1200 \times 0.20 = \$240\).

Answer

They budget \(\$360\) for groceries and \(\$240\) for entertainment.
Find part given percent and whole 31
An apartment building collected \(1500\,\text{kg}\) of waste in one week. The circle graph shows the percent in each category. Find the mass of the paper and compost categories.
Figure for problem 535621

Hints

- Multiply the total mass by each percent written as a decimal. - You can use \(10\%\) of \(1500\,\text{kg}\) to estimate whether each result is reasonable.

Solution

1. Paper: \(1500\,\text{kg} \times 0.42 = 630\,\text{kg}\). 2. Compost: \(1500\,\text{kg} \times 0.30 = 450\,\text{kg}\).

Answer

Paper: \(630\,\text{kg}\) Compost: \(450\,\text{kg}\)
Find part given percent and whole 32
A club of \(120\) members holds an election. The circle graph shows each member's response. Members marked “Abstained” submitted a blank ballot, while members marked “Did not participate” submitted no ballot. 1. How many members voted for Candidate A? 2. What percent of the members took part by submitting a ballot? 3. How many members submitted no ballot?
Figure for problem 535636

Hints

- Find the candidate's sector on the graph. - Think about which of the other groups actually turned in a ballot. - The circle accounts for all club members.

Solution

1. Candidate A received \(60\%\) of \(120\), which is \(72\) members. 2. Members who voted for A, voted for B, or submitted blank ballots participated: \(60\%+25\%+10\%=95\%\). 3. The remaining \(5\%\) submitted no ballot: \(0.05\times120=6\) members.

Answer

1. \(72\) members 2. \(95\%\) 3. \(6\) members
Find part given percent and whole 33
A family used \(3600\,\text{kWh}\) of electricity last year. The circle graph shows how the electricity was used. How many kilowatt-hours were used for heating, and how many were used for lighting?
Figure for problem 535639

Hints

- The total electricity use represents \(100\%\). - Write each percent as a decimal and multiply by \(3600\,\text{kWh}\).

Solution

1. Heating represents \(70\%\) of the total: \(3600\,\text{kWh} \times 0.70 = 2520\,\text{kWh}\). 2. Lighting represents \(5\%\) of the total: \(3600\,\text{kWh} \times 0.05 = 180\,\text{kWh}\).

Answer

Heating: \(2520\,\text{kWh}\); lighting: \(180\,\text{kWh}\)
Find part given percent and whole 34
A library has \(40\) extra books. It gives \(20\%\) of them to a nearby library. How many of the extra books does it keep?

Hints

- The donated books are not the same as the books remaining. - Find the number moved first. - Compare what was moved with the starting amount.

Solution

1. The number donated is \(20\%\) of \(40\), or \(\frac{1}{5}\times40=8\) books. 2. The books left are \(40-8=32\).

Answer

\(32\) books are kept.
Find part given percent and whole 35
The shaded sections show the part of a set of \(60\) stickers that is metallic. How many metallic stickers are in the set?
Figure for problem 559318

Hints

- Read the shaded fraction from the strip. - The whole set is sixty, not ten. - Use the fraction shown to find the corresponding part of sixty.

Solution

1. The strip has \(10\) equal sections, with \(3\) shaded. 2. The metallic fraction is \(\frac{3}{10}=30\%\). 3. The number of metallic stickers is \(\frac{3}{10}\times60=18\).

Answer

\(18\) stickers
Find part given percent and whole 36
A school needs \(\$1200\) for a “Green Schoolyard” project. Students have already raised \(40\%\) of the cost through a fun run. A community grant will cover \(\frac{1}{3}\) of the total cost. The PTA will donate another \(\$350\). Is there enough money for the project? Find how many dollars the total is below or above the goal.

Hints

- Find the dollar amounts for the fun run and the community grant. - Add all three contributions. - Compare the total with the project cost.

Solution

1. Find the amount raised by the fun run: \(40\%\) of \(\$1200\) is \(0.40 \times \$1200 = \$480\). 2. Find the community grant: \(\frac{1}{3}\) of \(\$1200\) is \(\$400\). 3. Add all contributions: \(\$480 + \$400 + \$350 = \$1230\). 4. Compare the total with the goal: \(\$1230 - \$1200 = \$30\). 5. The project has enough money and is \(\$30\) above the goal.

Answer

Yes. The total is \(\$1230\), which is \(\$30\) above the \(\$1200\) goal.
Find part given percent and whole 37
A class has \(25\) students. Of the students, \(60\%\) have a pet. Of the students who have a pet, \(40\%\) have a dog. How many students in the class have a dog?

Hints

- Break the problem into two steps. - How many students have any pet? - Use that result as the whole for the second percent.

Solution

1. Find the number of students who have a pet: \(25 \times 0.60 = 15\). 2. Use \(15\) as the new whole. Find \(40\%\) of \(15\): \(15 \times 0.40 = 6\). 3. Therefore, \(6\) students have a dog.

Answer

\(6\) students have a dog.
Find part given percent and whole 38
A school has \(1200\) students. Of the students, \(35\%\) play a musical instrument. Exactly one third of those students play piano. How many students at the school play piano?

Hints

- First find how many students play any instrument. - How do you find one third of a number? - You can write the percent as a decimal or fraction.

Solution

1. Find the number of students who play an instrument: \(1200 \times 0.35 = 420\). 2. Find one third of that group: \(420 \times \frac{1}{3} = 140\).

Answer

\(140\) students play piano.
Find part given percent and whole 39
A region uses \(5{,}000{,}000\,\text{m}^3\) of water each year. Industry uses \(18\%\) of that amount. Of the water used by industry, \(\frac{2}{5}\) is used for cooling. How many cubic meters of water are used for cooling each year?

Hints

- First find industry's portion of the total water use. - Then find \(\frac{2}{5}\) of that amount. - Keep track of the zeros in the large numbers.

Solution

1. Find the water used by industry: \(5{,}000{,}000\,\text{m}^3 \times 0.18 = 900{,}000\,\text{m}^3\). 2. Find the portion used for cooling: \(900{,}000\,\text{m}^3 \times \frac{2}{5} = 360{,}000\,\text{m}^3\).

Answer

\(360{,}000\,\text{m}^3\) of water are used for cooling each year.
Find part given percent and whole 40
Two sports clubs compare their memberships. Club A has \(400\) members, and \(65\%\) are children. Club B has \(520\) members, and \(55\%\) are children. Lucas says, “Because \(65\%\) is greater than \(55\%\), Club A must have more children than Club B.” Check his claim with calculations and explain the result.

Hints

- Find the actual number of children in each club. - Does the number depend only on the percent, or also on the whole? - Compare your two results with Lucas's claim.

Solution

1. Find the number of children in Club A: \(400 \times 0.65 = 260\). 2. Find the number of children in Club B: \(520 \times 0.55 = 286\). 3. Compare: \(286 > 260\). 4. Lucas is incorrect. Club B has a larger total membership, so its smaller percent still represents a greater number of children.

Answer

Lucas is incorrect. Club A has \(260\) children, and Club B has \(286\) children. The larger whole in Club B makes \(55\%\) of its membership greater than \(65\%\) of Club A's membership.
Find part given percent and whole 41
A school library has \(800\) books. Of the books, \(40\%\) are novels and \(25\%\) are nonfiction. All remaining books are children’s books. a) How many children’s books are there, and what percent of the collection do they represent? b) In a circle graph of the collection, what central angle would represent the novels?

Hints

- Subtract the two known percentages from \(100\%\). - Use the remaining percent to find the number of children’s books. - A full circle is \(360^\circ\).

Solution

1. Children’s books represent \(100\% - 40\% - 25\% = 35\%\). 2. Their number is \(800 \times 0.35 = 280\). 3. The novel sector has an angle of \(360^\circ \times 0.40 = 144^\circ\).

Answer

a) \(280\) children’s books, representing \(35\%\) b) \(144^\circ\)
Find part given percent and whole 42
Lucas and Sarah are saving for a new game. Lucas has saved \(\$120\) and plans to spend \(30\%\) of it. Sarah has saved \(\$150\) and plans to spend \(25\%\) of it. Who plans to spend more money? Justify your answer by comparing the amounts.

Hints

- Find the dollar amount for each person separately. - The two whole amounts are different. - Compare the two dollar amounts at the end.

Solution

1. Lucas plans to spend \(30\%\) of \(\$120\): \(\$120 \times 0.30 = \$36\). 2. Sarah plans to spend \(25\%\) of \(\$150\): \(\$150 \times 0.25 = \$37.50\). 3. Since \(\$37.50 > \$36\), Sarah plans to spend more.

Answer

Sarah plans to spend more: \(\$37.50\), compared with Lucas's \(\$36\).
Find part given percent and whole 43
Find the missing part in each row. <table><tr><th>Whole</th><th>Percent</th><th>Part</th></tr><tr><td>\(80\,\text{kg}\)</td><td>\(15\%\)</td><td>\(\square\)</td></tr><tr><td>\(\$250\)</td><td>\(120\%\)</td><td>\(\square\)</td></tr><tr><td>\(16\,\text{m}\)</td><td>\(2.5\%\)</td><td>\(\square\)</td></tr><tr><td>\(0.5\) metric ton</td><td>\(40\%\)</td><td>\(\square\)</td></tr></table>

Hints

- Write each percent as a decimal or a fraction. - The part can be larger than the whole when the percent is over one hundred. - Keep the unit of each whole in its part.

Solution

1. \(15\%\) of \(80\,\text{kg}\) is \(0.15\times80=12\,\text{kg}\). 2. \(120\%\) of \(\$250\) is \(1.2\times\$250=\$300\). 3. \(2.5\%\) of \(16\,\text{m}\) is \(0.025\times16=0.4\,\text{m}\). 4. \(40\%\) of \(0.5\) metric ton is \(0.4\times0.5=0.2\) metric ton.

Answer

<table><tr><th>Whole</th><th>Percent</th><th>Part</th></tr><tr><td>\(80\,\text{kg}\)</td><td>\(15\%\)</td><td>\(12\,\text{kg}\)</td></tr><tr><td>\(\$250\)</td><td>\(120\%\)</td><td>\(\$300\)</td></tr><tr><td>\(16\,\text{m}\)</td><td>\(2.5\%\)</td><td>\(0.4\,\text{m}\)</td></tr><tr><td>\(0.5\) metric ton</td><td>\(40\%\)</td><td>\(0.2\) metric ton</td></tr></table>
Find part given percent and whole 44
Find both percent amounts in each comparison. Then write \(<\), \(>\), or \(=\). a) \(20\%\) of \(\$60\) ___ \(15\%\) of \(\$80\) b) \(110\%\) of \(40\,\text{kg}\) ___ \(8\%\) of \(500\,\text{kg}\) c) \(0.5\%\) of \(2000\,\text{m}\) ___ \(200\%\) of \(5\,\text{m}\)

Hints

- Find the amount on the left first. - Then find the amount on the right. - Compare the two results. - Remember that \(0.5\%\) is half of \(1\%\).

Solution

1. For a), \(\$60 \times 0.20 = \$12\) and \(\$80 \times 0.15 = \$12\). Therefore, the amounts are equal. 2. For b), \(40\,\text{kg} \times 1.10 = 44\,\text{kg}\) and \(500\,\text{kg} \times 0.08 = 40\,\text{kg}\). Therefore, the left side is greater. 3. For c), \(2000\,\text{m} \times 0.005 = 10\,\text{m}\) and \(5\,\text{m} \times 2.00 = 10\,\text{m}\). Therefore, the amounts are equal.

Answer

a) \(=\) b) \(>\) c) \(=\)
Find part given percent and whole 45
Write \(<\), \(>\), or \(=\) in each blank. Justify each choice with a short calculation. a) \(40\%\) of \(50\,\text{kg}\) ___ \(50\%\) of \(40\,\text{kg}\) b) \(25\%\) of \(120\,\text{min}\) ___ \(35\,\text{min}\) c) \(9\) out of \(20\) ___ \(50\%\)

Hints

- Write both sides of each comparison in the same form. - Recall the percents for simple fractions such as one fourth and one half. - In a percent problem, “of” indicates multiplication.

Solution

1. For a), \(50\,\text{kg} \times 0.40 = 20\,\text{kg}\), and \(40\,\text{kg} \times 0.50 = 20\,\text{kg}\). The amounts are equal. 2. For b), \(25\%\) is one fourth, and \(120\,\text{min} \div 4 = 30\,\text{min}\). Since \(30 < 35\), the left side is less. 3. For c), \(\frac{9}{20} = \frac{45}{100} = 45\%\). Since \(45\% < 50\%\), the left side is less.

Answer

a) \(=\), because both amounts are \(20\,\text{kg}\) b) \(<\), because \(30\,\text{min} < 35\,\text{min}\) c) \(<\), because \(45\% < 50\%\)
Find part given percent and whole 46
Compare the two amounts. Is \(30\%\) of \(\$60\) greater or less than \(25\%\) of \(\$70\)? Justify your answer with calculations.

Hints

- Calculate both amounts before comparing them. - Keep track of the cents when working with money. - Write each percent as a decimal.

Solution

1. Find \(30\%\) of \(\$60\): \(\$60 \times 0.30 = \$18\). 2. Find \(25\%\) of \(\$70\): \(\$70 \times 0.25 = \$17.50\). 3. Since \(\$18 > \$17.50\), the first amount is greater.

Answer

\(30\%\) of \(\$60\), or \(\$18\), is greater than \(25\%\) of \(\$70\), or \(\$17.50\).
Find part given percent and whole 47
A school library has \(500\) books, and \(12\%\) are science-fiction novels. After a large donation, the library has \(600\) books, and science-fiction novels make up \(15\%\) of the collection. a) How many science-fiction novels were in the library before the donation? b) How many science-fiction novels are in the library after the donation? c) How many of the \(100\) donated books are science-fiction novels?

Hints

- Find the number of science-fiction books at each time. - The whole number of books changes after the donation. - Subtract the earlier number from the later number.

Solution

1. Before the donation, \(12\%\) of \(500\) is \(500 \times 0.12 = 60\). There were \(60\) science-fiction novels. 2. After the donation, \(15\%\) of \(600\) is \(600 \times 0.15 = 90\). There are now \(90\) science-fiction novels. 3. Subtract to find the number added: \(90 - 60 = 30\).

Answer

a) \(60\) science-fiction novels b) \(90\) science-fiction novels c) \(30\) of the donated books are science-fiction novels.
Find part given percent and whole 48
Compare these three expressions: A: \(12\%\) of \(250\) B: \(24\%\) of \(125\) C: \(6\%\) of \(500\) a) Without calculating each product, predict whether one expression is greatest or whether all three are equal. b) Calculate all three values to check your prediction. c) Explain the relationship among the results. What happens to the part when the percent is doubled and the whole is cut in half?

Hints

- Look for relationships among the three percents and among the three whole numbers. - Make a prediction before calculating the values. - What happens to a product when one factor is doubled and the other factor is halved?

Solution

1. From A to B, the percent doubles while the whole is halved, so the product stays the same. From A to C, the percent is halved while the whole doubles, so all three expressions should be equal. 2. Calculate each value: A: \(250 \times 0.12 = 30\) B: \(125 \times 0.24 = 30\) C: \(500 \times 0.06 = 30\) 3. All three expressions have the same value, \(30\). 4. Doubling one factor and halving the other keeps the product unchanged. For example, from A to B, the percent doubles from \(12\%\) to \(24\%\), while the whole is halved from \(250\) to \(125\).

Answer

a) All three expressions should have the same value. b) A: \(30\); B: \(30\); C: \(30\) c) The part stays the same when the percent is doubled and the whole is halved because the product of the two factors does not change.
Find part given percent and whole 49
Two angles, \(\alpha\) and \(\beta\), are compared. Angle \(\alpha\) is \(35\%\) of a straight angle. Angle \(\beta\) is \(75\%\) of a right angle. Find the measure of each angle in degrees. Then determine which angle is greater and support your answer with calculations.

Hints

- Recall the degree measures of a straight angle and a right angle. - Write each percent as a decimal and multiply by the corresponding angle measure. - Compare the two degree measures after calculating them.

Solution

1. A straight angle measures \(180^\circ\). 2. Therefore, \(\alpha = 180^\circ \times 0.35 = 63^\circ\). 3. A right angle measures \(90^\circ\). 4. Therefore, \(\beta = 90^\circ \times 0.75 = 67.5^\circ\). 5. Since \(67.5^\circ > 63^\circ\), angle \(\beta\) is greater than angle \(\alpha\).

Answer

\(\alpha = 63^\circ\) and \(\beta = 67.5^\circ\). Angle \(\beta\) is greater.
Find part given percent and whole 50
A sporting-goods store has \(450\) soccer balls in stock. Of these, \(40\%\) are made by one major brand. Of the remaining balls, \(20\%\) are designed for indoor play. a) How many balls are made by the major brand? b) How many of the balls not made by that brand are designed for indoor play?

Hints

- Use the full inventory as the whole in part a. - In part b, the \(20\%\) is taken from the balls that remain. - Find the number of non-brand balls before finding \(20\%\) of them.

Solution

1. For part a, the number of major-brand balls is \(450 \times 0.40 = 180\). 2. The number of balls from other brands is \(450 - 180 = 270\). 3. For part b, the number of indoor balls among the remaining balls is \(270 \times 0.20 = 54\).

Answer

a) \(180\) balls are made by the major brand. b) \(54\) of the remaining balls are designed for indoor play.
Find part given percent and whole 51
An orchard harvests \(1200\,\text{lb}\) of apples. Of the harvest, \(15\%\) has minor blemishes and is pressed into cider. Of the remaining apples, \(80\%\) is sold at the farm stand, and the rest is delivered to a grocery store. a) How many pounds of apples are pressed into cider? b) What percent of the entire harvest is delivered to the grocery store?

Hints

- First find how much of the harvest remains after the blemished apples are removed. - If \(80\%\) is sold at the farm stand, what percent of the remaining apples goes to the grocery store? - Compare the grocery-store amount with the original harvest.

Solution

1. For part a, the amount pressed into cider is \(1200\,\text{lb} \times 0.15 = 180\,\text{lb}\). 2. The amount of unblemished apples is \(1200\,\text{lb} - 180\,\text{lb} = 1020\,\text{lb}\). 3. Since \(80\%\) is sold at the farm stand, \(20\%\) of the unblemished apples goes to the grocery store: \(1020\,\text{lb} \times 0.20 = 204\,\text{lb}\). 4. As a percent of the entire harvest, this is \(\frac{204}{1200} \times 100\% = 17\%\).

Answer

a) \(180\,\text{lb}\) of apples are pressed into cider. b) \(17\%\) of the entire harvest is delivered to the grocery store.
Find part given percent and whole 52
A software developer plans an update with \(200\) changes. Of the changes, \(24\%\) are classified as critical. Of the remaining changes, \(37.5\%\) are classified as useful. All other changes are cosmetic. Find the number of cosmetic changes. Then determine whether cosmetic changes make up more than half of all changes.

Hints

- Find the number in each category one step at a time. - The useful changes are a percent of the changes remaining after the critical changes. - Compare the number of cosmetic changes with half of the original total.

Solution

1. The number of critical changes is \(200 \times 0.24 = 48\). 2. The number of changes remaining is \(200 - 48 = 152\). 3. The number of useful changes is \(152 \times 0.375 = 57\). 4. The number of cosmetic changes is \(152 - 57 = 95\). 5. Half of \(200\) is \(100\). Since \(95 < 100\), cosmetic changes do not make up more than half.

Answer

There are \(95\) cosmetic changes. They do not make up more than half of the \(200\) changes.
Find part given percent and whole 53
Two saltwater samples have different salt concentrations. Sample A is a \(15\,\text{kg}\) saltwater sample that is \(0.8\%\) salt. Sample B is a \(4\,\text{kg}\) saltwater sample that is \(3.5\%\) salt. Which sample contains the greater total mass of salt? Justify your answer with calculations.

Hints

- Find the salt mass in each sample separately. - The sample masses are different, so do not compare only the percentages. - Convert the results to the same unit before comparing.

Solution

1. Sample A contains \(15\,\text{kg} \times 0.008 = 0.12\,\text{kg} = 120\,\text{g}\) of salt. 2. Sample B contains \(4\,\text{kg} \times 0.035 = 0.14\,\text{kg} = 140\,\text{g}\) of salt. 3. Since \(140\,\text{g} > 120\,\text{g}\), Sample B contains more salt.

Answer

Sample B contains more salt: \(140\,\text{g}\), compared with \(120\,\text{g}\) in Sample A.
Find part given percent and whole 54
In a school survey, \(40\%\) of the students said they own a pet. Of those pet owners, \(30\%\) own a dog. a) What percent of all students own a dog? b) The school has \(450\) students. How many own a pet but do not own a dog? c) Kanya says, “Since \(40\%\) own a pet and \(30\%\) of them own a dog, \(70\%\) of all students must be connected with pets.” Explain Kanya's error.

Hints

- Identify the whole used by the second percentage. - You can test the situation by imagining \(100\) students. - A percent of a subgroup is not automatically a percent of the entire group.

Solution

1. For part a, the percent of all students who own a dog is \(40\% \times 30\% = 0.40 \times 0.30 = 0.12 = 12\%\). 2. The number of pet owners is \(450 \times 0.40 = 180\). 3. The number of dog owners among the pet owners is \(180 \times 0.30 = 54\). 4. For part b, the number who own a pet but not a dog is \(180 - 54 = 126\). 5. For part c, the two percentages use different wholes. The \(30\%\) applies only to the pet owners, not to the entire school, so the percentages cannot be added.

Answer

a) \(12\%\) of all students own a dog. b) \(126\) students own a pet but not a dog. c) Kanya's addition is incorrect because \(30\%\) refers only to the pet-owner group, not to all students.
Find part given percent and whole 55
A chess club starts with \(40\) members in Year \(0\). The table compares each later year's membership with that starting number. <table><tr><th>Year</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th><th>\(4\)</th></tr><tr><th>Percent of starting membership</th><td>\(100\%\)</td><td>\(150\%\)</td><td>\(225\%\)</td><td>\(300\%\)</td><td>\(275\%\)</td></tr></table> a) How many members did the club have in each year? b) Between which consecutive years did its membership increase the most, measured in members?

Hints

- All percentages in the table refer to the same starting group. - A percent above \(100\%\) represents more than the original amount. - Compare the changes in membership between neighboring years.

Solution

1. Use the same starting group of \(40\) for each percent: years \(0\) through \(4\) have \(40\), \(1.5\times40=60\), \(2.25\times40=90\), \(3\times40=120\), and \(2.75\times40=110\) members. 2. The consecutive changes are \(60-40=20\), \(90-60=30\), \(120-90=30\), and a decrease of \(120-110=10\). 3. The greatest increase, \(30\) members, occurred twice: Year \(1\) to Year \(2\) and Year \(2\) to Year \(3\).

Answer

a) Year \(0\): \(40\); Year \(1\): \(60\); Year \(2\): \(90\); Year \(3\): \(120\); Year \(4\): \(110\) b) Years \(1\)–\(2\) and \(2\)–\(3\), tied at \(30\) members
Find part given percent and whole 56
A family's monthly budget is \(\$3200\). Next year, it plans to move \(5\%\) of the full monthly budget from recreation to savings. The total budget stays the same. a) How much does the family currently spend each month on housing and utilities? b) How much does it currently save each month? c) By how many dollars will monthly savings increase? d) How much will be available for recreation each month after the change?
Figure for problem 531912

Hints

- Read the original categories and shares from the graph. - Each percentage is taken from the full monthly budget. - The moved amount is a share of the whole monthly budget, not a share of the old savings.

Solution

1. Housing and utilities are \(35\%\) of the budget, or \(0.35\times\$3200=\$1120\). 2. Current savings are \(10\%\), or \(0.10\times\$3200=\$320\). 3. The extra savings are \(5\%\) of the full budget: \(0.05\times\$3200=\$160\). 4. Recreation falls from \(15\%\) to \(10\%\), so its new amount is \(0.10\times\$3200=\$320\).

Answer

a) \(\$1120\) b) \(\$320\) c) \(\$160\) d) \(\$320\)
Find part given percent and whole 57
A study measured the recycling rates of five cities. The bar graph shows the percent of each city’s total waste that is recycled. Use the graph to answer each question. a) Pine Harbor produces \(800\,\text{t}\) of waste. How much is recycled? b) Maple Junction produces \(1500\,\text{t}\) of waste. How much is recycled? c) Redwood Falls produces \(250\,\text{t}\) of waste. How much is recycled? d) Willow Bend produces \(400\,\text{t}\) of waste. How much is recycled? e) Silver Creek produces \(2000\,\text{t}\) of waste. How much is not recycled?
Figure for problem 535089

Hints

- Read each city’s recycling rate from the graph. - Multiply each city’s total waste by its recycling rate written as a decimal. - For part e), pay attention to whether the question asks for the recycled or unrecycled amount. - You can find the unrecycled percent by subtracting the recycling rate from \(100\%\).

Solution

1. Read the recycling rates from the graph: Pine Harbor, \(45\%\); Maple Junction, \(62\%\); Redwood Falls, \(28\%\); Willow Bend, \(75\%\); Silver Creek, \(54\%\). 2. For a), \(800\,\text{t} \times 0.45 = 360\,\text{t}\). 3. For b), \(1500\,\text{t} \times 0.62 = 930\,\text{t}\). 4. For c), \(250\,\text{t} \times 0.28 = 70\,\text{t}\). 5. For d), \(400\,\text{t} \times 0.75 = 300\,\text{t}\). 6. For Silver Creek, the percent not recycled is \(100\% - 54\% = 46\%\). Then \(2000\,\text{t} \times 0.46 = 920\,\text{t}\).

Answer

a) \(360\,\text{t}\) b) \(930\,\text{t}\) c) \(70\,\text{t}\) d) \(300\,\text{t}\) e) \(920\,\text{t}\)
Find part given percent and whole 58
A transportation survey included \(400\) households from the Oakwood and Pine Hill neighborhoods. Graph 1 shows what percent of all surveyed households came from each neighborhood. Graph 2 shows, within each neighborhood, the percent of surveyed households that own at least two bicycles. a) How many surveyed households came from Oakwood and from Pine Hill? b) Tim says, “Pine Hill has more surveyed households with at least two bicycles because more households were surveyed there.” Check his claim with calculations. c) What percent of all \(400\) surveyed households own at least two bicycles?
Figure for problem 535111

Hints

- Apply each percentage in Graph 1 to the total of \(400\). - Then apply the percentages in Graph 2 to the neighborhood totals. - Compare the resulting numbers of households. - Add the two counts and divide by \(400\) to find the overall percent.

Solution

1. Oakwood accounts for \(45\%\) of \(400\): \(0.45 \times 400 = 180\) households. Pine Hill accounts for \(55\%\): \(0.55 \times 400 = 220\) households. 2. In Oakwood, \(60\%\) of \(180\) is \(0.60 \times 180 = 108\) households. In Pine Hill, \(40\%\) of \(220\) is \(0.40 \times 220 = 88\) households. 3. Tim's claim is false because \(108 > 88\); Oakwood has more surveyed households with at least two bicycles. 4. Altogether, \(108 + 88 = 196\) households own at least two bicycles. The overall percent is \(196 \div 400 = 0.49 = 49\%\).

Answer

a) Oakwood: \(180\) households; Pine Hill: \(220\) households b) Tim's claim is false. Oakwood has \(108\) such households, while Pine Hill has \(88\). c) \(49\%\)
Find part given percent and whole 59
A class poster will include a circle graph of students’ favorite subjects. The results are math, \(20\%\); physical education, \(30\%\); art, \(15\%\); and English, \(35\%\). Find the central angle needed for each sector so the graph can be drawn accurately.

Hints

- A full circle is \(360^\circ\). - Find how many degrees correspond to \(1\%\). - Multiply each percentage by the number of degrees per percent.

Solution

1. A full circle is \(360^\circ\), so \(1\%\) corresponds to \(3.6^\circ\). 2. Math: \(20 \times 3.6^\circ = 72^\circ\). 3. Physical education: \(30 \times 3.6^\circ = 108^\circ\). 4. Art: \(15 \times 3.6^\circ = 54^\circ\). 5. English: \(35 \times 3.6^\circ = 126^\circ\). 6. Check: \(72^\circ + 108^\circ + 54^\circ + 126^\circ = 360^\circ\).

Answer

Math: \(72^\circ\) Physical education: \(108^\circ\) Art: \(54^\circ\) English: \(126^\circ\)
Find part given percent and whole 60
A school festival has a planned budget of \(\$600\). The pie chart shows how the money will be divided. a) How much money is planned for Music & Tech? b) Find the central angle of the Food & Drinks sector.
Figure for problem 535526

Hints

- Multiply the total budget by the percent for Music & Tech. - A full circle measures \(360^\circ\). - The central angle is proportional to the sector's percent of the circle. - You may first find the number of degrees represented by \(1\%\).

Solution

1. Music & Tech receives \(20\%\) of the budget: \(\$600 \times 0.20 = \$120\). 2. A full circle measures \(360^\circ\). The Food & Drinks sector is \(55\%\) of the circle, so its central angle is \(360^\circ \times 0.55 = 198^\circ\).

Answer

a) \(\$120\) b) \(198^\circ\)
Find part given percent and whole 61
Find the central angle needed in a circle graph to represent each amount. a) \(15\%\) b) \(\frac{2}{9}\) c) \(7\) out of \(20\)

Hints

- A full circle is \(360^\circ\). - Write each amount as a fraction or decimal part of the whole. - Multiply that part by \(360^\circ\).

Solution

1. For part a), \(360^\circ \times 0.15 = 54^\circ\). 2. For part b), \(360^\circ \times \frac{2}{9} = 80^\circ\). 3. For part c), the fraction is \(\frac{7}{20}\), so \(360^\circ \times \frac{7}{20} = 126^\circ\).

Answer

a) \(54^\circ\) b) \(80^\circ\) c) \(126^\circ\)
Find part given percent and whole 62
A class survey asked students to choose their favorite color. The results are shown in the table. Find the central angle for each sector in a circle graph. <table><tr><th>Favorite color</th><th>Share</th></tr><tr><td>Blue</td><td>\(40\%\)</td></tr><tr><td>Red</td><td>\(25\%\)</td></tr><tr><td>Green</td><td>\(20\%\)</td></tr><tr><td>Other</td><td>\(15\%\)</td></tr></table>

Hints

- A full circle is \(360^\circ\). - Write each percent as a decimal part of the whole. - Multiply each decimal by \(360^\circ\).

Solution

1. Blue: \(360^\circ \times 0.40 = 144^\circ\). 2. Red: \(360^\circ \times 0.25 = 90^\circ\). 3. Green: \(360^\circ \times 0.20 = 72^\circ\). 4. Other: \(360^\circ \times 0.15 = 54^\circ\). 5. Check: \(144^\circ + 90^\circ + 72^\circ + 54^\circ = 360^\circ\).

Answer

Blue: \(144^\circ\) Red: \(90^\circ\) Green: \(72^\circ\) Other: \(54^\circ\)
Find part given percent and whole 63
A city has an annual arts and culture budget of \(\$750{,}000\). The pie chart shows how the budget is allocated. Find the dollar amount planned for each category.
Figure for problem 535541

Hints

- Use the total budget as the whole. - Find each category’s percent of the total separately. - Add the four results to check that they equal the full budget.

Solution

1. Library: \(\$750{,}000 \times 0.28 = \$210{,}000\). 2. Music programs: \(\$750{,}000 \times 0.32 = \$240{,}000\). 3. Theater: \(\$750{,}000 \times 0.25 = \$187{,}500\). 4. Festivals: \(\$750{,}000 \times 0.15 = \$112{,}500\). 5. Check: \(\$210{,}000 + \$240{,}000 + \$187{,}500 + \$112{,}500 = \$750{,}000\).

Answer

Library: \(\$210{,}000\) Music programs: \(\$240{,}000\) Theater: \(\$187{,}500\) Festivals: \(\$112{,}500\)
Find part given percent and whole 64
A hiking group completed a \(24\,\text{km}\) trail. The circle graph shows the type of terrain, but one percent label is missing. a) What percent of the trail was downhill? b) How many kilometers of the trail were uphill?
Figure for problem 535561

Hints

- The sectors of a circle graph must total \(100\%\). - What fraction is equivalent to \(25\%\)?

Solution

1. All sectors total \(100\%\), so the downhill portion is \(100\% - 50\% - 25\% = 25\%\). 2. The uphill portion is \(25\%\) of \(24\,\text{km}\): \(24\,\text{km} \times 0.25 = 6\,\text{km}\).

Answer

a) \(25\%\) b) \(6\,\text{km}\)
Find part given percent and whole 65
A family shares a \(40\,\text{GB}\) monthly mobile-data plan. The pie chart shows the percent used by each family member last month. a) How many gigabytes did Lucas use? b) Mia plans to use an additional \(2\,\text{GB}\) next month. What percent of the current \(40\,\text{GB}\) plan would her total usage then represent? c) The family is considering increasing the plan to \(50\,\text{GB}\). If everyone keeps the same percent share, how many gigabytes would Dad use?
Figure for problem 535619

Hints

- Convert each percent share to gigabytes using the total plan size. - For part b, find Mia’s current usage before adding \(2\,\text{GB}\). - When the total changes but a percent share stays fixed, the corresponding number of gigabytes changes.

Solution

1. For part a, Lucas used \(40\,\text{GB} \times 0.30 = 12\,\text{GB}\). 2. Mia currently uses \(40\,\text{GB} \times 0.10 = 4\,\text{GB}\). With an additional \(2\,\text{GB}\), she would use \(6\,\text{GB}\). 3. For part b, \(\frac{6}{40} \times 100\% = 15\%\). 4. For part c, Dad would use \(50\,\text{GB} \times 0.35 = 17.5\,\text{GB}\).

Answer

a) Lucas used \(12\,\text{GB}\). b) Mia’s total usage would be \(15\%\) of the current plan. c) Dad would use \(17.5\,\text{GB}\).
Find part given percent and whole 66
A town covers a total area of \(5400\) acres. The pie chart shows how the land is used. 1. How many acres are forest? 2. How many more acres are farmland than residential land? 3. How many acres are used for roads?
Figure for problem 535637

Hints

- Treat the total area as \(100\%\). - You can find the difference in two areas by using the difference in their percentages. - The category areas should add to the total area.

Solution

1. Forest: \(5400 \times 0.40 = 2160\) acres. 2. Farmland exceeds residential land by \(35\% - 15\% = 20\%\). The difference is \(5400 \times 0.20 = 1080\) acres. 3. Roads: \(5400 \times 0.10 = 540\) acres.

Answer

1. \(2160\) acres are forest. 2. Farmland is \(1080\) acres larger than residential land. 3. \(540\) acres are used for roads.
Find part given percent and whole 67
Nadia makes \(1\,\text{L}\) of fruit drink. 1) What percent of the drink is sugar syrup? 2) How many milliliters of orange juice does she use? Use \(1\,\text{L}=1000\,\text{mL}\).
Figure for problem 535728

Hints

- The sectors of the circle add to one whole drink. - Use the missing share for the first question. - Use the orange-juice share, not the missing share, for the volume.

Solution

1. The labeled percentages total \(55\%+20\%+10\%=85\%\), leaving \(100\%-85\%=15\%\) sugar syrup. 2. Orange juice is \(20\%\) of \(1000\,\text{mL}\), or \(0.20\times1000=200\,\text{mL}\).

Answer

1) \(15\%\) 2) \(200\,\text{mL}\)
Find part given percent and whole 68
A bronze bell weighs \(250\,\text{lb}\). The pie chart shows the percentages of the metals used to make the bell. 1) How many pounds of copper are in the bell? 2) How many pounds of tin are in the bell? 3) How many times as much copper as tin is in the bell?
Figure for problem 535729

Hints

- To find each mass, multiply the total weight by the metal’s percent written as a decimal. - To compare the masses, divide the copper mass by the tin mass.

Solution

1. Copper: \(250\,\text{lb} \times 0.78 = 195\,\text{lb}\). 2. Tin: \(250\,\text{lb} \times 0.20 = 50\,\text{lb}\). 3. Compare the two masses: \(\frac{195}{50} = 3.9\). The bell contains \(3.9\) times as much copper as tin.

Answer

1) \(195\,\text{lb}\) 2) \(50\,\text{lb}\) 3) \(3.9\) times as much
Find part given percent and whole 69
The grid represents a garden with a total area of \(120\,\text{m}^2\). The shaded cells represent a sandbox. a) What percent of the garden is the sandbox? b) What is the area of the sandbox?
Figure for problem 535825

Hints

- Count the total number of cells and the number of shaded cells. - Write the shaded fraction as a percent. - Use that percent to find the sandbox’s actual area.

Solution

1. The grid has \(5 \times 8 = 40\) equal cells, and \(14\) are shaded. 2. The shaded portion is \(\frac{14}{40} = \frac{35}{100} = 35\%\). 3. Find \(35\%\) of the total area: \(120\,\text{m}^2 \times 0.35 = 42\,\text{m}^2\).

Answer

a) \(35\%\) b) \(42\,\text{m}^2\)
Find part given percent and whole 70
A city has two schools. School A has \(500\) students, and \(10\%\) of them belong to a soccer club. School B has \(100\) students, and \(20\%\) of them belong to a soccer club. A local headline says, “Students at School B are twice as athletic because twice as many students there play soccer.” Calculate the actual number of soccer-club members at each school. Then explain the error in the headline.

Hints

- Find each percentage of its school’s total enrollment. - Compare both the percentages and the actual numbers. - Decide whether soccer-club membership alone measures how athletic all students are.

Solution

1. At School A, \(10\%\) of \(500\) is \(0.10\times500=50\) students. 2. At School B, \(20\%\) of \(100\) is \(0.20\times100=20\) students. 3. School B has twice the percentage of soccer-club members, but School A has more soccer-club members because \(50>20\). 4. The headline confuses a percentage with an actual count. It also uses soccer-club membership to make an unsupported claim about overall athletic ability.

Answer

School A has \(50\) soccer-club members, and School B has \(20\). School B has twice the percentage, not twice the number of students. The claim that its students are “twice as athletic” is not supported by these data.
Find part given percent and whole 71
In a science experiment, a powder sample must contain no more than \(0.005\%\) dye by mass. A \(2\,\text{kg}\) sample has \(120\,\text{mg}\) of dye. Is that above the limit? Use \(1\,\text{kg}=1{,}000{,}000\,\text{mg}\).

Hints

- Compare masses expressed in the same unit. - Remember that this percent is much smaller than one percent. - Find the largest allowed contaminant mass, then compare it with the sample.

Solution

1. Convert the sample mass: \(2\,\text{kg}=2{,}000{,}000\,\text{mg}\). 2. The permitted amount is \(0.00005\times2{,}000{,}000=100\,\text{mg}\). 3. Since \(120>100\), the sample exceeds the limit by \(20\,\text{mg}\).

Answer

Yes; the sample contains \(120\,\text{mg}\), more than the \(100\,\text{mg}\) limit.

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