Disclaimer: this is really minor stuff as far as how often it appears on the SAT, so if you’re looking for quick tips to really raise your score, I suggest you start elsewhere. This kind of question is pretty rare.
I trust you already know the very basics of absolute value: that |5| = 5, and |-5| = 5, etc. If you don’t, leave this page open and view a quick tutorial here before continuing. Ok, all caught up? Let’s do this.
Absolute values and inequalities
Remember that |x| = 5 means that x = 5, or x = -5. You can draw similarly simple conclusions with inequalities. If I told you that |y| < 3, and y is an integer, then what are the possible values for y? There aren’t many: y could equal 2, 1, 0, -1, or -2. In other words, y has to be less than 3, and greater than -3. Like so:
|x| < 5
x < 5 and x > -5
-5 < x < 5
Here’s an example of the kind of question you’ll usually get on the SAT:
- In order to be considered “good for eating” by the La’Urthg Orcs of Kranranul, a human must weigh between 143 and 181 pounds. Which of the following inequalities gives all the possible weights, w, that a human in Kranranul should NOT want to be?
(A) |w – 143| < 38
(B) |w – 162| < 19
(C) |w + 38| < 181
(D) |w – 181| < 22
(E) |w – 162| < 24
Answer, explanation, and so much more below the cut:
First of all, it’s possible to plug in here, although it matters greatly what numbers you choose. Picking a number right in the middle of the range (like 165) probably won’t help you out much. To plug in successfully, choose weights that should just barely be within the range (like 144 or 180) and then, if you still haven’t eliminated all the answers, choose weights that should just barely be outside the range (like 142 or 182). Be careful with this second part! You’re looking to eliminate any choices that WORK when you know you picked a number that SHOULDN’T WORK. Let’s see what happens when we pick w = 144.
A person who weighs 144 pounds should try to lose weight, pronto, lest he/she become a meal for the orcs. So we’re looking for choices that will give us TRUE inequalities when we plug in 144:
(A) |144 – 143| < 38
|1| < 38
1 < 38 — TRUE
(B) |144 – 162| < 19
|-18| < 19
18 < 19 — TRUE
(C) |144 + 38| < 181
|182| < 181
182 < 181 — FALSE
(D) |144 – 181| < 22
|-37| < 22
37 < 22 — FALSE
(E) |144 – 162| < 24
|-18| < 24
18 < 24 — TRUE
So, bummer. our first plug-in only eliminated 2/5 choices. What happens when we plug in a number that shouldn’t work, like 141? Remember, now we’re looking to eliminate anything that gives us TRUE, because a 141 pound person is NOT considered delectable by the orcs. Note that I’m not bothering with (C) and (D) since we’ve already eliminated them.
(A) |141 – 143| < 38
|-2| < 38
2 < 38 — TRUE
(B) |141 – 162| < 19
|-21| < 19
21 < 19 — FALSE
(E) |141 – 162| < 24
|-21| < 24
21 < 24 — TRUE
Only (B) was true when we needed it to be true, and false when we needed it to be false, so that’s our answer. But if plugging in here seems cumbersome to you, you’re not alone. I actually prefer to do this question another way.
Let’s have a look at (B), our correct answer, and convert it like we did at the beginning of this post:
|w – 162| < 19
-19 < w – 162 < 19
Here’s where things get awesome. Add 162 to all 3 sections of the inequality now:
-19 + 162 < w – 162 + 162 < 19 +162
143 < w < 181
Wow, that’s exactly what we were looking for. I mean, I knew that was coming, and I’m still amazed. I can’t even imagine how you must feel.
So on a question like this, you can just convert every answer choice in this way to see which one gives you what you want, or you can even take it one step further: in the correct answer, the part inside the absolute value brackets will be the middle of the range, and the bit on the other side of the less-than sign will be the distance from the middle to the ends of the range, or half the range.
|w – 162| < 19
|variable – middle of range| < half of range
Not bad, right? This question, as I said above, is pretty rare, but if it appears on your test you know exactly what to do. GET SOME.
Absolute values and functions
Of course, if you have absolute value brackets around an expression, you don’t perform the absolute value operation until you’ve completed all the operations inside. For example:
GOOD: |-8 + 5| = |-3| = 3
BAD: |-8 + 5| = 8 + 5 = 13
Seriously, don’t do that second one. Don’t.
The same is true if you have absolute value brackets around a function, like |f(x)|. The absolute value brackets don’t take effect until the function has done its thing inside. If the function comes out positive on its own, the brackets have no effect. If the function comes out negative, it becomes positive:
See? Now here’s the important part: what happens to the graph of a function when you take the absolute value of the function? Well, when it’s positive, nothing. When it’s negative, it reflects off the x-axis. In other words, it bounces. BOING! Observe:
ZOMG I can’t wait to practice!
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For Q16, D and E have the same answers.
Nope. One says 223 and one says 233. Two-TWENTY-three and two-THIRTY-three. They do look similar, though…you gave me a scare for a minute there. 🙂
Question 18 is a trick of reading comprehension. It asks for the Weight a human shouldn’t want to be (that is, the Weight of the ‘can-be-eaten’ human). How come you come up with that interesting question? I first thought it asks for the range the Beast wouldn’t want from a person and chose B, which is definitely wrong. Does this kind of trick appear in SAT?
No, the SAT would not have fun with a question the way I do here. I do this to keep people interested. This kind of question does appear on the test, but never with man-eating monsters. 🙂
Can you please explain #15 ?
f(1) = 2(1) – 10 = –8. But |f(1)| = |2(1) – 10| = |–8| = 8.
Since f(1) = –8 and |f(1)| = 8, f(1) ≠ |f(1)|.
All the other choices work.
Q20 Drill #2 !!! Please explain!
You should be able to take the formula in this post (|variable – middle of range| < half of range) and nail that one. Plug-in also works, if you're careful.
I’ll try plugging in… where is that formula from?
That formula is from this post. Just scroll up! 🙂
Oh. Well that’s embarrassing.
I solved Q15 by plugging in! Is there any other way to solve it?
You pretty much have to plug in those function arguments. f(x) < |f(x)| for some values of x, but not for x = 3. 🙂
Aw, well doesn’t hurt to ask eh?
Tried finding it somewhere else and it looks original, kudos for the help.