Free math worksheets

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

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

a) \(35\,\text{m}\) b) \(\$12\)

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

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.

Drink A contains more fruit juice: \(80\,\text{mL}\), compared with \(75\,\text{mL}\) in Drink B.

- 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 \(10\%\).

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.

\(448\,\text{mAh}\) of charge remains.

- 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?

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: \(\$120 + \$240 = \$360\).

\(1\%: \$4.80\); \(10\%: \$48\); \(25\%: \$120\); \(50\%: \$240\); \(75\%: \$360\). The \(75\%\) amount is the sum of the \(25\%\) and \(50\%\) amounts.

- 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?

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 \(60 + 30 = 90\).

a) \(60\) students b) \(30\) students c) \(120\) students d) \(90\) students

- 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.

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) \(5000\,\text{m}^2 \times 0.008 = 40\,\text{m}^2\)

a) \(36\,\text{kg}\) b) \(\$99\) c) \(90\,\text{m}\) d) \(9\,\text{L}\) e) \(40\,\text{m}^2\)

- Write each percent as a decimal before multiplying. - You can also start by finding \(1\%\) or \(10\%\). - Read the letters in the requested reverse order.

1. Find the percent amounts: (1) \(0.19 \times 300 = 57\); (2) \(0.51 \times 80 = 40.8\); (3) \(1.24 \times 50 = 62\); (4) \(0.05 \times 140 = 7\). 2. Match each result to the closest card value: (1) \(57\), card H; (2) \(40.8\), closest to \(41\), card T; (3) \(62\), card A; (4) \(7\), card M. 3. Read the letters from problem 4 to problem 1: M, A, T, H.

(1) \(57\) (2) \(40.8\) (3) \(62\) (4) \(7\) The word is MATH.

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

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, \(75\).

a) The amounts are equal; both are \(9\). b) The amount is \(90\). It is greater than \(75\) because \(120\%\) is greater than \(100\%\).

- 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.

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: \(25\% = \frac{25}{100} = \frac{1}{4}\).

a) \(1500\,\text{mL}\) b) \(\frac{1}{4}\), or \(25\%\)

- 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 \(50\%\).

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 \(360^\circ\).

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.

- A full circle is \(360^\circ\). - Multiply \(360^\circ\) by each percent written as a decimal. - Compare the bicycle angle with the car angle.

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.

a) Walking: \(144^\circ\); bicycle: \(72^\circ\); bus: \(108^\circ\); car: \(36^\circ\) b) The statement is correct because \(72^\circ=2\times36^\circ\).

- 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.

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.

a) \(60\) students b) \(80\) students c) \(36\) students d) The musical-instrument group is largest, with \(80\) students.

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

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

The spoon contains \(9\,\text{g}\) of copper.

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

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

The scooter's cold-weather range is \(18\) miles.

- 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.

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, \(25\% = \frac{25}{100} = \frac{1}{4}\).

a) \(\$12\) b) \(\$14\) c) \(\frac{1}{4}\)

- 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.

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: \(120 \times 0.15 = 18\).

a) \(48\) students chose sports. b) \(18\) students chose making music.

- 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 \(800\).

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: \(800 \times 0.05 = 40\).

Sports: \(280\) students Video games: \(200\) students Listening to music: \(160\) students Reading: \(120\) students Other: \(40\) students

- 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 \(200\).

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.

Soccer: \(80\) students Gymnastics: \(50\) students Swimming: \(40\) students Other: \(30\) students

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

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: \(\$4000 \times 0.10 = \$400\).

Lodging and meals: \(\$1800\) Transportation: \(\$1200\) Admission fees: \(\$600\) Other: \(\$400\)

- 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 \(350\,\text{g}\).

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: \(350\,\text{g} \times 0.10 = 35\,\text{g}\).

Peanuts: \(140\,\text{g}\) Raisins: \(105\,\text{g}\) Almonds: \(70\,\text{g}\) Hazelnuts: \(35\,\text{g}\)

- 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.

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 \(40 \times 0.25 = 10\).

a) \(18\) students b) \(12\) students c) \(10\) students

- 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.

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

a) \(96\) students b) \(108\) students

- 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.

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: \(250 \times 0.20 = 50\).

a) \(100\) people b) Italy: \(75\) people; Spain: \(50\) people

- 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.

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 \(4500\,\text{kg} \times 0.50 = 2250\,\text{kg}\).

a) \(1800\,\text{kg}\) b) \(450\,\text{kg}\) c) \(2250\,\text{kg}\)

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

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.

a) \(180\,\text{kg}\) b) \(60\,\text{kg}\) c) Food and yard waste, paper, and plastic

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

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: \(\$6 + \$10 + \$16 + \$8 = \$40\).

Snacks: \(\$6\); movies: \(\$10\); games: \(\$16\); savings: \(\$8\)

- 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.

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

They budget \(\$360\) for groceries and \(\$240\) for entertainment.

- 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.

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

Paper: \(630\,\text{kg}\) Compost: \(450\,\text{kg}\)

- Match each question with the relevant sector or sectors in the graph. - To find a number of members, multiply \(120\) by the percent written as a decimal. - For participation, include everyone who cast a vote or abstained.

1. Candidate A received \(60\%\) of the members’ responses: \(120 \times 0.60 = 72\). 2. Members who voted for either candidate or abstained still participated: \(60\% + 25\% + 10\% = 95\%\). 3. The nonparticipating group is \(5\%\) of the members: \(120 \times 0.05 = 6\).

1. \(72\) members 2. \(95\%\) 3. \(6\) members

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

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

Heating: \(2520\,\text{kWh}\); lighting: \(180\,\text{kWh}\)

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

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.

Yes. The total is \(\$1230\), which is \(\$30\) above the \(\$1200\) goal.

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

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.

\(6\) students have a dog.

- 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.

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

\(140\) students play piano.

- 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.

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: \(900{,}000\,\text{m}^3 \times \frac{2}{5} = 360{,}000\,\text{m}^3\).

\(360{,}000\,\text{m}^3\) of water are used for cooling each year.

- 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.

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.

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.

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

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 \(360^\circ \times 0.40 = 144^\circ\).

a) \(280\) children’s books, representing \(35\%\) b) \(144^\circ\)

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

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.

Sarah plans to spend more: \(\$37.50\), compared with Lucas's \(\$36\).

- Each row gives the whole and the percent. - Multiply the whole by the percent written as a decimal. - Be careful when writing \(2.5\%\) as a decimal. - The part has the same unit as the whole.

1. Row 1: \(80\,\text{kg} \times 0.15 = 12\,\text{kg}\). 2. Row 2: \(\$250 \times 1.20 = \$300\). 3. Row 3: \(16\,\text{m} \times 0.025 = 0.4\,\text{m}\). 4. Row 4: \(0.5\,\text{metric ton} \times 0.40 = 0.2\,\text{metric ton}\). 5. Complete the table. <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>\(0.5\,\text{metric ton}\)</td><td>\(40\%\)</td><td>\(0.2\,\text{metric ton}\)</td></tr></table>

<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>\(0.5\,\text{metric ton}\)</td><td>\(40\%\)</td><td>\(0.2\,\text{metric ton}\)</td></tr></table>

- 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 \(1\%\).

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.

a) \(=\) b) \(>\) c) \(=\)

- 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.

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.

a) \(=\), because both amounts are \(20\,\text{kg}\) b) \(<\), because \(30\,\text{min} < 35\,\text{min}\) c) \(<\), because \(45\% < 50\%\)

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

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.

\(30\%\) of \(\$60\), or \(\$18\), is greater than \(25\%\) of \(\$70\), or \(\$17.50\).

- 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.

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: \(90 - 60 = 30\).

a) \(60\) science-fiction novels b) \(90\) science-fiction novels c) \(30\) of the donated books are science-fiction novels.

- 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?

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 \(125\).

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.

- 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.

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 \(\alpha\).

\(\alpha = 63^\circ\) and \(\beta = 67.5^\circ\). Angle \(\beta\) is greater.

- 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.

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 \(270 \times 0.20 = 54\).

a) \(180\) balls are made by the major brand. b) \(54\) of the remaining balls are designed for indoor play.

- 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.

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 \(\frac{204}{1200} \times 100\% = 17\%\).

a) \(180\,\text{lb}\) of apples are pressed into cider. b) \(17\%\) of the entire harvest is delivered to the grocery store.

- 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.

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.

There are \(95\) cosmetic changes. They do not make up more than half of the \(200\) changes.

- 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.

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.

Sample B contains more salt: \(140\,\text{g}\), compared with \(120\,\text{g}\) in Sample A.

- 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.

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.

a) \(12\%\) of all students own a dog. b) \(126\) students own a pet but not a dog. c) The addition is incorrect because \(30\%\) refers only to the pet-owner group, not to all students.

- First find the value represented by \(1\%\) of the baseline. - How do you find an amount when \(100\%\) is known? - What does a value greater than \(100\%\) mean here? - Subtract consecutive membership values to find each change.

1. Find each membership using the baseline of \(40\): Year \(0\): \(40\) members. Year \(1\): \(1.5 \times 40 = 60\) members. Year \(2\): \(2.25 \times 40 = 90\) members. Year \(3\): \(3 \times 40 = 120\) members. Year \(4\): \(2.75 \times 40 = 110\) members. 2. Find the consecutive changes: Year \(0\) to Year \(1\): \(60 - 40 = 20\). Year \(1\) to Year \(2\): \(90 - 60 = 30\). Year \(2\) to Year \(3\): \(120 - 90 = 30\). Year \(3\) to Year \(4\): \(110 - 120 = -10\). 3. The greatest increase was \(30\) members, occurring from Year \(1\) to Year \(2\) and from Year \(2\) to Year \(3\).

a) Year \(0\): \(40\); Year \(1\): \(60\); Year \(2\): \(90\); Year \(3\): \(120\); Year \(4\): \(110\) b) The greatest increase occurred from Year \(1\) to Year \(2\) and from Year \(2\) to Year \(3\). Each increase was \(30\) members.

- Treat the full monthly budget as the whole. - Multiply the budget by each category’s percent. - A change of \(5\) percentage points represents \(5\%\) of the full budget here. - Use the new recreation percent for part d.

1. For part a, housing and utilities are \(35\%\) of the budget: \(\$3200 \times 0.35 = \$1120\). 2. For part b, current savings are \(10\%\) of the budget: \(\$3200 \times 0.10 = \$320\). 3. For part c, the savings share increases by \(5\) percentage points, so the dollar increase is \(\$3200 \times 0.05 = \$160\). 4. For part d, the new recreation share is \(10\%\), giving \(\$3200 \times 0.10 = \$320\).

a) \(\$1120\) b) \(\$320\) c) \(\$160\) d) \(\$320\)

- 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 \(100\%\).

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 \(2000\,\text{t} \times 0.46 = 920\,\text{t}\).

a) \(360\,\text{t}\) b) \(930\,\text{t}\) c) \(70\,\text{t}\) d) \(300\,\text{t}\) e) \(920\,\text{t}\)

- 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.

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 \(196 \div 400 = 0.49 = 49\%\).

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) \(49\%\)

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

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: \(72^\circ + 108^\circ + 54^\circ + 126^\circ = 360^\circ\).

Math: \(72^\circ\) Physical education: \(108^\circ\) Art: \(54^\circ\) English: \(126^\circ\)

- 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 \(1\%\).

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 \(360^\circ \times 0.55 = 198^\circ\).

a) \(\$120\) b) \(198^\circ\)

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

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 \(360^\circ \times \frac{7}{20} = 126^\circ\).

a) \(54^\circ\) b) \(80^\circ\) c) \(126^\circ\)

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

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: \(144^\circ + 90^\circ + 72^\circ + 54^\circ = 360^\circ\).

Blue: \(144^\circ\) Red: \(90^\circ\) Green: \(72^\circ\) Other: \(54^\circ\)

- 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.

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: \(\$210{,}000 + \$240{,}000 + \$187{,}500 + \$112{,}500 = \$750{,}000\).

Library: \(\$210{,}000\) Music programs: \(\$240{,}000\) Theater: \(\$187{,}500\) Festivals: \(\$112{,}500\)

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

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

a) \(25\%\) b) \(6\,\text{km}\)

- 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.

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 \(50\,\text{GB} \times 0.35 = 17.5\,\text{GB}\).

a) Lucas used \(12\,\text{GB}\). b) Mia’s total usage would be \(15\%\) of the current plan. c) Dad would use \(17.5\,\text{GB}\).

- 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.

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.

1. \(2160\) acres are forest. 2. Farmland is \(1080\) acres larger than residential land. 3. \(540\) acres are used for roads.

- All sectors of the circle graph total \(100\%\). - Write the orange-juice percent as a decimal or fraction before finding its volume.

1. The known percentages total \(75\% + 15\% + 5\% = 95\%\). Therefore, sugar represents \(100\% - 95\% = 5\%\). 2. Orange juice represents \(15\%\) of \(1000\,\text{mL}\): \(1000\,\text{mL} \times 0.15 = 150\,\text{mL}\).

1) \(5\%\) 2) \(150\,\text{mL}\)

- 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.

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.

1) \(195\,\text{lb}\) 2) \(50\,\text{lb}\) 3) \(3.9\) times as much

- 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.

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: \(120\,\text{m}^2 \times 0.35 = 42\,\text{m}^2\).

a) \(35\%\) b) \(42\,\text{m}^2\)

- 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.

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.

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.

- Convert the sample mass and contaminant mass to the same unit. - Find the allowed contaminant mass as a percent of the total sample mass. - Compare the measured amount with the allowed amount.

1. Convert the sample mass: \(2\,\text{kg} = 2{,}000{,}000\,\text{mg}\). 2. The maximum allowed contaminant mass is \(2{,}000{,}000\,\text{mg} \times 0.00005 = 100\,\text{mg}\). 3. Since \(120\,\text{mg} > 100\,\text{mg}\), the sample exceeds the safety limit.

Yes. The sample contains \(120\,\text{mg}\), which is greater than the \(100\,\text{mg}\) limit.