E X A M P L E S
Begin with Choice (C)
Begin with Choice (C)
Level 1: Number Theory
Three consecutive integers are listed in increasing order. If their sum is 531, what is the second integer in the list?
(A)176
(B)177
(C)178
(D)179
(E)180
Begin by looking at choice (C). If the second integer is 178, then the first integer is 177 and the third integer is 179. Therefore we get a sum of 177 + 178 + 179 = 534. This is a little too big. So we can eliminate choices (C), (D) and (E).
We next try choice (B). If the second integer is 177, then the first integer is 176 and the third integer is 178. So the sum is 176 + 177 + 178 = 531. Thus, the answer is choice (B). Remark 1: You should use your calculator to compute these sums. This will be quicker and you are less likely to make a careless error. Remark 2: This method is faster than solving the problem algebraically. You don't have to show your work on this test so it's usually best to avoid algebra when possible. * A quick, clever solution: A really sharp math student might realize that you can get the answer to this problem very quickly by dividing 531 by 3. An algebraic solution: This method is not recommended for the SAT! We include it for the more advanced student that wants an actual solution to the problem that doesn't involve any tricks. If we name the least integer x, then the second and third integers are x +1 and x + 2, respectively. So we have1, 2, 3 these are three consecutive integers
-3, -2, -1, 0, 1 these are five consecutive integers
In general, if x is an integer, then x, x + 1, x +2, x + 3, … are consecutive integers. Interesting fact: In a set of consecutive integers, the average (arithmetic mean) and median are equal.
Take A Guess
Level 3: Algebra
What is one possible value of x for which 
2 .5
1 1
.5 2
.1 10
We see that .1 < 5 < 10. So we can grid in .1. Note: When using the symbols < and >, the symbol always points to the smaller number. Choosing Guesses: If guesses of a certain kind aren't working, try a number of a different kind. For example, in this problem positive integers weren't working, so we switched to decimals less than 1. In general, trying extreme cases is always good. In this case a large integer and a small decimal would get the answer quickly. In other examples, extreme cases might consist of a positive integer, a negative integer, a positive fraction and a negative fraction. Note that attempting negative numbers would be a waste of time in this problem since we can't grid in a negative number. An algebraic solution for the advanced student: If 5 < 1/x, then we have 1/5 > x, or equivalently, x < 1/5, or as a decimal x< .2. So we can grid in any answer strictly between 0 and .2. Note that an answer of .2 will be marked wrong (and 0 will be marked wrong as well).
Pick A Number
Level 3: Number Theory
Which of the following is equal to 
?
(B)
(C)
(D)
(E)
Let’s choose a value for k, say k = 3. We first substitute a 3 in for k into the given expression and use our calculator.
We type in the following: (3 + 60)/12 and we get k = 5.25. Put a nice big, dark circle around this number so that you can find it easily later. We now substitute a 3 into each answer choice and use our calculator.
(A) 3/12 + 5 = 5.25 (B) 3 + 5 = 8 (C) 5*3 = 15 (D) (3 + 30)/6 = 5.5 (E) (3 + 5)/6 ~ 1.33 (~ means “is approximately”)We now compare each of these numbers to the number that we put a nice big, dark circle around. Since (B), (C), (D) and (E) are incorrect we can eliminate them. Therefore the answer is choice (A).
Important note: (A) is not the correct answer simply because it is equal to 5.25. It is correct because all 4 of the other choices are not 5.25. You absolutely must check all five choices! As an example of how things could go wrong with incorrect reasoning, suppose we choose k = 0. Then the given expression becomes 5, and the answer choices become (A) 5(B) 5
(C) 0
(D) 5
(E) 5/6
In this case we have eliminated (C) and (E), but (A), (B) and (D) are all potential solutions. A common error is to choose the first answer to come out correct.
This is certainly a good guessing strategy, especially if you’re running out of time, but it can potentially lead to the wrong answer. This is also why we generally try not to pick numbers to be too simple. It often leads to multiple answer choices coming out to the correct number.
A quick algebraic solution: A discussion of this problem wouldn’t be complete without showing how to solve this problem very quickly using an algebraic method. Most students have no trouble at all adding two fractions with the same denominator. For example, k/12 + 60/12 = (k + 60)/12 But these same students have trouble reversing this process. (k + 60)/12 = k/12 + 60/12 Note that these two equations are identical except that the left and right hand sides have been switched. Note also that to break a fraction into two (or more) pieces, the original denominator is repeated for each piece. * An algebraic solution to the above problem consists of the following quick computation (k + 60)/12 = k/12 + 60/12 = k/12 + 5 This is choice (A).