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More Help To Ace Your SAT Math Test, Tip #6

Updated on November 29, 2014

What to do, what to do, when one of those crazy math questions shows up your exam? You know the kind. Either you look at it and think, “This must be the kind of math I never got”; or you look at it and think, “What in the world are they talking about?” For example:

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This SAT question isn’t even kind enough to give you some answer choices; you have to come up with an answer yourself. Those of you who are comfortable with absolute values and inequalities, congratulations! The rest of us – in the past – might have thrown our hands up and moved on to the next question, or worse, spiraled into a flare of negative thoughts that compromise our ability to think about this or any more SAT questions! But no more! At least, not for anyone who’s been practicing and preparing to use the strategy of applying actual numbers.

A Puzzling Condition

The confusing part of many such questions is found in the condition, the statement following the word “if.” Number property problems sometimes evoke the same kind of response to their “if” statements. Not surprisingly, the same “trial and error” approach works pretty well with both number property problems and these other puzzling problems.

In the problem above, as long as I know what absolute value is, I can start picking negative numbers for x (since x < 0) until I find one that works, i.e. one for which the expression falls between 6 and 7:

-1 – 3 = -4 (and the absolute value of -4 is 4, which is not between 6 and 7)
-2 – 3 = -5
-3 – 3 = -6
-4 – 3 = -7

Since -3 and -4 fall on either side of range, any number between -3 and -4 satisfies the “if” statement:

-3.5 – 3 = -6.5 (and the absolute value of -6.5 is 6.5, which is between 6 and 7)

This means that -3.5 is one possible value of x, and therefore 3.5 is one possible absolute value of x. So the answer I would put into the gridded response on my answer sheet is 3.5.

More Puzzlements

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Here is a problem that might either stump some test takers or cause them to spend an inordinate amount of time trying to get their programmable calculator to figure it out. Instead, I’ll do more trial and error, starting with p = 2 and n = 1, and working up the scale:

p2n2 =

4 – 2 = 3
9 – 4 = 5
16 – 9 = 7
25 – 16 = 9
36 – 25 = 11
49 – 36 = 13

Since I just passed 12, I know that p and n can’t be consecutive; p2n2 will just keep getting bigger (did you notice the pattern?). This means that pn cannot equal 1. I can now cross off answer choices (A), (C) and (E), because they all include Roman numeral I.

Here’s a cool thing: The remaining answer choices both include Roman numeral II, so I don’t even have to think about it! All that is left is to figure out is whether p n could be 4. So I’ll start with p = 5 and n = 1, and work my way up:

25 – 1 = 24
36 – 4 = 32
49 – 9 = 40

That’s far enough to satisfy me; the differences are getting bigger and are never as small as 12. So I’ll select answer choice (B).

Let’s do one more of these crazy problems and call it a day. How about this monstrosity?

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My test prep students have heard me say over and over, “Exponent rules don’t apply when there’s a plus or minus sign between terms.” Believe me, plenty of test takers incorrectly apply an exponent rule, and since 3yy = 2y, they’ll pick answer choice (A). Wrong!

Knowing that I can’t apply exponent rules doesn’t tell me what I CAN apply. So I’ll get myself out of this pickle by applying actual numbers. Since y is the exponent, I’ll make y small. So I’m going with y = 2 and x = 3. This is not one of my trial and error problems. Like problems involving equations, I’m coming up with an answer to the question using the values I’ve picked, then applying those values to the answer choices and seeing which one gives me the same answer:

Now I work out the answer choices with my calculator – taking care to follow the order of operations but otherwise just doing straightforward arithmetic – and I find which answer choice comes out to 46,620. Oh, and by the way, since tests like the SAT generally (though admittedly not always) list their answer choices from least to greatest, I’ll start by calculating (C), and move up or down the list depending on how big (C) is.

Lo and behold, (C) comes out to 46,620. Had it been too small, I would have tried (D) next; and had it been too big, I would have tried (B) next.

Apply, Apply, Apply!

Let me finish with a reminder. I trust you were able to follow what I did with the problems I’ve presented in this series of hubs. But following someone else’s efforts is only the first step in making this strategy work for you. Grab some SAT math sections, look for problems to which you can apply actual numbers, and practice, practice, practice!

And along the way, let me know how you’re doing!

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    • Karen Lanigan profile image
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      Karen Lanigan 2 years ago from Belmont, NC

      Yes, Rajesh, a good point. In fact, finding a common factor is often the key to solving exponent problems in which the terms are being added or subtracted. I consider finding a common factor to be a multiplication rule, so that's why I say that exponent rules can't solve this problem -- at least, not exclusively.

    • profile image

      RAJESH CHANDRA PANDEY 2 years ago from India

      @Karen Lenigan

      The answer 3 would have been easier had we applied the exponent rules. The first term in the expression could be written as 2x raised to the power y and whole of this raised by cube power. In the next step we could take 2x raised to power y as a common factor and could have reached the answer easily.

      However I understand that your method is more general and would apply everywhere. Thanks a lot.