It turns out that on some of the problems in the SAT math sections, you don’t really need to do the math (at least not the math they want you to do). Crazy, right? One of the most important strategies that you should master if you want to score higher on SAT math is “picking numbers.” If a student were to come to me a week before the test and say “I know it’s late, but I’m desperate. Show me something!” this technique is what we’d work on.
Consider this example:
If a is an odd integer and b is an even integer, which of the following must be odd?
(B) ab + b
(C) (a + b)2
(D) ab + 2
(E) ab – 2
The person who wrote this question would love for you to start considering the properties of even and odd numbers (“Lets see …. if I multiply an even times an odd, that must be even.”). Instead let’s just choose numbers for a and b. I’ll let a = 3 and b = 4. Now plug these into the answer choices:
(A) 3(4)2 = 48
(B) 3(4) + 4 = 16
(C) (3 + 4)2 = 49
(D) 3(4) + 2 = 14
(E) 3(4) – 2 = 10
Of these, the only answer that turns out odd is Choice (C), so that’s the right answer.
Here’s another one. This was the Official SAT Question of the DayTM for November 6th.
The population of Norson, the largest city in Transitania, is 50 percent of the rest of the population of Transitania. The population of Norson is what percent of the entire population of Transitania?
(A) 20 %
(B) 25 %
(C) 30 %
(D) 33 1/3 %
(E) 50 %
(Source: College Board SAT Question of the Day, 11/6/13)
Of the more than 105,000 students who have tried this question, only 34 % have gotten it correct. So it must be difficult, right? Let’s try picking a number for the population of Norson. Let’s say that population is 100 people. Since that is 50%, or half, of the rest of the population, we know the rest of the population is 200 people. Here’s what we have:
Population of Norson: 100
Rest of Population: 200
Total Population: 300
Percent of Population that Norson is: 100/300 = .333333…. = 33 1/3%
The correct answer to this problem is, therefore, Choice (D).
Let’s look at one more. This problem would be considered “hard” and would be one of the last few problems in a section.
After the first term, each term in a sequence is found by adding 5 to 3/4 of the previous term. If the first term of the sequence is n and n ≠ 0, what is the ratio of the second term to the first term?
(A) (n + 5)/4
(B) (n + 20)/4
(C) (3n + 5)/4n
(D) (3n + 20)/4n
(E) (5n + 3)/4n
Let’s start by picking a first term, n. Since we know that we’ll have to take 3/4 of this number, it makes sense to choose a multiple of 4, so how about we try n = 8. The rest of our work would look something like this:
First term: 8
Second term: 5 + (3/4)(8) = 11
Ratio of second term to first term: 11/8
Now we’ll plug our value for n into each of the answer choices until one of them is equal to 11/8.
(A) (8 + 5)/4 = 13/4
(B) (8 + 20)/4 = 28/4 = 7
(C) (3·8 + 5)/4·8 = 29/32
(D) (3·8 + 20)/4·8 = 44/32 = 11/8 ← Eureka!
(E) (5·8 + 3)/4·8 = 43/32
Choice (D) works out to be 11/8 and that’s the correct answer!
Some Other Important Points:
- If you see variables in the answer choices, there’s a really good chance that problem will be a good one to try picking numbers.
- Percent increase/decrease problems also lend themselves well to picking numbers. You’ve seen them before. They start something like “The width of a rectangle is increased by 30 % …..” For these problems a good starting value is 100 since “percent” literally means “per hundred.”
- Avoid choosing 0 or 1 as these two numbers have special properties. For instance, multiplying a number by 1 and dividing that same number by 1 leaves you with the exact same value.
- Make sure you test all of the answer choices. It doesn’t happen often but there’s a chance that the numbers you choose will lead more than one of the answer choices to equal the value you’re looking for. When that happens, choose a new number to break the “tie.”
To try a few more of these problems, go here.
Questions about this strategy or anything else related to the SAT? Leave a comment or send me an email at email@example.com