| 2 ≤ z ≤ 4 | |
| Quantity A | Quantity B |
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The most important thing you have to figure out on QC Algebra questions is when you are allowed to plug in a number, and when you are not. In other words, when is a variable not a variable?
Consider this example. Can you plug in numbers?
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x − 3 = 12 |
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| Quantity A | Quantity B |
| y | 9 |
Pay attention to any constraints that have been placed on variables. In this question, the first equation gives you enough information to find the value of x. And because you have enough information to find x, you also have enough information to find y through the second equation. In fact, x = 15, which means that y will equal 10, and thus the answer to this question is (A). Although y is a variable, it actually has a definite value.
Problem Recap: When variables have a unique value, you must solve for the value of the variable.
On this test, variables can assume a variety of forms. They can:
On any question that involves variables, you should identify which situation you are dealing with. Take this problem, for example:
| 2 ≤ z ≤ 4 | |
| Quantity A | Quantity B |
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In this problem, z doesn’t have one specific value, but its range is well-defined. In a situation such as this, you should examine the upper and lower bounds of z.
Start with the lower bound. Plug in 2 for z in both quantities:
| 2 ≤ z ≤ 4 | |
| Quantity A | Quantity B |
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When z = 2, Quantity B is bigger.
Now try the upper bound. Plug in 4 for z in both quantities:
| 2 ≤ z ≤ 4 | |
| Quantity A | Quantity B |
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When z = 4, Quantity A is bigger. The correct answer is (D).
The way in which variables are constrained (or not) can tell you a lot about efficient ways to approach that particular problem.
Another way variables can be defined on this test is in terms of another variable. Take the following example:
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| Quantity A | Quantity B |
| 6x | 5y |
In this problem, you are given an equation that contains two variables: x and y. You won’t be able to solve for the value of either variable, but that doesn’t mean the answer will be (D). For this type of problem, the best course of action is to make a direct comparison of the variables. You can do this by simplifying the equation so that all unnecessary terms have been eliminated. Begin by cross-multiplying:
Now you have a 30 on each side that should be eliminated:
6x = 5y
You still don’t know the value of either variable, but you do have enough information to answer the question. The answer is (C).
Problem Recap: If a variable is defined in terms of another variable, simplify and find a direct comparison.
Sometimes, you will not be given any information about a variable. If there are no constraints on the variable, then your goal is to prove (D). For example:
| Quantity A | Quantity B |
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2x |
No information about x has been given. If x is positive, Quantity B will be bigger. For instance, if x = 1, Quantity
and Quantity B = 2.
However, there is no reason x must be positive. Remember, one way to try to prove (D) is to check negative possibilities. If x is negative, then Quantity A will be bigger. For instance, if x = −1, then Quantity
and Quantity B = −2.
The correct answer is (D).
Finally, variables may be defined as having certain properties. The most common include a variable being positive or negative or an integer. The strategy for this type of problem is identical to the strategy for problems in which variables have no constraints: prove (D). The only difference is that the types of numbers you can use are restricted. This type is also similar to Range of Values, in that you should test extreme values of the possible range.
| x is positive. | |
| Quantity A | Quantity B |
| x(x + 1) | x(x2 + 1) |
Begin by distributing both quantities:
| x is positive. | |
| Quantity A | Quantity B |
| x(x + 1) = x2 + x | x(x2 + 1) = x3 + x |
Both sides have an x, which you can cancel out.
| x is positive. | |
| Quantity A | Quantity B |
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You know x is positive, so x can’t be negative or 0. If x = 2, then Quantity A = 4 and Quantity B = 8. 
To be thorough, however, make sure that you try numbers that have a chance of behaving differently. You can’t try negatives, but you can try 1 and fractions between 0 and 1. If x = 1, then Quantity A = 1 and Quantity B = 1. Also, if x were a positive fraction (e.g., 1/2), then Quantity A would be greater than Quantity B. The correct answer is (D).