SAT Test Problem Solving (MC)

1.)  D  To solve an averages problem, you should multiply for the products of the average. In this case, a 3.0 GPA after 3 semesters gives us 9. (3 times 3.) We also know that after the fourth semester, the cumulative GPA is supposed to be a 3.1. Hence, this multiplier's product is 12.4. (3.1 times 4.) Subtracting 9 from 12.4 gives you choice D, 3.4. We hope you were able to eliminate choice A by noting that a 2.7 in the fourth semester would make the GPA less than 3.0 and would not increase it to 3.1.

2.)  D  Using a brief picture such as the one below can be helpful with this type of question. (Remember not to spend too much time on the accuracy and neatness of the picture. You can follow our example here!) Hopefully, you were able to eliminate choice A outright.

The algebraic equation for question 2 would be 2x + y = 24 where x equals the number of students in each club and 24 equals the total number of members for both the chess team and the science club. Inserting 5 for x yields a y value of 14, which is the number of students who belong to only one of the organizations.

3.)  A  We certainly hope the terribly-drawn fraction problem wasn't too much of a distraction, but hey we aren't guru web designers and this is free! Simplification is the key to this question. Remember that dividing a fraction is the same as multiplying by its inverse. The "2's" and the "8's" will cancel out and that will leave you with (1 x 3 x 1) in the numerator and (2 x 4 x 4) in the denominator.

4.)  E  All of these fractions have a numerator that is one greater than the denominator. Hence, they all equal 1 + 1/denominator. The larger the denominator, the smaller the number.

5.)  D  The key to solving this question is to multiply the second term's numerator and denominator by -1. Since 7/(a-b) is not an answer choice, you must re-examine the answers, multiply this solution by -1 and you will see it equals -7/(b-a).

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