Aimathic – Percent increase and decrease

Percent increase and decrease

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Difficulty: 1 – 5

5115427

A skateboard has an original price of $\$120$. Find the new price in each case. a) The price is decreased by $15\%$. b) The price is increased by $5\%$. c) The price is reduced to $80\%$ of the original price.

Hints

- Decide what percent of the original remains after each change. - Distinguish between decreasing by $15\%$ and reducing to $80\%$.

Solution

1. a) After a $15\%$ decrease, the price is $85\%$ of the original: $\$120 \cdot 0.85 = \$102$. 2. b) After a $5\%$ increase, the price is $105\%$ of the original: $\$120 \cdot 1.05 = \$126$. 3. c) The new price is $80\%$ of the original: $\$120 \cdot 0.80 = \$96$.

Answer

a) $\$102$ b) $\$126$ c) $\$96$

5128547

A small leak causes a rainwater tank to lose $0.25\,\text{gal}$ of water per minute. After a partial repair, the leak rate decreases by $60\%$. Find the new leak rate in gallons per minute.

Hints

- What percent of the original rate remains after a $60\%$ decrease? - Find that percent of $0.25\,\text{gal/min}$.

Solution

A decrease of $60\%$ means $40\%$ of the original rate remains. Therefore, $0.25\,\text{gal/min} \cdot 0.40 = 0.10\,\text{gal/min}$.

Answer

The new leak rate is $0.10\,\text{gal/min}$.

5349567

Two companies, Alpha and Beta, record their daily production of machine parts. The graph shows the number of parts each company produced on four days. On which days did Beta produce exactly $20\%$ more parts than Alpha?

Hints

- Write “$20\%$ more” as a multiplication factor. - Read both companies’ values for each day. - Test whether Beta’s value is $1.20$ times Alpha’s value.

Solution

1. On Day 1, Alpha produced $50$ parts and Beta produced $60$. Since $50 \cdot 1.20 = 60$, Beta produced $20\%$ more. 2. On Day 2, $100 \cdot 1.20 = 120 \ne 110$, so the condition is not met. 3. On Day 3, $150 \cdot 1.20 = 180$, so Beta produced $20\%$ more. 4. On Day 4, $200 \cdot 1.20 = 240 \ne 210$, so the condition is not met.

Answer

Day 1 and Day 3.

5100537

Linda chooses a number. She increases it by $10\%$, then increases the new result by $20\%$, and finally increases that result by $50\%$. By what percent has the original number increased?

Hints

- Successive percent increases are each applied to a new value, so the percents should not simply be added. - Try starting with $100$ and applying the increases one at a time. - Write each percent increase as a decimal multiplier.

Solution

1. The multipliers for the three increases are $1.10$, $1.20$, and $1.50$. 2. Multiply the factors: $1.10 \cdot 1.20 \cdot 1.50 = 1.98$. 3. The final value is $198\%$ of the original value, so the overall increase is $198\% - 100\% = 98\%$.

Answer

$98\%$

5101267

A forest contains $24{,}000$ trees. Find the number of trees after one year for each growth rate: $0.25\%$, $0.5\%$, $0.75\%$, and $k\%$.

Hints

- Work through the growth rates one at a time. - Find each increase as a percent of $24{,}000$. - Add the increase to the original number of trees. - Use the same process for the variable rate and leave $k$ in the expression.

Solution

1. At $0.25\%$, the increase is $24{,}000 \cdot 0.0025 = 60$, so the new total is $24{,}060$. 2. At $0.5\%$, the increase is $24{,}000 \cdot 0.005 = 120$, so the new total is $24{,}120$. 3. At $0.75\%$, the increase is $24{,}000 \cdot 0.0075 = 180$, so the new total is $24{,}180$. 4. At $k\%$, the increase is $24{,}000 \cdot \frac{k}{100} = 240k$, so the new total is $24{,}000 + 240k$.

Answer

At $0.25\%$: $24{,}060$ At $0.5\%$: $24{,}120$ At $0.75\%$: $24{,}180$ At $k\%$: $24{,}000 + 240k$

5101287

A city has a population of $45{,}000$, and the population decreases during one year. Find the remaining population for each annual decrease rate: $0.2\%$, $0.5\%$, $0.8\%$, and $m\%$.

Hints

- Work with each decrease rate separately. - Find the portion of the population that is lost. - Subtract each loss from the original population. - Use the same method for the variable rate and leave $m$ in the expression.

Solution

1. At a $0.2\%$ decrease, the loss is $45{,}000 \cdot 0.002 = 90$, so the remaining population is $44{,}910$. 2. At a $0.5\%$ decrease, the loss is $45{,}000 \cdot 0.005 = 225$, so the remaining population is $44{,}775$. 3. At a $0.8\%$ decrease, the loss is $45{,}000 \cdot 0.008 = 360$, so the remaining population is $44{,}640$. 4. At an $m\%$ decrease, the loss is $45{,}000 \cdot \frac{m}{100} = 450m$, so the remaining population is $45{,}000 - 450m$.

Answer

At a $0.2\%$ decrease: $44{,}910$ At a $0.5\%$ decrease: $44{,}775$ At a $0.8\%$ decrease: $44{,}640$ At an $m\%$ decrease: $45{,}000 - 450m$

5115367

Because of high demand, the price of a mountain bike was increased by $15\%$. The new price is $\$414$. Find the original price. State the percent of the original price represented by the new price and the decimal multiplier you use.

Hints

- Determine what percent of the original price the new price represents. - Is the new price equal to, greater than, or less than $100\%$ of the original? - To reverse a percent increase, divide by the decimal multiplier.

Solution

1. After a $15\%$ increase, the new price is $100\% + 15\% = 115\%$ of the original price. 2. The corresponding decimal multiplier is $1.15$. 3. Divide the new price by the multiplier: $\$414 \div 1.15 = \$360$.

Answer

The new price is $115\%$ of the original price, so the multiplier is $1.15$. The original price was $\$360$.

5115387

Two classes are collecting donations for an animal shelter. Class A exceeded its fundraising goal by $25\%$ and collected $\$250$. Class B collected $10\%$ less than its goal and collected $\$225$. Which class originally had the greater fundraising goal? Find both original goals.

Hints

- Find two different original amounts. - For each class, determine what percent of the goal the collected amount represents. - An amount above the goal represents more than $100\%$; an amount below the goal represents less than $100\%$.

Solution

1. Class A collected $125\%$ of its goal. Its goal was $\$250 \div 1.25 = \$200$. 2. Class B collected $90\%$ of its goal. Its goal was $\$225 \div 0.90 = \$250$. 3. Since $\$250 > \$200$, Class B had the greater original goal.

Answer

Class A's goal was $\$200$. Class B's goal was $\$250$. Class B had the greater fundraising goal.

5115547

At a middle school, the percent of students who ride bicycles to school increased from $20\%$ last year to $30\%$ this year. a) By how many percentage points did the percent increase? b) By what percent did the share of bicycle riders increase compared with last year?

Hints

- Distinguish between subtracting two percent values and finding a relative percent increase. - Use last year's percent as the base for part b). - Convert the ratio of the change to the original value into a percent.

Solution

1. The percentage-point increase is $30\% - 20\% = 10$ percentage points. 2. The relative increase is based on the original $20\%$: $\frac{30 - 20}{20} = \frac{10}{20} = 0.5$. 3. Convert to a percent: $0.5 = 50\%$.

Answer

a) $10$ percentage points b) $50\%$

5115557

A sports club has a constant total of $200$ members. The percent of members who are teenagers increased from $15\%$ to $24\%$. a) Find the number of teenage members before and after the increase. b) Find the percent increase in the number of teenage members. c) A board member says, “The number of teenage members increased by $9\%$.” Explain why this statement is mathematically imprecise and correct it.

Hints

- Use the total membership and each percent to find the corresponding number of teenagers. - For the relative increase, compare the change with the original number of teenagers. - Distinguish between percent and percentage points.

Solution

1. Before the increase, the number of teenage members was $200 \cdot 0.15 = 30$. 2. After the increase, the number was $200 \cdot 0.24 = 48$. 3. The number increased by $48 - 30 = 18$. 4. Relative to the original $30$ teenagers, the percent increase was $\frac{18}{30} = 0.60 = 60\%$. 5. The membership share rose from $15\%$ to $24\%$, a change of $9$ percentage points. The number of teenage members increased by $60\%$, not $9\%$.

Answer

a) Before: $30$ teenagers; after: $48$ teenagers b) $60\%$ c) The share increased by $9$ percentage points, while the number of teenage members increased by $60\%$.

5115567

Two classes compare their results in a Math Kangaroo competition. Class A: The percent of students who earned recognition increased from $10\%$ to $20\%$. Class B: The percent of students who earned recognition increased from $40\%$ to $50\%$. In both classes, the rate increased by $10$ percentage points. Which class had the greater relative increase? Support your answer with calculations.

Hints

- For each class, compare the increase with the starting percent. - Imagine each class has $100$ students and compare the before-and-after counts. - The same percentage-point increase can represent different relative increases.

Solution

1. For Class A, the relative increase was $\frac{20 - 10}{10} = \frac{10}{10} = 1 = 100\%$. 2. For Class B, the relative increase was $\frac{50 - 40}{40} = \frac{10}{40} = 0.25 = 25\%$. 3. Since $100\% > 25\%$, Class A had the greater relative increase. Its rate doubled, while Class B's rate increased by one-fourth of its original value.

Answer

Class A had the greater relative increase: $100\%$, compared with $25\%$ for Class B.

5115907

One square has side length $20\,\text{cm}$. A second square has side length $25\,\text{cm}$. a) By what percent is the second square's side length greater than the first square's side length? b) Find the area of each square. c) By what percent is the second square's area greater than the first square's area? Compare this result with your answer to part a).

Hints

- For each percent increase, use the first square's value as the base. - Recall the formula for the area of a square. - Use the two calculated areas for part c).

Solution

1. The side-length increase is $\frac{25 - 20}{20} = \frac{5}{20} = 0.25 = 25\%$. 2. The areas are $A_1 = 20\,\text{cm} \cdot 20\,\text{cm} = 400\,\text{cm}^2$ and $A_2 = 25\,\text{cm} \cdot 25\,\text{cm} = 625\,\text{cm}^2$. 3. The area increase is $\frac{625 - 400}{400} = \frac{225}{400} = 0.5625 = 56.25\%$. 4. The area increases by $56.25\%$, which is greater than the $25\%$ increase in side length.

Answer

a) $25\%$ b) $400\,\text{cm}^2$ and $625\,\text{cm}^2$ c) $56.25\%$

5115917

A cube has an edge length of $10\,\text{cm}$. A second cube has an edge length that is $20\%$ greater. a) Find the edge length of the second cube. b) Find the volume of each cube. c) By what percent is the second cube's volume greater than the first cube's volume?

Hints

- Use a multiplier of $1.20$ to find the new edge length. - Recall the formula for the volume of a cube. - Compare the change in volume with the original volume.

Solution

1. The second edge length is $10\,\text{cm} \cdot 1.20 = 12\,\text{cm}$. 2. The volumes are $V_1 = 10\,\text{cm} \cdot 10\,\text{cm} \cdot 10\,\text{cm} = 1000\,\text{cm}^3$ and $V_2 = 12\,\text{cm} \cdot 12\,\text{cm} \cdot 12\,\text{cm} = 1728\,\text{cm}^3$. 3. The percent increase in volume is $\frac{1728 - 1000}{1000} = \frac{728}{1000} = 0.728 = 72.8\%$.

Answer

a) $12\,\text{cm}$ b) $1000\,\text{cm}^3$ and $1728\,\text{cm}^3$ c) $72.8\%$

5116137

Two sports clubs compare their membership growth. Club A grew from $200$ members to $224$ members. Club B grew from $500$ members to $555$ members. Which club had the greater percent increase? Support your answer with calculations.

Hints

- Comparing only the numbers of new members is not enough. - For each club, compare the increase with that club's starting membership. - Compare the two percent increases.

Solution

1. Club A added $224 - 200 = 24$ members. 2. Its percent increase was $\frac{24}{200} = 0.12 = 12\%$. 3. Club B added $555 - 500 = 55$ members. 4. Its percent increase was $\frac{55}{500} = 0.11 = 11\%$. 5. Since $12\% > 11\%$, Club A had the greater percent increase.

Answer

Club A had the greater percent increase: $12\%$, compared with $11\%$ for Club B.

5117387

A city currently has $15{,}000$ residents. Its population grows by $2\%$ each year. a) What will the population be after one year? b) By how many people will the population increase during the second year? Remember that the second year's $2\%$ growth is based on the population at the end of the first year.

Hints

- Find the first-year increase and add it to the starting population. - Use the first-year result as the base for the second-year calculation. - The number of people added is not exactly the same each year.

Solution

1. The first-year increase is $15{,}000 \cdot 0.02 = 300$. 2. After one year, the population is $15{,}000 + 300 = 15{,}300$. 3. The second-year increase is $15{,}300 \cdot 0.02 = 306$.

Answer

a) $15{,}300$ residents b) $306$ people

5117607

A full rain barrel holds $50$ gallons of water. 1. First, $30\%$ of the water is used for tomato plants. 2. Then $20\%$ of the remaining water is used for flower boxes. How many gallons remain? What percent of the original amount is this?

Hints

- Find the amount remaining after the first use. - The second percent is based on the remaining amount, not on the full barrel. - Compare the final amount with the original $50$ gallons.

Solution

1. The first amount used is $50\,\text{gal} \cdot 0.30 = 15\,\text{gal}$, leaving $50 - 15 = 35\,\text{gal}$. 2. The second amount used is $35\,\text{gal} \cdot 0.20 = 7\,\text{gal}$. 3. The final amount is $35 - 7 = 28\,\text{gal}$. 4. As a percent of the original amount, $\frac{28}{50} = 0.56 = 56\%$.

Answer

$28$ gallons remain, which is $56\%$ of the original amount.

5118507

The Miller family reduced its water use by installing water-saving showerheads. Last year, the family used an average of $12{,}000$ gallons of water per month. This year, the monthly average decreased to $10{,}500$ gallons. Find the percent decrease. First identify the original amount and calculate the amount of water saved.

Hints

- Use the amount before the change as the original value. - Subtract to find the number of gallons saved. - Compare the savings with the original monthly use.

Solution

1. The original amount is $12{,}000$ gallons. 2. The amount saved is $12{,}000 - 10{,}500 = 1500$ gallons. 3. The percent decrease is $\frac{1500}{12000} = \frac{1}{8} = 0.125 = 12.5\%$.

Answer

The original amount was $12{,}000$ gallons, and the family saved $1500$ gallons per month. The water use decreased by $12.5\%$.

5118517

A water tank has a capacity of $500$ gallons. At the beginning of a rainy period, it is $20\%$ full. After the rain, it is $35\%$ full. a) By how many percentage points did the fill level increase? b) How many gallons of water were added? c) By what percent did the actual amount of water in the tank increase compared with the starting amount?

Hints

- Distinguish between the percent of the tank's capacity and the number of gallons in the tank. - A change in percentage points is found by subtracting the two fill percentages. - For part c), use the starting amount of water as the base.

Solution

1. The fill level increased by $35 - 20 = 15$ percentage points. 2. At first, the tank held $500\,\text{gal} \cdot 0.20 = 100\,\text{gal}$. After the rain, it held $500\,\text{gal} \cdot 0.35 = 175\,\text{gal}$. 3. The amount added was $175 - 100 = 75\,\text{gal}$. 4. Relative to the starting amount, the percent increase was $\frac{75}{100} = 0.75 = 75\%$.

Answer

a) $15$ percentage points b) $75$ gallons c) $75\%$

5127327

The Weber and Schmidt families live in the same apartment building. After an $8\%$ rent increase, the Weber family now pays $\$918.00$ per month. Because of construction noise, the Schmidt family received a temporary $10\%$ rent reduction and now pays $\$837.00$ per month. Which family had the higher monthly rent before these changes? Support your answer with calculations.

Hints

- Find each family's original rent separately. - One new amount reflects an increase, while the other reflects a decrease. - Compare the two original amounts after reversing the changes.

Solution

1. The Weber family's new rent is $108\%$ of the original rent. Their original rent was $\$918.00 \div 1.08 = \$850.00$. 2. The Schmidt family's reduced rent is $90\%$ of the original rent. Their original rent was $\$837.00 \div 0.90 = \$930.00$. 3. Since $\$930.00 > \$850.00$, the Schmidt family had the higher rent before the changes.

Answer

The Schmidt family had the higher original monthly rent: $\$930.00$, compared with $\$850.00$ for the Weber family.

5127347

An electronic device loses $24\%$ of its value during the first year after purchase. At the end of the year, it is worth $\$418$. a) What was the original purchase price? b) What would the device have been worth after one year if it had lost only $15\%$ of its value?

Hints

- Identify the original value in this situation. - After a percent loss, subtract that percent from $100\%$ to find the percent that remains. - Find the original price before answering part b).

Solution

1. After a $24\%$ loss, $100\% - 24\% = 76\%$ of the original value remains. 2. The original price was $\$418 \div 0.76 = \$550$. 3. With a $15\%$ loss, $85\%$ of the original value would remain. 4. The hypothetical value would be $\$550 \cdot 0.85 = \$467.50$.

Answer

a) $\$550$ b) $\$467.50$

5127397

Rice gains mass while cooking because it absorbs water. A $4\,\text{oz}$ portion of uncooked rice becomes $12\,\text{oz}$ of cooked rice. The uncooked portion contains $9\,\text{g}$ of protein. a) How many grams of protein are in $4\,\text{oz}$ of the cooked rice? b) By what percent did the amount of protein per $4\,\text{oz}$ decrease during cooking? c) Explain why the protein per $4\,\text{oz}$ decreases even though no protein is lost during cooking.

Hints

- Track the total amount of protein in the portion being cooked. - Compare the cooked mass with the original mass. - Use the change from $9\,\text{g}$ to the new amount to find the percent decrease.

Solution

1. The original $9\,\text{g}$ of protein remains in the full $12\,\text{oz}$ of cooked rice. 2. Since $4\,\text{oz}$ is one-third of $12\,\text{oz}$, the protein in $4\,\text{oz}$ of cooked rice is $9\,\text{g} \div 3 = 3\,\text{g}$. 3. The decrease is $9 - 3 = 6\,\text{g}$. Relative to the original $9\,\text{g}$, the percent decrease is $\frac{6}{9} = \frac{2}{3} \approx 66.7\%$. 4. The protein amount stays the same, but the rice gains water and mass, so the protein is spread through a larger total mass.

Answer

a) $3\,\text{g}$ b) Approximately $66.7\%$ c) Water increases the rice's mass, so the same protein is distributed through more cooked rice.

5127407

A $4\,\text{oz}$ portion of uncooked spaghetti contains $12\,\text{g}$ of protein. Assume no protein is lost during cooking. After cooking, $4\,\text{oz}$ of the spaghetti contains $4.8\,\text{g}$ of protein. a) What is the mass of the cooked spaghetti made from the $4\,\text{oz}$ uncooked portion? b) By what percent did the spaghetti's mass increase during cooking?

Hints

- The total $12\,\text{g}$ of protein remains after cooking. - Use the protein amount per $4\,\text{oz}$ of cooked spaghetti to find the total cooked mass. - For the percent increase, compare the increase in mass with the original mass.

Solution

1. The cooked spaghetti still contains a total of $12\,\text{g}$ of protein. 2. Since $4\,\text{oz}$ of cooked spaghetti contains $4.8\,\text{g}$ of protein, the cooked mass containing $12\,\text{g}$ is $\frac{12}{4.8} \cdot 4\,\text{oz} = 10\,\text{oz}$. 3. The mass increase is $10 - 4 = 6\,\text{oz}$. 4. Relative to the original $4\,\text{oz}$, the percent increase is $\frac{6}{4} = 1.5 = 150\%$.

Answer

a) $10\,\text{oz}$ b) $150\%$

5128677

A student is training for a 2-mile race and records a best time each week. <table> <thead> <tr> <th>Week</th> <td>1</td> <td>2</td> <td>3</td> <td>4</td> <td>5</td> </tr> </thead> <tbody> <tr> <th>Best time (min:sec)</th> <td>$12{:}40$</td> <td>$12{:}15$</td> <td>$12{:}20$</td> <td>$11{:}50$</td> <td>$11{:}35$</td> </tr> </tbody> </table> a) Between which two consecutive weeks did the student have the greatest improvement? Give the improvement in seconds. b) Find the total improvement from Week 1 to Week 5 in seconds. c) By what percent did the time improve from Week 1 to Week 5? Use the Week 1 time as the base and round to the nearest tenth of a percent.

Hints

- Convert all times to seconds before comparing them. - A faster time is a smaller time. - For the percent improvement, use the Week 1 time as the base.

Solution

1. Convert each time to seconds: Week 1: $12 \cdot 60 + 40 = 760\,\text{s}$ Week 2: $12 \cdot 60 + 15 = 735\,\text{s}$ Week 3: $12 \cdot 60 + 20 = 740\,\text{s}$ Week 4: $11 \cdot 60 + 50 = 710\,\text{s}$ Week 5: $11 \cdot 60 + 35 = 695\,\text{s}$ 2. The consecutive changes are an improvement of $25\,\text{s}$, a slowdown of $5\,\text{s}$, an improvement of $30\,\text{s}$, and an improvement of $15\,\text{s}$. The greatest improvement was from Week 3 to Week 4. 3. The total improvement was $760 - 695 = 65\,\text{s}$. 4. The percent improvement was $\frac{65}{760} \cdot 100\% \approx 8.6\%$.

Answer

a) Week 3 to Week 4; $30$ seconds b) $65$ seconds c) Approximately $8.6\%$

5133997

Each side of a square lot is reduced by $15\%$. By what percent does the area of the lot decrease?

Hints

- What multiplier represents a $15\%$ decrease? - How does that multiplier affect the square of the side length? - Express the new area as a percent of the original area.

Solution

1. Let the original side length be $s$, so the original area is $s^2$. 2. Reducing the side length by $15\%$ multiplies it by $1 - 0.15 = 0.85$. 3. The new area is $(0.85s)^2 = 0.7225s^2$, which is $72.25\%$ of the original area. 4. Therefore, the area decreases by $100\% - 72.25\% = 27.75\%$.

Answer

The area decreases by $27.75\%$.

5139407

Determine whether each statement is true or false. Correct each false statement. a) $0.2\%$ is equal to $\frac{1}{5}$. b) $\frac{2}{3}$ is less than $67\%$. c) An increase of $150\%$ means the new value is $1.5$ times the original value. d) One-fourth of $20\%$ is $5\%$.

Hints

- Pay close attention to the difference between increasing a value by a percent and changing it to a given percent of the original. - Convert percents to decimals by dividing by $100$. - Write $\frac{2}{3}$ as a percent before comparing it with $67\%$. - For d), find one-fourth of the number $20$.

Solution

1. For a), $0.2\% = \frac{0.2}{100} = \frac{2}{1000} = \frac{1}{500}$, while $\frac{1}{5} = 0.2 = 20\%$. The statement is false. 2. For b), $\frac{2}{3} = 66.\overline{6}\% \approx 66.67\%$. Since $66.\overline{6}\% < 67\%$, the statement is true. 3. For c), an increase of $150\%$ gives $100\% + 150\% = 250\%$ of the original value, or $2.5$ times the original. The statement is false. A new value equal to $150\%$ of the original would be $1.5$ times the original. 4. For d), $\frac{1}{4} \cdot 20\% = 5\%$, so the statement is true.

Answer

a) False. $0.2\% = \frac{1}{500}$; equivalently, $20\% = \frac{1}{5}$. b) True. c) False. An increase of $150\%$ produces a value that is $2.5$ times the original. d) True.

5141917

A square has side length $20\,\text{cm}$. One side length is increased by $25\%$, while the perpendicular side length is decreased by $25\%$, creating a rectangle. a) Find the area of the original square and the new rectangle. b) By what percent did the area change overall?

Hints

- Use the area formulas for a square and a rectangle. - Apply the increase and decrease to the original side length. - Compare the change in area with the original area.

Solution

1. The square's area is $20\,\text{cm} \cdot 20\,\text{cm} = 400\,\text{cm}^2$. 2. The new side lengths are $20\,\text{cm} \cdot 1.25 = 25\,\text{cm}$ and $20\,\text{cm} \cdot 0.75 = 15\,\text{cm}$. 3. The rectangle's area is $25\,\text{cm} \cdot 15\,\text{cm} = 375\,\text{cm}^2$. 4. The area decreased by $400 - 375 = 25\,\text{cm}^2$. 5. The percent decrease is $\frac{25}{400} = 0.0625 = 6.25\%$.

Answer

a) Square: $400\,\text{cm}^2$; rectangle: $375\,\text{cm}^2$ b) A decrease of $6.25\%$

5141927

A square has a perimeter of $48\,\text{cm}$. Each side length is increased by $50\%$. a) By what percent does the perimeter increase? b) By what percent does the area increase?

Hints

- Find the original side length from the perimeter. - Perimeter depends directly on side length, while area depends on the product of two side lengths. - Compare each new measurement with its original value.

Solution

1. The original side length is $48\,\text{cm} \div 4 = 12\,\text{cm}$. 2. The original area is $12\,\text{cm} \cdot 12\,\text{cm} = 144\,\text{cm}^2$. 3. The new side length is $12\,\text{cm} \cdot 1.50 = 18\,\text{cm}$. 4. The new perimeter is $4 \cdot 18\,\text{cm} = 72\,\text{cm}$. Its percent increase is $\frac{72 - 48}{48} = \frac{24}{48} = 0.5 = 50\%$. 5. The new area is $18\,\text{cm} \cdot 18\,\text{cm} = 324\,\text{cm}^2$. Its percent increase is $\frac{324 - 144}{144} = \frac{180}{144} = 1.25 = 125\%$.

Answer

a) $50\%$ b) $125\%$

5142587

A town tracked the share of electric vehicles among new car registrations. The share was $8\%$ in 2022 and $14\%$ in 2023. a) Find the increase in percentage points. b) By what percent did the electric-vehicle share increase compared with 2022? c) Briefly explain the difference between your answers to parts a) and b).

Hints

- Subtract the two percent values for part a). - For part b), compare the change with the starting value of $8\%$. - A percentage-point change and a relative percent change answer different questions.

Solution

1. The percentage-point increase is $14\% - 8\% = 6$ percentage points. 2. The relative increase is based on the original $8\%$: $\frac{14 - 8}{8} = \frac{6}{8} = 0.75 = 75\%$. 3. Percentage points measure the direct difference between two percent values. A percent increase compares that difference with the original percent.

Answer

a) $6$ percentage points b) $75\%$ c) The percentage-point change is the direct difference, while the percent increase is relative to the original $8\%$.

5143607

A car-sharing company changes its pricing plan. For the same fixed charge, a customer could previously drive $50$ miles but can now drive only $40$ miles. a) By what percent did the distance available for the fixed charge decrease? b) By what percent did the effective price per mile increase?

Hints

- For part a), use the original distance as the base. - For part b), compare the old and new price per mile. - You may use any convenient fixed charge because it cancels in the comparison.

Solution

1. The available distance decreased by $50 - 40 = 10$ miles. 2. Relative to the original $50$ miles, the decrease was $\frac{10}{50} = 0.20 = 20\%$. 3. Let the fixed charge be $F$. The old price per mile was $\frac{F}{50}$, and the new price per mile is $\frac{F}{40}$. 4. The ratio of the new unit price to the old unit price is $\frac{F/40}{F/50} = \frac{50}{40} = 1.25$, so the price per mile increased by $25\%$.

Answer

a) A decrease of $20\%$ b) An increase of $25\%$

5154227

An apartment's monthly rent is adjusted over two years. It increases by $5\%$ in the first year and then increases by another $2\%$ in the second year. By what percent has the rent increased overall compared with the original rent?

Hints

- Apply the two increases one after the other. - Write each percent increase as a decimal multiplier. - Convert the combined multiplier back to an overall percent increase.

Solution

1. The multipliers for the two increases are $1.05$ and $1.02$. 2. The combined multiplier is $1.05 \cdot 1.02 = 1.071$. 3. The final rent is $107.1\%$ of the original rent, so the overall increase is $107.1\% - 100\% = 7.1\%$.

Answer

The rent increased by $7.1\%$ overall.

5222147

A square poster has side length $s$. It is redesigned as a rectangle whose length is $10\%$ greater than $s$ and whose width is $20\%$ less than $s$. 1) Write expressions for the new length $l$ and width $w$ in terms of $s$. 2) Write and simplify an expression for the perimeter $P$ of the new poster. 3) Compare the new perimeter with the perimeter of the original square. By what percent did the perimeter change?

Hints

- Convert each percent change to a multiplier. - Use the rectangle perimeter formula $P=2(l+w)$. - The original square has perimeter $4s$. - Divide the amount of change by the original perimeter to find the percent change.

Solution

1. Increasing $s$ by $10\%$ gives $l=1.10s$. Decreasing $s$ by $20\%$ gives $w=0.80s$. 2. The new perimeter is $P=2(l+w)=2(1.10s+0.80s)=2(1.90s)=3.80s$. 3. The original perimeter is $4s$. The decrease is $4s-3.80s=0.20s$. 4. The percent decrease is $\frac{0.20s}{4s}\cdot100\%=5\%$.

Answer

1) $l=1.10s$; $w=0.80s$ 2) $P=3.80s$ 3) The perimeter decreased by $5\%$.

5233197

A region tracks annual electricity generation from wind power. The table shows the generation in gigawatt-hours ($\text{GWh}$) for selected years. <table> <tr><td>Year</td><td>2016</td><td>2019</td><td>2022</td></tr> <tr><td>Electricity generation in $\text{GWh}$</td><td>$80.0$</td><td>$104.0$</td><td>$145.6$</td></tr> </table> a) Find the percent increase in electricity generation from 2016 to 2019. b) Find the percent increase from 2019 to 2022. c) How many times as large was the 2022 generation as the 2016 generation?

Hints

- For each percent increase, use the value at the beginning of that interval as the base. - Compare the change with the starting value. - To find how many times as large the final value is, divide the final value by the initial value.

Solution

1. From 2016 to 2019, the increase was $104.0 - 80.0 = 24.0\,\text{GWh}$. The percent increase was $\frac{24.0}{80.0} = 0.30 = 30\%$. 2. From 2019 to 2022, the increase was $145.6 - 104.0 = 41.6\,\text{GWh}$. The percent increase was $\frac{41.6}{104.0} = 0.40 = 40\%$. 3. The overall factor was $\frac{145.6}{80.0} = 1.82$. Therefore, the 2022 generation was $1.82$ times the 2016 generation.

Answer

a) $30\%$ b) $40\%$ c) $1.82$ times as large

5240147

A sports club has $120$ members. The number of adults is $40\%$ greater than the number of children. How many children and how many adults belong to the club?

Hints

- Which group is the base for the phrase “$40\%$ greater”? - If the number of children is $100\%$, what percent represents the number of adults? - Write an equation in which the number of children is the unknown.

Solution

1. Let $x$ be the number of children. 2. Since the number of adults is $40\%$ greater, the number of adults is $x + 0.40x = 1.40x$. 3. The total gives $x + 1.40x = 120$, so $2.40x = 120$. 4. Solve for $x$: $x = \frac{120}{2.40} = 50$. 5. The number of adults is $120 - 50 = 70$.

Answer

The club has $50$ children and $70$ adults.

5240207

The number of solar-panel systems in a community increased by $35\%$ last year. The community now has $1026$ systems. a) How many systems were there before the increase? b) How many systems were added during the year?

Hints

- What percent of the previous number does the current number represent? - After finding the previous number, subtract it from the current number. - You can also find the increase directly from the original number and $35\%$.

Solution

1. The current number is $100\% + 35\% = 135\%$ of the previous number. 2. Let $x$ be the previous number. Then $1.35x = 1026$. 3. Solve: $x = 1026 \div 1.35 = 760$. 4. The number added was $1026 - 760 = 266$.

Answer

a) $760$ systems b) $266$ systems

5251977

A mountain bike originally cost $\$400$. After one year, its price increased by $10\%$. The following year, the price increased again, reaching $\$462$. By what percent did the price increase during the second year, based on the price after the first year?

Hints

- First find the price after the $10\%$ increase. - Use that new price as the whole for the second increase. - Find how many dollars the price increased during the second year. - Compare that increase with the price after the first year.

Solution

1. After the first increase, the price was $\$400 \cdot 1.10 = \$440$. 2. The second increase was $\$462 - \$440 = \$22$. 3. Relative to the $\$440$ price after the first year, the percent increase was $\frac{22}{440} \cdot 100\% = 5\%$.

Answer

The price increased by $5\%$ during the second year.

5279417

Bread dough loses $15\%$ of its weight during baking as water evaporates. A baker wants a loaf that weighs exactly $34\,\text{oz}$ after baking. How much dough should the baker weigh before baking?

Hints

- What percent of the original weight remains after baking? - Express the baked weight as a percent of the starting weight. - The starting weight must be greater than the baked weight. - Reverse the percent decrease by dividing by the remaining-value multiplier.

Solution

1. After losing $15\%$, the loaf retains $100\% - 15\% = 85\%$ of its original weight. 2. Let $x$ be the original dough weight. Then $0.85x = 34\,\text{oz}$. 3. Solve: $x = 34\,\text{oz} \div 0.85 = 40\,\text{oz}$.

Answer

The baker should weigh $40\,\text{oz}$, or $2.5\,\text{lb}$, of dough.

5318287

A local newspaper reports on a new grocery store with the headline **“Market Share Doubled in Just Two Years!”** The bar graph shows the store’s market share from $2020$ through $2022$. a) Read the market shares for $2020$ and $2022$. By how many percentage points did the market share increase? b) Find the percent increase from $2020$ to $2022$. c) Explain why the headline can sound especially impressive even though the increase is only $5$ percentage points.

Hints

- Read the two values from the y-axis. - Percentage-point change is the difference between the two percentages. - Percent increase compares the change with the original value. - Distinguish the relative increase from the final market share.

Solution

1. The graph shows $5\%$ in $2020$ and $10\%$ in $2022$. 2. The increase is $10\%-5\%=5$ percentage points. 3. The percent increase relative to the initial value is $\frac{10-5}{5}\cdot100\%=100\%$. 4. The market share did double, so the headline is mathematically correct. However, the store gained only $5$ percentage points and still has a market share of just $10\%$.

Answer

a) $2020: 5\%$; $2022: 10\%$; increase: $5$ percentage points b) $100\%$ c) “Doubled” emphasizes the large relative increase, while the actual share rose by only $5$ percentage points to $10\%$.

5349557

Two savings plans, $A$ and $B$, are shown on the graph over $10$ years. Each plan grows at a constant yearly rate. By what percent is Plan $B$’s yearly savings rate greater than Plan $A$’s?

Hints

- Use the graph to find how much each plan gains in one year. - Compare the two yearly dollar amounts. - Use Plan $A$’s rate as the base for the percent increase.

Solution

1. Plan $A$ increases from $\$0$ to $\$1000$ in $10$ years, so its yearly savings rate is $\$1000 \div 10 = \$100$ per year. 2. Plan $B$ increases from $\$0$ to $\$1250$ in $10$ years, so its yearly savings rate is $\$1250 \div 10 = \$125$ per year. 3. The percent increase from $\$100$ to $\$125$ is $\frac{125-100}{100} \cdot 100\% = 25\%$.

Answer

Plan $B$’s yearly savings rate is $25\%$ greater than Plan $A$’s.

5351057

The graph shows the percentage of households in a city that have high-speed fiber internet. a) By how many percentage points did the share increase from $2021$ to $2024$? b) Between which consecutive years was the increase in percentage points greatest? c) A newspaper says, “The percentage of households with fiber internet increased to more than six times its $2021$ level by $2024$.” Determine whether this statement is correct and justify your answer using the graph.

Hints

- Read each annual percentage from the graph. - Percentage-point change is found by subtraction. - Compare the changes for each pair of consecutive years. - For part c), multiply the $2021$ value by $6$.

Solution

1. The total increase is $65\%-10\%=55$ percentage points. 2. The annual increases are $25\%-10\%=15$ percentage points from $2021$ to $2022$, $40\%-25\%=15$ percentage points from $2022$ to $2023$, and $65\%-40\%=25$ percentage points from $2023$ to $2024$. The last interval has the greatest increase. 3. Six times the $2021$ percentage is $6\cdot10\%=60\%$. Since $65\%>60\%$, the newspaper statement is correct.

Answer

a) $55$ percentage points b) From $2023$ to $2024$, with an increase of $25$ percentage points c) Yes. Six times $10\%$ is $60\%$, and the $2024$ value is $65\%$.

5351367

The bar graph shows the percentage of a city’s household waste that was recycled each year from $2018$ through $2023$. a) By how many percentage points did the recycling rate increase over the entire period? b) Between which consecutive years was the increase greatest? How many percentage points was it? c) In which year did the recycling rate decrease from the previous year?

Hints

- Compare the first and last percentages for the overall change. - Subtract each year’s value from the following year’s value. - A negative difference identifies a decrease.

Solution

1. The overall increase was $55\%-30\%=25$ percentage points. 2. The yearly changes were $35-30=5$, $32-35=-3$, $42-32=10$, $50-42=8$, and $55-50=5$ percentage points. 3. The greatest increase was $10$ percentage points from $2020$ to $2021$. 4. The only decrease occurred in $2020$, when the rate fell from $35\%$ to $32\%$.

Answer

a) $25$ percentage points b) From $2020$ to $2021$, by $10$ percentage points c) $2020$

5100427

Taylor stores building bricks in a rectangular box. Taylor wants a new box that is $10\%$ longer and $20\%$ wider than the old box, with the same height. By what percent is the new box's volume greater than the old box's volume?

Hints

- Write the volume formula for a rectangular prism. - A $10\%$ increase corresponds to a scale factor of $1.10$. - Multiply the scale factors for the dimensions that change.

Solution

1. Let the old box have length $L$, width $W$, and height $H$, so its volume is $V=LWH$. 2. The new dimensions are $1.1L$, $1.2W$, and $H$. 3. The new volume is $V'=(1.1L)(1.2W)H=1.32LWH=1.32V$. 4. The increase is $1.32V-V=0.32V$, which is $32\%$ of the original volume.

Answer

The new box's volume is $32\%$ greater.

5115927

A rectangular billboard is $8\,\text{ft}$ wide and $12\,\text{ft}$ high. A designer compares two ways to enlarge it. Option 1: Increase only the width by $50\%$. Option 2: Increase both the width and the height by $25\%$. Determine which option produces the greater billboard area. For that option, find the percent increase in area compared with the original billboard.

Hints

- Find the original area first. - Calculate the new dimensions for each option. - Compare the two new areas. - When both dimensions change, both changes affect the area.

Solution

1. The original area is $8\,\text{ft} \cdot 12\,\text{ft} = 96\,\text{ft}^2$. 2. For Option 1, the new width is $8\,\text{ft} \cdot 1.50 = 12\,\text{ft}$. The new area is $12\,\text{ft} \cdot 12\,\text{ft} = 144\,\text{ft}^2$. 3. For Option 2, the new dimensions are $8\,\text{ft} \cdot 1.25 = 10\,\text{ft}$ and $12\,\text{ft} \cdot 1.25 = 15\,\text{ft}$. The new area is $10\,\text{ft} \cdot 15\,\text{ft} = 150\,\text{ft}^2$. 4. Option 2 produces the greater area. Its increase is $150 - 96 = 54\,\text{ft}^2$, and $\frac{54}{96} = 0.5625 = 56.25\%$.

Answer

Option 2 produces the greater area, $150\,\text{ft}^2$. The area increases by $56.25\%$.

5127257

A city recorded $5000$ new car registrations in 2020. In 2021, the number decreased by $8\%$. In 2022, the number then increased by $12\%$ compared with 2021. a) Find the number of new registrations in 2021 and 2022. b) Find the overall percent change from 2020 to 2022. c) Explain why the overall change is not simply $+4\%$, found by calculating $12\% - 8\%$.

Hints

- Apply each year's percent change in order. - Identify the base value for each percent calculation. - Compare the 2022 value directly with the 2020 value.

Solution

1. In 2021, the number was $5000 \cdot 0.92 = 4600$. 2. In 2022, the number was $4600 \cdot 1.12 = 5152$. 3. The overall increase was $5152 - 5000 = 152$. Relative to the 2020 value, $\frac{152}{5000} = 0.0304 = 3.04\%$. 4. The $12\%$ increase was applied to the reduced 2021 value, not to the original 2020 value. Therefore, the two percent changes cannot be combined by simple subtraction.

Answer

a) 2021: $4600$; 2022: $5152$ b) An increase of $3.04\%$ c) The $12\%$ increase uses the lower 2021 value as its base.

5127357

In a small town, the number of trees in a city park increases by $20\%$ during the first year because of new plantings. During the next year, $15\%$ of the trees then in the park must be removed because of drought. After the two years, the park has $1224$ trees. a) How many trees were in the park originally? b) By what percent did the number of trees change overall compared with the original number?

Hints

- The second percent change is based on the number after the first change, not on the original number. - Represent the two changes with decimal multipliers and multiply them. - Explain why simply subtracting $15\%$ from $20\%$ does not give the correct overall change.

Solution

1. Let $x$ be the original number of trees. After the first year, the number is $1.20x$. After the second year, the number is $1.20x \cdot 0.85$. 2. The combined multiplier is $1.20 \cdot 0.85 = 1.02$. 3. Solve $1.02x = 1224$: $x = 1224 \div 1.02 = 1200$. 4. A multiplier of $1.02$ means the final number is $102\%$ of the original, so the overall change is a $2\%$ increase.

Answer

a) $1200$ trees b) An overall increase of $2\%$

5127417

Beef jerky is made by drying fresh beef. During drying, the beef loses $60\%$ of its original mass as water evaporates. Assume no protein is lost. A $4\,\text{oz}$ portion of fresh beef contains about $22\,\text{g}$ of protein. a) How many grams of protein are in $4\,\text{oz}$ of the finished beef jerky? b) By what percent did the protein concentration, measured as protein per $4\,\text{oz}$, change during drying? c) Compare this process with cooking pasta. Why does the protein per $4\,\text{oz}$ increase for jerky but decrease for cooked pasta?

Hints

- Find the mass that remains after a $60\%$ loss. - The protein remains while the water leaves. - Scale the protein in the smaller jerky portion to a $4\,\text{oz}$ portion. - Compare losing water with absorbing water.

Solution

1. After losing $60\%$ of its mass, the $4\,\text{oz}$ portion becomes $4\,\text{oz} \cdot 0.40 = 1.6\,\text{oz}$ of jerky. 2. The $22\,\text{g}$ of protein remains in the $1.6\,\text{oz}$ of jerky. 3. Scale to $4\,\text{oz}$: $\frac{22\,\text{g}}{1.6\,\text{oz}} \cdot 4\,\text{oz} = 55\,\text{g}$. 4. The protein amount per $4\,\text{oz}$ rises from $22\,\text{g}$ to $55\,\text{g}$, an increase of $33\,\text{g}$. The percent increase is $\frac{33}{22} = 1.5 = 150\%$. 5. Drying removes water and concentrates the protein, while cooking pasta adds water and spreads the protein through a greater mass.

Answer

a) $55\,\text{g}$ b) An increase of $150\%$ c) Jerky loses water and mass, while pasta gains water and mass.

5134007

A square poster is enlarged so that its area increases by $125\%$. By what percent must each side length increase?

Hints

- What total percent of the original area results from an increase of $125\%$? - How are the area scale factor and side-length scale factor related for a square? - Convert the side-length scale factor into a percent increase.

Solution

1. An increase of $125\%$ means the new area is $225\%$ of the original area, so the area scale factor is $2.25$. 2. If the side-length scale factor is $k$, then $k^2 = 2.25$. 3. Since $1.5^2 = 2.25$ and a side-length scale factor is positive, $k = 1.5$. 4. A scale factor of $1.5$ means each side length increases by $50\%$.

Answer

Each side length must increase by $50\%$.

5134017

A square sheet of metal has side length $a$. a) Explain mathematically how the area changes when the side length is doubled. b) Find the factor $k$ by which the side length must be multiplied to exactly double the area. Round $k$ to the nearest hundredth. c) A student says, “If the side length increases by $50\%$, the area also increases by $50\%$.” Evaluate and correct the claim.

Hints

- Substitute the changed side length into the area formula for a square. - What happens to a factor inside a squared expression? - Distinguish between linear growth and area growth.

Solution

1. The original area is $A = a^2$. Doubling the side gives $(2a)^2 = 4a^2$, so the area is multiplied by $4$. 2. To double the area, solve $(ka)^2 = 2a^2$. Then $k^2 = 2$. Since $1.41^2 = 1.9881 < 2$ and $1.415^2 = 2.002225 > 2$, the positive value of $k$ lies between $1.41$ and $1.415$. Therefore, $k \approx 1.41$ to the nearest hundredth. 3. A $50\%$ increase multiplies the side length by $1.5$. The area is multiplied by $1.5^2 = 2.25$, so the area increases by $125\%$, not $50\%$.

Answer

a) The area is multiplied by $4$. b) $k \approx 1.41$ c) The claim is false. A $50\%$ increase in side length produces a $125\%$ increase in area.

5141937

Two rectangles each have an area of $120\,\text{cm}^2$. Rectangle A has side lengths $10\,\text{cm}$ and $12\,\text{cm}$. Rectangle B has side lengths $8\,\text{cm}$ and $15\,\text{cm}$. Both side lengths of each rectangle are increased by exactly $2\,\text{cm}$. Does the area of each rectangle increase by the same percent? Support your answer with calculations.

Hints

- Add $2\,\text{cm}$ to both dimensions of each rectangle. - Find each new area. - Compare each area increase with the original area of $120\,\text{cm}^2$.

Solution

1. Rectangle A's new dimensions are $12\,\text{cm}$ and $14\,\text{cm}$, so its new area is $12\,\text{cm} \cdot 14\,\text{cm} = 168\,\text{cm}^2$. 2. Its percent increase is $\frac{168 - 120}{120} = \frac{48}{120} = 0.40 = 40\%$. 3. Rectangle B's new dimensions are $10\,\text{cm}$ and $17\,\text{cm}$, so its new area is $10\,\text{cm} \cdot 17\,\text{cm} = 170\,\text{cm}^2$. 4. Its percent increase is $\frac{170 - 120}{120} = \frac{50}{120} \approx 0.4167 = 41.67\%$. 5. Since $41.67\% \ne 40\%$, the areas do not increase by the same percent.

Answer

No. Rectangle A's area increases by $40\%$, while Rectangle B's area increases by approximately $41.67\%$.

5350487

The bar graph shows the monthly attendance at a new amusement park from May through September. a) In which month was the greatest numerical increase in attendance from the previous month? b) In which month was the greatest percent increase from the previous month?

Hints

- Read each monthly attendance value from the graph. - Numerical increase is the difference between consecutive months. - Percent increase compares that difference with the previous month’s value. - The greatest numerical increase does not have to be the greatest percent increase.

Solution

1. The graph shows May, $1000$; June, $2000$; July, $3500$; August, $5500$; and September, $8000$. 2. The numerical increases are: June, $2000-1000=1000$; July, $3500-2000=1500$; August, $5500-3500=2000$; September, $8000-5500=2500$. The greatest numerical increase occurred in September. 3. The percent increases are: June, $\frac{1000}{1000}\cdot100\%=100\%$; July, $\frac{1500}{2000}\cdot100\%=75\%$; August, $\frac{2000}{3500}\cdot100\%\approx57.1\%$; September, $\frac{2500}{5500}\cdot100\%\approx45.5\%$. 4. The greatest percent increase occurred in June.

Answer

a) September, with $2500$ more visitors than in August b) June, with a $100\%$ increase from May

5134137

Let $p$ be the decimal rate by which the side length of a square is increased. An increase by $p$ makes the area increase by $44\%$. Determine whether doubling the side-length rate to $2p$ also doubles the original area increase from $44\%$ to $88\%$. Show your calculations.

Hints

- First find the value of $p$ that produces a $44\%$ area increase. - What side-length factor corresponds to $2p$? - Square that factor to find the new area factor. - Compare the new increase with $88\%$.

Solution

1. An area increase of $44\%$ gives the equation $(1 + p)^2 = 1.44$. 2. Since $1.2^2 = 1.44$ and the side-length factor is positive, $1 + p = 1.2$, so $p = 0.2$, or $20\%$. 3. Doubling the rate gives $2p = 0.4$, or $40\%$, so the new side-length factor is $1.4$. 4. The new area factor is $1.4^2 = 1.96$, which is an area increase of $96\%$. 5. Since $96\% \neq 88\%$, doubling the side-length rate does not double the area increase.

Answer

No. The original decimal rate is $p = 0.2$, or $20\%$. Doubling it to $0.4$, or $40\%$, makes the area increase by $96\%$, not $88\%$.

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