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Ace Your SAT Math Test with This Strategy

Updated on November 29, 2014

If I had to select the one strategy that could improve your SAT or ACT score the most, I know just what it would be: applying actual numbers. Before I tell you how this strategy works, let me tell you a few things about what it does and does not do:

Applying actual numbers:

  • Does not use the regular algebra and geometry methods your math teachers taught you
  • Does give you an alternative when the regular methods escape you
  • Does not avoid math concepts all together
  • Does get you away from variables and back to basic arithmetic
  • Does not protect you from every silly math error you could make (though it does eliminate some)
  • Does give you a way to check your answers

Sound good? Then let me demonstrate how it works. Here is a SAT practice question from the SAT website:

SAT Math Problem

Source

Your math teacher might expect you to explain your answer, build a factor tree, discuss prime factors, and so on—those regular things you learn in math class. But your math teacher isn’t grading your SAT or ACT. In fact, on a standardized test, your job is not to demonstrate your math mastery; it’s to get the question right! And there is always more than one way to get the right answer to a multiple-choice math problem. Always!

I’m betting that anyone reading this hub who is good at math has already selected an answer to this problem. And you may have done so by thinking the problem through abstractly; that is, you worked with the variable k while you solved it. If you can do that—great! Keep that skill in your arsenal for Test Day! But you can still benefit from applying actual numbers. Sit tight and I’ll return to your situation in a moment.

Many others of you looked at that question and felt that familiar pit in your stomach. “I have to work out problems like this?” you might be thinking, “And getting into college depends on it?” Let’s relieve that awful feeling as quickly as we can! Here is how you’ll apply numbers to make this a simple arithmetic problem.

The first step is to come up with a value for k that works. That is, I’m going to pick a number that follows the rules in this problem – it has to be divisible by 2, 3 and 15. So I’m not picking 10 since neither 3 nor 15 divides evenly into 10. One easy way to find a number divisible by all three is simply to multiply them together: 2 x 3 x 15 = 90. So we’re going to work with k = 90. Did I have to pick 90? Nope. There are lots of other numbers that are divisible by 2, 3 and 15. So to be clear: I can pick any number I want, as long as it works with the rules laid out in the question.

Once I’ve picked a number for the variable, I move on to the second step, which is to apply that number to each of the answer choices:

(A) 90 + 5 = 95

(B) 90 + 15 = 105

(C) 90 + 20 = 110

(D) 90 + 30 = 120

(E) 90 + 45 = 135

Now I just select the answer choice that answers the question correctly. And I can do so without punching a bunch of buttons on my calculator. (Calculating only when you have to is another general test-taking principle that I’ll address in a future hub.) I can cross off (A), (B) and (E) immediately, since odd numbers are not divisible by 2. That leaves (C) and (D), and (C) is not divisible by 3. Some of you would punch a few buttons on your calculator to eliminate (C). That’s fine! And you might want to divide 120 by 3 on the calculator just to be sure that (D) works. Bottom line: the only answer choice that answers the question correctly is (D), so I’m extremely confident that (D) is the right answer.


Applying Actual Numbers

Pick numbers for variables

Apply these numbers to the answer choices

Select the answer choice that correctly answers the question

A few more things, then we’ll wrap up this hub, and in future hubs demonstrate some variations on this strategy. First, for those of you who already knew that the answer was (D) before we went through this process: I’m sure it took you a lot less time to select (D) than it took me to apply actual numbers before selecting (D). That’s why I recommend that you keep the regular algebra technique in your arsenal. It is often fastest to use the regular technique on SAT or ACT math problems.

So why bother applying actual numbers? Two reasons: (1) You may encounter other problems for which you can’t remember the regular technique or how to use it, which makes applying actual numbers a great alternative that allows you to still get the problem right. And, (2) when you do use the regular technique, you can apply actual numbers to ensure that you haven’t let a careless error cost you. In this case, you can quickly apply numbers only to the answer choice you’ve selected, rather than going through the full strategy.

Finally, for all of you, but especially for those of you who had that pit in your stomach: Applying actual numbers provides not only a great performance boost, but also a great psychological boost. Imagine on Test Day, instead of having your heart sink a little lower with each tough math problem, you can instead ask yourself, “Could I apply actual numbers to solve this problem?” Since the answer to this question is frequently “Yes,” you’ll feel better and better about your math performance as the test progresses. So start developing this habit today.

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    • Karen Lanigan profile image
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      Karen Lanigan 3 years ago from Belmont, NC

      Yes, the regular approach taught in math classes is usually the faster method. Those who remember the concept of lowest common multiples would therefore apply an actual number only to check their work. But those who don't remember LCMs, or were never comfortable working with this concept, can nevertheless get the right answer to this problem. Good news for all, I hope.

    • rajeshcpandey profile image

      RAJESH CHANDRA PANDEY 3 years ago from India

      @Karen Lanigan

      The LCM of 2,3,15 is 30. So the minimum number greater than k and divisible by 2,3,15 will be k+30. The next bigger number in the given choices is k+45 which is not a multiple of the LCM. The answer hence is the choice 'D'. I think this is a short method.

      But yeah I agree that thinking as you did helps in some problems

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