problem solving involving percent

• Member login
• Pre-algebra lessons
• Pre-algebra word problems
• Algebra lessons
• Algebra word problems
• Algebra proofs
• Advanced algebra
• Geometry lessons
• Geometry word problems
• Geometry proofs
• Trigonometry lessons
• Consumer math
• Baseball math
• Math for nurses
• Statistics made easy
• High school physics
• Basic mathematics store
• SAT Math Prep
• Math skills by grade level
• Ask an expert
• Other websites
• K-12 worksheets
• Worksheets generator
• Algebra worksheets
• Geometry worksheets
• Free math problem solver
• Pre-algebra calculators
• Algebra Calculators
• Geometry Calculators
• Math puzzles
• Math tricks

Percentage word problems

Before you take a look at the percentage word problems in this lesson and their solutions, it may help to review the lesson about  formula for percentage or you can use the different techniques that I use here.

Different types of percentage word problems

There are three different types of percentage word problems. We will show how to solve them using proportions. 

• What is 80% of 20? ( example #1 )
• 50 is 25% of what number? ( example #2 )
• 18 is what percent of 24? What percent of 2000 is 3500? ( example #3 and example #4 )

Solving percentage word problems using proportions

You can solve problems involving percents using the proportion you see in the figure above:   ( n% / 100% = Part / Whole )

First, study the figure carefully! Then, we will show how to use the proportion to solve percentage word problems by creating diagrams to visualize relationships.

Example #1: A test has 20 questions. If peter gets 80% correct, how many questions did peter miss?

First, you need to find the number of correct answers by looking for 80% of 20.

When the problem involves looking for the part or the problem says something like, "Find 80% of 20" or "Find 30% of 50," just change the percent to a decimal and multiply.

80% of 20 = (80 / 100) × 20 = 0.80 × 20 = 16

Since the test has 20 questions and he got 16 correct answers, the number of questions Peter missed is 20 − 16 = 4

Recall that 16 is called the percentage. It is the answer you get when you take the percent of a number.

Percentage  =   Part

Example #2: In a school, 25% of the teachers teach basic math. If there are 50 basic math teachers, how many teachers are there in the school?

Once again set the problem up as shown in the figure below. Notice that the question is, " How many teachers are in the school?"

Therefore, the whole is missing this time!

Method #2 I shall help you reason the problem out!

When we say that 25% of the teachers teach basic math, we mean 25% of all teachers in the school equals number of teachers teaching basic math.

Since we don't know how many teachers there are in the school, we replace this with x or a blank. However, we know that the number of teachers teaching basic math is equal to the percentage = part =  50 Putting it all together, we get the following equation: 25% of ____ = 50 or 25% × ____ = 50 or 0.25 × ____ = 50 Thus, the question is 0.25 times what gives me 50? A simple division of 50 by 0.25 will get you the answer 50 / 0.25 = 200 Therefore, we have 200 teachers in the school In fact, 0.25 × 200 = 50

More percentage word problems

Example #3: 24 students in a class took an algebra test. If 18 students passed the test, what percent do not pass?

Solution First, find out how many student did not pass. Number of students who did not pass is 24 − 18 = 6

Then, write down the following equation: x% of 24 = 6 or x% × 24 = 6

To get x%, just divide 6 by 24 6 / 24 = 0.25 = 25 / 100 = 25% Therefore, 25% of students did not pass.

Example #4: A fundraising company would like to raise $2000 for a cause. The fundraiser was so successful that they ended up raising $3500. What percent of their goal did they raise?

Notice that the whole is 2000 since this is the whole money they expect to raise. The part is the amount that the fundraiser ended with and it usually lower than the amount they expect to raise. However, in this particular case, the part ended up being bigger than the whole. Keeping this in mind, here is how to set it up and solve it!

The fundraising company was able to raise 175% of the expected amount.

Example #5:

A department has a total of 22,000 units of stock. 25% of the garments are black and 10% of the garments are size 14.

a) How many black garments are there? 
b) How many size 14 garments are there? 
c) If 10% of the black garments are size 14,how many garments are black and size 14?

Note that the solution we show below for example #5 use a completely different approach or technique. Read it carefully and try to learn it as well!

25% = 25 per 100 = 250 per 1000 For 22,000 just multiply 250 by 22 250    ×    22   =  250 × (10 + 10 + 2)

                       =  2500 + 2500 + 500                        =  5000 + 500                        =  5500

So, there are 5500 black garments.

10% = 10 per 100 = 100 per 1000 For 22,000 just multiply 100 by 22 100 × 22 = 2200 So, 2200 of the garments are size 14.

If 10% or 10 per 100 of the black garments are size 14, then 100 per 1000 of the black garments are size 14.

500 per 5000 are size 14. However, you need to find it for 5500 black garments.

Then, what is 10% of 500? 10% = 10 per 100, so 50 per 500. So 550 of the black garments are size 14.

If you really understand the percentage word problems above, you can solve any other similar percentage word problems. If you still do not understand them, I strongly encourage you to study them again and again until you get it. The end result will be very rewarding!

Finding a percentage

percentage worksheets

Applied math

Calculators.

100 Tough Algebra Word Problems. If you can solve these problems with no help, you must be a genius!

 Recommended

About me :: Privacy policy :: Disclaimer :: Donate   Careers in mathematics  

Copyright © 2008-2021. Basic-mathematics.com. All right reserved

Real Life Problems on Percentage

Real life problems on percentage will help us to solve different types of problems related to the real-life situations. To understand the procedures follow step-by-step explanation so that you can solve any other similar type of percentage problems.

Solved real life problems on percentage:

1.  Mike needs 30% to pass. If he scored 212 marks and falls short by 13 marks, what was the maximum marks he could have got?

If Mike had scored 13 marks more, he could have scored 30%

Therefore, Mike required 212 + 13 = 225 marks

Let the maximum marks be m.

Then 30 % of m = 225

(30/100) × m = 225

m = (225 × 100)/30

m = 22500/30

2. A number is increased by 40 % and then decreased by 40 %. Find the net increase or decrease per cent.

Solution:            

Let the number be 100.

Increase in the number = 40 % = 40 % of 100

                               = (40/100 × 100)

                               = 40

Therefore, increased number = 100 + 40 = 140

This number is decreased by 40 %

Therefore, decrease in number = 40 % of 140

                                             = (40/100 × 140)

                                             = 5600/100

                                             = 56

Therefore, new number = 140 - 56 = 84

Thus, net decreases = 100 - 84 = 16

Hence, net percentage decrease = (16/100 × 100) %

                                               = (1600/100) %

                                               = 16 %

3. Max scored 6 marks more than what he did in the previous examination in which he scored 30. Maria scored 30 marks more than she did in the previous examination in which she scored 60. Who showed less improvement?

Max percentage improvement in the first exam = (6/30 × 100) %

                                                                     = (600/30) %

                                                                     = 20 %

Maria percentage improvement in the first exam = (30/60 × 100) %

                                                                       = (3000/60) %

                                                                       = 50 %

Hence, 20 % < 50 %

Therefore, Max showed less improvement.

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

Word Problems on Percentage

Application of Percentage

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

Didn't find what you were looking for? Or want to know more information about Math Only Math . Use this Google Search to find what you need.

New! Comments

Share this page: What’s this?

• Preschool Activities
• Kindergarten Math
• 1st Grade Math
• 2nd Grade Math
• 3rd Grade Math
• 4th Grade Math
• 5th Grade Math
• 6th Grade Math
• 7th Grade Math
• 8th Grade Math
• 9th Grade Math
• 10th Grade Math
• 11 & 12 Grade Math
• Concepts of Sets
• Probability
• Boolean Algebra
• Math Coloring Pages
• Multiplication Table
• Cool Maths Games
• Math Flash Cards
• Online Math Quiz
• Math Puzzles
• Binary System
• Math Dictionary
• Conversion Chart
• Homework Sheets
• Math Problem Ans
• Free Math Answers
• Printable Math Sheet
• Funny Math Answers
• Employment Test
• Math Patterns
• Link Partners
• Privacy Policy

Recent Articles

Bisecting an angle | bisecting an angle by using a protractor | rules.

Aug 21, 24 01:22 AM

Construction of Angles by using Compass, Construction of Angles

Aug 21, 24 01:14 AM

Construction of an Angle using a Protractor | Drawing an Angle

Aug 21, 24 01:04 AM

5th Grade Math Problems | Table of Contents | Worksheets |Free Answers

Aug 20, 24 03:42 PM

Construction of Perpendicular Bisector of a Line Segment|Steps-by-Step

Aug 20, 24 03:41 PM

Worksheet on Fraction into Percentage

Worksheet on Percentage into Fraction

Worksheet on Percentage into Ratio

Worksheet on Ratio into Percentage

Worksheet on Percentage into Decimal

Worksheet on Percentage of a Number

Worksheet on Finding Percent

Worksheet on Finding Value of a Percentage

Worksheet on Percentage of a Given Quantity

Worksheet on Word Problems on Percentage

Worksheet on Increase Percentage

Worksheet on Decrease Percentage

Worksheet on increase and Decrease Percentage

Worksheet on Expressing Percent

Worksheet on Percent Problems

Worksheet on Finding Percentage

© and ™ math-only-math.com. All Rights Reserved. 2010 - 2024.

A free service from Mattecentrum

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?"

• Pre-Algebra
• The mean, the median and the mode
• Stem-and-Leaf Plots and Box-and-Whiskers Plot
• Calculating the outcome
• Combinations and permutations
• Finding the odds
• Probability of events
• Geometry – fundamental statements
• Circle graphs
• Angles and parallel lines
• Quadrilaterals, polygons and transformations
• Measure areas
• Pyramids, prisms, cylinders and cones
• Square roots and real numbers
• The Pythagorean Theorem
• Trigonometry
• Algebra 1 Overview
• Algebra 2 Overview
• Geometry Overview
• SAT Overview
• ACT Overview

Percent Maths Problems

Math 7: Unit 8- Problems Involving Percents

PLEASE NOTE: Percents are not difficult, but people do struggle with them. One of the main reasons for the struggle is there are so many different ways of thinking about each percent problem. If there are lots of ways to think about percent problems, then there are lots of different ways to teach about percent problems.

The explanations in this unit rely on equations to solve percent problems. They also emphasize that we are always making comparisons to 100%. Tax, tip, markup, discount, and other types of percent problems are computed by always recognizing that we are either going up or down from 100%. The idea of always using 100% as our starting point can make topics in higher math easier to understand.

In short, these lessons show one way to think about percent problems, but it is not the only way. Each way of method of approaching percent problems has its advantages and disadvantages.

Unit "I CAN" Checklist

8.1A Solving Percent Problems

8.1B Solving Percent Problems

8.2A Taxes, Tips, & Commissions

8.2b taxes, tips, & commissions, 8.3a markups & discounts, 8.3b markups & discounts.

8.4A Percent Change (Increase & Decrease)

8.4b percent change (increase & decrease), 8.5 simple interest, 8.6 problems involving percents.

See Practice Worksheet

Practice Test #1

Practice Test #2

Practice Test #2 solutions can only be found by watching the video.

If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

Course: 6th grade   >   Unit 3

• Percent word problem: recycling cans

Percent word problems

• Rates and percentages FAQ

• Your answer should be
• an integer, like 6 ‍  
• a simplified proper fraction, like 3 / 5 ‍  
• a simplified improper fraction, like 7 / 4 ‍  
• a mixed number, like 1   3 / 4 ‍  
• an exact decimal, like 0.75 ‍  
• a multiple of pi, like 12   pi ‍   or 2 / 3   pi ‍  

PERCENTAGE INCREASE AND DECREASE WORD PROBLEMS

Formula to find percentage increase/decrease

Problem 1 :

The price of a TV is $260. In a sale the price is decreased by 20%. Work out the price of the TV sale.

Price of the TV = $260

It is given that the price of the TV is decreased by 20%.

Then, the selling price of the TV :

= 80% of 260

= 0.80(260)

Problem 2 :

The value of a painting rises from $120000 to $192000. Work out the percentage increase in the value of painting.

Original price of the painting = $120000

After increase, the new price = $192000.

Price increase = 192000 - 120000

Percentage increase :

= ⁷²⁰⁰⁰⁄₁₂₀₀₀₀   ⋅ 100%

=  ⅗   ⋅ 10 0%

The value of the paiting is increased by 60%.

Problem 3 :

A puppy weighed 2 kg. Eight weeks later the puppy weighed 3.5 kg. What was the percentage increase in the Puppy's weight?

Weight of puppy = 2 kg.

Puppy's weight after eight weeks = 3.5 kg.

Increase in weight in eight weeks :

= (1.5/2) ⋅ 100% 

= 0.75 ⋅ 100%

Puppy's weight was increased by 75%.

Problem 4 :

Peter's weight decreases from 80 kg. to 64 kg. Calculate the percentage decrease in Peter's weight.

Old weight = 80 kg.

New weight = 64 kg.

Decrease in weight :

Percentage decrease :

=  ¹⁶⁄₈₀   ⋅ 100%

=  ⅕   ⋅ 100%

Peter has reduced 20% of his weight.

Problem 5 :

Alice buys a book for $19.80 A year later she sells the book for $12.87 Calculate the percentage decrease in the value of the book.

Cost price of the book = $19.80

Selling price = $12.87

Decrease in the price :

= 19.80 - 12.87

= (6.93/19.80) ⋅ 100%

The value of the book is decaresed by 35%.

Problem 6 :

The volume of juice in a can is increased from 250 ml. to 330 ml. Work out the percentage increase.

Original volume of juice in a can = 250 ml.

Volume of juice in can after increase = 330 ml.

Increase in volume :

= 330 - 250

= ⁸⁰⁄₂₅₀   ⋅ 100%

=  ⁸⁄₂₅ ⋅ 100%

Volume of juice in can is increased by 32%.

Problem 7 :

Sarah bought a TV for $250. Three years later she sold it for $180. Work out her percentage loss.

Cost price of the TV = $250

Selling price = $180

Cost price > Selling price ----> Loss

Loss = Cost price - Selling price

= 250 - 180

Percentage loss :

= ⁷⁰⁄₂₅₀   ⋅ 100%

= ⁷⁄₂₅  ⋅ 100%

= 7  ⋅ 4%

Problem 8 :

A car is travelling at 40 kilometers per hour. The car increases its speed to 56 kilometers per hour. Calculate the percentage increase in the speed of the car.

Initial speed of car = 40 km. per hour

Increased speed = 56 km. per hour

Increase in speed :

= 16 km. per hour

Percentage increase in speed of the car :

= ¹⁶⁄₄₀   ⋅ 100%

= ⅖  ⋅ 100%

Problem 9 :

Susan buys an antique for $120 and sells it for $216. Work out her percentage profit.

Cost price of an item = $120

Selling price of an item = $216

Selling price > Cost price ----> Profit

Profit  = Sleeing price - Cost price

= 216 - 120

Percentage profit :

= ⁹⁶⁄₁₂₀   ⋅ 100%

=  ⅘  ⋅ 100%

Problem 10 :

Holly bought a table for $80 She sold the table for $108 Find the percentage profit

Cost price = $80

Selling price = $108

Profit = Selling price - Cost price

=  ²⁸⁄₈₀ ⋅ 100%

=  ⁷⁄₂₀ ⋅ 100%

= 7 ⋅ 5%  

Kindly mail your feedback to   [email protected]

We always appreciate your feedback.

© All rights reserved. onlinemath4all.com

• Sat Math Practice
• SAT Math Worksheets
• PEMDAS Rule
• BODMAS rule
• GEMDAS Order of Operations
• Math Calculators
• Transformations of Functions
• Order of rotational symmetry
• Lines of symmetry
• Compound Angles
• Quantitative Aptitude Tricks
• Trigonometric ratio table
• Word Problems
• Times Table Shortcuts
• 10th CBSE solution
• PSAT Math Preparation
• Privacy Policy
• Laws of Exponents

Recent Articles

Digital sat math problems and solutions (part - 31).

Aug 20, 24 12:09 PM

SAT Math Resources (Videos, Concepts, Worksheets and More)

Aug 20, 24 01:35 AM

Geometry Problems with Solutions (Part - 5)

Aug 19, 24 10:56 PM

Lesson Percentage word problems (Type 3 problems, Finding the Base)

• Texas Go Math
• Big Ideas Math
• Engageny Math
• McGraw Hill My Math
• enVision Math
• 180 Days of Math
• Math in Focus Answer Key
• Math Expressions Answer Key
• Privacy Policy

Problems Involving Percentage | Percentage Word Problems with Solutions

The percentage of the whole number is calculated by dividing the value by the total value and then multiply by 100. The percentage is nothing but “per 100”. The students of 5th grade can learn the relationship between fractions and percentages with the help of this article. By learning the concept of percent the students can solve different types of problems. We have shown percentage problems with answers in the below sections so that you can verify if you are stuck at some point in percentage problem-solving.

Percentage = (Value/Total Value) × 100

• To Convert a Percentage into a Fraction
• Worksheet on Problems Involving Percentage

Real World Problems Involving Percentage

Learn the concept of percentage in-depth by referring to some of the percentage word problems with solutions.

Example 1. In an exam, Preethi secured 278 marks. If she secured 81% Find the maximum marks? Solution: Given, Total number of marks = 278 Preethi secured marks = 81% Percentage formula = P% × X = Y Where P = 81% Y = 278 Then 81% × X = 278 81/100 × X = 278 X = 278 × 100/81 X = 27800/81 = 343.2 Thus Preethi got 278 marks out of 343.2 marks.

Example 2. A container contains 40% of milk. what quantity of container is required to get 160 l of milk? Solution: Given, Let the container contains a milk = 40% Quality of container is required to get = 160 l Percentage formula = P% × X = Y Where P = 40% Y = 160 Then 40% × X = 160 40/100 × X = 160 X = 160 × 100/40 X = 16000/40 = 400

Example 3. There are 200 students in a class. If 20% are absent on a Saturday. Find the number of students present in the class? Solution: Given, Number of students absent on Saturday = 200 Then 20/100 × 200 = 40 Therefore the number of students present = 200 – 40 = 160 students.

Example 4. A box contains of oranges .6% of them are spoiled and 48 are good. find the total number of oranges in the box. Solution: Given, Let the total number of boxes = m 6% of oranges are spoiled and 48 are good Therefore 6% of m = 48 Percentage formula = P% × X = Y 6/100 × m = 66 Then P = 6/100 X = m Y = 66 m = 66 × 100/6 m = 6600/6 = 1100

Example 5. Find the decimal 0.6 into a percentage? Solution: Given, First, we convert the decimal number into a fraction 0.6 = 6/100 = 6% Thus the percentage of the decimal 0.6 is 6%

Example 6. Find the fraction 3/25 into a percentage? Solution: In order to convert the fraction into a percentage, we have to multiply the given fraction by 100. 3/25 × 100 = 12%

Example 7.  Navya scores 68 marks out of 90 in her maths exam. Convert her Marks into percentages? Solution: Given, Navya scores 68 marks out of 90 in her exam 68 × 100/90 = 75% Thus Navya scored 75% on her maths exam.

Example 8. A donkey gives 2l of milk each day. If the milkman sells 50% of the milk, how many litres of milk is left with him Solution: Given, A donkey gives milk = 2l Milk man sells = 50% Percentage formula = P% × X = Y Where P = 50% Y = 2l 50/100 × X = 2 X = 2 × 100/50 X = 200/50 = 4

Example 9. Arjun was able to cover 10% of the 20 km walk in the morning. what is the percentage of the journey is still left to be covered? Solution: Given, Arjun was able to cover 10% of 20 km Percentage formula = P% × X = Y Where P = 10% Y = 20 10/100 × X = 20 X = 20 × 100/10 X = 200/10 = 200

Example 10. In a class 10% of the students are boys. If the total number of students in a class is 80. What is the number of girls? Solution: Given, Total number of boys in a class = 10% Total number of students in a class = 80 Percentage formula = P% × X = Y Where P = 10% Y = 80 10/100 × X = 80 X = 80 × 100/10 X = 800/10 = 80

Leave a Comment Cancel Reply

You must be logged in to post a comment.