Here’s  a question from the May 2006 SAT exam:

19. A container in the shape of a right circular cylinder has an inside base radius of 4 inches and an inside height of 9 inches.  This cylinder is completely filled with water. All of the water is then poured into a second right circular cylinder with a larger inside base radius of 9 inches. What must be the minimum inside height, in inches, of the second cylinder.

(A) 4/3

(B) 16/9

(C) 9/4

(D) 4

(E) 6

Don’t worry about the terminology — it’s not important that you know what a right circular cylinder is. It is, however, important to remember that any time you’re asked about a cylinder, you’re most likely being asked about volume. It’s not necessary to even memorize the volume formula since it’s given to you on the reference information portion of the section (right below the directions).  However, we at Stylus suggest you memorize the formulas anyway because it will save you the time of flipping back and forth.  Cylinder volume is fairly easy to memorize if you already know the formula for the area of a circle.  Think of a cylinder as being a circle with the added dimension of height.  Thus, the formula for volume of a cylinder is the same as that of a circle, but with height added: V = πr²h.

Since this question appears toward the end of the section, we can bet on needing to do at least two separate calculations.  One way the makers of the SAT love to make geometry equations more complex is by asking you to solve one equation and plug that solution into a second (or even a third) equation to solve for a different variable — to plug and plug again.  This equation is no different.  Eventually, we’ll need to solve for the height of the second cylinder, but before we can do that, we need to solve for the volume of the original cylinder.

Since the first cylinder is completely full, we only need to plug the given numbers into the formula to find the original volume of water: V = π(4)²(9) = π(16)(9).  Don’t bother solving this equation yet — doing so won’t give us the answer we’re looking for, and any time you avoid using a calculator to do messy calculations, you save yourself time.

Onto the second equation: If the water is poured into the second cylinder, it needs to have at least the volume of the original cylinder.  So we need to use the volume formula once again, this time using the volume from equation 1 and solving for the height of cylinder 2.

V = πr²h

π(16)(9) = π(9)²h

Divide both sides by 9π to get

16 = 9h (see why I told you not to bother with the calculator?) and then divide by 9 again to get

16/9 = h, choice B.

Here is a question that appeared at the end of the grid-ins section of the May 2006 SAT examination.

18. In the integer 3,589 the digits are all different and increase from left to right. How many integers between 4,000 and 5,000 have digits that are all different and increase from left to right?

When confronted with this problem, many students will decide they don’t know how to approach it mathematically and, as a result, skip the problem altogether. However, there is really no reason not to attempt this problem; because it’s a grid in, you won’t be penalized for guessing the wrong answer. Although knowledge of combinations might help you solve this problem, it can also be solved by simply listing the possibilities in an organized manner.

Since the integers must be between 4,000 and 5,000, the first digit of the number must be 4. Because the integer must increase from left to right, the remaining digits must be either 5, 6, 7, 8, or 9. List possibilities one by one, starting with the smallest possibility, and moving to the largest.

4567 4568 4569
4578 4579
4589
4678 4679
4689
4789

As you can see, there are 10 integers altogether.

It is very important to be organized and systematic when listing possibilities for a problem like this. Otherwise, you are apt to miss one or two. Begin by listing all numbers that start with 456, then that start with 457, then 458, then move on to all that start with 46, and so on. Work slowly and meticulously; rushing will likewise lead to errors.

Although this problem can be solved mathematically, it would involve more advanced permutations than high school students are typically expected to know. The test-makers do not intend for you to solve this mathematically, but for all interested parties, the mathematical solution would be (5!)/(3!)(5-3!).

Here is a tricky SAT question from the October 2006 exam.

In an election, 2.8 million votes were cast and each vote was for Candidate I or Candidate II.  Candidate I received 28,000 more votes than Candidate II.  What percent of the 2.8 million votes were cast for Candidate I?

(A) 50.05%

(B) 50.1%

(C) 50.5%

(D) 51%

(E) 55%

Begin by asking yourself a) what do I already know? and b) what do I need to know?  In this case, we already know the number of total votes cast = 2.8 million, and we know that Candidate I received 28,000 more votes than Candidate II.  We need to know the percent of votes cast for Candidate I.  The percent of votes cast for Candidate I is the number of votes cast for him/her divided by the total number of votes cast.  We already know the total number of votes cast, so we only need to know the number of votes cast for Candidate I.

Can we write an equation to figure out the number of votes cast for Candidate I?  Let’s call that number x.

Let x = the number of votes cast for Candidate I

x – 28,000 = the number of votes cast for Candidate II (since candidate I received 28,000 more).

x + (x – 28,000) = 2,800,000

2x – 28,000 = 2,800,000

2x = 2,828,000

x = 1,414,000

Now we only need to figure out what percent of the total 1,414,000 represents.  Remember that percent is part/whole, so the percent of votes Candidate I receives is 1,414,000/2,800,000, which is 0.505, or 50.5%

Don’t make the mistake of oversimplifying this problem (as many students did) and figuring out that 28,000 votes is 1% of the total.  This is true, but it does not mean that Candidate I received 51% of the votes.  If you are performing only one step to solve a problem near the end of the section, you are definitely oversimplifying it.

Here’s a question from the May 2007 exam to demonstrate the usefulness of “inventing numbers” when it comes to math problems with variables in the answer choices.

19. The toll for trucks to cross a certain bridge is a total of $2 for the first two axles plus $2 for each additional axle. Which of the following functions gives the toll T(n), in dollars, for a truck with n axles?

(A) T(n) = 2n – 2

(B) T(n) = 2n – 1

(C) T(n) = 2n

(D) T(n) = 2n + 1

(E) T(n) = 2n + 2

Don’t be intimidated by the use of the word “function” in this problem.   A function is just an equation that, when x (or in this case n) is plugged in, will yield only one y (or in this case, T(n)).  In other words, this problem asks you to find an equation to determine the total toll paid by a truck, which varies based upon the number of axles.   N is the number of axles.  T(n) is the total amount paid in tolls.

Start by inventing a number of axles (n).  Make it a number that’s easy to work with.  Let’s say that our truck has 3 axles (n = 3).  How much money would we pay in tolls?  Well, the first two axles cost $2, and the third axle costs another $2, so that’s $4 in total.  Circle that answer.

Now plug your variable into the answer choices (n = 3) until you find the one that yields a value of $4.  We can eliminate B, D, and E.  But how do we determine whether A or C is correct?

Invent a different number.  Say the truck has only two axles (n = 2).   That truck would pay only $2 in tolls.  Plugging n = 2 into the answer choices now eliminates choice C because 2n = 2(2)  = $4 in tolls.

Therefore, the answer is choice A: T(n) = 2n – 2.

Make sure to always test all answer choices when you are using the “invent numbers” strategy.  Do not stop just because you found one right answer, especially if the question appears toward the end of the section (as this one did).   If you find duplicate correct answers, invent another number and try again.