First of all, watch out for the trap answer choice “The answer cannot be determined from the information given in the question”!! NEVER GUESS THIS ANSWER CHOICE if you are not completely confident that it MUST be true! Most often the SAT uses this answer choice as a trap for students who don’t know the advanced method that does exist and is necessary to find the correct answer, which can be determined from the information given in the question if you know the correct method. If you don’t understand how to answer the question and you have to guess, guess one of the other three answer choices!

In this question, the SAT is trying to confuse you with the extra numbers and algebra and parentheses in the first equation they give you. Don’t be fooled or confused! There is no trick to the parentheses: You can just add all the terms together and get . Now just subtract 10 from both sides and you get .

Now you can see the connection to a question about tangent values: Since ,  and  can be the two smaller angles of a right triangle!

Pro Tip: At this point the most experienced math students will already see the answer instantly: Since you know from SOH-CAH-TOA that , and switching from one smaller angle of a right triangle to the other smaller angle simply swaps the angles’ opposite sides and adjacent sides, the tangent of one angle is always the reciprocal of the tangent of the other angle! So, since , the most experienced math students know right away that , which is the correct answer choice.

If you don’t see this right away, the other way to finish solving the question is to draw your own diagram:

Now you can see that , so you can assume the simplest case that side  and side . (Keep in mind that they can actually be any side lengths that have a ratio of 2:3.) Now switching to look at the triangle from the perspective of angle , you see that now the opposite side is  and the adjacent side is . So , which is the correct answer choice.

The question tells you that the tangent of angle a is 1. We’ll want to keep in mind that a tangent is a ratio: it’s the length of the opposite side divided by the length of the adjacent side. When does a ratio equal 1? When both figures are the same. So, side XY = side YZ, meaning that you’re dealing with an isosceles right triangle (and that’s where the “not drawn to scale” notation really comes in handy since you wouldn’t have thought that to be true from the picture). Since it’s an isosceles right triangle, you know the side ratio is: . At this point, even though you don’t now know the actual lengths, all you need to do is apply that ratio. The sine of b is the opposite over the hypotenuse, which is a ratio of . From here you have to “rationalize the denominator” by multiplying by 1 (in this case, ). Once simplified, this leaves you with .

In this example, we’re being tasked to apply our understanding of the ratio-driven relationships we express with SOH-CAH-TOA, our sin, cosine, and tangent. Keep in mind that these ratios are as follows:

From this, if we’re told that the tangent (the ) of  is , then the tangent of  must be the reciprocal since the opposite of a° is the adjacent of b°, and vice versa. In many cases, recognizing the way SOH-CAH-TOA manipulates the same relationships and sides of a triangle in SAT questions can simplify the extent to which we really end up needing to “do the math.” Here, by recognizing that the tangent of  is simply the reciprocal of the tangent of , this question becomes extremely efficient, and allows us to save time for more step-by-step questions elsewhere on the exam.

This question is a great example of a case where the SAT might try to trap you into doing more work than is necessary. Since the problem asks you whether that mean is exactly 3.0, it is possible to determine this without tedious work. If the average were 3.0, the values on either side of the twelve B (3.0) grades would all even out. And they ALMOST do: there are 6 As (4.0 each) and 6 Cs (2.0 each, so those grades all average out to 3.0); there are 8 A-s and 8 C+s, so those grades (3.7 and 2.3) all average out to 3.0; and there are 11 B+s and 11 B-s, so those all average out to 3. On the list from A on down to C, all those grades average out to 3.0…but then there are the 3 C− grades, and those do not get “averaged out” by anything above an A. So without fully calculating, you can tell that the mean is lower than 3.0. So, we can eliminate “The mean math grade of the 65 students was 3.0.”

Since two of our options deal with the median, we can focus there next. The middle of the 65 terms will be the 33rd term (32 terms above it and 32 terms below it). So when you go from the top down and see that there are 25 grades from A through B+ and then 12 grades of B, you should see that the 33rd term will fall in that group of Bs. This means that the median is 3.0. 

This knowledge allows us to both eliminate “The median math grade of the 65 students is less than a 3.0” and select “The median math grade of the 65 students is higher than the mean math grade of the 65 students.​” Since the mean is somewhat less than 3.0, and the median *is* 3.0, the mean is less than the median.

“If three new students were added to the class and each scored an A, the new class average would be greater than or equal to 3.0” can be eliminated on the logic of the first paragraph: The 62 grades from A down to C all average out to a 3.0, leaving the three C− grades of 1.7 to weigh down that average. In order to balance three 1.7s you’d need three 4.3s; three 4.0s, as this option presents, aren’t quite enough, so the average would still be below 3.0.

This question is an excellent example of a case where the SAT might try to trap us into doing more math than needed. If Henry’s current grades are 3.7, 3.7, 3.3, and 3.0, his current average GPA is lower than needed for him to achieve his goal. (3.7 and 3.3 average to 3.5, so 3.7 and 3.0 will average to lower than 3.5, making his current GPA lower than needed to reach his goal.) So, his final grade needs to bring his average *up* and bridge the gap between his current, below 3.5, average, and his goal. Only “A” has the potential to do so. Keep in mind - A- wasn’t even in the running! So, before you go to tackle complex calculations, make sure you keep in mind that the answer choices are a part of the question, and you’ll want to take a look at what options are even possible before doing the math.

You only care about the freshmen who have declared a major. That means that you need to look at Freshmen-Sciences; Freshmen-Arts; and FreshmenProfessional. As you approach table problems on the SAT, it is important to focus on the question first and then on the table; often the table has much more information than you actually need.

Here, of the freshmen who have declared a major, 1875 are science majors, 1476 are arts majors, and 284 are majoring in a professional degree program. This should allow you to get by with an estimate: the number of “non-science majors” declared is going to be between 1700 and 1800 and the number of science majors is 1875. That should tell you that freshman science majors make up just a little more than half of all freshman declared majors, so 51% is the only answer choice that makes sense.

As you’ve learned by now, you should look specifically at the question that you’re being asked before you get too concerned with the numbers in each cell. Since you’re asked about the amount raised by September 30, you’re really only concerned with the rightmost column. The sum of the cells in that column is just above 900 (you can use your calculator to add them up, or you can look at the answer choices and realize that you can get away with an estimate).

But what does that ∼900 figure represent? Twice in the problem – in the title of the table at the top and in the question stem below the table – the phrase “in thousands of dollars” appears. This means that the figure 900 actually represents approximately $900,000. So the percentage of the $1,000,000 that the charity has raised is just above 90% of the goal, making “91%” the only plausible option.

The two columns that matter are the arts column and the science column, and you’re looking for the highest ratio of arts to sciences. Furthermore, with five rows and only four answer choices, a quick scan should tell you that you don’t need to worry about the last “5th year” row, because it doesn’t match an answer choice. Since arts is less than science for freshmen but greater for the next three rows, you can quickly eliminate “freshman.” Additionally, since the totals are far apart (by more than a thousand) for sophomores and seniors but quite close together for juniors, you should focus your attention on the sophomores and seniors rows and can eliminate “junior” as well. Here, again, a relatively quick estimate should show you that the ratio will be bigger for sophomores than for seniors: for sophomores the difference between the two categories is higher and the number of science majors is lower, suggesting that the ratio will be much higher. Thus, our correct answer is “sophomore”.

The SAT will frequently test you on your ability to recognize concepts with word problems, graphs, and equations. Here the algebra in each answer choice – and the assignment of variables in the problem – looks messy, but the SAT is essentially testing whether you can recognize what type of change is occurring within each city. From the graphs you should see that City A is increasing in population and that City B is decreasing. So you’ll want the rate of change (x) to be added for City A and to be subtracted for City B. And here is where the algebraic terms “linear” and “exponential” really come to life in graphical form: you should see that City A’s graph forms a straight line, meaning that it’s a linear increase. City B’s graph is curved, suggesting a nonlinear, exponential decrease. A linear increase means that you add the same amount each period. That means that the same amount (x% of the starting value) is added each year. For an exponential change, as with City B, the exponent shows that the change is compounded. So you need addition for City A and subtraction for City B, with a linear (times y) function for City A and an exponential function (to the y power) for City B. The only option that satisfies these conditions is:

To express a positive correlation, as one component (our “x” axis) goes up, the other (our “y” axis) should also increase. In this case, both of the below options accomplish this to some extent:

However, of these options, the first clearly carries a closer positive relationship that appears more linear and consistent in the increasing nature of both components. Thus, it is our correct answer.