Picking Numbers In Percent Problems

Yesterday we looked at the basic but important strategy of picking numbers. Click the following link to view that post: Picking Numbers

At the end of that post, I did say that we would be applying this strategy to percent problems.

When picking a number in percent problems, the best choice is usually the number 100. After all, the word percent literally translates to “out of 100.”

Example 1:

Let’s take a look at an SAT problem involving percents:

There are b bricks that need to be stacked. After k of them have been stacked, then in terms of b and k, what percent of the bricks have not yet been stacked?

Since this is a percent problem let’s choose 100 for the total number of bricks. So we have b = 100. For k, let’s choose 25, so that 25 bricks have been stacked, and so that means 100 – 25 = 75 have not been stacked. Since we started with 100 as our total, 75% of the bricks have not been stacked. Remember to put a big, dark circle around 75%. We make the substitutions b = 100 and k = 25 into each answer choice.

            (A)   100/7500 ~ 0.0133% (~ means “is approximately”)
            (B)   7500/100 = 75%
            (C)   10,000/25 = 400%
            (D)   2500/100 = 25%

We now compare each of these percents to the percent that we put a nice big, dark circle around. Since A, C and D are incorrect we can eliminate them. Therefore the answer is choice B.

As I always like to stress: B is not the correct answer simply because it is equal to 75%. It is correct because all of the other choices are not 75%. You absolutely must check every choice!

Example 2:

Now let’s look at a percent problem without answer choices. Even though there is no variable in the problem to pick a number for, we can still use the number 100 to solve the problem quickly and easily.

If Matt’s weight increased by 30 percent and Lisa’s weight decreased by 20 percent during a certain year, the ratio of Matt’s weight to Lisa’s weight at the end of the year was how many times the ratio at the beginning of the year?

Again, 100 is the magic number – let’s choose 100 pounds for both Matt’s weight and Lisa’s weight at the beginning of the year. Matt’s weight at the end of the year was then 100 + 30 = 130 pounds and Lisa’s weight at the end of the year was 100 – 20 = 80 pounds. We then have that the ratio of Matt’s weight to Lisa’s weight at the beginning of the year was 100/100 = 1, and the ratio of Matt’s weight to Lisa’s weight at the end of the year was 130/80 = 13/8. We can therefore grid in 13/8.

Side note: 13/8 is equal to 1.625 as a decimal. Thus, we can also grid in 1.62 or 1.63. We get 1.62 by truncating the decimal, and 1.63 by rounding the decimal. Truncating is better because less thought is involved. Note that if you grid in 1.6 the answer will be marked wrong.

So remember the magic number ‘100’ whenever you see a problem with the word “percent” in it… but as always, practice makes perfect – so find more of these problems, such as those found in my 28 SAT Math Lessons Series. Click on the picture below for more information about these books.

Speak to you soon!

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