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# Percent Word Problems Help

based on 5 ratings
By McGraw-Hill Professional
Updated on Sep 27, 2011

## Introduction to Percent Word Problems

To many algebra students, applications (word problems) seem impossible to solve. You might be surprised how easy solving many of them really is. If you follow the program in this chapter, you will find yourself becoming a pro at solving word problems. Mastering the problems in this chapter will also train you to solve applied problems in science courses and in more advanced mathematics courses.

### Percents

A percent is a decimal number in disguise. In fact, the word “percent” literally means “per hundred.” Remember that “per” means to divide, so 16% means 16 ÷ 100 or 16/100 = 0.16. Then 16% of 25 will be translated into (0.16)(25). (Remember that “of” means “multiply.”) So, 16% of 25 is (0.16)(25) = 4.

#### Examples

82% of 44 is (0.82)(44) = 36.08

150% of 6 is (1.50)(6) = 9

of 24 is (0.0875)(24) = 2.1

0.65% of 112 is (0.0065)(112) = 0.728

Find practice problems and solutions at Percent Word Problems Practice Problems - Set 1.

## Increasing/Decreasing by a Percent Word Problems

As consumers, we often see quantities being increased or decreased by some percentage. For instance, a cereal box boasts “25% More.” An item might be on sale, saying “Reduced by 40%.” When increasing a quantity by a percent, first compute what the percent is, then add it to the original quantity. When decreasing a quantity by a percent, again compute the percent then subtract it from the original quantity.

#### Example 1:

80 increased by 20%

(0.20)(80) = 16

So, 80 increased by 20% is 80 + 16 = 96.

#### Example 2:

24 increased by 35%

(0.35)(24) = 8.4

24 increased by 35% is 24 + 8.4 = 32.4

#### Example 3:

36 increased by 250%

(2.50)(36) = 90

36 increased by 250% is 36 + 90 = 126

#### Example 4:

64 decreased by 27%

(0.27)(64) = 17.28

64 decreased by 27% is 64 – 17.28 = 46.72

### Formulas for Solving Percent Increase/Decrease Word Problems

Many word problems involving percents fit the above model—that is, a quantity being increased or decreased. Often you can solve these problems using one of the following formats:

x + .____ x (for a quantity being increased by a percent)

or

x – .___ x (for a quantity being decreased by a percent).

#### Examples

A \$100 jacket will be reduced by 15% for a sale. What will the sale price be?

Let x = sale price.

Then, 100 − (0.15)(100) = x .

100 – 15 = x

85 = x

The sale price is \$85.

Find practice problems and solutions at Percent Word Problems Practice Problems - Set 2.

## Solving for an Unknown Variable in Percent Word Problems

More often, the sale price will be known and the original price is not known.

#### Example

The sale price for a computer is \$1200, which represents a 20% markdown. What is the original price?

Let x represent original price. Then the sale price is x – 0.20 x = 0.80 x . The sale price is also 1200. This gives us the equation 0.80 x = 1200.

The original price is \$1500.

In the first example, the percent was multiplied by the number given; and in the second, the percent was multiplied by the unknown. You must be very careful in deciding of which quantity you take the percent. Suppose the first problem was worded, “An item is marked down 20% for a sale. The sale price is \$80, what is the original price?” The equation to solve would be x − 0.20 x = 80 x , where x represents the original price. A common mistake is to take 20% of \$80 and not 20% of the original price

The data used in the following examples and practice problems are taken from the 117th edition of Statistical Abstract of the United States . Many quantities and percentages are approximate.

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