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# Algebra: Praxis I Exam (page 3)

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Updated on Jul 5, 2011

### Problems Involving Ratio

A ratio is a comparison of two quantities measured in the same units. It is symbolized by the use of a colon—x:y. Ratios can also be expressed as fractions or using words (x to y).

Ratio problems are solved using the concept of multiples.
Example

A bag contains 60 screws and nails. The ratio of the number of screws to nails is 7:8. How many of each kind are there in the bag?

From the problem, it is known that 7 and 8 share a multiple and that the sum of their product is 60. Whenever you see the word ratio in a problem, place an "x" next to each of the numbers in the ratio, and those are your unknowns.
Let 7x = the number of screws.
Let 8x = the number of nails.
Write and solve the following equation:

Therefore, there are (7)(4) = 28 screws and (8)(4) = 32 nails.

Check: 28 + 32 = 60 screws, .

### Problems Involving Variation

Variation is a term referring to a constant ratio in the change of a quantity.

• Two quantities are said to vary directly if their ratios are constant. Both variables change in an equal direction. In other words, two quantities vary directly if an increase in one causes an increase in the other. This is also true if a decrease in one causes a decrease in the other.
Example
If it takes 300 new employees a total of 58.5 hours to train, how many hours of training will it take for 800 employees?
Because each employee needs about the same amount of training, you know that they vary directly. Therefore, you can set the problem up the following way:

Cross multiply to solve:

Therefore, it would take 156 hours to train 800 employees.

• Two quantities are said to vary inversely if their products are constant. The variables change in opposite directions. This means that as one quantity increases, the other decreases, or as one decreases, the other increases.
Example

If two people plant a field in six days, how many days will it take six people to plant the same field? (Assume each person is working at the same rate.)

As the number of people planting increases, the days needed to plant decreases. Therefore, the relationship between the number of people and days varies inversely. Because the field remains constant, the two products can be set equal to each other.
2 people × 6 days = 6 people × x days

Thus, it would take six people two days to plant the same field.

### Rate Problems

In general, there are three different types of rate problems likely to be encountered in the workplace: cost per unit, movement, and work output. Rate is defined as a comparison of two quantities with different units of measure.

Example

### Cost per Unit

Some problems will require the calculation of unit cost.

Example
If 100 square feet cost \$1,000, how much does 1 square foot cost?

### Movement

In working with movement problems, it is important to use the following formula:

(rate)(time) = distance

Example

A courier traveling at 15 mph traveled from his base to a company in of an hour less than it took when the courier traveled at 12 mph. How far away was his drop-off?

First, write what is known and unknown.
Unknown: time for courier traveling 12 mph = x
Known: time for courier traveling 15 mph = x
Then, use the formula (rate)(time) = distance to find expressions for the distance traveled at each rate:
12 mph for x hours = a distance of 12x miles
15 mph for x mph = a distance of 15x miles.
The distance traveled is the same; therefore, make the two expressions equal to each other:

Be careful: 1.25 is not the distance; it is the time. Now you must plug the time into the formula (rate)(time) = distance. Either rate can be used.

12x = distance
12(1.25) = distance
15 miles = distance

### Work Output

Work-output problems are word problems that deal with the rate of work. The following formula can be used on these problems:

(rate of work)(time worked) = job or part of job completed
Example
Danette can wash and wax two cars in six hours, and Judy can wash and wax the same two cars in four hours. If Danette and Judy work together, how long will it take to wash and wax one car?
Because Danette can wash and wax two cars in six hours, her rate of work is , or one car every three hours. Judy's rate of work is , or one car every two hours. In this problem, making a chart will help:
Because they are both working on only one car, you can set the equation equal to one: Danette's part + Judy's part = 1 car:
Solve by using 6 as the LCD for 3 and 2 and clear the fractions by multiplying by the LCD:

Thus, it will take Judy and Danette hours to wash and wax one car.