The SAT® is fast approaching, and you’re stressing about that dreaded math section. You’ve done everything you can to study and prepare, but still can’t seem to get that extra edge in defeating the math demons and raise your score. Have no fear, SupertutorTV is here to rescue you with some simple hacks, tricks and strategies to overcome those pesky problems and save you time on exam day.
Essentially, these SAT® Math Hacks are backdoor shortcuts to solving problems on the math section. Sometimes these hacks will save you time on the exam, other times they will actually be a little bit more time consuming, but they will always be really great problem-solving tools to keep in your arsenal of test-taking strategies. If you’re getting bogged down on certain problems, if you’re having trouble seeing the problems from another angle, these tips will be your way of hacking down into that problem and getting the right answer choice so you can squeeze out a few more points on the test.
So let’s get started!
Hack #1: Use The Answers
A lot of tutors have probably already told you this, or you may have seen it in other videos or on other blogs, but I’m going to give you one particular way to use it on the New SAT® because of the nature of this test. This test is different from the old SAT® and the ACT® in the sense that SAT® Math has only four answer choices per question. Why is this a big deal? Well, obviously, one less answer to eliminate!
Elimination matters because these answers frequently have a lot of repetition in them. When you see that kind of repetition, you have a HUGE opportunity to speed up the solving of that particular problem. Essentially, you’ll be skipping solving the entire problem itself and deriving your answer from only small parts of the problem that will let you eliminate certain answers. This will really start to matter the closer you get to the end of the section.
Tip: This trick REALLY comes in handy in the “No Calculator” portion of the Math Section.
Let’s take a look at a problem:
If y=x3+2x+5 and z=x2+7x+1, what is 2y+z in terms of x?
This is what I like to call an Equation/Expression question. Whenever I see this pattern, I know my best solving option is SUBSTITUTION. Substitute in a value from the equation into the expression you need. In this case, we have two equations (y=x3+2x+5 and z=x2+7x+1) and an expression (2y+z), and conveniently, y and z are already isolated in the equations, so I’ll plug in for y and z into the expression using the equations at left.
2(x3+2x+5) + x2+7x+1
Obviously, I could foil and combine all terms– but I want to be a little lazy and save 20 seconds so that I can move a bit faster and finish the section before resorting to super careful double checking mode.
Let’s look at the answer options:
The easiest terms in FOIL are the first and last terms– so those are what I’ll focus on. The last term is almost always easiest to deal with, so that’s where I’ll start.
Rather than FOIL everything, I just worry about the last term– 2(x3+2x+5) + x2+7x+1 — I multiply the 2 times the 5 in the first set and get 10 and then add that to 1 from the second expression and get 11. Therefore, I can eliminate B and D. Between A and C, I see that the coefficient of the x3 term differs– so then I look for that term. x2+7x+1 has no x3 term so I know all I’ll have is 2 times the x3 or 2x3. Thus C is correct.
You might think I’m not saving that much time by skipping the middle of my FOIL– and it’s true that explaining this method takes longer than doing the problem with the whole FOIL — but the concept still does save time if you can apply it quickly without getting too confused. It also works wonderfully on problems like this with far more polynomial terms to combine– imagine if this had a degree of 7 instead of 3– you get the picture. This also works on many other questions– look at the answers and use some common sense if you see repetition or distinct pieces of an answer– that can speed up process of elimination at times and help you get an answer more quickly.
Hack #2: Make Up Numbers
This trick is also a tutoring classic: when you don’t understand a word problem, or encounter a word problem with variables in the answer choice, MAKE UP NUMBERS that could work in place of variable(s) and plug them in to see how everything is working together.
The gas mileage for Peter’s car is 21 miles per gallon when the car travels at an average speed of 50 miles per hour. The car’s gas tank has 17 gallons of gas at the beginning of a trip. If Peter’s car travels at an average speed of 50 miles per hour, which of the following functions f models the number of gallons of gas remaining in the tank t hours after the trip begins?
This problem can seem pretty confusing at first– so to better understand what’s going on, make up a number for the time, t, solve for the gallons left and then look at your work and try to match to the answer choices.
Here, I’m going to let t equal 2. That means at 50 mph Peter will travel 100 miles.
Now if he travels 100 miles, the next question is how much gas does he use?
If he gets 21 miles per gallon, that means for every 21 miles he uses one gallon of gas– he would use 2 gallons if he drove 42 miles, etc. — 100 is about 5 times 21– so he can go about 5 stretches that are 21 miles each, right? What am doing here to get to the number of gallons from that 100 and 21? I’m dividing 100 by 21 — but making up numbers makes it easier for me to see and understand how the numbers work together so I don’t divide the wrong numbers.
What I have now:
Amount of gas initially – Amount of gas used = gas left in tank
17 gallons of gas – 100/21 gallons of gas = gas left in the tank
Now I look at my answer choices, knowing t=2.
Clearly B is the answer– 100 came from 2(50), right? I can also plug in 2 for t and I see that B works and gives me the pattern I need.
Hack #3: Plug In Zero or One
This is a tip for impossible seeming exponential functions, polynomials or quadratic equations — anything with an exponent that is in function form. When you get stuck on a problem like this, besides simplifying, substituting, foiling or factoring in any way you can, you can always plug in zero or one and ask yourself “what does it mean if this variable is one or zero?”
For quadratics, I’m often finding the “zeros” of the equation and then asking this question (i.e. I let y=0 )– for polynomials it could be anything– but I often aim to create an exponent of one or a one that “disappears” (if it multiplies times something else). In exponential word problems, I try to step forward one “step” or generation and plug out the problem one cycle at a time to understand the problem more.
Let’s look at a problem:
A biology class at Central High School predicted that a local population of animals will double in size every 12 years. The population at the beginning of 2014 was estimated to be 50 animals. If P represents the population n years after 2014, then which of the following equations represents the class’s model of the population over time?
Notice the exponents in choices C and D? They indicate this could be exponential. Now you might be savvy enough to know that this problem is exponential growth just from the nature of it (doubling / halving in size problems are exponential). If so, you’re likely to know we’re down to C and D already. But if not, you can plug in zero for n and at least get rid of one other answer.
If n (the number of years since 2014) is zero– what does that mean? It means it’s still 2014 — so the population in 2014 is the population denoted by P when I plug in zero. I can thus plug in n=0 and expect to get 50 — the initial population amount. Go to your answers and plug in zero for n and you’ll find quickly that A doesn’t work– P does not equal 12!
The rest do work so zero’s helpfulness stops there.
Now I’m going to think– what happens one generation down the line? What if I step the situation forward once (think “1” — one generation, one more step forward, etc.)? Here “once” is actually 12 years– every 12 years the population doubles. So I know that when n=12 (12 years after 2014 in 2026) my population of animals (once 50) is going to double and become 100 (twice 50). This gives me another set of points to “plug in.” If you plug these two numbers into the answers, you’ll find that only choice D gives us a P=100 — everything else results in outrageously large populations.
We could also plug in “1” into n, and then would expect a result between 50 and 100 right? Because it takes 12 whole years to double– even with that thinking we could find the answer.
Are these going to work for every problem and for every situation? No. But what you can get from these hacks is a better way, or rather an easier way, to get the correct answer. After all, correct answers are what give you a good score. That’s a simple concept, right?
Are you taking the SAT® soon? Leave us a comment below! If you want more tips, tricks, & strategies for the SAT®, check out some of our other videos and blog posts at the bottom of this page. Also, subscribe to our mailing list to stay up to date on all our strategy guides, informational blogs, promotions, coupons, and upcoming giveaways right here on SupertutorTV!