Part 2: The Good
The ACT and SAT Math sections have a similar purpose. They both are designed to test juniors and seniors on the maths they learned in high school. In practice, the tests are very different. As we compare the easiest level of problem on the ACT and SAT, you’ll see that they approach the same topics in different ways.
I’ve selected two problems from each test so that we can compare the tests’ way of assessing your knowledge of the similar topics.
For our “easy” ACT problems, notice that the problem is number 10 out of 60 total questions. Both tests are kind in that they place the problems into approximately increasing order of difficulty. We can use this fact to strategize timing and maximize scores for students at every target score. Unfortunately, the ACT is also cruel in that even the first of these “easy” problems has multiple steps.
In this problem you use two concepts to calculate a variety of angles until you eventually solve for the unknown angle. The angles inside a triangle add up to 180 degrees. You need to know that the check marks on segments AD, BD, and BC mean that all of those segments are the same length; that the angles opposite those hash marks within each individual triangle have equal measures; and that collinear means that points A, D, C, and E are all on the same line, and you can therefore assume that angles ADB and BDC together add to get 180 degrees.

We find the measure of <BDA using the knowledge of isosceles triangle and the triangle interior angle sum theorem. <BDA is 130 degrees.

We determine that <BDC is 50 degrees using the info we are given about collinear points.

The hash marks tell us that triangle BDC is isosceles so <BDC is also 50 degrees.

We use find the measure of the indicated angle by subtracting that from 180 degrees to get that <BCE is also 130 degrees.
These 4 steps require very simple calculations: subtractions from 180 or multiplying by two, but the number of steps leaves even this “easy” problem with plenty of room for mistakes.
ACT Math Easy #1 Answer: J  130
The first “easy” SAT problem covers topics that some students may not remember: what an arc or the circumference even is. Students who recognize this type of problem can quickly solve it by creating a ratio equation comparing the central angle and unknown arc to the sum of central angles and the circumference. If the previous sentence doesn’t make a lot of sense to you, that’s OK. It’s problems like this that drive the content we review.
Students who don’t immediately see a path to the solution should start with what they know: that square corner mark indicates that an angle has a measure of 90 degrees. Properties of intersecting straight lines inform us that each central angle within the circle has a measure of 90 degrees. If those four angles are equal, then we can assume that the arcs, parts of the circumference, must be equal as well. We can divide the total circumference into four to solve for the unknown arc.
SAT Math Easy #1 Answer: A  9
Our second “easy” ACT problem requires us to correctly use the common factoring skill on the expression 7a+7b. Higher level students will breeze past this problem recognizing that they can use common factoring to transform the expression without changing its value into 7(a+b). The final two options are a bit tempting, but the principles of combining like terms tell us that we cannot add the coefficients of two expressions unless the variables are the same. Notice that this problem can also be solved by looking at the answer choices and simply distributing the first option. This problem is an inch deep. It requires one skill, albeit not the one you might think of first, and can be backsolved very easily by working the first answer option.
ACT Math Easy #2 Answer: F  7(a+b)
Our second “easy” SAT problem is a bit more conceptual. In general, the SAT tests deep understanding of topics, so during our tutoring, we focus on training conceptual knowledge and deep learning of the topics we review. The SAT is also not as specific about how they will test your knowledge of math coursework. They will ask you many different facets of the same material. As such, you need to prepare deeply on the most important topics.
Like ACT Easy #2, SAT Easy number 2 also requires us to do common factoring. However, this SAT problem requires common factoring on both the numerator and denominator, then requires you to correctly cancel out terms that are present in both. There are both more steps and more concepts incorporated into this lowlevel problem.

x3+x2x4+x3x2(x+1)x3(x+1)

Cancel the term (x+1) in both numerator and denominator.

Simplify x2x3 to get 1x.

Recognize that is not one your answer choices and that

x1=1x
SAT Math Easy #2 Answer: D  x1
Phew. “Easy” problems out of the way. If you struggled with these problems, don’t worry! That is what test preparation is for. We teach strategy and content so that these problems will be familiar and easily solved!