## Overview

These questions are your classic standardized test math questions—think SAT but more difficult. These multiple-choice problems (five answer choices) test your quantitative reasoning ability by presenting questions that incorporate arithmetic, basic algebra, and elementary geometry. Some problems will be direct mathematical calculations while others will be word problems.

Problem solving questions are designed to test:

• Mathematical skills
• Understanding of elementary mathematical concepts
• The ability to reason quantitatively and solve quantitative problems

This section requires knowledge of the following:

• Arithmetic
• Elementary algebra
• Widely-known concepts of geometry

## Question 1

If a sandbox in the shape of a right triangle has a hypotenuse of h feet, an area of a square feet, and one leg of length x feet, which of the following must be true?

We could solve this question using algebra, but since the algebra is relatively complicated, and since there are variables in both the answer choices and the question prompt, it is a great candidate for the plugging in numbers strategy.

The easiest right triangle to work with is a 3-4-5 right triangle. In this case, the hypotenuse is 5, so h = 5. The area of the triangle is , so a = 6. Finally, since the triangle has legs of two different lengths, we could choose x = 4 or x = 3. We will choose x = 3, because smaller numbers are generally easier to work with.

Now that we have chosen values for the variables, we plug those values into the answer choices in order to identify the equation that must be true. Before we start, notice that answer choices B and E have only addition (no subtraction). We know that each expression in is positive, and we know that there is no way that three positive quantities can sum to zero, so there is no way the equation in B can be true. The same goes for the expression in answer choice E.

Therefore, we will plug the values above into answers A, C, and D.

A: . Since , we can write this as . Since this is not equal to zero, A is incorrect.

C: . Since , we can write this as . C is a possible answer, but we cannot stop here, since it is possible that answer D is also equal to zero.

D: . Since , we can write this as . Since this is not equal to zero, D is incorrect.

We can also solve the problem algebraically. Let y be the length of the second leg of the triangle. Then, we can use the Pythagorean Theorem to write x2 + y2 = h2, and we can use the formula for area of a triangle to write . In order to arrive at a statement that does not have the variable y in it, we solve for y in the second equation and then substitute that value of y into the first equation. Since , we can multiply by 2 and divide by x to find .

We then substittue this into the first equation. . This simplifies to . If we subtract h2 from both sides, we get . This is answer C, so C is correct.

## Question 2

Which of the following is equal to the average (arithmetic mean) of , , and ?

This question asks us for the average of three algebraic expressions. Since there are variables in the question prompt and in the answer choices, we can plug in a value for x and then compare the output from the question prompt with the output from the answer choices.

Let x = 4. The three expressions in the question prompt are then equal to (x + 2)2 = 62 = 36, (x + 4)2 = 82 = 64, and x2 – 11 = 42 – 11 = 16 – 11 = 5. We use the average formula to find the average of these three terms: . In this case the average is . Now we plug x = 4 into the answer choices to see which ones yield the same output value, 35.

A: (x + 2)(x + 3) = (6)(7) = 42

B: (x – 1)(x – 2) = (3)(2) = 6

C: (x – 4)(x + 1) = (0)(5) = 0

D: (x + 4)(x – 2) = (8)(2) = 16

E: (x + 3)(x + 1) = (7)(5) = 35

Only answer E gives the desired output of 35. Therefore, E is the right answer.

We can also solve this question algebraically. We compute the average of algebraic expressions in EXACTLY the same way as we compute the average of three numbers. In this case, the average is: . Since this is not one of the answer choices, we must simplify the numerator.

First, multiply out each expression in the numerator. This gives us . Then, combine like terms: . Next, perform the division: x2 + 4x + 3. This, too, is not an answer choice. It looks like we need to factor the expression. x2 + 4x + 3 = (x + 3)(x + 1), because 3 + 1 = 4 and 3 × 1 = 3.

(x + 3)(x + 1) is answer choice E, so E is correct.

## Question 3

In a certain box of sports drinks, of the bottles contain Orange Drink and of the remaining bottles contain Lemon Drink. If b bottles in the box contain Lemon Drink, how many bottles in the box contain Orange Drink?

This question is a classic candidate for the plugging in numbers strategy, because it contains fractions of an unknown (and irrelevant) whole. We can pick the total number of bottles in the box and work from there. It is a good idea to pick a number that works as a common denominator for the fractions in the question. In this case, let’s assume that there are 42 bottles in the box, since the denominators of the fractions in the question are 6 and 7, and 6 × 7 = 42.

Orange Drink bottles make up of the box, so bottles of Orange Drink. The remaining 42 – 7 = 35 bottles contain something other than Orange Drink.

of the remaining bottles are Lemon Drink, so bottles have Lemon Drink. This means that b = 10.

The question asks how many bottles contain Orange Drink. Since the answer in this case is 7, we need to plug in b = 10 and find the answer choice that yields 7.

A: since 42 × 7 = 294.

B: . This choice has the desired output of 7.

C: since 7 × 7 = 49.

D: 7b = 7 × 10 = 70 ≠ 7.

E: since 14 × 7 is much bigger than 30.

Only B yields 7, so the correct answer is B.
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