Posts Tagged ‘SAT math’

SAT Math: Counting Problems

Posted in Standardized Test Questions on December 10th, 2009 by admin – Comments Off

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!).

SAT Math: Percent Word Problems

Posted in Standardized Test Questions, Uncategorized on November 9th, 2009 by admin – Comments Off

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.

SAT Math: Inventing Numbers

Posted in Standardized Test Questions on July 22nd, 2009 by admin – Comments Off

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.