Nailing an 800 on the math portion of the SAT can be a tricky feat, even if you are steadfastly familiar with all of the requisite formulas and rules. A difficult problem can overwhelm even the most prepared individual come test day. Time constraints,
test surroundings, and the overall weight of the exam can unnerve the most grounded students.

So what do you do when panic strikes and your mind draws a blank? How do you re-center yourself and charge forward with ferocity and confidence? What you do is this: write everything down from the problem. This is the most important part of the problem solving process. As you peruse the question, write down the pertinent data and establish relationships by setting up equations. This exercise will help you see solutions that were previously difficult to decipher.

As you work on practice tests and sample problems, you must work diligently to form a solid habit of writing down important bits of information as you plow through the SAT math section. To give you an example of what it means to “write everything down from the problem,” I will explore the following three math questions in great detail. These in-depth explanations will give you an idea of what should be going through your brain every time you see a math problem. With practice, these thoughts and processes will manifest faster and faster until solving problems in this fashion becomes a reflexive response.

1. The average of 4 different integers is 75. If the largest integer is 90, what is the least possible value of the smallest integer?

a. 1

b. 19

c. 29

d. 30

e. 33

So what do you do when panic strikes and your mind draws a blank? How do you re-center yourself and charge forward with ferocity and confidence? What you do is this: write everything down from the problem. This is the most important part of the problem solving process. As you peruse the question, write down the pertinent data and establish relationships by setting up equations. This exercise will help you see solutions that were previously difficult to decipher.

As you work on practice tests and sample problems, you must work diligently to form a solid habit of writing down important bits of information as you plow through the SAT math section. To give you an example of what it means to “write everything down from the problem,” I will explore the following three math questions in great detail. These in-depth explanations will give you an idea of what should be going through your brain every time you see a math problem. With practice, these thoughts and processes will manifest faster and faster until solving problems in this fashion becomes a reflexive response.

1. The average of 4 different integers is 75. If the largest integer is 90, what is the least possible value of the smallest integer?

a. 1

b. 19

c. 29

d. 30

e. 33

Right off the bat, the problem states that we have four different integers. We can begin the problem by creating variables to represent the four integers:

W X Y Z

We also know that the average of the integers is 75. This means that we can set up another equation based on this relationship:

(W + X + Y + Z)/4 = 75

Isolating the variables, we get:

W + X + Y + Z = 300

We also know that the largest integer is 90. So:

W + X + Y + 90 = 300

The question then asks “what is the least possible value of the smallest integer?” This detail is a bit tricky to interpret, but we can reason this out fairly quickly. To get the smallest possible number, what needs to be true about the other two integers? They need to be as large as possible. Since 90 is the highest value for the integers, it makes sense to assign the other two variables to 90, right?

Not so fast. If we read the question carefully, it says that there are “four different integers.” This restricts us from using 90 for the other two values. Instead, we must use 89 and 88. We now have an equation to represent the four integers (where W = the smallest integer):

W + 88 + 89 + 90 = 300

Solving algebraically, we get:

W + 267 = 300

W = 33

Therefore, the final answer is

**e**.

2. Solution X is 10 percent alcohol by volume, and solution Y is 30 percent alcohol by volume. How many milliliters of solution Y must be added to 200 milliliters of solution X to create a solution that is 25 percent alcohol by volume?

a. 250/3

b. 500/3

c. 400

d. 480

e. 600

Let’s start writing down the relevant information:

.1X = AX

.3Y = AY

The above equations denote the amount of alcohol given a certain number of milliliters of solution (where AX = alcohol for X, AY = alcohol for Y, X = milliliters of solution X, and Y = milliliters of solution Y). The next part of the question asks how many milliliters of Y must be added to 200 milliliters of X to create a solution that is 25% alcohol? To answer this, we can represent the facts as an equation:

.3Y + .1X = .25(X + Y)

Once again, we have a two variable equation. Translation: we cannot solve it. But, we have a value for X: 200. So, plugging in 200 for X, we get the equation down to one variable:

.3(Y) + .1(200) = .25(Y + 200)

Perfect. Solving for Y algebraically, we get:

.3Y + 20 = .25Y + 50

.3Y - .25Y = 50 – 20

.05Y = 30

Y = 600

Therefore, the answer is

3. On a certain multiple-choice test, 9 points are awarded for each correct answer, and 7 points are deducted for each incorrect or unanswered question. Sally received a total score of 0 points on the test. If the test has fewer than 30 questions, how many questions are on the test?

a. Cannot be determined

b. 16

c. 19

d. 21

e. 24

.1X = AX

.3Y = AY

The above equations denote the amount of alcohol given a certain number of milliliters of solution (where AX = alcohol for X, AY = alcohol for Y, X = milliliters of solution X, and Y = milliliters of solution Y). The next part of the question asks how many milliliters of Y must be added to 200 milliliters of X to create a solution that is 25% alcohol? To answer this, we can represent the facts as an equation:

.3Y + .1X = .25(X + Y)

Once again, we have a two variable equation. Translation: we cannot solve it. But, we have a value for X: 200. So, plugging in 200 for X, we get the equation down to one variable:

.3(Y) + .1(200) = .25(Y + 200)

Perfect. Solving for Y algebraically, we get:

.3Y + 20 = .25Y + 50

.3Y - .25Y = 50 – 20

.05Y = 30

Y = 600

Therefore, the answer is

**e**.3. On a certain multiple-choice test, 9 points are awarded for each correct answer, and 7 points are deducted for each incorrect or unanswered question. Sally received a total score of 0 points on the test. If the test has fewer than 30 questions, how many questions are on the test?

a. Cannot be determined

b. 16

c. 19

d. 21

e. 24

The first step is to write down what we know and assign variables:

+9 points = correct (X)

-7 points = incorrect (Y)

Sally scored a total of 0 points

We can set up an equation with this information:

9X – 7Y = 0

Since we have two variables, this is not a solvable problem. Unfortunately, we do not have another relationship that we can reference to simplify this further. What can we do in this situation? When all else fails, try to isolate the variables:

9X = 7Y

X/Y = 7/9

What this tells you is that the ratio of questions answered correctly and incorrectly must be 7 correct (X) to 9 incorrect (Y). This is very useful information. According to this ratio, the number of questions on the test must be some multiple of 16 (so that the 7 to 9 ratio can be preserved). For example, 7 right and 9 wrong would work, as would 14 right and 18 wrong.

Now comes the critical piece of information: the total number of questions must be less than 30. With this helpful tidbit, the only possibly choice is 16 questions.

Therefore, the answer is

**b**.