# 5 Strategic Approaches To SAT MATHS Questions

The SAT Math, in the latest (2016 revised) form, split into two sections – the calculator section and the non-calculator section, isn’t as much about tricks as it is about conceptual clarity. However, the tricks that used to work in the previous version of the SAT, still work in favor of the test-takers’ pace in solving the questions.

We shall not only try revisiting the old tricks but also check the validity of the above using some math questions from the Official (CollegeBoard’s) SAT practice papers.

Example 1:

The above question is picked from the Practice Test 8 provided by the CollegeBoard.

Subtracting 4 from both sides of the equation, we get

Squaring on both sides, we get 2x + 6 = x2 + 1 -2x

Solving the quadratic equation, we get x = 5, -1.

However, the trick lies here. One of the above solutions fits into the given equation in the question while the other doesn’t.

“The traditional math approach was vulnerable to traps!”

If we now try the “Substituting the Answers” method and verify whether the answer choice A (-1) is a valid solution to the given equation. We may be surprised to find that it is not. Now, we substitute the answer choice B (5) and find that it is a valid solution. We simply choose B as the right answer choice.

The Substitution Strategy is much more powerful now. It saved us from falling prey to the trap. It also saved a lot of time.

Example 2:

The above question is again picked from the same test paper from College Board.

Once you read the question carefully, you will notice that the last sentence in it has “the total amount of energy cost savings will exceed the installation cost”.

The question also gives information that the installation cost is $25000. Now, if something has to exceed the installation cost, then something has to be greater than$25000.

The only answer choice which has “> 25000” or “25000 <” is the answer choice B, which is also the right answer in this case.

Generally, the questions which are too wordy such as the above one are easily solved. In this example, what works best is the “Process of Elimination”.

Strategic Elimination to solve questions is here to stay and is going to save a lot of time.

“Elimination eliminates a lot of work!”

Example 3:

The above question picked up from the same test falls under the area “Additional Topics in Math” tested on the SAT Math. The underlying concept is basic trigonometry. There are no answer choices given making it a bit more challenging. One needs to be good at geometry and basic trigonometry to solve the above question.

However, the following approach may make the solution a lot simpler.

The point ‘W’ mentioned in the question but (not shown) can be assumed to be lying anywhere on RT, nearer to R or nearer to T or in fact on R or on T as well. Yes, if we assume it to be lying on the point T itself, the solution turns out to be pretty simple.

If we assume W to lie on T itself, then the required value   is going to be cos (angle RST) – sin (angle TST) = cos 90 – sin 0 = 0-0 = 0.

The traditional math approach would have been to apply the concept cos(90-x) = sin x and/or sin(90-x) = cos x and get the answer as 0.

The strategy demonstrated in the above simpler solution is to “exploit the design of the question”.

Since the question gave us the freedom to assume the point W anywhere on RT, we choose the easier way out.

“Exploit the information in the question”

Example 4:

The above question is from the same paper.

When solving the above question, it is very much possible that one tries solving for x and then may solve for x+1. Sometimes, the time pressure during the test may make test takers choose the wrong answer (the value of x instead of x+1) too. These type of errors are very common on the test.

If one chooses to relax and take it easy, he/she may not only save a lot of time and effort but also get the right answers. In this example, it is simply about multiplying with (x+1) on both sides and applying square root on both sides. The value of (x+1) from the equation, (x+1)2 = 2 can be easily derived as + √2. Hence, the right answer would be C.

The strategy demonstrated in the above method is “Reading the Question Carefully” which is a very important task in not just the SAT but also in any other test.

Example 5:

In the above question, equating the values of y, we get a quadratic equation in x, which generally is assumed to be having 2 solutions. However, when we equate the values of y, we get x2-2x+1 = 0, which implies (x-1)2 = 0 and then x = 1 only. There are high chances that people choose the right answer as B and fall prey to the trap. The right answer obviously has to be C since the roots of the quadratic equation turned out to be equal. Hence, only 1 solution exists for the pair of equations given.

This question is in the calculator section of the test. If one used the graphical calculator to plot the graphs of the given equations, it would be clear that the parabola and the line intersect at only 1 point and hence the have only 1 solution.

The point illustrated in the above question is that, though the calculator may not be a necessary option for all the questions on the calculator section of the test, it may sometimes save you from answering questions wrongly.
Use the Calculator whenever necessary and know when it is necessary!

For more such strategies to score high on the SAT, you may visit fabmarks.com and access a lot of FREE learning RESOURCES available apart from the PAID Premium Services from fabmarks.

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