| 0 < p < q < r | ||
| Quantity A | Quantity B | |
|
(?) |
|
Inequalities are a common theme in Quantitative Comparisons, and can take many forms. As was noted earlier in this section, one thing inequalities can do is restrict the range of a variable. Another way they are used is in combination with absolute values. For example:
| −2 ≤ x ≤ 3 −3 ≤ y ≤ 2 |
|
| Quantity A | Quantity B |
| The maximum value of | x − 4| | The maximum value of | y + 4| |
Once again, inequalities are used to bound a variable. As before, you should test the boundaries of the range. But now, there’s the added twist of absolute values. On QC, it is important to understand how to maximize and minimize values. The smallest possible value of any absolute value will be 0. It is impossible for an absolute value to have a value less than 0.
This question asks you to maximize the absolute values in Quantities A and B. In Quantity B, the maximum value of | y + 4| will be when y = 2, because that is the largest number you can add to positive 4. The absolute value of | y + 4| will equal 6.
To maximize the absolute value of |x − 4| in Quantity A, however, you have to do the opposite. There is a negative 4 already in the absolute value. If you try to increase the value by adding a positive number to −4, you will only make the absolute value smaller. For instance, if x is 3, then the absolute value is:
|3 − 4| = |−1| = 1
You can actually maximize the absolute value by making x = −2. Then the absolute value becomes:
|−2 − 4| = |−6| = 6
The maximum value in each quantity is the same (6), therefore, the answer is (C).
The GRE often uses inequalities to show much more than the range of possible values for a variable. For instance, the common information may tell you that 0 < p < q < r.
This inequality tells you two crucial things: 1) p, q, and r are all positive, and 2) p, q, and r are in order from least to greatest.
Questions that provide this type of information will often use different combinations of these variables in each quantity and perform some kind of mathematical operation on them (e.g., +, −, ×, ÷). You now have to look for the pattern.
If there is a pattern, the answer will be (A), (B), or (C). If there is no pattern, the answer will be (D). Make use of the Invisible Inequality to discern the pattern, if one is present. Take a look at four basic examples, one for each of the four basic mathematical operations (+, −, ×, ÷).
| 0 < p < q < r | |
| Quantity A | Quantity B |
| p + q | q + r |
Pretend there is an unknown inequality between the two quantities, designated by a (?).
| 0 < p < q < r | ||
| p + q | (?) | q + r |
Both sides contain a q, so subtract the q:
| 0 < p < q < r | ||
| Quantity A | Quantity B | |
|
|
(?) |
|
From the common information (0 < p < q < r), you know that r is bigger than p, so Quantity B is definitely bigger.
| 0 < p < q < r | ||
| Quantity A | Quantity B | |
| pq | (?) | qr |
Once again, both sides have a q. Because you know that q is positive, you can divide both sides by q without changing the Invisible Inequality:
| 0 < p < q < r | ||
| Quantity A | Quantity B | |
|
(?) |
|
Once again, from the common information, you know that r is definitely greater than p. The correct answer is again (B).
In both of the last two examples, you were able to successfully eliminate common terms to arrive at a definite conclusion.
However, take a look at the next example.
| 0 < p < q < r | ||
| Quantity A | Quantity B | |
| q − p | (?) | r − q |
Both sides contain a q, but notice that their signs are different. You can’t actually eliminate q altogether. If you try adding q to both sides, here’s what you get:
| 0 < p < q < r | ||
| Quantity A | Quantity B | |
|
(?) |
|
Quantity A still contains q. Likewise, if you try subtracting q from both sides, you just push q into Quantity B:
| 0 < p < q < r | ||
| Quantity A | Quantity B | |
|
(?) |
|
Either way, you cannot arrive at a definite conclusion.
You can also pick numbers to show that there is no pattern. Remember to pick numbers satisfying 0 < p < q < r. If p = 1, q = 3, and r = 6, then:
| 0 < p < q < r | ||
| Quantity A | Quantity B | |
| q − p = 3 − 1 = 2 | (?) | r − q = 6 − 3 = 3 |
With these numbers, Quantity B is bigger. 
Now space the numbers differently. To visualize the possibilities, imagine placing the variables p, q, and r to the right of 0 on a number line, in that order from left to right, like beads on a wire. Now slide the beads, keeping their order but changing their spacing. In the previous case, q was closer to p than to r. Try putting q closer to r than to p.
If p = 2, q = 7, and r = 8, then:
| 0 < p < q < r | ||
| Quantity A | Quantity B | |
| q − p = 7 − 2 = 5 | (?) | r − q = 8 − 7 = 1 |
Now Quantity A is bigger. The correct answer is (D).
There’s a similar dilemma with this fourth example.
| 0 < p < q < r | ||
| Quantity A | Quantity B | |
|
(?) |
|
Because all the variables are positive, you can cross-multiply:
| 0 < p < q < r | ||
| Quantity A | Quantity B | |
| q2 | (?) | pr |
It’s impossible to know for sure which quantity will be bigger. For extra practice, use numbers satisfying 0 < p < q < r to prove the answer is (D). (Hint: Space the numbers differently, as in the previous example.)