problem solving of percentage

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How to Solve Percent Problems? (+FREE Worksheet!)

Learn how to calculate and solve percent problems using the percent formula.

Related Topics

• How to Find Percent of Increase and Decrease
• How to Find Discount, Tax, and Tip
• How to Do Percentage Calculations
• How to Solve Simple Interest Problems

Step by step guide to solve percent problems

• In each percent problem, we are looking for the base, or part or the percent.
• Use the following equations to find each missing section. Base \(= \color{black}{Part} \ ÷ \ \color{blue}{Percent}\) \(\color{ black }{Part} = \color{blue}{Percent} \ ×\) Base \(\color{blue}{Percent} = \color{ black }{Part} \ ÷\) Base

Percent Problems – Example 1:

\(2.5\) is what percent of \(20\)?

In this problem, we are looking for the percent. Use the following equation: \(\color{blue}{Percent} = \color{ black }{Part} \ ÷\) Base \(→\) Percent \(=2.5 \ ÷ \ 20=0.125=12.5\%\)

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Percent problems – example 2:.

\(40\) is \(10\%\) of what number?

Use the following formula: Base \(= \color{ black }{Part} \ ÷ \ \color{blue}{Percent}\) \(→\) Base \(=40 \ ÷ \ 0.10=400\) \(40\) is \(10\%\) of \(400\).

Percent Problems – Example 3:

\(1.2\) is what percent of \(24\)?

In this problem, we are looking for the percent. Use the following equation: \(\color{blue}{Percent} = \color{ black }{Part} \ ÷\) Base \(→\) Percent \(=1.2÷24=0.05=5\%\)

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Percent problems – example 4:.

\(20\) is \(5\%\) of what number?

Use the following formula: Base \(= \color{black}{Part} \ ÷ \ \color{blue}{Percent}\) \(→\) Base \(=20÷0.05=400\) \( 20\) is \(5\%\) of \(400\).

Exercises for Calculating Percent Problems

Solve each problem..

• \(51\) is \(340\%\) of what?
• \(93\%\) of what number is \(97\)?
• \(27\%\) of \(142\) is what number?
• What percent of \(125\) is \(29.3\)?
• \(60\) is what percent of \(126\)?
• \(67\) is \(67\%\) of what?

Download Percent Problems Worksheet

• \(\color{blue}{15}\)
• \(\color{blue}{104.3}\)
• \(\color{blue}{38.34}\)
• \(\color{blue}{23.44\%}\)
• \(\color{blue}{47.6\%}\)
• \(\color{blue}{100}\)

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by: Effortless Math Team about 4 years ago (category: Articles , Free Math Worksheets )

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4.2: Percents Problems and Applications of Percent

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• Page ID 142718

• Morgan Chase
• Clackamas Community College via OpenOregon

You may use a calculator throughout this module.

Recall: The amount is the answer we get after finding the percent of the original number. The base is the original number, the number we find the percent of. We can call the percent the rate.

When we looked at percents in a previous module, we focused on finding the amount. In this module, we will learn how to find the percentage rate and the base.

\(\text{Amount}=\text{Rate}\cdot\text{Base}\)

\(A=R\cdot{B}\)

We can translate from words into algebra.

• “is” means equals
• “of” means multiply
• “what” means a variable

Solving Percent Problems: Finding the Rate

Suppose you earned \(56\) points on a \(60\)-point quiz. To figure out your grade as a percent, you need to answer the question “\(56\) is what percent of \(60\)?” We can translate this sentence into the equation \(56=R\cdot60\).

Exercises \(\PageIndex{1}\)

1. \(56\) is what percent of \(60\)?

2. What percent of \(120\) is \(45\)?

1. \(93\%\) or \(93.3\%\)

2. \(37.5\%\)

Be aware that this method gives us the answer in decimal form and we must move the decimal point to convert the answer to a percent.

Also, if the instructions don’t explicitly tell you how to round your answer, use your best judgment: to the nearest whole percent or nearest tenth of a percent, to two or three significant figures, etc.

Solving Percent Problems: Finding the Base

Suppose you earn \(2\%\) cash rewards for the amount you charge on your credit card. If you want to earn $ \(50\) in cash rewards, how much do you need to charge on your card? To figure this out, you need to answer the question “\(50\) is \(2\%\) of what number?” We can translate this into the equation \(50=0.02\cdot{B}\).

3. $ \(50\) is \(2\%\) of what number?

4. \(5\%\) of what number is \(36\)?

3. $ \(2,500\)

5. An \(18\%\) tip will be added to a dinner that cost $ \(107.50\). What is the amount of the tip?

6. The University of Oregon women’s basketball team made \(13\) of the \(29\) three-points shots they attempted during a game against UNC. What percent of their three-point shots did the team make?

7. \(45\%\) of the people surveyed answered “yes” to a poll question. If \(180\) people answered “yes”, how many people were surveyed altogether?

5. $ \(19.35\)

6. \(44.8\%\) or \(45\%\)

7. \(400\) people were surveyed

Solving Percent Problems: Percent Increase

When a quantity changes, it is often useful to know by what percent it changed. If the price of a candy bar is increased by \(50\) cents, you might be annoyed because it’s it’s a relatively large percentage of the original price. If the price of a car is increased by \(50\) cents, though, you wouldn’t care because it’s such a small percentage of the original price.

To find the percent of increase:

• Subtract the two numbers to find the amount of increase.
• Using this result as the amount and the original number as the base, find the unknown percent.

Notice that we always use the original number for the base, the number that occurred earlier in time. In the case of a percent increase, this is the smaller of the two numbers.

8. The price of a candy bar increased from $ \(0.89\) to $ \(1.39\). By what percent did the price increase?

9. The population of Portland in 2010 was \(583,793\). The estimated population in 2019 was \(654,741\). Find the percent of increase in the population. [1]

8. \(56.2\%\) increase

9. \(12.2\%\) increase

Solving Percent Problems: Percent Decrease

Finding the percent decrease in a number is very similar.

To find the percent of decrease:

• Subtract the two numbers to find the amount of decrease.

Again, we always use the original number for the base, the number that occurred earlier in time. For a percent decrease, this is the larger of the two numbers.

10. During a sale, the price of a candy bar was reduced from $ \(1.39\) to $ \(0.89\). By what percent did the price decrease?

11. The number of students enrolled at Clackamas Community College decreased from \(7,439\) in Summer 2019 to \(4,781\) in Summer 2020. Find the percent of decrease in enrollment.

10. \(36.0\%\) decrease

11. \(35.7\%\) decrease

Relative Error

In an earlier module, we said that a measurement will always include some error, no matter how carefully we measure. It can be helpful to consider the size of the error relative to the size of what is being measured. As we saw in the examples above, a difference of \(50\) cents is important when we’re pricing candy bars but insignificant when we’re pricing cars. In the same way, an error of an eighth of an inch could be a deal-breaker when you’re trying to fit a screen into a window frame, but an eighth of an inch is insignificant when you’re measuring the length of your garage.

The expected outcome is what the number would be in a perfect world. If a window screen is supposed to be exactly \(25\) inches wide, we call this the expected outcome, and we treat it as though it has infinitely many significant digits. In theory, the expected outcome is \(25.000000...\)

To find the absolute error , we subtract the measurement and the expected outcome. Because we always treat the expected outcome as though it has unlimited significant figures, the absolute error should have the same precision (place value) as the measurement , not the expected outcome .

To find the relative error , we divide the absolute error by the expected outcome. We usually express the relative error as a percent. In fact, the procedure for finding the relative error is identical to the procedures for finding a percent increase or percent decrease!

To find the relative error:

• Subtract the two numbers to find the absolute error.
• Using the absolute error as the amount and the expected outcome as the base, find the unknown percent.

Exercisew \(\PageIndex{1}\)

12. A window screen is measured to be \(25\dfrac{3}{16}\) inches wide instead of the advertised \(25\) inches. Determine the relative error, rounded to the nearest tenth of a percent.

13. The contents of a box of cereal are supposed to weigh \(10.8\) ounces, but they are measured at \(10.67\) ounces. Determine the relative error, rounded to the nearest tenth of a percent.

12. \(0.1875\div25\approx0.8\%\)

13. \(0.13\div10.8\approx1.2\%\)

The tolerance is the maximum amount that a measurement is allowed to differ from the expected outcome. For example, the U.S. Mint needs its coins to have a consistent size and weight so that they will work in vending machines. A dime (10 cents) weighs \(2.268\) grams, with a tolerance of \(\pm0.091\) grams. [2] This tells us that the minimum acceptable weight is \(2.268-0.091=2.177\) grams, and the maximum acceptable weight is \(2.268+0.091=2.359\) grams. A dime with a weight outside of the range \(2.177\leq\text{weight}\leq2.359\) would be unacceptable.

A U.S. nickel (5 cents) weighs \(5.000\) grams with a tolerance of \(\pm0.194\) grams.

14. Determine the lowest acceptable weight and highest acceptable weight of a nickel.

15. Determine the relative error of a nickel that weighs \(5.21\) grams.

A U.S. quarter (25 cents) weighs \(5.670\) grams with a tolerance of \(\pm0.227\) grams.

16. Determine the lowest acceptable weight and highest acceptable weight of a quarter.

17. Determine the relative error of a quarter that weighs \(5.43\) grams.

14. \(4.806\) g; \(5.194\) g

15. \(0.21\div5.000=4.2\%\)

16. \(5.443\) g; \(5.897\) g

17. \(0.24\div5.670\approx4.2\%\)

• www.census.gov/quickfacts/fact/table/portlandcityoregon,OR,US/PST045219 ↵
• https://www.usmint.gov/learn/coin-and-medal-programs/coin-specifications and https://www.thesprucecrafts.com/how-much-do-coins-weigh-4171330 ↵

Home / United States / Math Classes / 6th Grade Math / Solving Problems Based on Percentage

Solving Problems Based on Percentage

Percent is an alternate method of representing fractions and decimals. Here we will learn different methods of calculati ng the percent and the steps involved in each method. We will also look at some examples that will help you gain a better understanding of the concept. ...Read More Read Less

Table Of Contents

What is meant by percentage?

Solving problems based on percentages, finding the percentage of a number, finding the whole number from the percent, finding the whole using the ratio method, solved examples.

• Frequently Asked Questions

In mathematics, a percentage is a number or ratio that represents a fraction of 100. The symbol “ % ” is frequently used to represent it, and it has a few hundred years of history. While we are on the topic of percentages, one example will be, the decimal 0.35, or the fraction \(\frac{7}{20}\) , which is equivalent to 35 percent, or 35%.

By solving problems based on percentages, we can find the missing values and find the values of various unknowns in a given problem.

Find 40% of 200.

\(\frac{40}{100}\times 200\)              Write the percentage as a fraction

\(\frac{2}{5}\times 200=800\)            Simplify

First, write the percentage as a fraction or decimal. Then, divide the fraction or decimal by the part. This method applies to any situation in which a percentage and its value are given. 

If 2 percent equals 80, multiply 80 by 100 and divide it by 2 to get 4000.

Prove that 20% of 120 is 24. 

20% =\(\frac{20}{100}\)       Write the percent as a fraction or decimal.

Using multiplication equation:

\(\frac{20}{100}\times 120=24\)      Simplify

To prove the reverse of this solution we use the  division equation:

\(\frac{24}{\frac{20}{100}}\)      Simplify

\(\frac{2400}{20}=120\)    

A ratio table is the table that shows the comparison between two units and shows the relationship between them.

Example 1: What is 25% of 50?

We have 25% of 50.

So, 25% of 50 = \(\frac{1}{4}\times 50\) Write the percentage as a fraction or decimal.

                       = \(\frac{50}{4}\)     Simplify.

                       = 12.5  

Example 2: Using the ratio table, answer the following question:

What is 60% of 200?

We have 60% of 200.

Now, we have to use the ratio table to find the part. Let one row represent the part and the other row represent the whole row in the table and find the equivalent ratio of 200.

The first column represents the percentage = \(\frac{60}{100}\)

So, 60% of 200 is 120.

Example 3: Find the whole of the number.

                  50% of what number is 45.

We have: 50% of what number is 45?

Use division equation

\(\frac{45}{50%}\)    Write the percentage as a fraction or decimal

\(=\frac{45}{\frac{1}{2}}\)  Simplify

So, \(45\times 2=90\)

Hence, 50% of 90 is 45

Example 4: Find the whole of the number using the ratio table.

140% of what number is 84

We have to find 140% of what number is 84.

Use the ratio table to find the part. Let one be the part and the other be the whole row in the table. Now, find the equivalent ratio of 200.

So, 140% of 60 is 84.

Example 5: A rectangular hall’s width is 60 percent of its length.

What are the room’s dimensions?

Solution:  

Calculate the width of the room by taking 60% of 15 feet.

\(60%\times 15\) Write the percentage as a fraction or decimal.

= \(0.6\times 15\)      Simplify

We can al so understand it with the help of a diagram:

The width is 9 feet.

Area of the rectangle = \(\text{length}\times \text{width}\)

                                    = \(15\times 9\)

                                    = 135

Hence, the area of the given room is 135 \(feet^2\).

Example 6: You have won a camping trip at an auction at your school fair that cost $80. Your bid is 40% of your maximum bid for the price of the camping trip. How much more would you be willing to pay for the trip if you hadn’t already paid the full price?

You are given the winning camping bid that represents the maximum bid as well as the percentage of your maximum bid. You must calculate how much more you would have paid for the camping trip if you had known how much more you were willing to pay.

Your winning bid is the part, and your maximum bid is the whole.

Create a model based on the fact that 40% of the total is $80 to determine the highest bid. Then divide the winning bid by the maximum bid to find out how much more you were willing to pay.

The maximum bid is $200 and the winning bid is $80. So, you would be willing to bid $200 – $80 = $120 more for the tickets.

How do you calculate a percentage?

To calculate a percentage, divide the given value by the total value and multiply the result by 100. That is “(value/total value) x 100%”. This is the formula for calculating percentages.

In mathematics, a percentage is a number or ratio that represents a fraction of 100 in mathematics. Percentage is usually represented by the symbol “%”. It is also written simply as “percent” or “pct”. For example, the decimal 0.35, or the fraction \(\frac{35}{100}\), is equivalent to 0.35.

What is the purpose of percentages?

Percentages are used to figure out “how much” or “how many” of something is to be taken from a given value. Percentage makes it easier to calculate the exact amount or figure being discussed. In order to determine whether a percentage increase or decrease has occurred, a comparison of fractions is done. This aids in calculating percentages of profit and loss, for example in real life situations.

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Solving Percent Problems

Videos to help Grade 6 students solve percent problems. When given a part and the percent, students find the percent of a quantity and solve problems involving finding the whole.

New York State Common Core Math Module 1, Grade 6, Lesson 29

Lesson 29 outcome.

• Students find the percent of a quantity. • Given a part and the percent, students solve problems involving finding the whole.

Lesson 29 Summary

Claim: To find 10% of a number all you need to do is move the decimal to the left once.

Use at least one model to solve each problem (e.g., tape diagram, table, double number line diagram, 10x10 grid).

Claim: If an item is already on sale and then there is another discount taken off the sale price, this is the same as saving the sum of the two discounts from the original price.

Use at least one model to solve each problem (e.g., tape diagram, table, double number line diagram, 10 x 10 grid).

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Solving Percent Problems

Learning Objective(s)

·          Identify the amount, the base, and the percent in a percent problem.

·          Find the unknown in a percent problem.

Introduction

Percents are a ratio of a number and 100. So they are easier to compare than fractions, as they always have the same denominator, 100. A store may have a 10% off sale. The amount saved is always the same portion or fraction of the price, but a higher price means more money is taken off. Interest rates on a saving account work in the same way. The more money you put in your account, the more money you get in interest. It’s helpful to understand how these percents are calculated.

Parts of a Percent Problem

Jeff has a coupon at the Guitar Store for 15% off any purchase of $100 or more. He wants to buy a used guitar that has a price tag of $220 on it. Jeff wonders how much money the coupon will take off the original $220 price.

Problems involving percents have any three quantities to work with: the percent , the amount , and the base .

The percent has the percent symbol (%) or the word “percent.” In the problem above, 15% is the percent off the purchase price.

The base is the whole amount. In the problem above, the whole price of the guitar is $220, which is the base.

The amount is the number that relates to the percent. It is always part of the whole. In the problem above, the amount is unknown. Since the percent is the percent off , the amount will be the amount off of the price .

You will return to this problem a bit later. The following examples show how to identify the three parts, the percent, the base, and the amount.

The previous problem states that 30 is a portion of another number. That means 30 is the amount. Note that this problem could be rewritten: 20% of what number is 30?

Solving with Equations

Percent problems can be solved by writing equations. An equation uses an equal sign (= ) to show that two mathematical expressions have the same value.

Percents are fractions, and just like fractions, when finding a percent (or fraction, or portion) of another amount, you multiply.

The percent of the base is the amount.

Percent of the Base is the Amount.

Percent · Base = Amount

Once you have an equation, you can solve it and find the unknown value. To do this, think about the relationship between multiplication and division. Look at the pairs of multiplication and division facts below, and look for a pattern in each row.

Multiplication and division are inverse operations. What one does to a number, the other “undoes.”

When you have an equation such as 20% · n = 30, you can divide 30 by 20% to find the unknown: n =  30 ÷ 20%.

You can solve this by writing the percent as a decimal or fraction and then dividing.

n = 30 ÷ 20% =  30 ÷ 0.20 = 150

You can estimate to see if the answer is reasonable. Use 10% and 20%, numbers close to 12.5%, to see if they get you close to the answer.

10% of 72 = 0.1 · 72 = 7.2

20% of 72 = 0.2 · 72 = 14.4

Notice that 9 is between 7.2 and 14.4, so 12.5% is reasonable since it is between 10% and 20%.

This problem is a little easier to estimate. 100% of 24 is 24. And 110% is a little bit more than 24. So, 26.4 is a reasonable answer.

Using Proportions to Solve Percent Problems

Let’s go back to the problem that was posed at the beginning. You can now solve this problem as shown in the following example.

You can estimate to see if the answer is reasonable. Since 15% is half way between 10% and 20%, find these numbers.

10% of 220 = 0.1 · 220 = 22

20% of 220 = 0.2 · 220 = 44

The answer, 33, is between 22 and 44. So $33 seems reasonable.

There are many other situations that involve percents. Below are just a few.

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Solving problems with percentages

• Price difference I
• Price difference II
• How many students?

To solve problems with percent we use the percent proportion shown in "Proportions and percent".

$$\frac{a}{b}=\frac{x}{100}$$

$$\frac{a}{{\color{red} {b}}}\cdot {\color{red} {b}}=\frac{x}{100}\cdot b$$

$$a=\frac{x}{100}\cdot b$$

x/100 is called the rate.

$$a=r\cdot b\Rightarrow Percent=Rate\cdot Base$$

Where the base is the original value and the percentage is the new value.

47% of the students in a class of 34 students has glasses or contacts. How many students in the class have either glasses or contacts?

$$a=r\cdot b$$

$$47\%=0.47a$$

$$=0.47\cdot 34$$

$$a=15.98\approx 16$$

16 of the students wear either glasses or contacts.

We often get reports about how much something has increased or decreased as a percent of change. The percent of change tells us how much something has changed in comparison to the original number. There are two different methods that we can use to find the percent of change.

The Mathplanet school has increased its student body from 150 students to 240 from last year. How big is the increase in percent?

We begin by subtracting the smaller number (the old value) from the greater number (the new value) to find the amount of change.

$$240-150=90$$

Then we find out how many percent this change corresponds to when compared to the original number of students

$$90=r\cdot 150$$

$$\frac{90}{150}=r$$

$$0.6=r= 60\%$$

We begin by finding the ratio between the old value (the original value) and the new value

$$percent\:of\:change=\frac{new\:value}{old\:value}=\frac{240}{150}=1.6$$

As you might remember 100% = 1. Since we have a percent of change that is bigger than 1 we know that we have an increase. To find out how big of an increase we've got we subtract 1 from 1.6.

$$1.6-1=0.6$$

$$0.6=60\%$$

As you can see both methods gave us the same answer which is that the student body has increased by 60%

Video lessons

A skirt cost $35 regulary in a shop. At a sale the price of the skirtreduces with 30%. How much will the skirt cost after the discount?

Solve "54 is 25% of what number?"

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PERCENTAGES

This section will explain how to apply algebra to percentage problems.

In algebra problems, percentages are usually written as decimals.

Example 1. Ethan got 80% of the questions correct on a test, and there were 55 questions. How many did he get right?

The number of questions correct is indicated by:

Ethan got 44 questions correct.

Explanation: % means "per one hundred". So 80% means 80/100 = 0.80.

Example 2. A math teacher, Dr. Pi, computes a student’s grade for the course as follows:

a. Compute Darrel's grade for the course if he has a 91 on the homework, 84 for his test average, and a 98 on the final exam.

Darrel’s grade for the course is an 89.6, or a B+.

b. Suppose Selena has an 89 homework average and a 97 test average. What does Selena have to get on the final exam to get a 90 for the course?

The difference between Part a and Part b is that in Part b we don’t know Selena’s grade on the final exam.

So instead of multiplying 30% times a number, multiply 30% times E. E is the variable that represents what Selena has to get on the final exam to get a 90 for the course.

Because Selena studied all semester, she only has to get a 79 on the final to get a 90 for the course.

Example 3. Sink Hardware store is having a 15% off sale. The sale price of a toilet is $97; find the retail price of the toilet.

a. Complete the table to find an equation relating the sale price to the retail price (the price before the sale).

Vocabulary: Retail price is the original price to the consumer or the price before the sale. Discount is how much the consumer saves, usually a percentage of the retail price. Sale Price is the retail price minus the discount.

b. Simplify the equation.

Explanation: The coefficient of R is one, so the arithmetic for combining like terms is 1 - 0.15 = .85. In other words, the sale price is 85% of the retail price.

c. Solve the equation when the sale price is $97.

The retail price for the toilet was $114.12. (Note: the answer was rounded to the nearest cent.)

The following diagram is meant as a visualization of problem 3.

The large rectangle represents the retail price. The retail price has two components, the sale price and the discount. So Retail Price = Sale Price + Discount If Discount is subtracted from both sides of the equation, a formula for Sale Price is found. Sale Price = Retail Price - Discount

Percentages play an integral role in our everyday lives, including computing discounts, calculating mortgages, savings, investments, and estimating final grades. When working with percentages, remember to write them as decimals, to create tables to derive equations, and to follow the proper procedures to solve equations.

Study Tip: Remember to use descriptive letters to describe the variables.

CHAPTER 1 REVIEW

This unit introduces algebra by examining similar models. You should be able to read a problem and create a table to find an equation that relates two variables. If you are given information about one of the variables, you should be able to use algebra to find the other variable.

Signed Numbers:

Informal Rules:

Adding or subtracting like signs: Add the two numbers and use the common sign.

Adding or subtracting unlike signs: Subtract the two numbers and use the sign of the larger, (more precisely, the sign of the number whose absolute value is largest.)

Multiplying or dividing like signs: The product or quotient of two numbers with like signs is always positive.

Multiplying or dividing unlike signs: The product or quotient of two numbers with unlike signs is always negative.

Order of operations: P lease E xcuse M y D ear A unt S ally 1. Inside P arentheses, (). 2. E xponents. 3. M ultiplication and D ivision (left to right) 4. A ddition and S ubtraction (left to right)

Study Tip: All of these informal rules should be written on note cards.

Introduction to Variables:

Generate a table to find an equation that relates two variables.

Example 6. A car company charges $14.95 plus 35 cents per mile.

Simplifying Algebraic Equations:

Combine like terms:

Solving Equations:

1. Simplify both sides of the equation. 2. Write the equation as a variable term equal to a constant. 3. Divide both sides by the coefficient or multiply by the reciprocal. 4. Three possible outcomes to solving an equation. a. One solution ( a conditional equation ) b. No solution ( a contradiction ) c. Every number is a solution (an identity )

Applications of Linear Equations:

This section summarizes the major skills taught in this chapter.

Example 9. A cell phone company charges $12.50 plus 15 cents per minute after the first six minutes.

a. Create a table to find the equation that relates cost and minutes.

c. If the call costs $23.50, how long were you on the phone?

If the call costs $23.50, then you were on the phone for approximately 79 minutes.

Literal Equations:

A literal equation involves solving an equation for one of two variables.

Percentages:

Write percentages as decimals.

Example 11. An English teacher computes his grades as follows:

Sue has an 87 on the short essays and a 72 on the research paper. If she wants an 80 for the course, what grade does Sue have to get on the final?

Sue has to get a 78.36 in the final exam to get an 80 for the course.

Study Tips:

1. Make sure you have done all of the homework exercises. 2. Practice the review test on the following pages by placing yourself under realistic exam conditions. 3. Find a quiet place and use a timer to simulate the test period. 4. Write your answers in your homework notebook. Make copies of the exam so you may then re-take it for extra practice. 5. Check your answers. 6. There is an additional exam available on the Beginning Algebra web page. 7. DO NOT wait until the night before the exam to study.

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Word Problems on Percentage

Word problems on percentage will help us to solve various types of problems related to percentage. Follow the procedure to solve similar type of percent problems.

Word problems on percentage:

1.  In an exam Ashley secured 332 marks. If she secured 83 % makes, find the maximum marks.

Let the maximum marks be m.

Ashley’s marks = 83% of m

Ashley secured 332 marks

Therefore, 83% of m = 332

⇒ 83/100 × m = 332

⇒ m = (332 × 100)/83

⇒ m =33200/83

Therefore, Ashley got 332 marks out of 400 marks.

2. An alloy contains 26 % of copper. What quantity of alloy is required to get 260 g of copper?

Let the quantity of alloy required = m g

Then 26 % of m =260 g

⇒ 26/100 × m = 260 g

⇒ m = (260 × 100)/26 g

⇒ m = 26000/26 g

⇒ m = 1000 g

3. There are 50 students in a class. If 14% are absent on a particular day, find the number of students present in the class.

Solution:             

Number of students absent on a particular day = 14 % of 50

                                          i.e., 14/100 × 50 = 7

Therefore, the number of students present = 50 - 7 = 43 students.

4. In a basket of apples, 12% of them are rotten and 66 are in good condition. Find the total number of apples in the basket.

Solution:             

Let the total number of apples in the basket be m

12 % of the apples are rotten, and apples in good condition are 66

Therefore, according to the question,

88% of m = 66

⟹ 88/100 × m = 66

⟹ m = (66 × 100)/88

⟹ m = 3 × 25

Therefore, total number of apples in the basket is 75.

5. In an examination, 300 students appeared. Out of these students; 28 % got first division, 54 % got second division and the remaining just passed. Assuming that no student failed; find the number of students who just passed.

The number of students with first division = 28 % of 300

                                                             = 28/100 × 300

                                                             = 8400/100

                                                             = 84

And, the number of students with second division = 54 % of 300

                                                                        = 54/100 × 300

                                                                        =16200/100

                                                                        = 162

Therefore, the number of students who just passed = 300 – (84 + 162)

                                                                           = 54

Questions and Answers on Word Problems on Percentage:

1. In a class 60% of the students are girls. If the total number of students is 30, what is the number of boys?

2. Emma scores 72 marks out of 80 in her English exam. Convert her marks into percent.

Answer: 90%

3. Mason was able to sell 35% of his vegetables before noon. If Mason had 200 kg of vegetables in the morning, how many grams of vegetables was he able to see by noon?

Answer: 70 kg

4. Alexander was able to cover 25% of 150 km journey in the morning. What percent of journey is still left to be covered?

Answer:  112.5 km

5. A cow gives 24 l milk each day. If the milkman sells 75% of the milk, how many liters of milk is left with him?

Answer: 6 l

6.  While shopping Grace spent 90% of the money she had. If she had $ 4500 on shopping, what was the amount of money she spent?

Answer:  $ 4050

Fraction into Percentage

Percentage into Fraction

Percentage into Ratio

Ratio into Percentage

Percentage into Decimal

Decimal into Percentage

Percentage of the given Quantity

How much Percentage One Quantity is of Another?

Percentage of a Number

Increase Percentage

Decrease Percentage

Basic Problems on Percentage

Solved Examples on Percentage

Problems on Percentage

Real Life Problems on Percentage

Application of Percentage

8th Grade Math Practice From Word Problems on Percentage to HOME PAGE

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AusPost to halve letter delivery frequency from Monday

Australia Post will only deliver letters every second day from next week, as a generational shift in the nation's postal services takes effect.

From Monday, 98 per cent of locations will have letter delivery reduced to every second business day, and delivery times for ordinary letters will be extended by a business day.

That means within the same state, letter delivery times will be four or five business days, depending on location, and six to eight business days for interstate letters.

A daily priority letter service will remain available for customers who need fast delivery of mail, at a higher cost.

Requirements for Australia Post to maintain a minimum presence nationally and in rural, regional and remote areas will remain unchanged.

Separately, Australia Post is raising the price of ordinary letter stamps from $1.20 to $1.50, and large letter stamps from $3.60 to $4.50.

The federal government agreed to change the law to loosen Australia Post's requirements for letter delivery, acknowledging Australians send far fewer letters than they used to.

The postal service had warned without the change its "long-term viability" was at risk.

Australia Post says the changes will allow it to focus on expanding parcel posting and collection places, which have become its major business.

Trials of the change resulted in postal workers being able to carry 20 per cent more parcels on each delivery.

Four in five households bought goods online last year, while letter volumes have fallen by about two-thirds since their peak in 2008.

Communications Minister Michelle Rowland said the way Australians used the national postal service had changed, and so Australia Post needed to be modernised.

"[These] changes reflect the way Australians now interact with Australia Post, and will help improve the company’s financial sustainability," Ms Rowland said.

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