The SAT requires you to be able to manage your time and keep a consistent pace when completing the questions, so it is necessary to develop strategies and consider shortcuts to help you complete each section before the time runs out. Being able to utilize different tips and tricks will help you complete problems quicker, giving you more time to contemplate the harder questions without having to worry about time. Here is a list of shortcuts and tricks that can help you complete different types of questions quicker.
Determining the Characteristics of the Data (Calculator Active Math Section)
On the SAT calculator active math section, College Board tends to include questions regarding the median of data. The question will usually include this data out of order in a table, so you have to reorder the data in order to find the median. Doing so can be time-consuming and tedious, so utilizing the different functions on your calculator will help you solve these problems quicker, saving you time.
For example, if the data provided consists of the numbers 1, 5, 2, 9, 10, 3, 4, 7, 11, 12, 22, 24, 15, 19, 17, there is an easier way to reorder the data and get the precise central tendencies of the data. To do so, follow the steps below.
Turn on your TI-84 graphing calculator and press on “stat”
Press enter when “1:EDIT” is selected and enter the data provided (does not have to be in order)
Press “stat” again and navigate to the “Calc” tab
Press enter on “1:1-VAR Stats” until a list of stats shows up
The results will include important information regarding the data you entered. This includes the mean, median, mode, standard deviations, quartiles, and maximum/minimum values.
Solving for Original Price (Math Section)
Sometimes, solving for an original price before a discount on an item can be confusing for students. There is a simple way to utilize the data given and make use of a consistent rule to solve for either the original price, the price after the discount, or the discount itself. The rule is as follows:
100 - discount Price w/ Discount
---------- = ------------------------
100 Original Price
Example Problem: After receiving a 20% discount, Jonathan paid $200 for Airpod Pros. What is the original price of the Airpod Pros before the discount?
100-20 200
---------- = -------
100 x
Step 1: Cross multiply -> 80x = 20000
Step 2: Solve for x -> x = 250
The original price of the Airpod Pros was $250.
This trick is simple once you practice it more, and it can be used to solve for the discount and the price with the discount as well. All you have to do is substitute the information you are given into the proportion and solve.
Finding the Number of Solutions a Function Has
One common problem in the math section is finding the number of solutions an equation has. There are multiple ways to go about solving this problem, but there is one consistent method that always works and takes less than 15 seconds. Let’s say you have the function:
Y = 2X2 + 7x + 11
A simple way to solve for the number of solutions this function has is by using the formula
b2 - 4ac
In this example, a = 2, b = 7, and c = 11. All you have to do is substitute these values into the formula and you will end up with: (7)(7) - 4(2)(11) = 49 - 88 = -39. The rule for determining the number of solutions is that if the discriminant (the result) is negative, it has 0 solutions, one solution if it is 0, and two solutions if it is positive. In this case, the discriminant is negative, so it has zero solutions.
Finding the Value of “a” That Makes the System of Linear Equations Have No Solution
Oftentimes, the SAT will ask you to determine which value for “a” will make a system of linear equations have no solution. The general rule for these types of questions is to make the slopes equal to each other, assuming that they have the same “b” values according to the linear structure y=mx + b.
Example:
0.5x - 0.25y = 5
ax - 3y = 20
In the system of linear equations above, a is a constant. If the system has no solution, what is the value of a?
Step 1: Solve for y for both equations
Y = 2x - 20
Y = ax - 203= ax3- 203
Step 2: Set the slopes equal to each other
2 = a3
Step 3: Solve for a
A = 6
Step 4: Substitute and Check
2x - 20/3 = 2x-20
This final equation has no solution because for every x value, the left and right sides will never be equal as they have different “b” values.
Finding Maximum and Minimum Values
A common type of problem that the SAT may include is determining the minimum/maximum x and y values of a function given the equation. There are different ways to approach this problem. Some resort to graphing the equation and others might simply guess and check values. The most consistent way to solve these types of problems is by using the rule: -b/2a
This rule will help you find the x value of the vertex of the function, allowing you to substitute the x value back into the original equation to find the y value of the vertex.
Example: Given the function y = 5x2+ 10x + 9, find the coordinates of the vertex.
-10/2(5) = -1, so the x value of the vertex is -1. Substitute this value back into the original function to find the y value: 5(-1)(-1) + 10(-1) + 9 = 4.
Solving a System of Linear Equations (Math Calculator Active Section)
One of the types of problems that you will almost be guaranteed to encounter on the SAT revolves around solving a system of linear equations. This type of problem can either be given to you through a word problem or with two different equations, in which you will have to find a pair(s) of points that satisfy both equations.
One of the quickest and most accurate ways to go about solving these types of problems on the calculator active section is simply by solving the two equations for y and then graphing it on your calculator. To find the solutions, simply look for where the two functions intersect and those coordinate pairs will be your solutions. This method is just one of the few ways to go about solving these types of problems, with the second recommended strategy being solving through elimination. With this method, you simply have to multiply one of the equations by a constant to eliminate either “x” or “y” when adding the two equations together. This will then allow you to solve for one of the variables, then substitute that variable back in to find the other.
Comments