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Diagnostic Test 521

15.

8x2 – 18 =

 

 

2(4x2 – 9) =

by the distributive property ax + ay = a(x + y)

 

2(22x2 – 32) =

 

 

2([2x]2 – 32) =

by the difference of squares formula x2 – y2 = (x + y)(x – y)

 

2(2x + 3)(2x – 3)

The answer is (B).

It is common for students to wrongly apply the difference of squares formula to a perfect square: (x – y)2 ≠ (x + y)(x – y)

The correct formulas follow. Notice that the first formula is the square of a difference, and the second formula is the difference of two squares.

Perfect square trinomial:

(x – y)2 = x2 – 2xy + y2

Difference of squares:

x2 – y2 = (x + y)(x – y)

It is also common for students to wrongly distribute the 2 in a perfect square: (x – y)2 ≠ x2 – y2

Note, there is no factoring formula for a sum of squares: x2 + y2. It cannot be factored.

16. First, replace the inequality symbol with an equal symbol:

x2

= 2x

Subtracting 2x from both sides yields

x2

– 2x = 0

Factoring by the distributive rule yields

x(x – 2) = 0

Setting each factor to 0 yields

x = 0 and x – 2 = 0

Or

x = 0 and x = 2

Now, the only numbers at which the expression can change sign are 0 and 2. So 0 and 2 divide the number line into three intervals. Let’s set up a number line and choose test points in each interval:

Interval I

 

Interval II

 

Interval III

–1

0

1

2

3

When x = –1, x2 < 2x becomes 1 < –2. This is false. Hence, no numbers in Interval I satisfy the inequality. When x = 1, x2 < 2x becomes 1 < 2. This is true. Hence, all numbers in Interval II satisfy the inequality. That is, 0 < x < 2. When x = 3, x2 < 2x becomes 9 < 6. This is false. Hence, no numbers in Interval III satisfy the inequality. The answer is (B). The graph of the solution follows:

02

17.Since y is to be less than 1 and y = –3x + 7, we get

–3x + 7 < 1

 

–3x < –6

by subtracting 7 from both sides of the inequality

x > 2

by dividing both sides of the inequality by –3

(Note that the inequality changes direction when we divide both sides by a negative number. This is also the case if you multiply both sides of an inequality by a negative number.)

Since x is an integer and is to be as small as possible, x = 3. The answer is (C).

522 GRE Math Bible

18. Since the triangle is isosceles, with base AC, the base angles are congruent (equal). That is, A = C. Since the angle sum of a triangle is 180, we get

A + C + x = 180

Replacing C with A and x with 60 gives

A + A + 60 = 180 A + A + 60 = 180 2A + 60 = 180 2A = 120

A = 60

Hence, the triangle is equilateral (all three sides are congruent). Since we are given that side AB has length 4, side AC also has length 4. The answer is (C).

19. Since the unit square is circumscribed about the circle, the diameter of the circle is 1 and the radius of the circle is r = d/2 = 1/2. This is illustrated in the following figure:

 

1

1

1/2

1

1

Now, the circumference of a circle is given by the formula 2πr. For this circle the formula becomes 2πr = 2π(1/2) = π. We are told that the circumference of the circle is qπ. Setting these two expressions equal yields

 

 

 

π = qπ

 

 

Dividing both sides of this equation by π yields

 

 

 

 

 

1 = q

 

 

The answer is (A).

 

 

 

 

 

20. Let x be the unknown side of the triangle. Applying the Pythagorean Theorem yields

 

 

92 + x2 = 152

 

 

 

 

 

81+ x2 = 225

by squaring the terms

 

 

x2 = 144

by subtracting 81 from both sides of the equation

x = ±

144

 

by taking the square root of both sides of the equation

x = 12

since we are looking for a length, we take the positive root

In a right triangle, the legs are the base and the height of the triangle. Hence, A =

1

bh =

1

9 12 = 54 .

 

2

 

 

 

2

 

 

The answer is (D).

21. Since the average of 2x and 4x is 12, we get

2x + 4x = 12

2

62x = 12

3x = 12 x = 4

The answer is (D).

Diagnostic Test 523

22. Recall that the average of N numbers is their sum divided by N. That is, average = sum/N. Since the average of x, y, and z is 8 and the average of y and z is 4, this formula yields

x + y + z = 8 3

y +2 z = 4

Solving the bottom equation for y + z yields y + z = 8. Plugging this into the top equation gives x +3 8 = 8

x + 8 = 24 x = 16

The answer is (C).

23. Let the two numbers be x and y. Now, a ratio is simply a fraction. Forming the fraction yields x/y = 6, and forming the sum yields x + y = 21. Solving the first equation for x yields x = 6y. Plugging this into the second equation yields

6y + y = 21

7y = 21 y = 3

Plugging this into the equation x = 6y yields

x = 6(3) = 18

The answer is (E).

24. Let z% represent the unknown percent. Now, when solving percent problems, “of” means times. Translating the statement “What percent of 3x is 6y” into an equation yields

z%(3x) = 6y

Substituting x = 4y into this equation yields

z%(3 4y) = 6y z%(12y) = 6y

z% = 6y

12y

z% = 1/2 = .50 = 50%

The answer is (A).

25.Translating Column A into a mathematical expression yields .10y. Translating Column B into a mathematical expression yields .40x. Since y = 3x, Column A becomes .10y = .10(3x) = .30x. Since .40 is larger than .30 and x is positive, Column B is larger. The answer is (B).

26.Let x be the amount of water added. Since there is no alcohol in the water, the percent of alcohol in the water is 0%x. The amount of alcohol in the original solution is 40%(30), and the amount of alcohol in the final solution will be 25%(30 + x). Now, the concentration of alcohol in the original solution plus the concentration of alcohol in the added solution (water) must equal the concentration of alcohol in the resulting solution:

40%(30) + 0%x = 25%(30 + x)

Multiplying this equation by 100 to clear the percent symbol yields

40(30) + 0 = 25(30 + x) 1200 = 750 + 25x

450 = 25x

18 = x

The answer is (E).

524 GRE Math Bible

27. Except for the first term, each term of the sequence is found by adding 4 to the term immediately preceding it. In other words, we are simply adding 4 to the sequence 200 times. This yields

4 200 = 800

Adding the 2 in the first term gives 800 + 2 = 802. The answer is (C).

We can also solve this problem formally. The first term of the sequence is 2, and since each successive term is 4 more than the term immediately preceding it, the second term is 2 + 4, and the third term is (2 + 4) + 4, and the fourth term is [(2 + 4) + 4] + 4, etc. Regrouping yields (note that we rewrite the first term as 2 + 4(0). You’ll see why in a moment.)

2 + 4(0), 2 + 4(1), 2 + 4(2), 2 + 4(3), . . .

Notice that the number within each pair of parentheses is 1 less than the numerical order of the term. For instance, the first term has a 0 within the parentheses, the second term has a 1 within the parentheses, etc. Hence, the nth term of the sequence is

2 + 4(n – 1)

Using this formula, the 201st term is 2 + 4(201 – 1) = 2 + 4(200) = 2 + 800 = 802.

28. For the first model, there are 5 options. So there are 5 different types of cars in this model. For the second model, there are the same number of different types of cars. Likewise, for the other two types of models. Hence, there are 5 + 5 + 5 + 5 = 20 different types of cars. The answer is (E).

This problem illustrates the Fundamental Principle of Counting:

If an event occurs m times, and each of the m events is followed by a second event which occurs k times, then the first event follows the second event m k times.

29. This is considered to be a hard problem. However, it is actually quite easy. By the definition given, the function @ merely cubes the term on the left and then subtracts 1 from it (the value of the term on the right is irrelevant). In each column, the term on the left is x. Hence, in each case, the result is x3 – 1. This shows the two expressions are equal, and the answer is (C).

30.

((1 − x) *)* = (1− x )*

 

(1− (1− x))* = (1− x) *

 

(1− 1+ x)* = (1− x )*

 

(x )* = (1− x) *

 

1 − x = 1− (1− x)

 

1

– x = 1 – 1 + x

 

1

– x = x

 

1

= 2x

The answer is (A).

1/2 = x

 

 

Study Plan

Use the list below to review the appropriate chapters for any questions you missed.

Equations: Page 259

Factoring: Page 316

Percents: Page 330

Questions: 1, 2, 3

Questions: 14, 15

Questions: 24, 25

Algebraic Expressions: Page 322

Inequalities: Page 230

Word Problems: Page 376

Questions: 4, 5, 6

Questions: 16, 17

Question: 26

Number Theory: Page 37

Geometry: Page 78

Sequences & Series: Page 397

Questions: 7, 8

Questions: 18, 19, 20

Question: 27

Exponents & Roots: Page 304

Averages: Page 274

Counting: Page 405

Questions: 9, 10, 11

Questions: 21, 22

Question: 28

Fractions & Decimals: Page 243

Ratio & Proportion: Page 285

Defined Functions: Page 28

Questions: 12, 13

Question: 23

Questions: 29, 30

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