Multi Step Percentage Word Problems Worksheet PDF | airSlate SignNow - Free Printable

Let's solve each problem step by step:

---

Problem 1:

Hardy’s plant grows rapidly. It grows by 10 cm every 2 years. How much taller will Hardy’s plant be in 3 years?

#### Solution:
- The plant grows 10 cm every 2 years.
- In 3 years, we need to calculate how much it grows.

1. First, determine the growth rate per year:
\[
\text{Growth per year} = \frac{10 \text{ cm}}{2 \text{ years}} = 5 \text{ cm/year}
\]

2. Now, calculate the growth over 3 years:
\[
\text{Growth in 3 years} = 5 \text{ cm/year} \times 3 \text{ years} = 15 \text{ cm}
\]

#### Final Answer:
\[
\boxed{15 \text{ cm}}
\]

---

Problem 2:

A beaver burrows 10 holes every 5 minutes. How many holes will it burrow if it burrows for 25 minutes?

#### Solution:
- The beaver burrows 10 holes every 5 minutes.
- We need to find out how many holes it burrows in 25 minutes.

1. First, determine the burrowing rate per minute:
\[
\text{Burrowing rate per minute} = \frac{10 \text{ holes}}{5 \text{ minutes}} = 2 \text{ holes/minute}
\]

2. Now, calculate the total number of holes burrowed in 25 minutes:
\[
\text{Total holes in 25 minutes} = 2 \text{ holes/minute} \times 25 \text{ minutes} = 50 \text{ holes}
\]

#### Final Answer:
\[
\boxed{50}
\]

---

Problem 3:

Gregory drinks 6 liters of water over 2 days. How many liters does he drink in 7 days?

#### Solution:
- Gregory drinks 6 liters of water over 2 days.
- We need to find out how much he drinks in 7 days.

1. First, determine the drinking rate per day:
\[
\text{Drinking rate per day} = \frac{6 \text{ liters}}{2 \text{ days}} = 3 \text{ liters/day}
\]

2. Now, calculate the total amount of water he drinks in 7 days:
\[
\text{Total water in 7 days} = 3 \text{ liters/day} \times 7 \text{ days} = 21 \text{ liters}
\]

#### Final Answer:
\[
\boxed{21}
\]

---

Problem 4:

Ron’s factory is currently producing 38 chairs in a day. How many chairs will be produced if production is to be increased by 130%?

#### Solution:
- The factory currently produces 38 chairs per day.
- Production is to be increased by 130%.

1. Calculate the increase in production:
\[
\text{Increase} = 130\% \times 38 = 1.3 \times 38 = 49.4 \text{ chairs}
\]

2. Add the increase to the original production:
\[
\text{New production} = 38 + 49.4 = 87.4 \text{ chairs}
\]

#### Final Answer:
\[
\boxed{87.4}
\]

---

Problem 5:

Fred has a bakery. He sells 40 cakes in a day. If his sales are increased by 12%, how many cakes will he sell in a day?

#### Solution:
- Fred sells 40 cakes per day.
- Sales are to be increased by 12%.

1. Calculate the increase in sales:
\[
\text{Increase} = 12\% \times 40 = 0.12 \times 40 = 4.8 \text{ cakes}
\]

2. Add the increase to the original sales:
\[
\text{New sales} = 40 + 4.8 = 44.8 \text{ cakes}
\]

#### Final Answer:
\[
\boxed{44.8}
\]

---

Problem 6:

Henry eats 1/3 of a loaf of bread in 2 days. How many loaves of bread will he eat in 15 days?

#### Solution:
- Henry eats 1/3 of a loaf in 2 days.
- We need to find out how much he eats in 15 days.

1. Determine the eating rate per day:
\[
\text{Eating rate per day} = \frac{1/3 \text{ loaf}}{2 \text{ days}} = \frac{1}{3} \div 2 = \frac{1}{6} \text{ loaf/day}
\]

2. Calculate the total amount of bread eaten in 15 days:
\[
\text{Total bread in 15 days} = \frac{1}{6} \text{ loaf/day} \times 15 \text{ days} = \frac{15}{6} = 2.5 \text{ loaves}
\]

#### Final Answer:
\[
\boxed{2.5}
\]

---

Problem 7:

Veronica’s bike requires 5 liters of gas to run for 10 kilometers. How many liters of gas does Veronica need if she has to cover 24 kilometers?

#### Solution:
- Veronica’s bike uses 5 liters of gas for 10 kilometers.
- We need to find out how much gas is needed for 24 kilometers.

1. Determine the gas consumption rate per kilometer:
\[
\text{Gas consumption per kilometer} = \frac{5 \text{ liters}}{10 \text{ km}} = 0.5 \text{ liters/km}
\]

2. Calculate the total gas needed for 24 kilometers:
\[
\text{Total gas for 24 km} = 0.5 \text{ liters/km} \times 24 \text{ km} = 12 \text{ liters}
\]

#### Final Answer:
\[
\boxed{12}
\]

---

Problem 8:

Betty has a cow. The cow produces 5 liters of milk a day. If its diet is improved, the cow will produce 200% more milk. How much milk would the cow make in a day if that were to happen?

#### Solution:
- The cow currently produces 5 liters of milk per day.
- If the diet is improved, the cow will produce 200% more milk.

1. Calculate the additional milk produced due to the improvement:
\[
\text{Additional milk} = 200\% \times 5 = 2 \times 5 = 10 \text{ liters}
\]

2. Add the additional milk to the original production:
\[
\text{New production} = 5 + 10 = 15 \text{ liters}
\]

#### Final Answer:
\[
\boxed{15}
\]

---

Problem 9:

A machine makes 20 carpets in a day. After oiling the machine, its efficiency will go up by 150%. How many carpets would be produced at this level?

#### Solution:
- The machine currently produces 20 carpets per day.
- After oiling, its efficiency increases by 150%.

1. Calculate the increase in production:
\[
\text{Increase} = 150\% \times 20 = 1.5 \times 20 = 30 \text{ carpets}
\]

2. Add the increase to the original production:
\[
\text{New production} = 20 + 30 = 50 \text{ carpets}
\]

#### Final Answer:
\[
\boxed{50}
\]

---

Problem 10:

Samantha can write 15 pages in 6 minutes. How many pages could she write in 12 minutes?

#### Solution:
- Samantha writes 15 pages in 6 minutes.
- We need to find out how many pages she can write in 12 minutes.

1. Determine the writing rate per minute:
\[
\text{Writing rate per minute} = \frac{15 \text{ pages}}{6 \text{ minutes}} = 2.5 \text{ pages/minute}
\]

2. Calculate the total number of pages written in 12 minutes:
\[
\text{Total pages in 12 minutes} = 2.5 \text{ pages/minute} \times 12 \text{ minutes} = 30 \text{ pages}
\]

#### Final Answer:
\[
\boxed{30}
\]

---

Final Answers:

1. \(\boxed{15}\)
2. \(\boxed{50}\)
3. \(\boxed{21}\)
4. \(\boxed{87.4}\)
5. \(\boxed{44.8}\)
6. \(\boxed{2.5}\)
7. \(\boxed{12}\)
8. \(\boxed{15}\)
9. \(\boxed{50}\)
10. \(\boxed{30}\)