# 5.2.1: Solving Percent Problems

Learning Objectives

• Identify the amount, the base, and the percent in a percent problem.
• Find the unknown in a percent problem.

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.

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.

Example

Identify the percent, amount, and base in this problem.

30 is 20% of what number?

Solution

Percent: The percent is the number with the % symbol: 20%.

Base: The base is the whole amount, which in this case is unknown.

Amount: The amount based on the percent is 30.

Percent=20%

Amount=30

Base=unknown

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?

Exercise

Identify the percent, base, and amount in this problem:

What percent of 30 is 3?

The percent is unknown, because the problem states "What percent?" The base is the whole in the situation, so the base is 30. The amount is the portion of the whole, which is 3 in this case.

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.

( ext { Percent } {color{red}cdot} ext { Base }{color{blue}=} ext { Amount })

In the examples below, the unknown is represented by the letter ( n). The unknown can be represented by any letter or a box ( square) or even a question mark.

Example

Write an equation that represents the following problem.

30 is 20% of what number?

Solution

 20% of what number is 30? Rewrite the problem in the form “percent of base is amount.” Percent is: 20%Base is: unknownAmount is: 30 Identify the percent, the base, and the amount. ( ext { Percent } cdot ext { Base }= ext { Amount })( 20 \% cdot n=30) Write the percent equation. using ( n) for the base, which is the unknown value.

( 20 \% cdot n=30)

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 Division ( 2 cdot 3=6) ( 6 div 2=3) ( 8 cdot 5=40) ( 40 div 8=5) ( 7 cdot 4=28) ( 28 div 7=4) ( 6 cdot 9=54) ( 54 div 6=9)

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

When you have an equation such as ( 20 \% cdot n=30), you can divide 30 by 20% to find the unknown: ( n=30 div 20 \%).

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

( n=30 div 20 \%=30 div 0.20=150)

Example

What percent of 72 is 9?

Solution

 Percent: unknownBase: 72Amount: 9 Identify the percent, base, and amount. ( n cdot 72=9) Write the percent equation: ( ext { Percent } cdot ext { Base }= ext { Amount }). Use ( n) for the unknown (percent). ( n=9 div 72) Divide to undo the multiplication of ( n) times 72. ( egin{array}{r}0.125 \72longdiv{9.000}end{array}) Divide 9 by 72 to find the value for ( n), the unknown. ( n=0.125)( n=12.5 \%) Move the decimal point two places to the right to write the decimal as a percent.

( 12.5 \% ext { of } 72 ext { is } 9).

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 \% ext { of } 72=0.1 cdot 72=7.2)

( 20 \% ext { of } 72=0.2 cdot 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%.

Example

What is 110% of 24?

Solution

 Percent: 110%Base: 24Amount: unknown Identify the percent, the base, and the amount. ( 110 \% cdot 24=n) Write the percent equation.( ext { Percent } cdot ext { Base }= ext { Amount }).The amount is unknown, so use ( n). ( 1.10 cdot 24=n) Write the percent as a decimal by moving the decimal point two places to the left. ( 1.10 cdot 24=26.4=n) Multiply 24 by 1.10 or 1.1.

( 26.4 ext { is } 110 \% ext { of } 24).

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.

Exercise

18 is what percent of 48?

1. ( 0.375 \%)
2. ( 8.64 \%)
3. ( 37.5 \%)
4. ( 864 \%)
1. ( 0.375 \%)

Incorrect. You may have calculated properly, but you forgot to move the decimal point when you rewrote your answer as a percent. The equation for this problem is ( n cdot 48=18). The corresponding division is ( 18 div 48), so ( n=0.375). Rewriting this decimal as a percent gives the correct answer, ( 37.5 \%).

2. ( 8.64 \%)

Incorrect. You may have used ( 18) or ( 48) as the percent, rather than the amount or base. The equation for this problem is ( n cdot 48=18). The corresponding division is ( 18 div 48), so ( n=0.375). Rewriting this decimal as a percent gives the correct answer, ( 37.5 \%).

3. ( 37.5 \%)

Correct. The equation for this problem is ( n cdot 48=18). The corresponding division is ( 18 div 48), so ( n=0.375). Rewriting this decimal as a percent gives ( 37.5 \%).

4. ( 864 \%)

Incorrect. You probably used 18 or 48 as the percent, rather than the amount or base, and also forgot to rewrite the percent as a decimal before multiplying. The equation for this problem is ( n cdot 48=18). The corresponding division is ( 18 div 48), so ( n=0.375). Rewriting this decimal as a percent gives the correct answer, ( 37.5 \%).

Percent problems can also be solved by writing a proportion. A proportion is an equation that sets two ratios or fractions equal to each other. With percent problems, one of the ratios is the percent, written as ( frac{n}{100}). The other ratio is the amount to the base.

( ext { Percent }=frac{ ext { amount }}{ ext { base }})

Example

Write a proportion to find the answer to the following question.

30 is 20% of what number?

Solution

 ( frac{20}{100}=frac{ ext { amount }}{ ext { base }}) The percent in this problem is 20%. Write this percent in fractional form, with 100 as the denominator. ( frac{20}{100}=frac{30}{n}) The percent is written as the ratio ( frac{20}{100}), the amount is 30, and the base is unknown. ( egin{array}{r}20 cdot n=30 cdot 100 20 cdot n=3,000 n=3,000 div 20 n=150end{array}) Cross multiply and solve for the unknown, ( n), by dividing 3,000 by 20.

30 is 20% of 150.

Example

What percent of 72 is 9?

Solution

 ( egin{array}{r} ext { Percent }=frac{ ext { amount }}{ ext { base }} frac{n}{100}=frac{9}{72}end{array}) The percent is the ratio of ( n) to 100. The amount is 9, and the base is 72. ( egin{array}{r}n cdot 72=9 cdot 100 n cdot 72=900 n=900 div 72 n=12.5end{array}) Cross multiply and solve for ( n) by dividing 900 by 72. ( 12.5 \% ext { of } 72 ext { is } 9) The percent is ( frac{12.5}{100}=12.5 \%).

Example

What is 110% of 24?

Solution

 ( egin{array}{l} ext { Percent }=frac{ ext { amount }}{ ext { base }} frac{110}{100}=frac{n}{24}end{array}) The percent is the ratio ( frac{110}{100}). The amount is unknown, and the base is 24. ( egin{array}{r}24 cdot 110=100 cdot n 2,640 div 100=n 26.4=nend{array}) Cross multiply and solve for ( n) by dividing 2,640 by 100. ( 26.4 ext { is } 110 \% ext { of } 24)

Exercise

18 is 125% of what number?

1. ( 0.144)
2. ( 14.4)
3. ( 22.5)
4. ( 694 frac{4}{9}) (or about ( 694.4))
1. ( 0.144)

Incorrect. You probably didn’t write a proportion and just divided 18 by 125. Or, you incorrectly set up one fraction as ( frac{18}{125}) and set this equal to the base, ( n). The percent in this case is 125%, so one fraction in the proportion should be ( frac{125}{100}). The base is unknown and the amount is 18, so the other fraction is ( frac{18}{n}). Solving the proportion ( frac{125}{100}=frac{18}{n}) gives ( n=14.4).

2. ( 14.4)

Correct. The percent in this case is 125%, so one fraction in the proportion should be ( frac{125}{100}). The base is unknown and the amount is 18, so the other fraction is ( frac{18}{n}). Solving the proportion ( frac{125}{100}=frac{18}{n}) gives ( n=14.4).

3. ( 22.5)

Incorrect. You probably put the amount (18) over 100 in the proportion, rather than the percent (125). Perhaps you thought 18 was the percent and 125 was the base. The correct percent fraction for the proportion is ( frac{125}{100}). The base is unknown and the amount is 18, so the other fraction is ( frac{18}{n}). Solving the proportion ( frac{125}{100}=frac{18}{n}) gives ( n=14.4).

4. ( 694 frac{4}{9}) (or about ( 694.4))

Incorrect. You probably confused the amount (18) with the percent (125) when you set up the proportion. The correct percent fraction for the proportion is ( frac{125}{100}). The base is unknown and the amount is 18, so the other fraction is ( frac{18}{n}). Solving the proportion ( frac{125}{100}=frac{18}{n}) gives ( n=14.4).

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.

Example

Jeff has a coupon at the Guitar Store for 15% off any purchase of $100 or more. Jeff wonders how much money the coupon will take off of the$220 original price.

Solution

 How much is 15% of $220? Simplify the problems by eliminating extra words. Percent: 15%Base: 220Amount: ( n) Identify the percent, the base, and the amount. ( 15 \% cdot 220=n) Write the percent equation.( ext { Percent } cdot ext { Base }= ext { Amount }) ( 0.15 cdot 220=33) Convert 15% to 0.15, then multiply by 220. 15% of$220 is $33. The coupon will take$33 off the original price.

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

( egin{array}{l}
10 \% ext { of } 220=0.1 cdot 220=22
20 \% ext { of } 220=0.2 cdot 220=44
end{array})

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. Example Evelyn bought some books at the local bookstore. Her total bill was$31.50, which included 5% tax. How much did the books cost before tax?

Solution

 What number +5% of that number is $31.50? In this problem, you know that the tax of 5% is added onto the cost of the books. So if the cost of the books is 100%, the cost plus tax is 105%. 105% of what number = 31.50?Percent: 105%Base: ( n)Amount: 31.50 Identify the percent, the base, and the amount. ( 105 \% cdot n=31.50) Write the percent equation. ( ext { Percent } cdot ext { Base }= ext { Amount }). ( 1.05 cdot n=31.50) Convert 105% to a decimal. ( n=31.50 div 1.05=30) Divide to undo the multiplication of ( n) times 1.05. The books cost$30 before tax.

Example

Susana worked 20 hours at her job last week. This week, she worked 35 hours. In terms of a percent, how much more did she work this week than last week?

Solution

 35 is what percent of 20? Simplify the problem by eliminating extra words. Percent: ( n)Base: 20Amount: 35 Identify the percent, the base, and the amount. ( n cdot 20=35) Write the percent equation.( ext { Percent } cdot ext { Base }= ext { Amount }). ( n=35 div 20) Divide to undo the multiplication of ( n) times 20. ( n=1.75=175 \%) Convert 1.75 to a percent.

Since 35 is 175% of 20, Susana worked 75% more this week than she did last week. (You can think of this as, “Susana worked 100% of the hours she worked last week, as well as 75% more.”)

Percent problems have three parts: the percent, the base (or whole), and the amount. Any of those parts may be the unknown value to be found. To solve percent problems, you can use the equation, ( ext { Percent } cdot ext { Base }= ext { Amount }), and solve for the unknown numbers. Or, you can set up the proportion, ( ext { Percent }=frac{ ext { amount }}{ ext { base }}), where the percent is a ratio of a number to 100. You can then use cross multiplication to solve the proportion.

## 5.2.1: Solving Percent Problems

THE THREE PERCENTAGE CASES

To explain the cases that arise in problems involving percents, it is necessary to define the terms that will be used. Rate (r) is the number of hundredths parts taken. This is the number followed by the percent sign. The base (b) is the whole on which the rate operates. Percentage (p) is the part of the base determined by the rate. In the example

5% is the rate, 40 is the base, and 2 is the percentage.

There are three cases that usually arise in dealing with percentage, as follows:

Case I-To find the percentage when the base and rate are known.

EXAMPLE: What number is 6% of 50?

Case II-To find the rate when the base and percentage are known.

EXAMPLE: 20 is what percent of 60?

Case III-To find the base when the percentage and rate are known.

EXAMPLE: The number 5 is 25% of what number ?

the "of" has the same meaning as it does in fractional examples, such as

In other words, "of" means to multiply. Thus, to find the percentage, multiply the base by the rate. Of course the rate must be changed from a percent to a decimal before multiplying can be done. Rate times base equals percentage.

The number that is 6% of 50 is 3.

FRACTIONAL PERCENTS.-A fractional percent represents a part of 1 percent. In a case such as this, it is sometimes easier to find 1 percent of the number and then find the fractional part. For example, we would find 1/4 percent of 840 as follows:

To explain case II and case III, we notice in the foregoing example that the base corresponds to the multiplicand, the rate corresponds to the multiplier, and the percentage corresponds to the product.

Recalling that the product divided by one of its factors gives the other factor, we can solve the following problem:

We are given the base (60) and percentage (20).

We then divide the product (percentage) by the multiplicand (base) to get the other factor (rate). Percentage divided by base equals rate. The rate is found as follows:

The rule for case II, as illustrated in the foregoing problem, is as follows: To find the rate when the percentage and base are known, divide the percentage by the base. Write the quotient in the decimal form first, and finally as a percent.

The unknown factor in case III is the base, and the rate and percentage are known.

We divide the product by its known factor to find the other factor. Percentage divided by rate equals base. Thus,

The rule for case III may be stated as follows: To find the boss when the rate and percentage are known, divide the percentage by the rate.

Practice problems. In each of the following problems, fact determine which case is involved then find the answer.

1. What is 3/4% of 740?
2. 7.5% of 2.75 = ?
3. 8 is 2% of what number ?
4. ?% of 18 = 15.
5. 12% of ? = 12.
6. 8 is what percent of 32?

1. Case I 5.55
2. Case I 0.20625
3. Case III 400
4. Case II 83 1/3%
5. Case III 100
6. Case II 25%

## Fraction Word Problems - Intermediate

Lessons on solving fraction word problems using visual methods like bar models, block diagrams or tape diagrams.

Here are some examples of fraction word problems. The videos will illustrate how to use the block diagrams (Singapore Math) method to solve word problems.

### Middle School Fraction Problem

Example:
Alex, Betty, and Chris are trading marbles. First, Alex gives Betty 1/2 of his marbles. Then Betty gives Chris 1/3 of her marbles. Finally, Chris gives Alex 2 marbles. If everyone ends up with 12 marbles, how many marbles did each person start with?

Example:
Iona started with a certain amount of money. She spent 2/5 of her money. She earned $5 the next day. Later, she lost 3/7 of her money, but she still had$20 left.

1. There are 96 children in a library. 5/8 of them are girls. How many of them are boys?
2. David had $40. She spent 1/5 of the money on a storybook and 3/10 on a calculator. How much did he spend altogether? 3. Scott had some eggs. He sold 5/8 of them. If he sold 300 eggs, how many eggs did he have at first? Examples: Al received some money from his grandmother. Al spent$50, and gave 1/3 of what was left to Bob. Bob spent $2, and gave 3/4 of what was left to Carl. Carl spent$6, and gave 3/5 of what was left to Dan. If Dan received $27 from Carl, how much money did Al receive from his grandmother? Examples: There were several pelicans basking in the sun. 2/5 of the pelicans were brown, and the rest were white. After some of the brown pelicans flew away, only 3/10 of the remaining pelicans were brown. After some of the white pelicans flew away, the fraction of the pelicans that were white turned to 2/3. If the difference between the number of brown pelicans that flew away and the number of white pelicans that flew away was 18, how many brown pelicans were basking in the sun originally? #### Sixth Grade Math Sum Fraction And Price Value PSLE math sum involving fraction and price value does using model method. Examples: Jack bought some erasers, pencils and rulers. 1/6 of them were erasers. The number of pencils he bought was 5 more 1/2 the total number of all items and the remaining were rulers. Each of the erasers, pencils and rulers costs$2.20, $3.45 and$1.70 respectively. He spent a total of $519.15 on all the items. How many more pencils than erasers did Jack buy? #### Division Of Fraction By Whole Number Word Problem Using Visual Models 6.NS.1 - Division of Fraction by Whole Number Word Problem (Singapore math). Example: 3/4 gallons of cake batter is poured equally into 2 bowls. How many gallons are in each bowl? #### Division Of Whole Number By Fraction Word Problem Using Visual Models 6.NS.1 - Division of Whole Number by Fraction Word Problem. Example: Benjamin has 9 cups of sugar. If this is 3/4 of the number he needs to cake, how many cups does he need? Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page. ## Managing Conflict: Solvable vs. Perpetual Problems Knowing the difference between the types of problems all couples have is the key to avoiding gridlock. Knowing the difference between the types of problems all couples have is the key to avoiding gridlock. Knowing the difference between the types of problems all couples have is the key to avoiding gridlock. When thinking about conflict in a relationship, it is important to ascertain whether a problem is solvable or perpetual. Sixty-nine percent of relationship conflict is about perpetual problems. All couples have them. These problems are grounded in the fundamental differences that any two people face. They are either fundamental differences in your personalities that repeatedly create conflict or fundamental differences in your lifestyle needs. Instead of solving perpetual problems, what seems to be important is whether or not a couple can establish a dialogue about them. If they cannot establish such a dialogue, the conflict becomes gridlocked, and gridlocked conflict eventually leads to emotional disengagement. In today’s post, we want to take the opportunity to explain the difference between a solvable problem, a perpetual problem, and a gridlocked perpetual problem. • Solvable problems can be about housecleaning, disciplining children, sex, and in-laws. Solvable problems for one couple can be about the exact same topics that could be perpetual problems for a different couple. A solvable problem within a relationship is about something situational. The conflict is simply about that topic, and there may not be a deeper meaning behind each partner’s position. A solution can be found and maintained. • Perpetual problems are problems that center on either fundamental differences in your personalities, or fundamental differences in your lifestyle needs. All couples have perpetual problems. These issues can seemingly be about the exact same topics as what for another couple might be solvable however, unlike a solvable problem, these are the problems that a couple will return to over and over and over again. • Gridlocked perpetual problems are perpetual problems that have been mishandled and have essentially calcified into something “uncomfortable.” When a couple tries to discuss a gridlocked issue, it can feel like they are “spinning their wheels” and getting nowhere. The nature of gridlock is that hidden agendas underlie the issue. The Gottman Method focuses on building emotional intelligence and developing skills for managing conflict and enhancing friendship to help couples create a system of shared meaning in your relationship. What matters is not solving perpetual problems, but rather the affect with which they are discussed. The goal should be to establish a dialogue about the perpetual problem that communicates acceptance of your partner with humor, affection, and even amusement, to actively cope with the unresolvable problem, rather than allowing it to fall into the condition of gridlock. Gridlocked discussions only lead to painful exchanges or icy silence, and almost always involve the Four Horsemen (criticism, contempt, stonewalling, and defensiveness). Learn how to recognize if a perpetual problem in your relationship has become gridlocked in our next blog post, which you can read here. If you want to build a deeply meaningful relationship full of trust and intimacy, then subscribe below to receive our blog posts directly to your inbox: ## Solving Mixture Problems We recommend using a table to organize your information for mixture problems. Using a table allows you to think of one number at a time instead of trying to handle the whole mixture problem at once. ### Replacing The Solution A tank has a capacity of 10 gallons. When it is full, it contains 15% alcohol. How many gallons must be replaced by an 80% alcohol solution to give 10 gallons of 70% solution? Set up a table for alcohol . The alcohol is replaced i.e. removed and added. Mixture Problems Some word problems using systems of equations involve mixing two quantities with different prices. To solve mixture problems, knowledge of solving systems of equations. is necessary. Most often, these problems will have two variables, but more advanced problems have systems of equations with three variables. Other types of word problems using systems of equations include rate word problems and work word problems. Solving a Mixture Problem using a system of equations. We set up and solve a mixture problem using a system of equations with two variables. Before solving the problem, a short introduction to what a -solution- with talking about a chemical mixture. Example: A chemist mixes a 12% acid solution with a 20% acid solution to make 300 milliliters of an 18% acid solution. How many milliliters of each solution does the chemist use? Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page. ## 5.2.1: Solving Percent Problems "Percent of" Word Problems: Markup and Markdown Examples (page 2 of 3) An important category of percentage exercises is markup and markdown problems. For these, you calculate the markup or markdown in absolute terms (you find by how much the quantity changed), and then you calculate the percent change relative to the original value. So they're really just another form of "increase - decrease" exercises. • A computer software retailer used a markup rate of 40%. Find the selling price of a computer game that cost the retailer$25.

The markup is 40% of the $25 cost, so the markup is: Then the selling price, being the cost plus markup, is: • A golf shop pays its wholesaler$40 for a certain club, and then sells it to a golfer for $75 . What is the markup rate? First, I'll calculate the markup in absolute terms: Then I'll find the relative markup over the original price, or the markup rate: ($35) is (some percent) of ($40) , or: Copyright © Elizabeth Stapel 1999-2011 All Rights Reserved . so the relative markup over the original price is: Since x stands for a percentage, I need to remember to convert this decimal value to the corresponding percentage. The markup rate is 87.5% . • A shoe store uses a 40% markup on cost. Find the cost of a pair of shoes that sells for$63.

This problem is somewhat backwards. They gave me the selling price, which is cost plus markup, and they gave me the markup rate, but they didn't tell me the actual cost or markup. So I have to be clever to solve this.

I will let " x " be the cost. Then the markup, being 40% of the cost, is 0.40x . And the selling price of $63 is the sum of the cost and markup, so: 63 = x + 0.40x 63 = 1x + 0.40x 63 = 1.40x 63 ÷ 1.40 = x= 45 The shoes cost the store$45.

First, I'll find the markdown. The markdown is 25% of the original price of $55 , so: By subtracting this markdown from the original price, I can find the sale price: The sale price is$41.25.

First, I'll find the amount of the markdown:

Then I'll calculate "the markdown over the original price", or the markdown rate: ($106.25) is (some percent) of ($425) , so:

. and the relative markdown over the original price is:

Since the " x " stands for a percentage, I need to remember to convert this decimal to percentage form.

The markdown rate is 25%.

This problem is backwards. They gave me the sale price ($127.46) and the markdown rate (15%) , but neither the markdown amount nor the original price. I will let " x " stand for the original price. Then the markdown, being 15% of this price, was 0.15x . And the sale price is the original price, less the markdown, so I get: x &ndash 0.15x = 127.46 1x &ndash 0.15x = 127.46 0.85x = 127.46 x = 127.46 ÷ 0.85 = 149.952941176. This problem didn't state how to round the final answer, but dollars-and-cents is always written with two decimal places, so: The original price was$149.95.

Note in this last problem that I ended up, in the third line of calculations, with an equation that said "eighty-five percent of the original price is \$127.46 ". You can save yourself some time if you think of discounts in this way: if the price is 15% off, then you're only actually paying 85% . Similarly, if the price is 25% off, then you're paying 75% if the price is 30% off, then you're paying 70% and so on.

## How to Solve for X

This article was co-authored by David Jia. David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math.

There are 7 references cited in this article, which can be found at the bottom of the page.

There are a number of ways to solve for x, whether you're working with exponents and radicals or if you just have to do some division or multiplication. No matter what process you use, you always have to find a way to isolate x on one side of the equation so you can find its value. Here's how to do it:

## Mass Percent Composition Problem

Bicarbonate of soda (sodium hydrogen carbonate) is used in many commercial preparations. Its formula is NaHCO3. Find the mass percentages (mass %) of Na, H, C, and O in sodium hydrogen carbonate.

First, look up the atomic masses for the elements from the Periodic Table. The atomic masses are found to be:

Next, determine how many grams of each element are present in one mole of NaHCO3:

• 22.99 g (1 mol) of Na
• 1.01 g (1 mol) of H
• 12.01 g (1 mol) of C
• 48.00 g (3 mole x 16.00 gram per mole) of O

The mass of one mole of NaHCO3 is:

22.99 g + 1.01 g + 12.01 g + 48.00 g = 84.01 g

And the mass percentages of the elements are

• mass % Na = 22.99 g / 84.01 g x 100 = 27.36 %
• mass % H = 1.01 g / 84.01 g x 100 = 1.20 %
• mass % C = 12.01 g / 84.01 g x 100 = 14.30 %
• mass % O = 48.00 g / 84.01 g x 100 = 57.14 %

When doing mass percent calculations, it's always a good idea to check to make sure your mass percents add up to 100% (helps catch math errors):

27.36 + 14.30 + 1.20 + 57.14 = 100.00

## Acid-Base Equilibria

Example: Consider the process by which we would calculate the H 3O + , OAc - , and HOAc concentrations at equilibrium in an 0.10 M solution of acetic acid in water.

We start this calculation by building a representation of what we know about the reaction.

 HOAc(aq) + H2O(l) H3O + (aq) + OAc - (aq) Ka = 1.8 x 10 -5 Initial: 0.10 M 0 0 Equilibrium: ? ? ?

We then compare the initial reaction quotient (Qa) with the equilibrium constant (Ka) for the reaction and reach the obvious conclusion that the reaction must shift to the right to reach equilibrium.

Recognizing that we get one H3O + ion and one OAc - ion each time an HOAc molecule dissociates allows us to write equations for the equilibrium concentrations of the three components of the reaction.

 HOAc(aq) + H2O(l) H3O + (aq) + OAc - (aq) Ka = 1.8 x 10 -5 Initial: 0.10 M 0 0 Equilibrium: 0.10 - C C C

Substituting what we know about the system at equilibrium into the Ka expression gives the following equation.

Although we could rearrange this equation and solve it with the quadratic formula, it is tempting to test the assumption that C is small compared with the initial concentration of acetic acid.

We then solve this approximate equation for the value of C.

C is small enough to be ignored in this problem because it is less than 5% of the initial concentration of acetic acid.

We can therefore use this value of C to calculate the equilibrium concentrations of H3O + , OAc - , and HOAc.

[H3O + ] = [OAc - ] = C 0.0013 M

We can confirm the validity of these results by substituting these concentrations into the expression for Ka.

Our calculation must be valid because the ratio of these concentrations agrees with the value of Ka for acetic acid, within experimental error.

When solving problems involving weak acids, it may appear that one assumption is madethat is small compared with the initial concentration of HOAc. In fact, two assumptions are made.

The second assumption is hidden in the way the problem is set up.

 HOAc(aq) + H2O(l) H3O + (aq) + OAc - (aq) Ka = 1.8 x 10 -5 Initial: 0.10 M 0 0 Equilibrium: 0.10 - C C C

The amount of H3O + ion in water is so small that we are tempted to assume that the initial concentration of this ion is zero, which isn't quite true.

It is important to remember that there are two sources of the H3O + ion in this solution. We get H3O + ions from the dissociation of acetic acid.

HOAc(aq) + H2O(l) H3O + (aq) + OAc - (aq)

But we also get H3O + ions from the dissociation of water.

2 H2O(l) H3O + (aq) + OH - (aq)

Because the initial concentration of the H3O + ion is not quite zero, it might be a better idea to write "0" beneath the H3O + term when we describe the initial conditions of the reaction, as shown below.

 HOAc(aq) + H2O(l) H3O + (aq) + OAc - (aq) Ka = 1.8 x 10 -5 Initial: 1.0 M 0 0 Equilibrium: 1.0 - C C C

Before we can trust the results of the calculation for acetic acid in the previous section, we have to check both of the assumptions made in this calculation.

• The assumption that the amount of acid that dissociates is small compared with the initial concentration of the acid.
• The assumption that enough acid dissociates to allow us to ignore the dissociation of water

We have already confirmed the validity of the first assumption. (Only 1.3% of the acetic acid molecules dissociate in this solution.) Let's now check the second assumption.

According to the calculation in the previous section, the concentration of the H3O + ion from the dissociation of acetic acid is 0.0013 M. The OH - ion concentration in this solution is therefore 7.7 x 10 -12 M.

All of the OH - ion in this solution comes from the dissociation of water. Since we get one H3O + ion for each OH - ion when water dissociates, the contribution to the total H3O + ion concentration from the dissociation of water must be 7.7 x 10 -12 M. In other words, only about 6 parts per billion of the H3O + ions in this solution come from the dissociation of water.

The second assumption is therefore valid in this calculation. For all practical purposes, we can assume that virtually none of the H3O + ion in this solution comes from the dissociation of water. As might be expected, this assumption only fails for dilute solutions of very weak acids.

The two assumptions that are made in weak-acid equilibrium problems can be restated as follows.

• The dissociation of the acid is small enough that the change in the concentration of the acid as the reaction comes to equilibrium can be ignored.
• The dissociation of the acid is large enough that the H3O + ion concentration from the dissociation of water can be ignored.

In other words, the acid must be weak enough that C is small compared with the initial concentration of the acid. But it also must be strong enough that the H3O + ions from the acid overwhelm the dissociation of water. In order for the approach taken to the calculation for acetic acid to work, the acid has to be "just right." If it's too strong, C won't be small enough to be ignored. If it's too weak, the dissociation of water will have to be included in the calculation.

Fortunately, many acids are "just right." To illustrate this point, the next section will use both assumptions in a series of calculations designed to identify the factors that influence the H3O + ion concentration in aqueous solutions of weak acids.

The following examples probe the relationship between the H3O + ion concentration at equilibrium and the acid-dissociation equilibrium constant for the acid.

Calculate the pH of 0.10 M solution of hypochlorous acid, HOCl, Ka = 2.9 x 10 -8

As expected, the H3O + ion concentration at equilibrium and therefore the pH of the solutiondepends on the value of Ka for the acid. The H3O + ion concentration decreases and the pH of the solution increases as the value of Ka becomes smaller. The next exercise shows that the H3O + ion concentration at equilibrium also depends on the initial concentration of the acid.

Calculate the H3O + ion concentration and the pH of acetic acid solutions with the following concentrations: 1.0 M, 0.10 M, and 0.01 M.

The concentration of the H3O + ion in an aqueous solution gradually decreases and the pH of the solution increases as the solution becomes more dilute.

The results of the previous two examples provide a basis for constructing a model that allows us to predict when we can ignore the dissociation of water in equilibrium problems involving weak acids. Two factors must be built into this model: (1) the strength of the acid as reflected by the value of Ka, and (2) the strength of the solution as reflected by the initial concentration of the acid.

We need to develop techniques to handle problems for which one or the other of our assumptions is not valid. (Either the acid is not weak enough to ignore the value of C, or the acid is so weak we have to include the dissociation of water in our calculations.)

In this section, we will consider acid solutions that aren't weak enough to ignore the value of C. Let's start by calculating the H3O + , HClO2, and ClO2 - concentrations at equilibrium in an 0.10 M solution of chlorous acid (Ka = 1.1 x 10 -2 ).

The first step, as always, involves building a representation of the problem.

 HClO2(aq) + H2O(l) H3O + (aq) + ClO2 - (aq) Ka = 1.1 x 10 -2 Initial: 0.10 M 0 0 Equilibrium: 0.10 - C C C

We then substitute this information into the Ka expression.

The value of Ka for this acid is close enough to 1 to make us suspicious of the assumption that C is small compared with the initial concentration of the acid. There is nothing wrong with trying this assumption, however, even if we suspect it isn't valid.

Solving this approximate equation gives a value for C that is 33% of the initial concentration of chlorous acid.

The assumption that C is small therefore fails miserably.

There are two ways out of this difficulty. We can expand the original equation and solve it by the quadratic formula. Or we can use successive approximations to solve the problem. Both techniques give the following value of C for this problem.

Using this value of C gives the following results.

Chlorous acid doesn't belong among the class of strong acids that dissociate more or less completely. Nor does it fit in the category of weak acids, which dissociate only to a negligible extent. Since the amount of dissociation in this solution is about 28%, it might be classified as a "not-so-weak acid."

It is more difficult to solve equilibrium problems when the acid is too weak to ignore the dissociation of water. Deriving an equation that can be used to solve this class of problems is therefore easier than solving them one at a time. To derive such an equation, we start by assuming that we have a generic acid, HA, that dissolves in water. We therefore have two sources of the H3O + ion.

HA(aq) + H2O(l) H3O + (aq) + A - (aq)

2 H2O(l) H3O + (aq) + OH - (aq)

Because we get one H3O + ion for each OH - ion when water dissociates, the concentration of the H3O + ion from the dissociation of water is always equal to the amount of OH - ion from this reaction.

The total H3O + ion concentration in an acid solution is equal to the sum of the H3O + ion concentrations from the two sources of this ion, the acid and water.

We now write three more equations that describe this system. The first equation is the equilibrium constant expression for this reaction.

The second equation summarizes the relationship between the total H3O + ion concentration in the solution and the OH - ion concentration from the dissociation of water.

The third equation summarizes the relationship between the positive and negative ions produced by the two reactions that occur in this solution.

(This equation simply states that the sum of the positive ions formed by the dissociation of the acid and water is equal to the sum of the negative ions produced by these reactions.)

We now substitute the second equation into the third.

We then solve this equation for the [A - ] term.

We then substitute this equation into the equilibrium constant expression.

Rearranging this equation by combining terms gives the following result.

We then solve this equation for the H3O + ion concentration and take the square root of both sides.

We can generate a more useful version of this equation by remembering that we are trying to solve equilibrium problems for acids that are so weak we can't ignore the dissociation of water. We can therefore assume that C is small compared with the initial concentration of the acid.

By convention, the symbol used to represent the initial concentration of the acid is Ca. If C is small compared with the initial concentration of the acid, then the concentration of HA when this reaction reaches equilibrium will be virtually the same as the initial concentration.

Substituting this approximation into the equation derived in this section gives an equation that can be used to calculate the pH of a solution of a very weak acid.

Calculate the H3O + concentration in an 0.0001 M solution of hydrocyanic acid (HCN).

This section compares the way in which the H3O + concentration is calculated for pure water, a weak acid, and a very weak acid.

The product of the concentrations of the H3O + and OH - ions in pure water is equal to Kw.

But the H3O + and OH - ion concentrations in pure water are the same.

Substituting the second equation into the first gives the following result.

The H3O + ion concentration in pure water is therefore equal to the square root of Kw.

The generic equilibrium constant expression for a weak acid is written as follows.

If the acid is strong enough to ignore the dissociation of water, the H3O + ion and A - ion concentrations in this solution are about equal.

Substituting this information into the acid-dissociation equilibrium constant expression gives the following result.

The concentration of the HA molecules at equilibrium is equal to the initial concentration of the acid minus the amount that dissociates: C.

If C is small compared with the initial concentration of the acid, we get the following approximate equation.

Rearranging this equation and taking the square root of both sides gives the following result.

When the acid is so weak that we can't ignore the dissociation of water, we use the following equation to calculate the concentration of the H3O + ion at equilibrium.

The equations used to calculate the H3O + ion concentration in these solutions are summarized below.

The first and second equations are nothing more than special cases of the third. When we can ignore the dissociation of the acidbecause there is no acid in the solutionwe get the first equation. When we can ignore the dissociation of water, we get the second equation. When we can't ignore the dissociation of either the acid or water, we have to use the last equation.

This discussion gives us a basis for deciding when we can ignore the dissociation of water. Remember our rule of thumb: we can ignore anything that makes a contribution of less than 5% to the total. Now compare the most inclusive equation for the H3O + ion concentration

with the equation that assumes that the dissociation of water can be ignored.

The only difference is the Kw term, which is under the square root sign.

As a rule: We can ignore the dissociation of water when KaCa for a weak acid is larger than 1.0 x 10 -13 . When KaCa is smaller than 1.0 x 10 -13 , the dissociation of water must be included in the calculation.

Calculate the pH of an 0.023 M solution of saccharin (HSc), if Ka is 2.1 x 10 -12 for this artificial sweetener.

## How to Use the 5 Whys

The model follows a very simple seven-step process:

### 1. Assemble a Team

Gather together people who are familiar with the specifics of the problem, and with the process that you're trying to fix. Include someone to act as a facilitator , who can keep the team focused on identifying effective counter-measures.

### 2. Define the Problem

If you can, observe the problem in action. Discuss it with your team and write a brief, clear problem statement that you all agree on. For example, "Team A isn't meeting its response time targets" or "Software release B resulted in too many rollback failures."

Then, write your statement on a whiteboard or sticky note, leaving enough space around it to add your answers to the repeated question, "Why?"

### 3. Ask the First "Why?"

Ask your team why the problem is occurring. (For example, "Why isn't Team A meeting its response time targets?")

Asking "Why?" sounds simple, but answering it requires serious thought. Search for answers that are grounded in fact: they must be accounts of things that have actually happened, not guesses at what might have happened.

This prevents 5 Whys from becoming just a process of deductive reasoning, which can generate a large number of possible causes and, sometimes, create more confusion as you chase down hypothetical problems.