1.      A     

If a statement is true, the contrapositive of the statement will also be true. To find the contrapositive, you need to take the opposite of both parts of the statement and then switch the order. The contrapositive of the original sentence is “If a student doesn’t do well on the SAT IIs, then that student didn’t study hard.”

2.      B     

Given two terms of a geometric sequence, it is possible to find the common ratio between consecutive terms. In any geometric series, g₃ = g₁ r^3–1 = g₁ r². (More generally, g_a = g_b r^a–b.) In this problem, ³/ ₄ = 3r². So, r² = ¹/₄, and therefore r = ¹ /₂. Now that you know the value of r and the value of at least one term in the geometric sequence, you can find any term. In this case, g₁₀ = g₁ r^10–1 = 3 ( ¹ /₂)⁹ = ³/₅₁₂.

3.      B     

The sum of an infinite geometric series is finite if |r| < 1 and infinite if |r| > 1. The formula for x_n is 8 (¹ /₂)ⁿ, where r = ¹ /₂ is the common ratio between consecutive terms. The sum of this series is therefore finite. The formula for the sum of an infinite series is ^x₁ /_{1 – r}, where r is the common ratio. In this problem, the sum is ⁴/_{1 – .5} = 8.

4.      D     

In order to find the limit as it approaches a value, plug the value into the expression for x. For this problem you immediately run into a problem, because plugging 3 into the expression produces a 0 in the denominator, and you cannot divide by 0. If this happens, see if the expression can be factored:

Now you can plug 3 into the simplified expression:

5.      A     

The algebra for complex numbers is the same as for real numbers. For this problem, use FOIL and keep in mind that i² = –1.

6.      C     

The magnitude of a complex number is the distance from the origin to that number in the complex plane. Using the Pythagorean theorem, the magnitude of the complex number a + bi is . In this problem the complex number is 3 – 7i, so its magnitude is = = ≈ 7.62.