percentage word problem solving examples

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Word Problems Involving Percent

Three Kinds of Percent Problems

Consider the statement " x percent of y is z ."

If any two of the variables are given, you can use algebra to find out the missing one. This results in three different kinds of problems. In each one, the unknown is in a different position.

•  problems where x is the unknown (e.g. "What percent of 44 is 11 ?")

•  problems where y is the unknown (e.g. " 58 is 25 % of what number?")

•  problems where z is the unknown (e.g. "What is 88 % of 5000 ?")

When given a word problem involving percents, then, the first job is to find out what the unknown is.

A concert hall has 400 seats, of which 325 are occupied. Express the attendance at a percent of capacity.

The question here can be stated as, "What percent of 400 is 325 ?" So, the unknown here is x .

Write a proportion.

325 400 = x 100

You could cross multiply. Or, you could notice that the denominator on the left is 4 times the denominator on the right.

So, to find x , just divide the numerator on the left by 4 .

x = 325 4 = 81.25

So, the hall is filled to 81.25 % capacity.

71 % of the earth's surface--that's about 362,100,000 square kilometers--is covered by water. Use this information to find the total surface area of the earth.

The question here can be stated as, " 71 % percent of what number is 362,100,000 ?" So, the unknown here is y .

362 , 100 , 000 y = 71 100

Cross multiply.

71 y = 36 , 210 , 000 , 000

Divide both sides by 71 .

y = 510 , 000 , 000

So, the earth's surface area is about 510,000,000 square kilometers.

The residents of a state were surveyed about their opinion of the public education system. 12 % said "Good", 54 % said "Bad", and the rest said "No opinion".

If 650 people were surveyed, how many said "No opinion"?

The total of the three percentages should be 100 . Find the percentage of people who said "No opinion."

12 + 54 + c = 100 66 + c = 100 c = 34

The question here can be stated as, "What is 34 % of 650 ?" So, the unknown here is z .

z 650 = 34 100

100 z = 22100 z = 221

So, 221 people said "No opinion".

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Percent Word Problems

In these lessons we look at some examples of percent word problems. The videos will illustrate how to use the block diagrams (Singapore Math) method to solve word problems.

Related Pages More Math Word Problems Algebra Word Problems More Singapore Math Word Problems

How to solve percent problems with bar models? Examples:

• Marilyn saves 30% of the money she earns each month. She earns $1350 each month. How much does she save?
• At the Natural History Museum, 40% of the visitors are children. There are 36 children at the museum. How many visitors altogether are at the museum?
• Bill bought cards to celebrate Pi day. He sent 60% of his cards to his friends. He sent 42 cards to his friends. How many cards did he buy altogether?
• Bruce cooked 80% of the pancakes at the pancake breakfast last weekend. They made 1120 pancakes. How many pancakes did Bruce cook?

Sales Tax and Discount An example of finding total price with sales tax and an example of finding cost after discount.

• Alejandro bought a TV for $900 and paid a sales tax of 8%. How much did he pay for the TV?
• Alice saved for a new bike. The bike was on sale for a discount of 35%. The original cost of the bike was $270. How much did she pay for the bike?

Percent Word Problems Example: There are 600 children on a field. 30% of them were boys. After 5 teams of boys join the children on the field, the percentage of children who were boys increased to 40%. How many boys were there in the 5 teams altogether?

Problem Solving - Choosing a strategy to solve percent word problems An explanation of how to solve multi-step percentage problems using bar models or choosing an operation. Example: The $59.99 dress is on sale for 15% off. How much is the price of the dress?

How to solve percent problems using a tape diagram or bar diagram? Example: An investor offers $200,000 for a 20% stake in a new company. What amount does the investor believe the toatl value of the business is worth at this time? How to use a tape diagram or bar diagram to find the answer?

• First draw a bar that represents the company’s whole value.
• Divide into 5 equal parts because 100%/20% = 5.
• Label one side with the percentages.
• Label the other side $200,000 across from 20% because that was given.
• Finish labeling the money side.
• Find solution.

Solve Percent Problems Using a Tape Diagram (Bar Diagram) Example: a) If $300 is increased by 25% what is the new amount? b) What is 19% of 120? c) Joe went to an athletic store to purchase new running shoes. To his surprise, the store was having a 20% off athletic shoes sale. He purchased a new pair of shoes that were regularly priced $60. How much did Joe pay for his shoes?

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Percent Word Problems - Examples & Practice - Expii

Percent word problems - examples & practice, explanations (3).

Solving Word Problems with Percents

We can break word problems with percents into smaller steps!

The following graphic breaks down how to solve word problems with percents into 3 simple steps. Following these steps will allow you have the confidence to tackle all future word problems with percents!

Image source: by Caroline Kulczycky

You can see that we started with the problem, identified what we knew, plugged in values, and solved the proportion. These same steps can be used for other word problems.

Let's do another example!

During a soccer tournament, referees gave a total of 124 yellow cards and 4 red cards. What percent of the total cards were red?

Related Lessons

(Videos) Percent Applications

by mathman1024

This video by mathaman1024 works through a percent word problem.

There are three steps you can take with any percentage-based word problem

• Identify the knowns and unknowns . Some of the information will be given to us, but not all of it.
• Plug the values into the proportion. We need to set up a proportion that incorporates the known values and the variables.
• Solve the proportion. Once we have a set proportion, we have to use our algebra skills to isolate the variable.

The example problem is: $109,570,000 of Lone Star College's $317,790,000 budget comes from property taxes. What percent of their budget comes from property taxes?

Step 1: Identify the knowns and unknowns from the problem

$109,570,000 of Lone Star College's $317,790,000 budget comes from property taxes. What percent of their budget comes from property taxes?

Step 2: Plug Values Into Proportion

PartWhole=Percent100 $109,570,000$317,790,000=Percent100

Step 3: Solve Proportion

109,750,000317,790,000⋅100=x0.348⋅100=x34.8=x34.8%

Word Problems with Percents

You need to translate problems with percentages into math and then solve them. You can break down this process into three steps.

• Identify the knowns and unknown from the problem.
• Convert values given in the problems into proportions.
• Solve the proportion.

Let’s try an example.

Suppose 60% of people in a math class have a dog. If the math class consist of 150 people, how many people in the math class has a dog?

Step 1: Identify the knowns and unknown from the problem

Knowns : We know that 60% of people in a math class have a dog. We also know that there are 150 people in the math class.

Unknowns : We want to determine how many people in the math class have a dog. We can call this unknown value x.

Step 2: Convert values given in the problems into proportions

Next, we can plug in our known and unknown values into the proportion.

Remember, this is our general set up for a proportion: PartWhole=Percent100 We are solving for x, which is a part of the class that has a dog, so we can plug x into “Part”. xWhole=Percent100 Next, we can substitute 60% and 150 into the proportion. Since 60% is a percent, we can plug 60 into ‘Percent”. 150 is the total number of students in the class, so 150 is the whole. x150=60100

Step 3: Solve the proportion using cross multiplication

x150=60100100x=150⋅60100x=9,000x=9,000100x=90 There are 90 students in the class of 150 that have a pet dog.

Let’s try another example.

Nick went to the mall and bought 2 shirts and a pair of jeans. Each shirt costs $50, and the pair of jeans costs 75. If Nick had a coupon for 25% off the entire purchase, how much did he pay?

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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 ‍  

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25 Percentage Word Problems For Year 5 To Year 8 With Tips On Supporting Pupils’ Progress

Emma johnson.

Percentage word problems and the concept of calculating percentages first appears in Upper Key Stage 2. As pupils progress through school from KS2 to KS3, the skills they need to solve percentage word problems develop.

It is important to expose students to percentage word problems alongside any fluency work on percentages, to help them understand how percentages are used in real-life. To help you with this, we have put together a collection of 25 percentage word problems which can be used by pupils from Year 5 to Year 8. Don’t miss our downloadable word problems worksheet to develop these skills further!

How pupils develop the necessary skills to solve percentage word problems

Percentage word problems in the national curriculum, why are word problems important for children’s understanding of percentage , how to teach solving percentage word problems in ks2 and early ks3, percentage word problems for year 5, percentage word problems for year 6, percentage word problems for key stage 3, more word problems resources, percentage word problems faqs, all kinds of word problems four operations.

Download this free, printable pack of word problems covering all four operations; a great way to build students' problem solving skills.

Initially, pupils are introduced to the per cent symbol (%) in Year 5. At this stage they are expected to understand that percent relates to ‘number of parts per hundred’ and should be able to solve problems requiring knowing percentage and decimal form equivalents of simple fractions.

As pupils progress into Year 6, they should be able to recall and use equivalences between simple fractions, mixed numbers, decimals and percentages. They also need to be able to solve word problems involving percentages of amounts and percentage increase and decrease.

Moving into Key Stage 3, pupils continue to build on the percentage work from primary, to solve percent problems, interpreting percentages and percentage change; expressing one quantity as a percentage of another; comparing two quantities using percentages and working with percentages greater than 100%.

Concrete resources (such as percentages cubes) and visual images (such as bar models) are important during the early stages of learning and understanding percentages. Word problems for year 3 and word problems for year 4 will often include visual aids. Upper Key Stage 2 teachers and pupils often have the mistaken belief that concrete resources are only for children who are struggling; however, with a new topic, such as percentages, it is important all children are initially introduced to the topic through the use of visual and concrete aids.

Children are first introduced to percentage problems in Year 5. The National Curriculum expectations for percentages are that children will be able to:

Percentage in Year 5

• Recognise the percent symbol (5)  and understand that per cent relates to ‘number of parts per hundred’.
• Write percentages as a fraction with a denominator 100, and in its decimal form.
• Solve problems which require knowing percentage and decimal equivalents of \frac{1}{2},\frac{1}{4},\frac{1}{5},\frac{2}{5},\frac{4}{5} , and fractions with a denominator of a multiple of 10 or 25.

Percentage in Year 6

• Recall and use equivalences between simple fractions, decimals and percentages, including in different contexts.
• Solve problems involving the calculation of percentages (for example, measures and such as 15% of 360) and use percentages for comparison.

Percentage in Key Stage 3

• Define percentage as ‘number of parts per hundred’
• Interpret percentages and percentage changes as a fraction or a decimal. Interpret these multiplicatively.
• Express one quantity as a percentage of another.
• Compare two quantities using percentages.
• Work with percentages greater than 100%.
• Interpret fractions and percentages as operators.

Percentage word problems will often include other skills, such as fraction word problems , multiplication word problems , addition word problems , subtraction word problems and division word problems .

Percentage word problems help children to develop their understanding of percentages and the different ways percentages are used in everyday life. Taken out of context, percentages can be quite an abstract concept, which some children can find quite difficult to understand. 

Real-life problems involving percentages enable students to see how they will make use of this key skill outside the classroom.

As with all word problems, students need to learn the skills required to solve percentage word problems. It’s important that children make sure they have read the questions carefully and thought about exactly what is being asked and whether they have fully understood this. They then need to identify what they will need to do to solve the problem and whether there are any concrete resources or pictorial representations which they can use to help them. Even pupils in Key Stage 3 can benefit from drawing a quick picture, to understand what a word problem is asking.

Third Space Learning’s online, one-to-one tutoring programmes work to build students’ maths fluency and reasoning skills. Personalised to the needs of each individual student, our programmes fill gaps and build students’ confidence in maths.

Percent word problem example :

A box of cupcakes sold by a bakery cost £3.40.

Due to the increased costs involved with running a bakery, the owner has decided to increase the price of everything sold by 20%.

How much will a box of cupcakes cost once the price has been increased?

How to solve step-by-step:

What do you already know?

–        We know that the original price of a box of cupcakes is £3.40.

–        If the price of the box is being increased by 20%, we need to work out how much 20% of £3.40 is.

–        To do this, we need to work out how much 10% of £3.40 is. We therefore need to divide £3.40 by 10 = £0.34

–        To calculate what 20% is, we need to multiply the £0.34 by 2 = £0.68

–        Finally, we need to add the 20% (£0.68) onto the original price.

–        £3.40 + £0.68 = £4.08

How can this be represented pictorially?

• We can draw a bar model to represent what 10% of £3.40 equals.
• Once we know what 10% of £3.40 is (34p), we can double it to calculate 20% of £3.40 (68p).
• We can then add this on to the original price of £3.40.
• £3.40 + 68p = £4.08.

To solve word problems for year 5 , children need to be able to convert fractions to percentages and calculate fractions of an amount.

Gemma saves \frac{1}{2} of her pocket money every week.

She receives £5 per week and is saving to buy a game costing £25.

• What percentage does she save each week?
• How long will it take her to save for the game?
• She saves 50% of her pocket money each week

Gemma saves £2.50 per week.

£2.50 x 10 = £25

Sam gives \frac{4}{10} of his sweets to Ahmed.

What percentage of the sweets does he keep for himself?

Answer : 60%

  \frac{4}{10} =  \frac{40}{100} = 40%

100 – 40 = 60%

A school football team has 11 players and 5 substitutes.

  \frac{3}{4} of the players are boys, the rest are girls.

What percentage are girls?

Answer: 25%

\frac{3}{4} of 16 are boys

  \frac{1}{4} are girls

  \frac{1}{4} = 25%

Children in Year 5 voted on their favourite food.

35% of children voted for pizza.

60 children took part in the survey.

How many voted for pizza?

Answer : 21 children 

10% of 60: 60 ÷ 10 = 6

5% of 60: Half of 10% (6) = 3

30%: 6 x 3 = 18

35% = 18 + 3 = 21

Ben was given a maths worksheet to complete for his homework.

He got  \frac{6}{10} of the maths problems correct

If there were 20 questions on the paper:

• How many questions did he get right?
• What percentage did he score?

  \frac{1}{10} of 20  = 2

  \frac{6}{10} of 20  = 12

• 60% correct

  \frac{12}{20} =  \frac{60}{100} .

An ice cream seller has been researching the most popular ice creams.

He knows the percentage of each flavour of ice cream sold, but wants to work out how many of each flavour were sold.

80 ice creams were sold in total.

40% vanilla

25% strawberry

20% cookie dough

15% mint choc chip

How many of each ice cream flavour were sold?

20 strawberry

16 cookie dough

12 mint choc chip

10% of 80 = 8 ice creams

5% of 80 = 4 ice creams

Vanilla: 4 x 8 = 32

Strawberry: 2 x 8 = 16. 16 + 4 = 20

Cookie dough: 2 x 8 = 16

Mint choc chip: 8 + 4 = 12

Pupils in Year 5 held a vote on where to go for their next school trip.

The vote was between the zoo and the aquarium.

90 children voted. 

40% voted for the zoo

How many pupils voted for the aquarium?

Answer : 54 children

60% voted for the aquarium

10% of 90 = 9 children

60% = 6 x 9 = 54 children

The price of burgers being sold by a burger van have increased by 25%

If the original price was £2 per burger. How much are the burgers now?

Answer : £2.50

25% of £2 = £2 ÷ 4 = 50p

£2 + 50p = £2.50

Word problems for year 6 involve solving problems involving equivalence between fractions, decimals and percentages; calculation of percentages and using percentages for comparison. Year 6 students will also tackle multi-step problems .

A rugby game lasts for 80 minutes.

A player is on the pitch for 85% of the game.

How long is he on the pitch for?

Answer : 68 minutes

10% of 80  = 8 minutes

5%  = Half of 8 minutes = 4 minutes

80% = 8 x 8 = 64 minutes

64 + 4 = 68 minutes

There are 480 pupils in a primary school.

15% in foundation

25% in Key Stage 1

How many pupils are in Key Stage 2?

Answer : 288 pupils

In foundation there are 15% + 25% (40% of the pupils)

Therefore, 60% of pupils are in KS2

10% of 480 = 48 pupils

60% = 6 x 48 = 288 pupils

A pizza restaurant decided to add a 15% increase to the cost of all their pizzas.

The cost of a meat feast pizza before the increase was £12.60

What is the new price of the pizza?

Answer : £14.49

10% of £12.60 = £1.26

5% = Half of £1.26 = £0.63

15% = £1.26 + £0.63 = £1.89

New price: £12.60 + £1.89 = £14.49

Oliver was shopping for a new pair of jeans.

The jeans were 15% off  in the sale, but the new sale price sticker had fallen off.

The original price of the jeans was £35.

How much did they cost in the sale?

Answer : £29.75

10% of £35 = £3.50

5% = Half of £3.50 = £1.75

35% = £3.50 + £1.75 = £5.25

New price: £35 – £5.25 = £29.75

200g of sugar is needed for a chocolate brownies recipe.

A 1kg bag of sugar is used.

25% of the remaining sugar is used to bake a cake too.

How much sugar was used to bake the cake?

Answer : 200g sugar

 200g sugar used for brownies, therefore 1000g – 200g = 800g remaining

25% of 800g = 800 ÷ 4 = 200g

Mr Jones bought a second hand car for £12,400

A year later, it had decreased in value by 15%

What was the value of the car after a year?

Answer : £10,540

10% of 12,400 = £1,240

5%  = half of £1240 = £620

15% = 1240 + 620 = £1,860

Value after a year = 12,400 – 1860 = £10,540

The number of visitors to a theme park in 2021 was 286,000.

The following year, there was a 24 percent increase in visitors.

How many visited the theme park in 2022?

Answer : 354,640 visitors

10% of 286,000 = 28,600

20% = 2 x 28,600 = 57,200

1% of 286,000 = 2,860

4% = 4 x 2860 = 11,440

24% = 57,200 + 11,440 = 68.640

Total number of visitors: 286,000 + 68,640 = 354,640

A library has 16,200 books

55% are fiction and 45% are non-fiction

968 non-fiction books are taken out in one week.

How many non-fiction books are left in the library, from the books which were there at the start of the week? 

Answer : 6,322 non-fiction books

10% of 16,200 = 1,620

40% = 1,620 x 4 = 6,480 books

5% = half of 1,620 = 810

45% = 6,480 + 810 = 7,290

7,290 – 968 = 6,322

In Key Stage 3, the work pupils carry out on percentages, builds upon the percentage skills developed in primary. Students need to be able to solve percent word problems involving interpreting percentages and percentage change as a fraction or decimal; expressing one quantity as a percentage of another; comparing two quantities using percentages; working with percentages greater than 100% and interpreting fractions and percentages as operators.

Jasmine wins £600

She gives 30% to her sister and 20% to her friend.

She keeps the rest.

How much does each person have?

Sister: £180

Friend: £120

Jasmine: £300

She gives 30% to her sister.

10% of £600 = £60

30% = 3 x 60 = £180

She gives 20% to her friend.

20% = 2 x 60 = £120

She must keep 50% for herself, if she has given 30% and 40% away.

50% of 600 =  \frac{1}{2} of 600 = £300

A car is reduced in the sale by 15%

If the original price was £18,500, what is the price of the car now?

Answer: £16,225

10% of 18500 = 1,850

5% = half of 1,850 = £925

15% = 1,850 + 925 = £2,275

New price: 18,500 – 2,275 = £16,225

Sales tax in Florida is 6%

Maisie has bought a pair of jeans, 3 T shirts and a jacket, which came to $150

How much will she have to pay, once she has added on the sales tax?

Answer: $159

1% of 150 = 1.5

6% = 6 x 1.5 = $9

150 + 9 = $159

A packet of biscuits is 300g

As a special offer, the biscuits currently have an extra 15% free.

How many grams of biscuit do you get with the special offer?

Answer : 345g

10% of 300 = 30g

5% = half of 30 = 15g

15% = 30 + 15 = 45g

New weight: 300 + 45 = 345g

Jason is travelling to Birmingham from Manchester.

His average speed is 62 miles per hour

On the return journey, the traffic on the M6 is terrible and his average speed it reduced by 35%

What is his average speed on the return journey?

Answer : 40.3mph

10% of 62 = 6.2mph

30% = 6.2 x 3 = 18.6mph

5% = half of 6.2 = 3.1mph

35% = 18.6 + 3.1 = 21.7mph

62 – 21.7 = 40.3 mph

Mr Andrews bought a car in January 2021 for £15000.

By January 2022 his car had depreciated in value by 20%.

By January 2023, his car had depreciated in value by another 30%.

What was the value of his car in January 2023?

Answer : £8400

January 2022

20% of £15000 = £3000, so the value of the car is £12000.

January 2023

30% of £12000 = £3600, so the value of the car is £8400.

Alternative method – using decimal multiplier

20% decrease means 80% of January 2021 value – decimal multiplier of 0.8

30% decrease means 70% of January 2022 value – decimal multiplier of 0.7

15000 x 0.8 x 0.7 = £8400.

In her half term test, Jasmine did a French test and scored 15 out of 30.

In her next half term test, Jasmine scored 21 out of 30 in her French test.

By what percentage did Jasmine improve?

Answer : 40% improvement

Percentage change

= 21 – 15/15  x 100

= 6/15  x 100 = 40% improvement.

In 2021, a company made a profit of $600000.

In 2022, the same company made a profit of $1350000.

By what percentage had their profit increased?

Answer : 125% improvement

= 1350000 – 600000/600000 x 100

= 750000/600000 x 100 = 125% improvement.

In 2022, Thomas earned £1800 a month for his job.

As part of his annual review in February 2023, he is going to ask for a pay rise of 3.5%.

If the pay rise is agreed, what will Thomas’ annual salary be?

Answer: £22356

3.5% of £1800 = £63

So new monthly salary would be £1863

£1863 x 12 = £22356.

Alternative method 1

£1800 x 12 = £21600

3.5% of £21600 = £756

£21600 + £756 = £22356.

Looking for more word problems practice questions? Take a look at our collection of addition and subtraction word problems , time word problems , money word problems and ratio word problems . Teaching percentages to KS3 or KS4? Check out our percentage worksheets here.

Do you have pupils who need extra support in maths? Every week Third Space Learning’s maths specialist tutors support thousands of pupils across hundreds of schools with weekly online 1-to-1 lessons and maths interventions designed to plug gaps and boost progress. Since 2013 we’ve helped over 150,000 primary and secondary school pupils become more confident, able mathematicians. Learn more or request a personalised quote for your school to speak to us about your school’s needs and how we can help.

There are different types of percentage problems. If you want to find the percentage of an amount, it can be calculated by writing the percentage as a decimal or a fraction and then multiplying it by the amount.

Pupils in Year 5 held a vote on where to go for their next school trip. The vote was between the zoo and the aquarium. 90 children voted.  40% voted for the zoo How many pupils voted for the aquarium?

1. Calculating a discount when shopping 2. Understanding bank interest rates 3. Understanding your grades in school

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• Percentage Word Problems

Introduction

Percentage is basically the ratio of the value of a particular quantity to its total value multiplied by 100. It is generally a way of expressing something as per 100. Further, in this article we will look at various scenarios for which we will find the percentage and learn the steps to solve word problems based on percentage.

What is the Percentage?

A percentage is something expressed part per hundred. It is generally a ratio which is given in terms of a fraction of 100. The word percent is taken from the Latin term per centrum which means “by the hundred” . It is denoted by the symbol “\[\% \]”.

What is Percentage

How to Calculate the Percentage?

Percentage is generally calculated by expressing the value whose percentage is to be calculated and the total value as a ratio which is multiplied by 100. The formula for calculating percentage is as follows :

Percentage \[ = \frac{{value}}{{Total\,Value}} \times 100\]

If in case, percentage of a number is to be calculated i.e. if % is given and the value of the quantity whose percentage is given with respect to the total is to be found we use -

For instance,\[{\rm{x\% }}\] of 450 \[ = \]Y and we wish to calculate the value of Y, we can do so by

\[{\rm{Y}} = \frac{{\rm{X}}}{{100}} \times 450\]

E.g. 20% of 800

So, \[\frac{{20}}{{100}} \times 800 = 20 \times 8 = 160\]

Question and Answer based on Percentage

This section contains basic problems based on the concept of percentage which can be solved very easily.

Raju scored 81 out of 90 in mathematics. Convert his marks into percentages.

Ans. Percentage \[ = \] Value \[/\]Total Value \[ \times \]100 \[ = \frac{{81}}{{90}} \times 100 = 9 \times 10 = 90\% \]

In a class of 300 students, there are 75 girls. Calculate the percentage of boys in the class.

Ans. Number of boys \[ = 300 - 75 = 225\]

Percentage of boys \[ = \frac{{225}}{{300}} \times 100 = 75\% \]

Prince spent 75 \[\% \] of the money he had to buy groceries. If he had 4000$ with him initially, what amount did he spend ?

Ans. Amount Spend \[ = \frac{{75}}{{100}} \times 4000 = 3000\]$

Compute 65% of 680

Ans. 65% of 680 \[ = \frac{{65}}{{100}} \times 680 = 442\]

Word Problems based on Percentage

The word problem based on percentage will have some scenarios in which we have to understand the requirements in the given problems and accordingly apply the formula for percentage and find the value of the quantity asked to be found.

1.If a brass article contains 72% of copper. What amount of brass will be required to get 360g of copper?

Ans : Let the quantity of brass required be x g.

Therefore, 72% of x = 360g

\[72\% \, \times \,x = 360g\]

\[0.72 \times x = 360g\]

Therefore, \[x = \frac{{360}}{{0.72}} = 500g\]

So, 500 g of brass is required to get 360g of copper.

2.Karishma appeared for a quiz in which she got 25 answers correct and 15 answers incorrect. What is the percentage of questions that she appeared correctly ?

Ans : Here, in this case 25 answers were correct and 15 were incorrect.

Therefore, total questions \[ = 25 + 15 = 40\]

So, Percentage of correct questions \[ = \frac{{25}}{{40}} \times 100 = 62.5\% \]

Therefore, 62.5\[\% \] of the questions were answered correctly by Karishma.

Solved Examples :

1. What is 30\[\% \] of 450?

Solution : 30 % of 450 \[ = \frac{{30}}{{100}} \times 450 = 0.3 \times 450 = 135\]

Therefore, 30% of 450 is 135.

2. The cost price of a bag is 1500 and the selling price of the same bag is 2100. At what profit is the article sold? What is the profit percentage?

Solution : As , Profit = Selling price - Cost price

Therefore, Profit earned on selling the bag \[ = 2100 - 1500 = 600\]

So, Profit Percentage \[ = \frac{{600}}{{1500}} \times 100 = 40\% \]

The seller earned 40\[\% \] profit on selling the bag.

3. Arun sells an object to Benny at a profit of 15%, Benny sells that object to Chandan for ₹1012 and makes a profit of 10%. At what cost did Arun purchase the object?

Solution: Let the actual cost price at which Arun bought the object be x

When Arun sells the object to Benny

Profit % = 15%

∴ selling price of object \[= \frac{{100 + 15}}{{100}} \times x = 1.15x\]

Now, this cost price of the object for Benny

When Benny sells the object to Chandan

Selling Price = ₹1012

Profit % = 10%

∴ Selling price = \[ = \frac{{100 + 10}}{{100}} \times 1.15x\]

\[ \Rightarrow 1012 = \frac{{100 + 10}}{{100}} \times 1.15x\]

\[ \Rightarrow x = \frac{{1012 \times 1000}}{{11 \times 115}}\]

Therefore, the price at which Arun bought the object is ₹800.

4. In an inter school aptitude test, 200 students appeared. Out of these students; 20 % got A grade, 50 % got B grade and the remaining got C grade. Assuming that no student got a D grade; find the number of students who got a C grade .

Solution : The number of students with A grade \[ = \] 20 \[\% \] of 200

\[ = \frac{{20}}{{100}} \times 200 = \frac{{4000}}{{100}}\]

And, the number of students with B grade \[ = \] 50 \[\% \] of 200

\[ = \frac{{50}}{{100}} \times 200\]

\[ = \frac{{10000}}{{100}} = 100\]

Therefore, the number of students who got C grade \[ = 200 - [40 + 100]\]

\[ = 200 - 140 = 60\]

So, 60 students scored C grade.

Conclusion:

Thus, we can say that percentage is a fraction of something expressed as part of 100. So, after understanding the method of calculating percentage and applying it according to the scenario we can easily solve the word problems based on percentage.

FAQs on Percentage Word Problems

1. Can the percentage be negative?

No, percentage can never be negative.

2. Is 0% of something valid ?

Yes, it is valid. For instance, if an alloy contains 0% of a certain element it simply means that the element is not present in that alloy.

Example: \[1250 \times 0\%  = 0\]

3. Can percentage be represented in decimals?

Yes, percentage can be represented in decimals. For example, 25% of something means \[\frac{{25}}{{100}} = 0.25\]\[\frac{{25}}{{100}} = 0.25\]

4. Can percentages be added?

Yes, percentages can be added if they are taken considering the same total value

Example: 5 % and 10% is 5 + 10 = 15%.

5. Can percentage be rounded off?

Yes, percentages can be rounded off to the nearest next natural number only if fine accuracy in decimals is not required in that scenario.

Example: 7.526% round off is 7.5%

Solved Examples on Percentage

The solved examples on percentage will help us to understand how to solve step-by-step different types of percentage problems. Now we will apply the concept of percentage to solve various real-life examples on percentage.

Solved examples on percentage:

1.  In an election, candidate A got 75% of the total valid votes. If 15% of the total votes were declared invalid and the total numbers of votes is 560000, find the number of valid vote polled in favour of candidate.

Total number of invalid votes = 15 % of 560000

                                       = 15/100 × 560000

                                       = 8400000/100

                                       = 84000

Total number of valid votes 560000 – 84000 = 476000

Percentage of votes polled in favour of candidate A = 75 %

Therefore, the number of valid votes polled in favour of candidate A = 75 % of 476000

= 75/100 × 476000

= 35700000/100

2. A shopkeeper bought 600 oranges and 400 bananas. He found 15% of oranges and 8% of bananas were rotten. Find the percentage of fruits in good condition.

Total number of fruits shopkeeper bought = 600 + 400 = 1000

Number of rotten oranges = 15% of 600

                                    = 15/100 × 600

                                    = 9000/100

                                    = 90

Number of rotten bananas = 8% of 400

                                   = 8/100 × 400

                                   = 3200/100

                                   = 32

Therefore, total number of rotten fruits = 90 + 32 = 122

Therefore Number of fruits in good condition = 1000 - 122 = 878

Therefore Percentage of fruits in good condition = (878/1000 × 100)%

                                                                 = (87800/1000)%

                                                                 = 87.8%

3. Aaron had $ 2100 left after spending 30 % of the money he took for shopping. How much money did he take along with him?

Solution:            

Let the money he took for shopping be m.

Money he spent = 30 % of m

                      = 30/100 × m

                      = 3/10 m

Money left with him = m – 3/10 m = (10m – 3m)/10 = 7m/10

But money left with him = $ 2100

Therefore 7m/10 = $ 2100          

m = $ 2100× 10/7

m = $ 21000/7

Therefore, the money he took for shopping is $ 3000.

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

Problems on Percentage

Real Life Problems on Percentage

Word Problems on Percentage

Application of Percentage

8th Grade Math Practice From Solved Examples on Percentage to HOME PAGE

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