SAT Hub # Miscellaneous Algebra There are a few additional algebra topics often tested on the SAT. Expect each of the following to show up at least once on each SAT Math Test.

## 1. Undefined Denominator

Math convention does not allow division by 0. When 0 appears in the denominator of an expression, then that expression is undefined. How does this convention affect quadratic equations?

#### Example: What are the solutions to the following equation?

$$\frac{x^{2} + x - 12}{x - 2} = 0$$

We notice a quadratic equation in the numerator. Since it is a good idea to start solving quadratic equations by factoring, we will factor this numerator as follows: $$\frac{(x-3)(x-4)}{x-2} = 0$$

If either of the factors in the numerator is 0, then the entire expression becomes 0. Thus, the solutions to this equation are x = 3 or x = - 4.

### Rationalizing of Denominator

The SAT generally does not include in its answer choices any mathematical expressions in which a radical ( $$\sqrt{}$$ ) appears in the denominator of a fraction. Accordingly, if your calculations yield a radical in the denominator, you should be prepared to take one last step: rationalize the denominator.

$$\frac{3}{\sqrt{2}}\rightarrow Multiply\: by \frac{\sqrt{2}}{\sqrt{2}}\: To \: get \: \frac{3\sqrt{2}}{2}$$

Example: What is the value of $$\frac{2}{2-\sqrt{2}}$$

In this problem, the multi-part denominator poses an algebraic challenge. Try to multiply both the numerator and denominator by the denominator, and you are still left with a messy radical in the denominator. Algebraically, the solution is to employ the difference of squares:

$$\frac{2}{2-\sqrt{2}}$$ x $$\frac{2+\sqrt{2}}{2+\sqrt{2}}$$

$$\frac{4 + 2\sqrt{2}}{2^{2}-\sqrt{2^{2}}}$$

$$\frac{4+2\sqrt{2}}{2}$$

$$2+\sqrt{2}$$

When rationalizing denominators with addition or subtraction in them, the difference of squares rule is a nearly magical way to transform the expression.

## 2. Direct and Inverse Proportions

Direct Variation: As one value increases, the other value increases. Decreases in one variable produce a decrease in the other.

#### Example:

• 10 doughnuts cost $6.00. How much will 15 doughnuts cost? • Set up proportion 6/10 = x/15 • x =$9.00 or use 6/10 as the constant of variation.
• The constant stays the same no matter what amount is purchased.
• If you want 15, then $0.6x15 =$9.00
• If you want 45, then $0.6 x 45 =$27.00

Inverse Variation: A change in one variable produces an inverse or opposite change in the other.

## 3. Exponential Growth and Decay

#### a = Constant specified in problem e = Numerical constant ≈ 2.718r = Rate of growth/decay t = Time

The key difference between growth and decay is that growth has an exponential increase, while decay has an exponential decrease.

## Rates and Work

One common type of word problem on the SAT Math Test is the rate problem. Rate problems come in a variety of forms, but all are marked by three primary components: rate, time, and distance/work.

#### These three elements are related by the following equations:

• RATE X TIME = DISTANCE (RT = D)
• RATE X TIME = WORK (RT = W)

The SAT makes rate situations more complicated by involving more than one person or vehicle traveling. To deal with this, you will need to consider multiple RT = D relationships.

Example: Harvey runs a 30-mile course at a constant rate of 4 miles per hour. If Clyde runsthe same track at a constant rate and completes the course in 90 fewer minutes, how fast did Clyde run?

4t = 30
t - 7.5

r x 6 = 30
r = 5

## 4. Relative Rates

#### Relative rate problems are a subset of multiple rate problems. The defining aspect of these problems is that two bodies are traveling at the same time. There are three possible scenarios:

1. The bodies move towards each other.
2. The bodies move away from each other.
3. The bodies move in the same direction on the same path.

Solved Example 5
Imagine that two people are 14 miles apart and begin walking towards each other. Person A walks 3 miles per hour, and Person B walks 4 miles per hour. How long will it take them to reach each other?
A. 1
B. 2
C. 3
D. 4

#### The rate at which they’re getting closer to each other is 3 + 4 = 7 miles per hour. In other words, after every hour they walk, they are 7 miles closer to each other. Now you can create one RT = D equation:

7t = 14
t = 2

The correct answer is B. You may very well have answered this question intuitively. But as the questions become more difficult, this method only becomes more valuable.

Solved Example 6

Car X is 40 miles west of Car Y. Both cars are traveling east, and Car X is going 50% faster than Car Y. If both cars travel at a constant rate and it takes Car X 2 hours and 40 minutes to catch up to Car Y, how fast is Car Y going?

A. 20 miles per hour
B. 30 miles per hour
C. 40 miles per hour
D. 50 miles per hour

#### Solution: The multiple RTD chart would look like this:

Instead, you can answer this question with one equation. If Car X is initially 40 miles behind Car Y, and they both travel until Car X catches up to Car Y, then the distance between them will have decreased by 40 miles. That is your distance. The distance between the two cars is decreasing at a rate of 1.5r - r = 0.5r, and the time they travel is 8/3 hours:

0.5r x 8/3 = 40
r = 30

You defined the rate Car Y was traveling as r, so if r = 30, then Car Y was going 30 miles per hour. The correct answer is B.

## 5. Work Problems

Work problems are just another type of rate problem. Instead of distances, however, these questions are concerned with the amount of work done.

Work takes the place of distance. Instead of RT = D, use the equation RT = W.

Be sure to express a rate as work per time (W/T), NOT time per work (T/W).

For example: if a machine produces pencils at a constant rate of 120 pencils every 30 seconds, the rate at which the machine works is 120 pencils/30 seconds = 4 pencils/second.

Example 7

Martha can paint 3/7 of a room in 4.5 hours. If Martha finishes painting the room at the same rate, how long will it have taken Martha to paint the room?

A. 8.3 hours
B. 9 hours
C. 9.5 hours
D. 10.5 hours

#### SOLUTION: The first step in this problem is to calculate the rate at which Martha paints the room. You can say that painting the entire room is completing 1 unit of work. Set up an RTW chart:

Now you can solve for the rate:

r x $$\frac{9}{2}=\frac{3}{7}$$

r = 2/21

#### The division would be messy, so leave it as a fraction. Martha paints 2/21 of the room every hour. Now you have what you need to answer the question. Remember, painting the whole room is the same as doing 1 unit of work. Set up another RTW chart:

$$\frac{(21)}{2}t = 1$$

$$t=\frac{21}{2}$$

The correct answer is D. Notice that the rate and the time in this case were reciprocals of each other. This will always be true when the amount of work done is 1 unit.

## 6. Working Together

More often than not, work problems will involve more than one worker. When two or more workers are performing the same task, their rates can be added together.

For instance, if Machine A can make 5 boxes in an hour, and Machine B can make 12 boxes in an hour, then working together the two machines can make 5 + 12 = 17 boxes per hour.

Likewise, if Lucas can complete 1/3 of a task in an hour and Serena can complete 1/2 of that task in an hour, then working together they can complete 1/3 + 1/2 = 5/6 of the task every hour.

If, on the other hand, one worker is undoing the work of the other, subtract the rates. For instance, if one hose is filling a pool at a rate of 3 gallons per minute, and another hose is draining the pool at a rate of 1 gallon per minute, the pool is being filled at a rate of 3 - 1 = 2 gallons per minute.

##### Example 9Alejandro, working alone, can build a doghouse in 4 hours. Betty can build the same doghouse in 3 hours. If Betty and Carmelo, working together, can build the doghouse twice as fast as Alejandro, how long would it take Carmelo, working alone, to build the doghouse?A. 6B. 7C. 8D. 9Solution: Let c represent the number of hours it takes Carmelo to build the doghouse.Alejandro can build 1/4 of the doghouse every hour, Betty can build 1/3 of the doghouse every hour, and Carmelo can build 1/C of the doghouse every hour.The problem states that Betty and Carmetlo, working together, can work twice as fast as Alejandro. That means that their rate is twice Alejandro’s rate:$$Rate_{B}+Rate_{C}=2(Rate_{n})$$$$\frac{1}{3}+\frac{1}{c}=2(\frac{1}{4})$$$$\frac{1}{c}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$$c=6The correct answer is A.     ## More Topics  