OK, so you know how I’m always saying that the SAT is not a math test? This is one of the primary reasons why. On the SAT, it’s often completely unnecessary to do the math that’s been so carefully laid out before you. A lot of the time (and on a lot of the most otherwise onerous problems), all you need to do is make up numbers.

Sounds crazy, right? Well it’s not. It would be crazy to just make up numbers on just about any other pain-in-the-ass task (for instance, it would be bad just to make up numbers on your taxes), but you’ll be dumbstruck by how often it works on the SAT. Of course, you have to practice doing it to get good at it, so that when an opportunity to do it on the real test pops up, you don’t panic and blow it. That’s what your old buddy Mike is here for.

I’m thinking we should start with a more obvious plug-in. If you would consider trying to solve this one with pure algebra, you’re probably out of your mind. Still, it’s a great illustration of the technique:

  1. If m and n are divided by 8, the remainders are 3 and 5, respectively. What is the remainder when mn is divided by 8?
     
    (A) 0
    (B) 1
    (C) 3
    (D) 5
    (E) 7

What we want to do with a question like this is plug in values for m and n so that we’re not dealing with abstracts. Of course, there are infinite possibilities for the values of both m and n, but we’re just going to pick values and stick with them.

Since the problem stipulates that m divided by 8 gives me a remainder of 3, and n divided by 8 gives me a remainder of 5, let’s pick m = 11 and n = 13 (because 8 + 3 = 11, and 8 + 5 = 13). That will keep our numbers nice and low, and make the division we’ll have to do in the next step less arduous.

Since I’ve plugged in 11 for m and 13 for n, I need to find the remainder when 11×13=143 is divided by 8. Remember long division? That’s going to be the easiest way to calculate a remainder, so let’s do it:

Bam. Remainder 7. That’s choice (E). I feel so alive right now.

Note that if we picked different numbers for m and n (like, say, 83 and 85), we’d still get the same answer (try it yourself to see). That’s the beautiful thing about plugging in!

Let’s do another, slightly tougher one:

  1. If the inequality above is true for integer constant k > 1, which of the following could be a value of x?
     
    (A) k – 3
    (B) k – 1
    (C) k
    (D) 3k – 4
    (E) 3k – 2

OK. Forget for a minute that this can be solved with algebra and think about how to solve it by plugging in. Remember, if you don’t practice plugging in on problems you know how to do otherwise, you won’t be able to plug in well when you come to a problem you don’t know how to solve otherwise!

We know k is a positive integer greater than 1, so let’s say it’s 2. If k = 2, then we can do a little manipulation to see that x has to be greater than or equal to 3:



Note that we don’t just make up a number for x! Once we’ve chosen a value for k, we’ve constrained the universe of possible x‘s. When we have an equation (or an inequality), we usually can’t plug in values for both sides; we have to choose one side on which to plug in, and then see what effects our choices have on the other side.

So, which answer choice, given our plugged in value of k = 2, gives us a number greater than or equal to 3 for x?

(A) k – 3 = 2 – 3 = -1 (too low!)
(B) k – 1 = 2 – 1 = 1 (nuh-uh)
(C) k = 2 (nope)
(D) 3k – 4 = 6 – 4 = 2 (still no good)
(E) 3k – 2 = 6 – 2 = 4 (yes!)

Rock. On. Note again that if we had picked a different number for k, we still would have been OK. Try running through this with k = 10 to see for yourself.

So…when do you plug in?
  • When you see variables in the question and the answer, you might want to try plugging in.
  • On percent questions, you’ll probably benefit from plugging in (and using 100 as your starting value).
  • On triangle questions where no angles are given, you might try plugging in 60 for all angles.
    • If you’re plugging in on a geometry question, just make sure that all the angles in your triangles and straight lines add up to 180°.
  • Anytime you don’t know something that you think it would be helpful to know, try making it up!
Anything else I should know?
  • As a general rule, DO NOT plug in 0 because when you multiply things by 0, you always get 0, and when you add 0 to anything, it stays the same. Usually, that will make too many answer choices work.
  • Similarly, DO NOT plug in 1, since when you multiply things by 1, they don’t change. This will also often make more than one choice seem correct.
  • DO try to keep your numbers small. There’s no need to plug in 2545 when 2 will do.
  • DO think for a minute before picking your numbers. Will the numbers you’re choosing result in messy fractions or negative numbers? We plug in to make our lives easier, so practice avoiding these scenarios!
  • You always have to check every single choice when there are variables in the answers and you plugged in, because there’s a small chance that more than one answer will work. If that happens, don’t panic…just try new numbers. You can greatly mitigate this by following rules #1 and 2 above.
Let’s try some more problems!

Note: all of these problems can be solved without plugging in, but you’re not here to do that right now, you’re here to practice plugging in. Don’t be intractable in your methods. If you’re amenable to change, it’s more feasible that you’ll improve your scores.

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Want to see more plugging in? Browse the “plug in” tag on my Tumblr for some recently posted examples!

Comments (40)

These questions are based on things I’ve seen on the SAT in the past. For #20, just pick one variable to get started, plug in, and carefully see where it leads you. Try plugging in for e, since that has to be greater than 2. See what happens when you make it 3. 🙂

wouldnt A work as well? i plugged in 3 for e and 8 for d. i got 9 on both A and E.

i tried plugging in 5 for e and 10 for d and STILL got 13 on both letter A and E.

This one is tough, no? Obviously, because of the name of this post, I think the best solution is to plug in.

Say e = 5, so there are 5 employees. That means there are 10 desks, since all but 5 desks are occupied — d = 10. And, since all but 2 employees have 2 chairs at their desks, that means 3 of our employees have 2 chairs at their desks. In other words, there is a chair at each of 10 desks, plus 3 more for the 3 employees who have 2 chairs. So c = 13. See which answer choice gives you 13 when d = 10 and e = 5 and you’re good to go!

(E) is the only one that works.

Thanks for the comment!

Yes, 11 – 8 = 3 and 13 – 8 = 5. That’s true. But what that means is that 11 is 3 more than 8 times 1. So if you divide 11 by 8, you get 1 remainder 3. Same deal for 13. Divide it by 8, and you get 1 remainder 5. You can pick larger numbers if you want, but that’s a shortcut I like to use.

For #14, make x = 2. 2^3 = 8, so y = 8. 2^6 = 64. 64 is how much greater than 8? 56 greater. Which answer choice gives you 56 when y = 8?

For #16, plug in 60 for each angle inside the triangle. Why can you do that? Well, when you’re given a triangle with NOTHING labeled in it, you can assume that whatever they’re testing you on must be true for ALL triangles–so make it easy on yourself. Each marked angle ends up being 300 (120 + 60 + 120), so the sum of the measures of the marked angles is 900.

[Link to drill]

Sure! This question is WAY easier if you plug in values for the sides of the original rectangle, and even more so if you’re clever about which numbers you plug in. Since a square is a special case of a rectangle, I recommend plugging in 10 for the length, and 10 for the width. The area of the original rectangle, then, is 100.

Your new rectangle, with adjusted sides, will then be 9 by 11, with an area of 99. That’s 99% of the area of the original rectangle.

How would you solve this question?
1. If sally drives m miles from her house to her office in f hours, and drives back to her house in g hours, what is her average speed of the entire trip, in miles per hour?
a) (f+g)/2
b)m/(f+g)
c)2m/(f+g)
d)fg/(f+g)
e)2fg/ (f+g)

I’d definitely plug in here, but before I get into it let me point out for everyone else that you made a typo in answer choice (C). That 9 shouldn’t be there…it probably came from a quick double tap when you were opening the parentheses. ::Spoiler alert:: This is pretty important, because that turns out to be the right answer.

Let’s say m = 60, f = 1, and g = 2. That means that her total travel time was 3 hours, and her total distance traveled is 120 miles. That’d make her average speed 120m/3hr = 40mph.

Now just plug the numbers you chose into the answer choices to see which one gives you 40!
(A) 3/2 ← nope
(B) 60/3 = 20 ← nope
(C) 120/3 = 40 ← yes!
(D) 2/3 ← nope
(E) 4/3 ← nope

Sure. When the SAT gives you a completely unmarked triangle like this, assume that they’re getting at something that must be true about ALL triangles, and then plug in angle values for an easy triangle. In this case, it’s totally fine to just assume the triangle is equilateral even though it isn’t, and say all the interior angles are 60º. That the supplementary angles all 120º, so each marked angle is 120º + 60º + 120º = 300º. Add ’em up, and you get 900º.

Now here’s the fun part: Try it with different angles. As long as the angles in your triangle add up to 180º, you’ll get 900º for the answer. Why is that??? 🙂

In #20, how come “all but two of the employees” means e-2 but ” all but five desks” means e+5? They have the same wording yet different signs. How do you differentiate between the two wordings when it comes to knowing which sign to use?

This is a great reminder than when you plug in you must test every answer choice (see #5 under “Anything else I should know?” above). If more than one answer choice works, that doesn’t mean plugging in didn’t work, it just means you need to plug in one more time with different numbers to eliminate more answers. So say 9 for d, which makes e = 4 and c = 11. (E), the correct answer, will still work. (A) will not.

In #20, how come “all but two of the employees” means e-2 but ” all but five desks” means e+5? They have the same wording yet different signs. How do you differentiate between the two wordings when it comes to knowing which sign to use?

I got confused by your other post which set up the equation d=e+5. How do you get d=e+5 from that wording instead of e=d-5? Maybe you did an extra step to get “d” by it self…
Questions about forming the algebraic equations:

1. does d-5=E because of the wording “occupied by EMPLOYEES?”
2. What would e-2 equal? I am not sure how to translate “two chairs at their desks” and “one chair” into numbers (or variables) to substitute in an equation. How does that information help me with solving the problem?

Plugging in:
1.If e=5, how do I know to ADD 5 to that to get “d” without setting up an equation?
2.Why add 3 to the number of desks?

Right. d = e + 5 and e = d – 5 are equivalent.

1. Yes.
2. I’m not sure what you’re asking here. e – 2 is just two less than the number of employees. If you’re looking for the number of chairs, we know all but 2 employees have 2 chairs at their desks, and all others have 1 chair. We also know that there are chairs at the empty desks. There are 5 empty desks which each have 1 chair, 2 employees who only have 1 chair, and then e – 2 employees who have 2 chairs. Another way to look at this is to say that each desk has at least one chair, but some (e –2) have a second chair. So one expression for the number of chairs could be c = d + e – 2. But since we know d = e + 5, we can substitute: c = e + 5 + e – 2 = 2e + 3.

1. Plugging in doesn’t mean you never set up small equations—it’s fine to set up a small equation to help you figure out what’s what.
2. We’re not adding 3 to the number of desks. In the numbers I chose in the comment above, there end up being 3 more chairs than desks, but if you chose, say, 100 employees, there would be many more chairs than desks.

Thank you for these tips! I’ve never thought of it this way. I feel so accomplished to be able to solve these questions although I missed some because of careless. Anyway, thanks again!

Here’s how I’d plug in for that one. Start by saying r = 20. Then r + 9, or 29, is 4 more than s, so s = 25. Now, r – 11, or 20 – 11 = 9, is how much less than 25? 16 less.

Just in case the algebra would help:
r + 9 = s + 4
  – 20     – 20
r – 11 = s – 16

The average of 3 numbers is x. If one of the numbers is y, what is the average of the remaining 2 numbers in terms of x and y?
FOR QUESTIONS LIKE THESE: do not use the same number for all the terms in the set. I plugged in 3 as the average of 3, 3, and 3. Though you can get away with this, look at two of the possible answer choices, one of which is incorrect.

(A) [3Y-X]/2
(B) [3X-Y]/2

I am glad I checked for all the answer choices instead of moving on immediately.

I suggest to avoid using 2 as well. I usually start with 3 or 5 just because.

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