Aimathic – Find percent given part and whole

Find percent given part and whole

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Difficulty: 1 – 5

5114766

Write each part as a percent. a) $15$ of $20$ students completed their homework. b) $9$ of $50$ people wear glasses. c) $12$ of $60$ animals at a shelter are dogs. d) $84$ of $400$ people surveyed enjoy hiking. e) $3$ of $4$ music students play a keyboard instrument.

Hints

- Write each part as a fraction of its whole. - Scale or simplify the fraction to an equivalent fraction with denominator $100$ when convenient.

Solution

1. a) $\frac{15}{20} = \frac{75}{100} = 75\%$ 2. b) $\frac{9}{50} = \frac{18}{100} = 18\%$ 3. c) $\frac{12}{60} = \frac{1}{5} = 20\%$ 4. d) $\frac{84}{400} = \frac{21}{100} = 21\%$ 5. e) $\frac{3}{4} = \frac{75}{100} = 75\%$

Answer

a) $75\%$ b) $18\%$ c) $20\%$ d) $21\%$ e) $75\%$

5114826

In a class of $20$ students, $13$ own a bicycle. What percent of the students do not own a bicycle?

Hints

- First find the number of students without a bicycle. - Write that number as a fraction of the class and convert the fraction to a percent.

Solution

1. The number without a bicycle is $20 - 13 = 7$. 2. The fraction without a bicycle is $\frac{7}{20}$. 3. Convert to a percent: $\frac{7}{20} = \frac{35}{100} = 35\%$.

Answer

$35\%$ of the students do not own a bicycle.

5114836

A fruit basket contains $80$ pieces of fruit. Of these, $16$ are bananas and the rest are apples. What percent of the fruit are apples?

Hints

- First find the number of apples. - Write the number of apples as a fraction of all the fruit, then convert it to a percent.

Solution

1. The number of apples is $80 - 16 = 64$. 2. The apple fraction is $\frac{64}{80} = \frac{8}{10}$. 3. Convert the fraction to a percent: $\frac{8}{10} = 80\%$.

Answer

$80\%$ of the fruit are apples.

5114976

A trail is $25\,\text{km}$ long. A family hikes $8\,\text{km}$ in the morning and another $7\,\text{km}$ after lunch. What percent of the trail remains?

Hints

- Find the distance already completed, then the distance remaining. - Write the remaining distance as a fraction of the entire trail and convert it to a percent.

Solution

1. The family has hiked $8\,\text{km} + 7\,\text{km} = 15\,\text{km}$. 2. The remaining distance is $25\,\text{km} - 15\,\text{km} = 10\,\text{km}$. 3. The remaining percent is $\frac{10}{25} \times 100\% = 40\%$.

Answer

$40\%$ of the trail remains.

5115346

Convert each part to a percent. a) A candidate received $7$ of $20$ votes. b) In a class of $25$ students, $6$ have a pet. c) Of $50$ parts produced by a factory, $3$ are defective.

Hints

- Write each part as a fraction of its whole. - Scale each fraction to an equivalent fraction with denominator $100$.

Solution

1. a) $\frac{7}{20} = \frac{35}{100} = 35\%$ 2. b) $\frac{6}{25} = \frac{24}{100} = 24\%$ 3. c) $\frac{3}{50} = \frac{6}{100} = 6\%$

Answer

a) $35\%$ b) $24\%$ c) $6\%$

5118486

In a class of $25$ students, $9$ have a pet. Identify the whole and the part, then find the percent of students who have a pet.

Hints

- Identify the total group and the subgroup being considered. - Write the subgroup as a fraction of the total and convert it to a percent.

Solution

1. The whole is the class of $25$ students, and the part is the $9$ students who have a pet. 2. Divide the part by the whole: $\frac{9}{25} = 0.36$. 3. Convert to a percent: $0.36 = 36\%$.

Answer

The whole is $25$ students, the part is $9$ students, and $36\%$ of the class has a pet.

5239426

A container holds $600\,\text{g}$ of a $5\%$ sugar solution. The solution is heated until some water evaporates, leaving $500\,\text{g}$ of solution. What is the new percent of sugar by mass?

Hints

- Does the amount of sugar change when only water evaporates? - Find the original mass of sugar. - Compare the unchanged sugar mass with the new total mass.

Solution

1. The original sugar mass is $600\,\text{g} \times 0.05 = 30\,\text{g}$. 2. Only water evaporates, so the sugar mass remains $30\,\text{g}$. 3. The new sugar portion is $\frac{30}{500} = 0.06 = 6\%$.

Answer

The new solution is $6\%$ sugar by mass.

5240616

A kitchen mixes $400\,\text{mL}$ of a concentrated vinegar solution that is $20\%$ acetic acid with $600\,\text{mL}$ of water. Assume the volumes add. What percent of the final mixture is acetic acid?

Hints

- Find the acid volume in the concentrate. - Adding water does not change the acid volume. - Find the total volume after mixing. - Compare the acid volume with the total volume.

Solution

1. The concentrate contains $400\,\text{mL} \times 0.20 = 80\,\text{mL}$ of acetic acid. 2. The total volume is $400\,\text{mL} + 600\,\text{mL} = 1000\,\text{mL}$. 3. The acid portion is $\frac{80}{1000} = 0.08 = 8\%$.

Answer

The final mixture is $8\%$ acetic acid by volume.

5318906

A survey asked all $400$ sixth-grade students at a school how they travel to school. The circle graph shows the number of students in each category. Find the percent for each form of transportation.

Hints

- The whole is $400$ students. - Write each category count as a fraction of $400$, then convert the fraction to a percent.

Solution

1. Bus: $\frac{160}{400} \times 100\% = 40\%$. 2. Bicycle: $\frac{120}{400} \times 100\% = 30\%$. 3. Walking: $\frac{80}{400} \times 100\% = 20\%$. 4. Car: $\frac{40}{400} \times 100\% = 10\%$.

Answer

Bus: $40\%$ Bicycle: $30\%$ Walking: $20\%$ Car: $10\%$

5320776

What percent of each figure is shaded? Find the shaded percent for figures a), b), and c).

Hints

- Count the total equal parts and the shaded parts. - Write each shaded portion as a fraction and convert it to a percent.

Solution

1. a) $7$ of $10$ equal parts are shaded: $\frac{7}{10} \times 100\% = 70\%$. 2. b) $6$ of $8$ equal sectors are shaded: $\frac{6}{8} \times 100\% = 75\%$. 3. c) $13$ of $20$ equal cells are shaded: $\frac{13}{20} \times 100\% = 65\%$.

Answer

a) $70\%$ b) $75\%$ c) $65\%$

5321116

A tile installer is designing three wall patterns. Each pattern is a grid of equal square tiles. Find the percent of tiles that are shaded in patterns a), b), and c).

Hints

- For each pattern, multiply rows by columns to find the total number of tiles. - Divide the number of shaded tiles by the total number and convert to a percent.

Solution

1. a) The grid has $4 \times 5 = 20$ tiles, and $8$ are shaded. Thus, $\frac{8}{20} \times 100\% = 40\%$. 2. b) The grid has $5 \times 6 = 30$ tiles, and $12$ are shaded. Thus, $\frac{12}{30} \times 100\% = 40\%$. 3. c) The grid has $5 \times 5 = 25$ tiles, and $15$ are shaded. Thus, $\frac{15}{25} \times 100\% = 60\%$.

Answer

a) $40\%$ b) $40\%$ c) $60\%$

5355946

A class has $25$ students. The diagram shows present students as shaded and absent students as unshaded. What percent of the class is absent today?

Hints

- Count the unshaded symbols. - Write the absent count as a fraction of the $25$ students and convert it to a percent.

Solution

1. The diagram shows $20$ present students and $5$ absent students. 2. The absent fraction is $\frac{5}{25}$. 3. Convert to a percent: $\frac{5}{25} = \frac{20}{100} = 20\%$.

Answer

$20\%$ of the class is absent.

5356536

A storage shelf has $20$ equal compartments. The occupied compartments are shaded in the diagram. What percent of the compartments are still available?

Hints

- Count the occupied compartments, then find the number that are available. - Write the available count as a fraction of $20$ and convert it to a percent.

Solution

1. The grid has $4 \times 5 = 20$ compartments. 2. Six compartments are occupied, so $20 - 6 = 14$ are available. 3. The available percent is $\frac{14}{20} \times 100\% = 70\%$.

Answer

$70\%$

5356946

A square is divided into $100$ equal cells. A $5 \times 5$ block is shaded. What percent of the square is shaded?

Hints

- Count the cells in the shaded square. - The entire grid already represents $100$ equal parts.

Solution

1. The shaded block contains $5 \times 5 = 25$ cells. 2. Since the whole square has $100$ cells, $25$ shaded cells represent $25\%$.

Answer

$25\%$

5357716

What percent of the grid is shaded blue?

Hints

- Multiply rows by columns to find the total number of cells. - Write the shaded cells as a fraction of the total and convert it to a percent.

Solution

1. The grid has $4 \times 5 = 20$ cells. 2. Seven cells are shaded. 3. The shaded percent is $\frac{7}{20} \times 100\% = 35\%$.

Answer

$35\%$ of the grid is shaded blue.

5357726

The diagram shows a group of circles. What percent of the circles are shaded orange?

Hints

- Count all the circles and then count the shaded circles. - Write the shaded amount as a fraction of the whole group.

Solution

1. There are $20$ circles in all and $9$ are shaded. 2. The shaded percent is $\frac{9}{20} \times 100\% = 45\%$.

Answer

$45\%$ of the circles are shaded orange.

5357756

What percent of the grid squares are shaded purple?

Hints

- Count the total grid squares and the shaded squares. - Scale the shaded fraction to an equivalent fraction with denominator $100$.

Solution

1. The grid has $5 \times 5 = 25$ squares. 2. Eleven squares are shaded. 3. The shaded percent is $\frac{11}{25} \times 100\% = 44\%$.

Answer

$44\%$

5358206

The diagram shows a chocolate bar after some pieces have been eaten. What percent of the bar remains, represented by the shaded pieces?

Hints

- Find the total number of pieces and the number that remain. - Convert the remaining fraction to a percent.

Solution

1. The bar has $4 \times 5 = 20$ pieces. 2. Thirteen pieces remain. 3. The remaining percent is $\frac{13}{20} \times 100\% = 65\%$.

Answer

$65\%$

5358666

A class has $20$ students. Each circle in the diagram represents one student, and the shaded circles represent students who forgot their gym clothes. What percent of the students forgot their gym clothes?

Hints

- Count the total circles and the shaded circles. - Write the shaded count as a fraction of the class and convert it to a percent.

Solution

1. There are $20$ students in all and $6$ shaded circles. 2. The percent is $\frac{6}{20} \times 100\% = 30\%$.

Answer

$30\%$ of the students forgot their gym clothes.

5103246

In Class A, $15$ of $25$ students have a pet. In Class B, $18$ of $30$ students have a pet. a) Compare the percentages of students who have a pet. b) Which class has more students without a pet?

Hints

- Write each part-to-whole relationship as a fraction. - Convert each fraction to a percent. - For part b), compare the actual numbers, not the percentages.

Solution

1. For Class A, $\frac{15}{25}=\frac{3}{5}=60\%$. For Class B, $\frac{18}{30}=\frac{3}{5}=60\%$. The percentages are equal. 2. Class A has $25-15=10$ students without a pet. Class B has $30-18=12$ students without a pet. Therefore, Class B has more students without a pet.

Answer

a) Both classes have $60\%$ of students with a pet. b) Class B has more students without a pet: $12$ compared with $10$.

5114746

A $20\,\text{cm}$ paper strip is divided into sections $A$, $B$, and $C$. Section $A$ is $5\,\text{cm}$ long, section $B$ is $40\%$ of the strip, and section $C$ is the remainder. What percent of the strip is section $C$? Also determine which section is longest.

Hints

- Write section $A$ as a fraction of the full strip, then convert it to a percent. - The three sections total $100\%$. - Compare all three percentages.

Solution

1. Section $A$ is $\frac{5}{20} \times 100\% = 25\%$ of the strip. 2. Sections $A$ and $B$ total $25\% + 40\% = 65\%$. 3. Section $C$ is $100\% - 65\% = 35\%$. 4. Since $40\% > 35\% > 25\%$, section $B$ is the longest.

Answer

Section $C$ is $35\%$ of the strip. Section $B$ is the longest.

5114786

A survey of $200$ students asked about their favorite hobby. Of the students, $60$ chose sports and $90$ chose gaming. The rest chose reading. Find the percent who chose each hobby.

Hints

- First find how many students chose reading. - Divide each category count by $200$ and convert to a percent. - Check that the percentages total $100\%$.

Solution

1. The number who chose reading is $200 - 60 - 90 = 50$. 2. Sports: $\frac{60}{200} \times 100\% = 30\%$. 3. Gaming: $\frac{90}{200} \times 100\% = 45\%$. 4. Reading: $\frac{50}{200} \times 100\% = 25\%$.

Answer

Sports: $30\%$ Gaming: $45\%$ Reading: $25\%$

5114846

A movie theater has $120$ seats, and $84$ tickets have been sold. a) What percent of the seats are occupied? b) An employee says, “More than one fourth of the seats are still available.” Is the statement correct? Explain.

Hints

- Find the occupied seats as a fraction of all seats. - Subtract from $100\%$ to find the available percent. - Convert one fourth to a percent before comparing.

Solution

1. The occupied-seat percent is $\frac{84}{120} \times 100\% = 70\%$. 2. The available-seat percent is $100\% - 70\% = 30\%$. 3. One fourth is $25\%$. Since $30\% > 25\%$, the statement is correct.

Answer

a) $70\%$ b) Yes. $30\%$ of the seats are available, which is more than one fourth, or $25\%$.

5114986

In one sixth-grade class, $12$ of $30$ students chose pizza as their favorite food. In another sixth-grade class, $9$ of $20$ students chose pizza. Which class has the greater percent of students who chose pizza? Justify your answer.

Hints

- Find the percent for each class separately. - Compare the two percentages rather than the two student counts.

Solution

1. First class: $\frac{12}{30} \times 100\% = 40\%$. 2. Second class: $\frac{9}{20} \times 100\% = 45\%$. 3. Since $45\% > 40\%$, the second class has the greater percent.

Answer

The second class has the greater percent: $45\%$, compared with $40\%$ in the first class.

5114996

A smartphone battery has a total capacity of $4000\,\text{mAh}$. Overnight, $1200\,\text{mAh}$ was used. During the morning, apps used another $25\%$ of the original capacity. a) What percent of the total capacity was used altogether? b) What percent remains? Is that more or less than half of the charge?

Hints

- Convert the overnight energy use to a percent of the full capacity. - Add the two usage percentages. - Subtract from $100\%$ and compare the result with $50\%$.

Solution

1. Overnight use was $\frac{1200}{4000} \times 100\% = 30\%$ of the capacity. 2. Total use was $30\% + 25\% = 55\%$. 3. The remaining charge is $100\% - 55\% = 45\%$. 4. Since $45\% < 50\%$, less than half remains.

Answer

a) $55\%$ b) $45\%$ remains, which is less than half.

5115056

A town uses $5000\,\text{MWh}$ of energy per year. In a circle graph, solar energy has a central angle of $144^\circ$, wind energy provides $25\%$ of the total, and the rest comes from biomass. For biomass, find: a) the percent of total energy, b) the central angle in the circle graph, c) the amount of energy in $\text{MWh}$.

Hints

- Convert the solar angle to a percent of $360^\circ$. - The three sources total $100\%$. - Use the biomass percent to find both its angle and energy amount.

Solution

1. Solar energy represents $\frac{144^\circ}{360^\circ} \times 100\% = 40\%$. 2. Biomass represents $100\% - 40\% - 25\% = 35\%$. 3. Its central angle is $360^\circ \times 0.35 = 126^\circ$. 4. Its energy amount is $5000\,\text{MWh} \times 0.35 = 1750\,\text{MWh}$.

Answer

a) $35\%$ b) $126^\circ$ c) $1750\,\text{MWh}$

5115086

A school environmental club recorded the waste collected during two weeks. Week 1: $80\,\text{kg}$ of waste in all, including $24\,\text{kg}$ of plastic. Week 2: $120\,\text{kg}$ of waste in all, including $42\,\text{kg}$ of plastic. A student says, “The percentage of the waste that was plastic increased in Week 2, even though the total amount of waste also increased.” Use percentages to determine whether the student is correct.

Hints

- For each week, divide the plastic amount by the total amount. - Convert both ratios to percentages. - Compare the percentages, not just the amounts in kilograms.

Solution

1. In Week 1, the percentage of plastic is $\frac{24}{80}=0.30=30\%$. 2. In Week 2, the percentage of plastic is $\frac{42}{120}=0.35=35\%$. 3. Since $35\%>30\%$, the percentage increased, so the student is correct.

Answer

The student is correct. Plastic was $30\%$ of the waste in Week 1 and $35\%$ in Week 2.

5115106

A circle graph has three sectors. The first sector has a central angle of $108^\circ$, and the second has a central angle of $54^\circ$. a) Find the central angle of the third sector. b) Find the percent of the whole represented by each sector. c) Which sector is largest? Does it represent more than half of the circle? Briefly justify your answer.

Hints

- The central angles of a circle graph total $360^\circ$. - To convert an angle to a percent, compare it with $360^\circ$. - Half of a circle is $180^\circ$, or $50\%$.

Solution

1. The third angle is $360^\circ-108^\circ-54^\circ=198^\circ$. 2. The first sector represents $\frac{108}{360}\times100\%=30\%$. 3. The second sector represents $\frac{54}{360}\times100\%=15\%$. 4. The third sector represents $\frac{198}{360}\times100\%=55\%$. 5. The third sector is largest and represents more than half because $55\%>50\%$, or equivalently because $198^\circ>180^\circ$.

Answer

a) $198^\circ$ b) First sector: $30\%$; second sector: $15\%$; third sector: $55\%$ c) The third sector is largest, and it represents more than half of the circle because $55\%>50\%$.

5115206

On a quiz, Taylor earned $18$ of $24$ possible points. a) What percent of the points did Taylor earn? b) Suppose the quiz had $30$ possible points and Taylor still earned $18$. What percent would that be? c) Compare the two percentages. Why does the percent decrease even though the earned score stays the same?

Hints

- Write each score as a fraction of the possible points. - Compare what changes in the denominator from part a) to part b).

Solution

1. a) $\frac{18}{24} = \frac{3}{4} = 75\%$. 2. b) $\frac{18}{30} = \frac{3}{5} = 60\%$. 3. c) The percent decreases from $75\%$ to $60\%$ because the same earned score is compared with a larger possible score.

Answer

a) $75\%$ b) $60\%$ c) The same number of points is a smaller part of a larger total.

5115326

A can of mixed vegetables has a total mass of $450\,\text{g}$. It contains $180\,\text{g}$ of peas, and the rest is carrots. a) What percent of the mixture is peas? b) How many grams of carrots are in the can? c) Find the percent of carrots in two different ways.

Hints

- Identify the part and the whole for the peas. - The two ingredient percentages must total $100\%$. - One method can use the carrot mass; the other can use the pea percent.

Solution

1. Peas: $\frac{180}{450} \times 100\% = 40\%$. 2. Carrots: $450\,\text{g} - 180\,\text{g} = 270\,\text{g}$. 3. First method: $\frac{270}{450} \times 100\% = 60\%$. 4. Second method: $100\% - 40\% = 60\%$.

Answer

a) $40\%$ b) $270\,\text{g}$ c) $60\%$

5115356

Solve each percent problem. a) A sports club has $200$ members, including $84$ teenagers. What percent of the members are teenagers? b) A pie was cut into $8$ equal slices. Three slices were eaten. What percent of the pie remains? c) Is $7$ out of $25$ greater than or less than $30\%$? Show a calculation.

Hints

- Pay attention to whether a question asks for the used part or the remaining part. - Simplify or scale each fraction before converting it to a percent. - For part c), convert both values to comparable forms.

Solution

1. a) $\frac{84}{200} = \frac{42}{100} = 42\%$. 2. b) There are $8 - 3 = 5$ slices left, so $\frac{5}{8} \times 100\% = 62.5\%$. 3. c) $\frac{7}{25} = \frac{28}{100} = 28\%$. Since $28\% < 30\%$, the given part is less than $30\%$.

Answer

a) $42\%$ b) $62.5\%$ c) Less than $30\%$, because $7$ out of $25$ is $28\%$

5115466

Two classes held elections. In one class, $18$ of $24$ students voted for Candidate A. In the other class, $19$ of $25$ students voted for Candidate B. In which class did the candidate receive the greater percent of votes? Justify your answer.

Hints

- Convert both fractions to percentages before comparing. - A common base of hundredths makes the comparison direct.

Solution

1. First class: $\frac{18}{24} \times 100\% = 75\%$. 2. Second class: $\frac{19}{25} \times 100\% = 76\%$. 3. Since $76\% > 75\%$, Candidate B received the greater percent in the second class.

Answer

The second class had the greater percent: $76\%$, compared with $75\%$ in the first class.

5115626

An after-school program has $20$ students. Of them, $5$ chose soccer as their main activity and $4$ chose music. The remaining students chose other activities. a) Find the percent of students who chose soccer and the percent who chose music. b) Find the corresponding central angles in a circle graph. c) How many degrees larger is the soccer sector than the music sector?

Hints

- Divide each activity count by the total number of students. - Convert each fraction to a percent. - Use $360^\circ$ as the whole circle. - Subtract the smaller angle from the larger angle.

Solution

1. Soccer represents $\frac{5}{20}=25\%$. 2. Music represents $\frac{4}{20}=20\%$. 3. The soccer angle is $360^\circ\times0.25=90^\circ$. 4. The music angle is $360^\circ\times0.20=72^\circ$. 5. The difference is $90^\circ-72^\circ=18^\circ$.

Answer

a) Soccer: $25\%$; music: $20\%$ b) Soccer: $90^\circ$; music: $72^\circ$ c) $18^\circ$

5116096

Two basketball players compare their free-throw accuracy. Player A made $12$ of $15$ attempts. Player B made $15$ of $20$ attempts. Find each player’s shooting percent and determine who was more accurate.

Hints

- For each player, divide the number made by the number attempted. - Convert both ratios to percentages before comparing.

Solution

1. Player A: $\frac{12}{15} \times 100\% = 80\%$. 2. Player B: $\frac{15}{20} \times 100\% = 75\%$. 3. Since $80\% > 75\%$, Player A was more accurate.

Answer

Player A: $80\%$; Player B: $75\%$. Player A was more accurate.

5117566

A drink is made with $100\,\text{mL}$ of apple juice and $300\,\text{mL}$ of sparkling water. a) What percent of the mixture is apple juice? b) Another $100\,\text{mL}$ of apple juice is added. What percent of the new mixture is apple juice?

Hints

- Find the total volume in each part. - After the addition, both the amount of juice and the total volume change.

Solution

1. The original mixture has $100\,\text{mL} + 300\,\text{mL} = 400\,\text{mL}$. The apple-juice percent is $\frac{100}{400} \times 100\% = 25\%$. 2. After the addition, there are $200\,\text{mL}$ of apple juice in $500\,\text{mL}$ total. 3. The new apple-juice percent is $\frac{200}{500} \times 100\% = 40\%$.

Answer

a) $25\%$ b) $40\%$

5117576

Two classes compare results from a charity walk. In Class A, $18$ of $24$ students met their lap goal. In Class B, $21$ of $28$ students met their goal. a) Find the percent for each class. What do you notice? b) How many students in Class A would need to meet their goal for the rate to be exactly $87.5\%$?

Hints

- Simplify both fractions before converting them to percentages. - For part b), write $87.5\%$ as a decimal and multiply by the class size.

Solution

1. Class A: $\frac{18}{24} \times 100\% = 75\%$. 2. Class B: $\frac{21}{28} \times 100\% = 75\%$. 3. The classes have the same success rate even though their class sizes differ. 4. For Class A, $24 \times 0.875 = 21$, so $21$ students would need to meet the goal.

Answer

a) Both classes have a rate of $75\%$. b) $21$ students

5123786

A full turn is divided into four angles. - The first angle is a straight angle. - The second angle is half of a right angle. - The third angle is half of the second angle. Find the fourth angle. What percent of the full turn does the fourth angle represent?

Hints

- Find the first three angle measures one at a time. - Subtract their sum from $360^\circ$. - To find the percent, divide the fourth angle by the full turn and multiply by $100\%$.

Solution

1. A full turn measures $360^\circ$. 2. The first angle is $180^\circ$. 3. The second angle is $90^\circ \div 2 = 45^\circ$. 4. The third angle is $45^\circ \div 2 = 22.5^\circ$. 5. The fourth angle is $360^\circ - (180^\circ + 45^\circ + 22.5^\circ) = 112.5^\circ$. 6. The percent is $\frac{112.5}{360} \times 100\% = 31.25\%$.

Answer

The fourth angle measures $112.5^\circ$ and represents $31.25\%$ of the full turn.

5127226

Two fruit drinks are prepared for a school event. Mixture A contains $150\,\text{mL}$ of apple juice and $350\,\text{mL}$ of water. Mixture B has a total volume of $400\,\text{mL}$, and $28\%$ of it is pure fruit juice. Which mixture has the greater percent of fruit juice?

Hints

- Find the total volume of Mixture A. - Compare the juice volume with the total volume of Mixture A. - Compare the calculated percent with the given percent for Mixture B.

Solution

1. Mixture A has a total volume of $150\,\text{mL} + 350\,\text{mL} = 500\,\text{mL}$. 2. The juice percent in Mixture A is $\frac{150}{500} \times 100\% = 30\%$. 3. Since $30\% > 28\%$, Mixture A has the greater fruit-juice percent.

Answer

Mixture A has the greater fruit-juice percent: $30\%$, compared with $28\%$ for Mixture B.

5239416

A chemist combines two salt solutions. The first solution has a mass of $150\,\text{g}$ and is $10\%$ salt. The second solution has a mass of $50\,\text{g}$ and is $22\%$ salt. What percent by mass of the combined mixture is salt?

Hints

- Find the mass of salt in each solution. - Add the salt masses and the total solution masses separately. - Compare the total salt mass with the total mixture mass.

Solution

1. The first solution contains $150\,\text{g} \times 0.10 = 15\,\text{g}$ of salt. 2. The second solution contains $50\,\text{g} \times 0.22 = 11\,\text{g}$ of salt. 3. The mixture contains $15\,\text{g} + 11\,\text{g} = 26\,\text{g}$ of salt and has a total mass of $150\,\text{g} + 50\,\text{g} = 200\,\text{g}$. 4. The salt percent is $\frac{26}{200} \times 100\% = 13\%$.

Answer

The combined mixture is $13\%$ salt by mass.

5319286

Three students—Maya, Ben, and Leo—ran for student council president. The pie chart shows the full election result, including invalid ballots and abstentions. A school newspaper later reported, “Maya received $44.4\%$ of the valid votes.” a) Verify whether the report is correct. b) What percent of the valid votes did Ben receive? Round to the nearest tenth of a percent.

Hints

- First find the percent of all ballots that were valid. - To find a candidate’s share of valid votes, divide the candidate’s share of all ballots by the valid-ballot share. - Convert the resulting ratio to a percent. - Round Ben’s result to the nearest tenth of a percent.

Solution

1. Valid votes make up $100\% - 19\% = 81\%$ of all ballots. 2. For part a, Maya’s share of the valid votes is $\frac{36\%}{81\%} = \frac{4}{9} \approx 0.4444 = 44.4\%$. The report is correct. 3. For part b, Ben’s share of the valid votes is $\frac{27\%}{81\%} = \frac{1}{3} \approx 0.3333 = 33.3\%$.

Answer

a) Yes. Maya received approximately $44.4\%$ of the valid votes. b) Ben received approximately $33.3\%$ of the valid votes.

5320636

For each figure, find the percent of the total area that is shaded. Round to the nearest hundredth of a percent when needed.

Hints

- Count the total equal parts and the shaded parts in each figure. - Divide the shaded count by the total count and convert to a percent. - Round only when the decimal does not terminate at the requested place.

Solution

1. a) $5$ of $8$ equal parts are shaded: $\frac{5}{8} \times 100\% = 62.5\%$. 2. b) $13$ of $20$ equal parts are shaded: $\frac{13}{20} \times 100\% = 65\%$. 3. c) $4$ of $6$ equal parts are shaded: $\frac{4}{6} \times 100\% \approx 66.67\%$.

Answer

a) $62.5\%$ b) $65\%$ c) $66.67\%$

5355156

A town survey asked residents how they usually travel to work. The circle graph shows the number of responses for each option. Find the percent of all responses represented by each form of transportation.

Hints

- Add all four counts to find the whole. - Divide each count by the total and convert to a percent.

Solution

1. The total number of responses is $480 + 240 + 120 + 160 = 1000$. 2. Car: $\frac{480}{1000} \times 100\% = 48\%$. 3. Bicycle: $\frac{240}{1000} \times 100\% = 24\%$. 4. Bus: $\frac{120}{1000} \times 100\% = 12\%$. 5. Walking: $\frac{160}{1000} \times 100\% = 16\%$.

Answer

Car: $48\%$ Bicycle: $24\%$ Bus: $12\%$ Walking: $16\%$

5355166

A sporting-goods store sold $40$ pairs of shoes on Saturday: $18$ pairs of running shoes, $12$ pairs of hiking shoes, $6$ pairs of soccer cleats, and $4$ pairs of court shoes. Find the percent for each type and match each type to the labeled sector in the circle graph.

Hints

- Divide each number of pairs by the total of $40$. - Match the largest percent to the largest labeled sector and continue in order.

Solution

1. Running shoes: $\frac{18}{40} \times 100\% = 45\%$, so they match the largest sector, A. 2. Hiking shoes: $\frac{12}{40} \times 100\% = 30\%$, so they match sector B. 3. Soccer cleats: $\frac{6}{40} \times 100\% = 15\%$, so they match sector C. 4. Court shoes: $\frac{4}{40} \times 100\% = 10\%$, so they match the smallest sector, D.

Answer

Running shoes: $45\%$, A Hiking shoes: $30\%$, B Soccer cleats: $15\%$, C Court shoes: $10\%$, D

5355176

A newsstand recorded its morning sales by category: newspapers, $\$105$; snacks, $\$75$; drinks, $\$90$; and baked goods, $\$30$. Find each category’s percent of the total morning sales.

Hints

- Add the four sales amounts to find total sales. - Divide each category’s sales by the total and convert to a percent.

Solution

1. Total sales were $\$105 + \$75 + \$90 + \$30 = \$300$. 2. Newspapers: $\frac{105}{300} \times 100\% = 35\%$. 3. Snacks: $\frac{75}{300} \times 100\% = 25\%$. 4. Drinks: $\frac{90}{300} \times 100\% = 30\%$. 5. Baked goods: $\frac{30}{300} \times 100\% = 10\%$.

Answer

Newspapers: $35\%$ Snacks: $25\%$ Drinks: $30\%$ Baked goods: $10\%$

5355256

An animal shelter currently cares for $80$ animals. The circle graph shows the number of animals in each category. Find each category’s percent of the total.

Hints

- The whole is the shelter’s $80$ animals. - Divide each category count by $80$ and convert to a percent. - Check that the percentages total $100\%$.

Solution

1. Dogs: $\frac{40}{80} \times 100\% = 50\%$. 2. Cats: $\frac{24}{80} \times 100\% = 30\%$. 3. Small animals: $\frac{12}{80} \times 100\% = 15\%$. 4. Birds: $\frac{4}{80} \times 100\% = 5\%$.

Answer

Dogs: $50\%$ Cats: $30\%$ Small animals: $15\%$ Birds: $5\%$

5355636

A $50\,\text{g}$ granola bar contains $20\,\text{g}$ of oats, $15\,\text{g}$ of nuts, $10\,\text{g}$ of dried fruit, and $5\,\text{g}$ of honey. Find the central angle for each ingredient in a circle graph of the bar’s composition.

Hints

- Find each ingredient’s fraction of the $50\,\text{g}$ total. - Convert each fraction to a percent or decimal. - Multiply each part of the whole by $360^\circ$.

Solution

1. Oats represent $\frac{20}{50}=40\%$, so their angle is $360^\circ\times0.40=144^\circ$. 2. Nuts represent $\frac{15}{50}=30\%$, so their angle is $360^\circ\times0.30=108^\circ$. 3. Dried fruit represents $\frac{10}{50}=20\%$, so its angle is $360^\circ\times0.20=72^\circ$. 4. Honey represents $\frac{5}{50}=10\%$, so its angle is $360^\circ\times0.10=36^\circ$.

Answer

Oats: $144^\circ$; nuts: $108^\circ$; dried fruit: $72^\circ$; honey: $36^\circ$

5355646

A $40\,\text{m}^2$ apartment has $24\,\text{m}^2$ of hardwood flooring, $10\,\text{m}^2$ of carpet, and $6\,\text{m}^2$ of tile. a) What percent of the floor area is hardwood? b) What central angle represents the hardwood sector in a circle graph?

Hints

- Write the hardwood area as a fraction of the total floor area. - A full circle measures $360^\circ$. Find the same percent of that angle.

Solution

1. The hardwood percent is $\frac{24}{40} \times 100\% = 60\%$. 2. The central angle is $60\%$ of $360^\circ$: $0.60 \times 360^\circ = 216^\circ$.

Answer

a) $60\%$ b) $216^\circ$

5355656

A school surveyed $200$ students about how they travel to school. The results are shown in the circle graph. a) Verify whether the graph correctly shows that the $40$ bicycle riders represent $20\%$ of the students. b) How many students walk to school?

Hints

- For part a), divide the number of bicycle riders by the total. - For part b), read the walking percent from the graph and find that percent of $200$.

Solution

1. The bicycle-rider percent is $\frac{40}{200} \times 100\% = 20\%$, so the graph is correct. 2. Walking represents $40\%$ of the students: $200 \times 0.40 = 80$.

Answer

a) Yes. $40$ of $200$ is $20\%$. b) $80$ students

5355676

A square mosaic has $25$ equal tiles. The diagram shows $9$ shaded tiles. a) What fraction of the tiles are shaded? b) What percent is shaded? c) What percent is not shaded? d) How many tiles would need to be shaded for exactly $40\%$ of the mosaic to be shaded?

Hints

- Use the number of shaded tiles over the total number of tiles. - Convert the fraction to a fraction with denominator $100$. - The shaded and unshaded percentages total $100\%$.

Solution

1. a) The shaded fraction is $\frac{9}{25}$. 2. b) $\frac{9}{25} = \frac{36}{100} = 36\%$. 3. c) The unshaded percent is $100\% - 36\% = 64\%$. 4. d) $25 \times 0.40 = 10$, so $10$ tiles would need to be shaded.

Answer

a) $\frac{9}{25}$ b) $36\%$ c) $64\%$ d) $10$ tiles

5356186

A school analyzed its annual electricity use. The table shows the amount used by each area. <table> <tr><th>Area</th><th>Electricity use (kWh)</th></tr> <tr><td>Lighting</td><td>1200</td></tr> <tr><td>Computers and technology</td><td>900</td></tr> <tr><td>Heat pumps</td><td>600</td></tr> <tr><td>Cafeteria</td><td>300</td></tr> </table> The school used a total of $3000\,\text{kWh}$. a) Find each area’s percent of the total electricity use. b) The circle graph labels its sectors A through D. Match each area to a letter.

Hints

- Divide each area’s electricity use by the total. - Match the largest percent to the largest sector, then continue in order.

Solution

1. Lighting: $\frac{1200}{3000} \times 100\% = 40\%$. 2. Computers and technology: $\frac{900}{3000} \times 100\% = 30\%$. 3. Heat pumps: $\frac{600}{3000} \times 100\% = 20\%$. 4. Cafeteria: $\frac{300}{3000} \times 100\% = 10\%$. 5. Match by sector size: A is lighting, B is computers and technology, C is heat pumps, and D is the cafeteria.

Answer

a) Lighting: $40\%$; computers and technology: $30\%$; heat pumps: $20\%$; cafeteria: $10\%$ b) A: lighting; B: computers and technology; C: heat pumps; D: cafeteria

5356416

A school snack bar sold a total of 200 items in one week. The pie chart shows the percent of each type of item sold. Beginning the next week, the snack bar will no longer sell candy. Assume that students who previously bought candy will not buy anything else instead. Under this assumption, what will be the new percent of sandwiches among all items sold? Round to the nearest tenth of a percent.

Hints

- First find how many sandwiches and candy items were included in the 200 items sold. - Find the new total number of items sold after the candy sales are removed. - Compare the number of sandwiches with the new total. - Convert the resulting fraction to a percent.

Solution

1. The number of sandwiches sold was $200 \times 0.40 = 80$. 2. The number of candy items sold was $200 \times 0.10 = 20$. 3. Without candy, the new total number of items sold would be $200 - 20 = 180$. 4. The new percent of sandwiches would be $\frac{80}{180} \times 100\% \approx 44.4\%$.

Answer

The new percent of sandwiches would be $44.4\%$.

5357126

A mosaic floor has $50$ equal square tiles. The shaded tiles are already installed. What percent of the floor is complete?

Hints

- Count the total tiles and the shaded tiles. - Write the shaded portion as a fraction of $50$ and convert it to a percent.

Solution

1. The grid has $5 \times 10 = 50$ tiles. 2. There are $13$ shaded tiles. 3. The completed percent is $\frac{13}{50} \times 100\% = 26\%$.

Answer

$26\%$ of the floor is complete.

5357246

The circle graph shows how a student spends free time after school. 1) What percent of the time is spent on screen activities, television and video games combined? 2) What percent is spent on active activities, exercise and time outdoors combined? 3) How many times as much time is spent on television and video games as on homework?

Hints

- Identify which sectors belong together in each question. - To find how many times as great one amount is, divide the two amounts.

Solution

1. Screen activities: $30\% + 20\% = 50\%$. 2. Active activities: $15\% + 10\% = 25\%$. 3. Compare screen time with homework time: $50\% \div 10\% = 5$. The student spends five times as much time on screen activities as on homework.

Answer

1) $50\%$ 2) $25\%$ 3) Five times as much

5357266

A circle graph shows how students travel to school. 1) What percent use an environmentally friendly option, riding a bicycle or walking? 2) How many times as many students ride the bus as travel by car? 3) Which option is used by exactly one fifth of the students?

Hints

- Add the two requested sectors in part 1). - Divide the bus percent by the car percent for part 2). - Convert one fifth to a percent.

Solution

1. Bicycle and walking together represent $30\% + 20\% = 50\%$. 2. Compare bus and car: $40\% \div 10\% = 4$. Four times as many students ride the bus. 3. One fifth is $\frac{1}{5} = 20\%$, which corresponds to walking.

Answer

1) $50\%$ 2) Four times as many 3) Walking

5357446

Find the percent of each figure that is shaded. Then order the percentages from least to greatest. The corresponding letters form a word.

Hints

- Write the shaded part of each figure as a fraction. - Convert each fraction to a percent, then order the percentages.

Solution

1. Figure F: $1$ of $4$ parts is shaded, so $\frac{1}{4} = 25\%$. 2. Figure O: $2$ of $5$ parts are shaded, so $\frac{2}{5} = 40\%$. 3. Figure R: $6$ of $10$ cells are shaded, so $\frac{6}{10} = 60\%$. 4. Figure M: $3$ of $4$ objects are shaded, so $\frac{3}{4} = 75\%$. 5. In order, $25\% < 40\% < 60\% < 75\%$, giving F–O–R–M.

Answer

F: $25\%$, O: $40\%$, R: $60\%$, M: $75\%$ Order: $25\% < 40\% < 60\% < 75\%$ Word: FORM

5357476

Find the percent of each figure that is shaded.

Hints

- Count the total parts and the shaded parts in each figure. - Write the shaded portion as a fraction and convert it to a percent.

Solution

1. a) $3$ of $6$ equal parts are shaded: $\frac{3}{6} \times 100\% = 50\%$. 2. b) $5$ of $25$ objects are shaded: $\frac{5}{25} \times 100\% = 20\%$. 3. c) $15$ of $50$ parts are shaded: $\frac{15}{50} \times 100\% = 30\%$.

Answer

a) $50\%$ b) $20\%$ c) $30\%$

5357506

Find the percent of each circle that is shaded.

Hints

- Count the equal sectors in each circle. - Look for familiar fractions such as one fourth, one half, or three fourths.

Solution

1. a) $9$ of $12$ sectors are shaded: $\frac{9}{12} \times 100\% = 75\%$. 2. b) $4$ of $16$ sectors are shaded: $\frac{4}{16} \times 100\% = 25\%$. 3. c) $10$ of $20$ sectors are shaded: $\frac{10}{20} \times 100\% = 50\%$.

Answer

a) $75\%$ b) $25\%$ c) $50\%$

5357516

Two pizzas were cut into different numbers of equal slices. Which pizza has the greater percent remaining? Show the percentages.

Hints

- Find the remaining fraction for each pizza separately. - Convert both fractions to percentages before comparing.

Solution

1. Pizza a) has $6$ of $8$ slices remaining: $\frac{6}{8} \times 100\% = 75\%$. 2. Pizza b) has $6$ of $10$ slices remaining: $\frac{6}{10} \times 100\% = 60\%$. 3. Since $75\% > 60\%$, pizza a) has the greater percent remaining.

Answer

Pizza a) has more remaining: $75\%$, compared with $60\%$ for pizza b).

5357736

What percent of the circle is shaded?

Hints

- Write the shaded sectors as a fraction of all sectors. - Consider the percent represented by one eighth.

Solution

1. The circle has $8$ equal sectors, and $3$ are shaded. 2. The shaded percent is $\frac{3}{8} \times 100\% = 37.5\%$.

Answer

$37.5\%$

5357746

What percent of the hexagon is shaded green? Round to the nearest tenth of a percent.

Hints

- Write and simplify the shaded fraction. - Convert the fraction to a decimal and use the hundredths digit to round to the nearest tenth.

Solution

1. The hexagon has $6$ equal triangular parts, and $4$ are shaded. 2. The shaded fraction is $\frac{4}{6} = \frac{2}{3}$. 3. Convert and round: $\frac{2}{3} \times 100\% \approx 66.7\%$.

Answer

Approximately $66.7\%$

5357766

What percent of the circles are shaded red?

Hints

- Count the total circles and the shaded circles. - Simplify the fraction before converting it to a percent.

Solution

1. There are $40$ circles, and $22$ are shaded. 2. The shaded fraction is $\frac{22}{40} = \frac{11}{20}$. 3. Convert to a percent: $\frac{11}{20} \times 100\% = 55\%$.

Answer

$55\%$ of the circles are shaded red.

5358246

Compare the shaded portions in figures A and B. Which figure has the greater shaded percent? Show your calculations.

Hints

- Write each shaded portion as a fraction. - Convert both fractions to percentages before comparing.

Solution

1. Figure A has $3$ of $8$ sectors shaded: $\frac{3}{8} \times 100\% = 37.5\%$. 2. Figure B has $2$ of $5$ sections shaded: $\frac{2}{5} \times 100\% = 40\%$. 3. Since $40\% > 37.5\%$, figure B has the greater shaded percent.

Answer

Figure B has the greater shaded percent: $40\%$, compared with $37.5\%$ for figure A.

5358266

A parking lot has $25$ spaces. The occupied spaces are shaded gray. What percent of the spaces are currently available?

Hints

- Pay attention to whether the question asks about occupied or available spaces. - Write the available spaces as a fraction of all $25$ spaces.

Solution

1. The grid represents $25$ parking spaces. 2. Seventeen spaces are occupied, so $25 - 17 = 8$ are available. 3. The available percent is $\frac{8}{25} \times 100\% = 32\%$.

Answer

$32\%$ of the spaces are available.

5103516

Anna earned $27$ of $36$ possible points on a math test. Lucas earned $32$ of $40$ possible points on a different test. Lucas says, “I did better because I earned more points.” Evaluate his claim and determine who performed better relative to the total possible points.

Hints

- Raw point totals are not directly comparable when the maximum scores differ. - Find the percent of possible points each student earned. - Simplify each score fraction before converting to a percent. - Decide what would need to be equal for raw points alone to be comparable.

Solution

1. Anna’s score is $\frac{27}{36}=\frac{3}{4}=75\%$. 2. Lucas’s score is $\frac{32}{40}=\frac{4}{5}=80\%$. 3. Since $80\%>75\%$, Lucas performed better. However, comparing only the number of points is not a valid justification because the tests had different totals.

Answer

Lucas performed better, with $80\%$ compared with Anna’s $75\%$. His conclusion is correct, but his reason is incomplete because the possible point totals were different.

5118446

A household uses $3000\,\text{kWh}$ of electricity in one year. A $10\,\text{cm}$-long strip diagram shows how the electricity is used: - Cooking: $1.5\,\text{cm}$ - Lighting: $1.0\,\text{cm}$ - Refrigeration: $2.5\,\text{cm}$ - Other appliances: $5.0\,\text{cm}$ a) Find the percent represented by each category. b) Find the actual electricity use in $\text{kWh}$ for each category. c) What would the central angle of the refrigeration sector be in a circle graph of the same data?

Hints

- Compare each segment length with the total strip length. - Use each percentage to find a part of $3000\,\text{kWh}$. - A full circle is $360^\circ$.

Solution

1. Compare each segment with the $10\,\text{cm}$ total: cooking, $\frac{1.5}{10}=15\%$; lighting, $\frac{1.0}{10}=10\%$; refrigeration, $\frac{2.5}{10}=25\%$; other appliances, $\frac{5.0}{10}=50\%$. 2. The electricity uses are: cooking, $3000\,\text{kWh}\times0.15=450\,\text{kWh}$; lighting, $3000\,\text{kWh}\times0.10=300\,\text{kWh}$; refrigeration, $3000\,\text{kWh}\times0.25=750\,\text{kWh}$; other appliances, $3000\,\text{kWh}\times0.50=1500\,\text{kWh}$. 3. The refrigeration angle is $360^\circ\times0.25=90^\circ$.

Answer

a) Cooking: $15\%$; lighting: $10\%$; refrigeration: $25\%$; other appliances: $50\%$ b) Cooking: $450\,\text{kWh}$; lighting: $300\,\text{kWh}$; refrigeration: $750\,\text{kWh}$; other appliances: $1500\,\text{kWh}$ c) $90^\circ$

5118536

At Pine School, $40\%$ of the $600$ students play an instrument. At neighboring City School, $50\%$ of the $400$ students play an instrument. a) Which school has more students who play an instrument? Show your calculation. b) If the two student groups were combined, what percent of all students would play an instrument?

Hints

- Convert each school’s percent to a number of students before comparing. - For the combined percent, add both the instrument players and the total enrollments.

Solution

1. Pine School: $600 \times 0.40 = 240$ students. 2. City School: $400 \times 0.50 = 200$ students. 3. Since $240 > 200$, Pine School has more students who play an instrument. 4. Together, $240 + 200 = 440$ of $600 + 400 = 1000$ students play an instrument. 5. The combined percent is $\frac{440}{1000} \times 100\% = 44\%$.

Answer

a) Pine School, with $240$ students compared with $200$ b) $44\%$

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