Tip #38 to Get a Top SAT Math Score

The expression |–3| means "the absolute value of –3." The | | bars mean absolute value. Like many SAT topics, this is really just vocab. If you don't know the meaning of the term, then it's very hard or impossible to get the question correct; but if you know the vocab, it's EASY!

Basically, absolute value just means, "Ditch the negative sign!" |–3| = 3 and |3| = 3. So it does not mean to switch the sign necessarily; it just means drop the (–) negative signs, which makes it positive. You do this only after you've done what is between the bars. For example, |–3 – 6| = |–9| = 9. You don't just drop negative signs right away. First you do the math between the bars, and when you've done all the math that you can, then you drop the negative sign.

Here's the key to absolute value on the SAT: usually you have to come up with the less obvious answer to the question. For example, of course 3 is one answer to |x – 1| = 2, but –1 is also an answer, and that's the one you'll need on the SAT. How do you find the less obvious answer? Drop the bars and switch the sign of the answer: x – 1 = –2; then solve for x.

That's it. That's absolute value. If you did not know this and you do by the end of the drills, then you will gain points!

Let's look at this question:

Solution: Great "Use the Answers" review. Try each answer choice in the absolute value equation in the question:

I. |2 + 3| = 5 correct II. |–2 + 3| ≠ 5 incorrect III. |–8 + 3 | = 5 correct

So I and III are correct and choice D is the answer.

Correct answer: D

Example Problems

Easy

If |6 – m| – 12, then m =
1. 6 or 0
2. 12 or –6
3. 18 or –6
4. 18 or 6
5. 0 or –6
Which of the following are solutions to |p – 6| = 5 ?
1. 11
2. 5
3. 1
1. I only
2. III only
3. II and III
4. I and III
5. I, II, and III

Medium

For negative integers m and n, if |m| + |n| = 8, what is the least possible value for m – n ?
1. 8
2. 0
3. –4
4. –6
5. –16
If in the equations above x < 0 and y < 0, what is the value of xy ?
1. –32
2. –24
3. 6
4. 12
5. 24
If |1 + 2a| < 1, what is one possible value of a ?

Hard

If m = |n|, then m could equal
1. 0
2. n
3. –n
1. I only
2. III only
3. I and II
4. II and III
5. I, II, and III

Answers

C Great "Use the Answers" review! Try each answer choice in the equation. Choice C is correct, since |6 –18| = |–12| = 12 and |6 –(–6)| = |12| = 12.
D Again, "Use the Answers." Try each answer choice in the question.
1. |2 + 3| = 5 correct
2. |–2 + 3| ≠ 5 incorrect
3. |–8 + 3| = 5 correct

So I and III are correct, and choice D is the answer.

D Remember to underline the vocab term "negative integers" so you don't forget about it. We must consider the possible values for m and n:

|–7| + |–1| = 8, m =–7, n =–1

|–6| + |–2| = 8, m =–6, n =–2

|–5| + |–3| = 8, m =–5, n =–3

|–4| + |–4| = 8, m =–4, n =–4

|–3| + |–5| = 8, m =–3, n =–5

|–2| + |–6| = 8, m =–2, n =–6

|–1| + |–7| = 8, m =–1, n =–7

So try each possible pair of m and n to see which gives the least value for m – n. –7 – (–1) =–6 and is the least value.

E Since x < 0, x must equal –12 since |4 + (– 12)| = |–8| = 8. And in the equation |3 – y| = 5, y equals –2 since |3 –(–2)| = 5. So xy = (–12)(–2) = 24.
–1 < a < 0. Start by just trying some numbers and see if you can get one that works. Any positive numbers yield an answer that is too large, and 0 is still too big since |1 + 2(1)| = 1. So try –1. |1 + 2(–1)| = |–1| = 1, so that doesn't work. What if a =–0.52. That works, so –0.5 is an answer; in fact any number between 0 and –1 works.
E Let's look at each choice. Could m equal
1. 0 Definitely, 0 = |0|.
2. n Sure, n = |n| could work if n > 0.
3. – n Sure, – n = |n|, if n < 0. Nice!

So all three work.

Go to: Tip #39

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