Sometimes, math problems on the SAT really are as straightforward as they seem. Here’s an example of a problem that should be fairly easy to solve.

In a sequence of numbers, the first number is 6, and each number after the first number is 2 more than 3 times the preceding number. What is the third number in the sequence?

Many students struggle with the SAT Multiple Choice Writing section, which is surprising given how coachable this section of the exam really is. The SAT Multiple Choice Writing section is nothing more than a basic test of your English grammar comprehension. In an effort to elucidate this theory, we have compiled a list of 14 example problems that highlight some of the more common grammatical mistakes students are expected to recognize on test day. Join us over the next several weeks as we review these fourteen questions, one-by-one, to help you improve your SAT score. Continue reading “Improve Your SAT Score – Multiple Choice Writing” »

To keep their negotiations —–, the leaders of rival groups in the country arranged meetings that were —-.

a.relaxed… complex

b.covert… prestigious

c.secret… clandestine

d.productive… unscheduled

e.diplomatic… illicit

Explanation: The framework of this sentence is such that the word in the second blank must be similar in meaning to the word in the first blank. Of the available answer choices only (c) has two answer choices with similar meanings.

You can never have enough vocabulary; you should add these words and their definitions to your flashcards.

Relaxed: To make lax or loose; to make less severe or strict; to reduce in intensity; slacken; to relieve from tension or strain.

Complex: Involved or intricate, as in structure; complicated.

Covert: Not openly practiced, avowed, engaged in, accumulated, or shown. Covered or covered over; sheltered.

Prestigious: Having prestige; esteemed.

Secret: kept hidden from knowledge or view; concealed. Dependably discreet. Operating in a hidden or confidential manner. Not expressed; inward. Not frequented; secluded. Known or shard only by the initiated. Beyond ordinary understanding; mysterious. Containing information, the unauthorized disclosure of which poses a grave threat to national security.

Clandestine: Held in secrecy; inconspicuous.

Productive: Performing the act of, or having the ability to, accomplish a task or activity in abundance or efficiency.

Unscheduled: Not planned; Spontaneous.

Diplomatic: Relating to the use of peaceful means of negotiation to accomplish political goals, especially in order to avoid conflict.

At the end of every SAT Math section, the test makers try to come up with an extremely difficult problem that will leave even the cleverest students scratching their heads. The really evil part, though, is that even these problems can be solved in under a minute without a calculator – if you know what to do. This means that once you “figure out the trick,” these difficult problems become easy. So, while those test makers are busy cackling with sadistic glee, let’s see if we can’t beat them at their own game.

Consider the following problem:

If the average (arithmetic mean) of four different positive integers is 5, then the least possible product of the four integers is:

A) 20

B) 84

C) 104

D) 480

E) 504

This one might seem like a hard nut to crack. After all, there are many possible sets of four distinct positive integers that have an average of 5. We could not possibly test them all one by one given the time constraints of the SAT. So how are we going to figure out which four integers are the ones that yield the smallest possible product?

Well, if we want the smallest product, then we want as many of the integers to be as small as possible. If the smallest of the three integers were equal to 1, then that would be great, since multiplying by one simply produces the same number and would thus not increase the product of the three numbers. Since the four integers must be distinct, we can make the second integer 2 so that it is also as small as possible, and by the same logic the third integer would be 3. All that remains is to determine the fourth integer, which we can find using algebra:

The fourth integer is 14, and the product is:

Logically, the answer must be 84, or choice B. If you are still feeling unsure, you could try a few other sets of four integers to see if their products are less than 84. What happens, for instance, when we increase the 3 to 4 and decrease the 14 to 13?

Already, the product begins to increase. Try any other set and you will see that its product is greater than 84. It doesn’t matter that we decreased 14 to 13, because what controls the outcome in this problem is how small we can make the smallest numbers in the set. Thus, our original reasoning was correct.

If you know what to do, it takes only about 30 seconds to solve this problem. So you see, with practice, even the hardest problems on the SAT become easy. Check back here each week for more extra hard problems and the tricks you need to solve them! Also, remember that you can find out all the tricks from experts like me with a Test Masters course or private tutoring. Until then, keep up the good work and happy studying!

At the end of every SAT Math section, the test makers try to come up with an extremely difficult problem that will leave even the cleverest students scratching their heads. The really evil part, though, is that even these problems can be solved in under a minute without a calculator – if you know what to do. This means that once you “figure out the trick,” these difficult problems become easy. So, while those test makers are busy cackling with sadistic glee, let’s see if we can’t beat them at their own game.

Consider the following problem:

If Aragorn can dispatch 5 orcs every 11 seconds and Legolas can dispatch 3 orcs every 9 seconds, then together, how many orcs they can dispatch in how many seconds?

A) 2 orcs every 5 seconds

B) 8 orcs every 20 seconds

C) 26 orcs every 33 seconds

D) half an orc every second

E) genocide

To solve this problem, all you need to realize is that the rates at which Aragorn and Legolas slaughter orcs can be represented as fractions and then combined to give their combined rate:

5 orcs every 11 seconds = 5/11

3 orcs every 9 seconds = 3/9 = 1/3

You will notice that we simplified 3/9 to 1/3. It’s always a good idea to simplify fractions whenever possible, as this will save time later. Next, we need to give the fractions like denominators so that we can add them together:

Thus, together they can kill 26 orcs every 33 seconds, and the answer is choice C. If you know what to do, it takes only about 30 seconds to solve this problem. So you see, with practice, even the hardest problems on the SAT become easy. Check back here each week for more extra hard problems and the tricks you need to solve them! Also, remember that you can find out all the tricks from experts like me with a Test Masters course or private tutoring. Until then, keep up the good work and happy studying!

Here is an example of a very classic SAT algebra problem. There are two ways to solve this problem; one is very intuitive and simple, but potentially very slow. The other is rooted in simple algebraic methodology (I promise you’ve done it a million times in school by now) and is extremely fast and precise. Let’s take a look at how basic algebra can help you save some major time on the math section. Continue reading “Intuition vs. Methodology: The Difference Between Spending 5 Minutes and 5 Seconds on an SAT Math Problem” »

At the end of every SAT Math section, the test makers try to come up with an extremely difficult problem that will leave even the cleverest students scratching their heads. The really evil part, though, is that even these problems can be solved in under a minute without a calculator – if you know what to do. This means that once you “figure out the trick,” these difficult problems become easy. So, while those test makers are busy cackling with sadistic glee, let’s see if we can’t beat them at their own game.

Consider the following problem:

Out of 50 space cadets, 30 speak English, 20 speak Klingon, and 5 speak neither English nor Klingon. What is the ratio of those who speak both English and Klingon to those who speak neither?

A) 1:1

B) 1:2

C) 2:3

D) 1:3

E) 5:1

We know that the total number of space cadets is 50, and that the number of cadets who only speak English plus the number of cadets who only speak Klingon plus the number of cadets who speak both plus the number of cadets who speak neither should equal 50. The trouble is in figuring out how many cadets speak both so that we don’t count them twice. If we just add the 30 who speak English to the 20 who speak Klingon, we will have counted the ones who speak both twice. So, how can we prevent this?

By subtracting them. If we add the cadets who speak English to those who speak Klingon and those who speak neither but subtract the number who speak both, then we will have made up for double counting and the result should equal the 50 total cadets. If we let b be the number of cadets who speak both, then:

50 = 30 + 20 – b + 5

50 = 55 – b

b = 5

Since the problem asked for the ratio of those who speak both English and Klingon to those who speak neither, the ratio will be 5 to 5, or simplified, 1:1, which is choice A. If you know what to do, it takes only about 30 seconds to solve this problem. So you see, with practice, even the hardest problems on the SAT become easy. Check back here each week for more extra hard problems and the tricks you need to solve them! Also, remember that you can find out all the tricks from experts like me with a Test Masters course or private tutoring. Until then, keep up the good work and happy studying!

At the end of every SAT Math section, the test makers try to come up with an extremely difficult problem that will leave even the cleverest students scratching their heads. The really evil part, though, is that even these problems can be solved in under a minute without a calculator – if you know what to do. This means that once you “figure out the trick,” these difficult problems become easy. So, while those test makers are busy cackling with sadistic glee, let’s see if we can’t beat them at their own game.

Consider the following problem:

The average of four numbers is five less than the average of the three numbers that remain after one has been eliminated. If the eliminated number is 2, what is the average of the four numbers?

A) 17

B) 22

C) 66

D) 68

E) 100

“What numbers?” you ask, “How am I supposed to find the averages of numbers that no one knows?!”

The SAT test makers love to make math questions harder by forcing you to work with variables instead of with actual numbers. Actual numbers are concrete, intuitive, and easy to understand, while variables are more abstract and difficult to grasp. At least that’s their theory. As you’ll see, variables aren’t actually harder than actual numbers. In fact, in some cases, problems that have many unknowns can actually be faster and easier to solve than problems with actual numbers, since there is less to figure out (because there’s less you can figure out). The main trick comes at the beginning, when we have to translate the words of a problem into a mathematical expression we can manipulate on our scratch paper.

First, the problem tells us that we have “the average of four numbers.” Well, how do you find the average of four numbers? You add them up and divide by 4. We can write this down if we pick variables to represent each of the four numbers. Let’s just go with a, b, c, and d.

Here, e represents the average of the four numbers (what we are trying to find). Next, the problem tells us that we take the average of 3 of the four numbers. This mean we have to leave out one of our variables. Since d is the last one, we’ll leave it out:

The new average is represented by f. Now the problem also tells us that the variable that was left out was equal to 2, so we know that:

The problem also tells us that the average of four numbers is five less than the average of the three numbers that remain after one has been eliminated. We have let e represent the average of the four numbers and f represent the average of the three numbers, so we could rephrase this as e is five less than f:

So, now we still have all of these variables. We want to substitute variables for each other until we have one equation with one variable, because then we can solve for that variable and hopefully get closer to solving for e. Here’s one possible way to do that:

We can now substitute 3f for a+b+c in the other equation:

We can also substitute f – 5 for e, since we know that e = f – 5:

Now we have one equation with only one variable, so we can solve for f:

Of course, we are trying to solve for e, so we should plug this into the equation e = f – 5:

Thus, the correct answer is choice A. Note, however, that if you had only solved for f, 22 was a choice as well. With this problem, it’s all about translating the words into variables and then moving the variables around until you can actually solve for one of them. This problem might be a little more time consuming than some of the others we’ve looked at in this series, but it can still be solved in under a minute without a calculator, if you know what to do. So you see, with patience and practice, even the hardest problems on the SAT become easy. As you do more practice problems you will get better and better at them – the test makers tend to use the same tricks over and over again. Check back here each week for more extra hard problems and the tricks you need to solve them! Also, remember that you can find out all the tricks from experts like me with a Test Masters course or private tutoring. Until then, keep up the good work and happy studying!

At the end of every SAT Math section, the test makers try to come up with an extremely difficult problem that will leave even the cleverest students scratching their heads. The really evil part, though, is that even these problems can be solved in under a minute without a calculator – if you know what to do. This means that once you “figure out the trick,” these difficult problems become easy. So, while those test makers are busy cackling with sadistic glee, let’s see if we can’t beat them at their own game.

Consider the following problem:

For all x and y where ,

A) 2(x – 2y)

B) 2y – x

C) 1

D) 0

E) -2

Well, as you know, you can’t add fractions unless their denominators are the same, so if you want to add these two fractions, then you would have to multiply the numerator and denominator of the first fraction by the denominator of the second fraction and the numerator and denominator of the second fraction by the denominator of the first fraction, like so:

So, the answer is E. But that’s the long way. By the time you get to the third line of all that awful math, you should realize that the numerator and denominator of each fraction are essentially the same, except that one has been multiplied by -1. In line 3 above, we factor out the -1 and realize that these fractions are just a glorified way of writing -1 + -1, but if we had stopped to think about it at the beginning, we could have realized it without multiplying all those binomials:

Granted, if you remember the FOIL method, multiplying the binomials shouldn’t take that long, but it’s still nice if you can do a 2 minute problem in 10 seconds, since that gives you more time for other questions you may need to think about. A method with fewer steps also leaves less room for careless errors. But how are you going to find these shortcuts on the day of the test? Ironically, if you are rushing through the test as fast as you can, you’re more likely to do problems the long way and miss the shortcuts. Before you dive into a problem and break out your calculator, pause and reflect: every SAT math problem can be solved in under a minute without a calculator. This means that for the harder problems, there is always some trick. If you ever find yourself scribbling down lines and lines of scratch work, that probably means that you’re doing a problem the long way. For this problem, just look at it. You will notice a lot of repetition: everything is made up of x and 2y. This is a red flag that there’s a trick to solving this problem. If everything is kind of the same, then if you arrange the expressions properly chances are things will start to cancel out. In this case, you have to factor out a -1 from either the numerators or the denominators. After that, it becomes clear that each fraction is equal to -1, and the problem becomes, as they say, easy as π!

So you see, with patience and practice, even the hardest problems on the SAT become easy. As you do more practice problems you will get better and better at spotting these shortcuts – the test makers tend to use the same tricks over and over again. Check back here each week for more extra hard problems and the tricks you need to solve them! Also, remember that you can find out all the tricks from experts like me with a Test Masters course or private tutoring. Until then, keep up the good work and happy studying!

At the end of every SAT Math section, the test makers try to come up with an extremely difficult problem that will leave even the cleverest students scratching their heads. The really evil part, though, is that even these problems can be solved in under a minute without a calculator – if you know what to do. This means that once you “figure out the trick,” these difficult problems become easy. So, while those test makers are busy cackling with sadistic glee, let’s see if we can’t beat them at their own game.

Consider the following problem:

If x and y are integers and y < 20, for exactly how many ordered pairs (x, y) will x^2 = y?

A) 4

B) 5

C) 7

D) 8

E) 9

This one actually doesn’t seem so bad, does it?

1^2 = 1

2^2 = 4

3^2 = 9

4^2 = 16

5^2 = 25 > 20

So we’ve got (1,1), (2,4), (3,9), and (4,16). Answer choice A, right? Not so fast! You forgot that the square of a negative number is also positive, so for ever y, there are two x values: one positive and one negative. So really our list should look like this:

(1,1) and (-1,1)

(2,4) and (-2,4)

(3,9) and (-3,9)

(4,16) and (-4,16)

So the answer is D, right? Wrong again! There’s one last square you forgot:

0^2 = 0

Thus, there are in fact 9 pairs: the eight already mentioned, plus (0,0). Thus, the correct answer is actually choice E.

Was there actually anything hard about this question? Not really. However, if you were going fast and running out of time, you might have easily made one of the careless errors above. Note that 4 and 8 are traps set for students who see this problem, think it’s easy, and then blow through it too fast without thinking carefully (if you forgot the negatives but remembered 0, there’s also choice B, 5). If you get toward the end of a math section and see a problem that looks really easy, be careful – there’s probably more to it than meets the eye. Sometimes it’s just as bad to spend too little time on a problem as it is to spend too much, so make sure you don’t go too fast through any “easy” problems at the end of a math section.

Check back here each week for more extra hard problems and the tricks you need to solve them! Also, remember that you can find out all the tricks from experts like me with a Test Masters course or private tutoring. Until then, keep up the good work and happy studying!