Algebra

For questions in the Quantitative Comparison format (“Quantity A” and “Quantity B” given), the answer choices are always as follows:

(A)Quantity A is greater.
(B)Quantity B is greater.
(C)The two quantities are equal.
(D)The relationship cannot be determined from the information given.

For questions followed by a numeric entry box , you are to enter your own answer in the box. For questions followed by a fraction-style numeric entry box , you are to enter your answer in the form of a fraction. You are not required to reduce fractions. For example, if the answer is , you may enter or any equivalent fraction.

All numbers used are real numbers. All figures are assumed to lie in a plane unless otherwise indicated. Geometric figures are not necessarily drawn to scale. You should assume, however, that lines that appear to be straight are actually straight, points on a line are in the order shown, and all geometric objects are in the relative positions shown. Coordinate systems, such as xy-planes and number lines, as well as graphical data presentations, such as bar charts, circle graphs, and line graphs, are drawn to scale. A symbol that appears more than once in a question has the same meaning throughout the question.

1.If 4(–3x – 8) = 8(–x + 9), what is the value of x²?

2.If 2x(4 – 6) = –2x + 12, what is the value of x?

3.If x ≠ 0 and = –6, what is the value of x?

4.If x ≠ 2 and = 17, what is the value of x?

–5 is 7 more than –z.

5.

Quantity A z

Quantity B –12

6.If (x + 3)² = 225, which of the following could be the value of x – 1?

(A)13

(B)12

(C)–12

(D)–16

(E)–19

x = 2

7.

Quantity A x2 – 4x + 3

Quantity B 1

p = 300c² – c
c = 100

8.

Quantity A p

Quantity B 29,000c

–(x)³ = 64

9.

Quantity A x4

Quantity B x5

10. If 3t³ – 7 = 74, what is the value of t² – t?

(A)–3

(B)3

(C)6

(D)9

(E)18

11. If y = 4x + 10 and y = 7x – 5, what is the value of y?

12. If x – y = 4 and 2x + y = 5, what is the value of x?

13. 4x + y + 3z = 34
4x + 3z = 21

What is the value of y?

14.

Quantity A (x + 2) (x – 3)

Quantity B x2 – x – 6

xy > 0

15.

Quantity A (2x – y)(x + 4y)

Quantity B 2x2 + 8xy – 4y2

x² – 2x = 0

16.

Quantity A x

Quantity B 2

17.

Quantity A d(d2 – 2d + 1)

Quantity B d(d2 – 2d) + 1

18.

Quantity A xy2z(x2z + yz2 – xy2)

Quantity B x3y2z2 + xy3z3 – x2y4z

a = 2b = 4c and a, b, and c are integers.

19.

Quantity A a + b

Quantity B a + c

k = 2m = 4n and k, m, and n are non-negative integers.

20.

Quantity A km

Quantity B kn

For the positive integers a, b, c, and d, a is half of b, which is one-third of c. The value of d is three times the value of c.

21.

Quantity A

Quantity B

22. If x² – y² = 0 and xy ≠ 0, which of the following must be true?

Indicate all such statements.

• x = y
• |x| = |y|
• = 1

3x + 6y = 27
x + 2y + z = 11

23.

Quantity A z + 5

Quantity B x + 2y – 2

24. If (x – y) = and (x + y) = , what is the value of x² – y²?

(A)3

(B)6

(C)9

(D)36

(E)It cannot be determined from the information given.

a ≠ b

25.

Quantity A

Quantity B 1

a =
c = 3b

26.

Quantity A a

Quantity B c

27. If xy ≠ 0 and x ≠ –y,

(A)1

(B)x² – y²

(C)x⁹ – y⁹

(D)x¹⁸ – y¹⁸

(E)

28. If x ≠ –y, what is the value of ?

(A)1

(B)

(C)

(D)xy

(E)2xy

x > y
xy ≠ 0

29.

Quantity A

Quantity B

30. If x + y = –3 and x² + y² = 12, what is the value of 2xy?

31. If x – y = and x² – y² = 3, what is the value of x + y?

32. If x² – 2xy = 84 and x – y = –10, what is the value of |y|?

33. Which of the following is equal to (x – 2)² + (x – 1)² + x² + (x + 1)² + (x + 2)²?

(A)5x²

(B)5x² + 10

(C)x² + 10

(D)5x² + 6x + 10

(E)5x² – 6x + 10

34. If a = (x + y)² and b = x² + y² and xy > 0, which of the following must be true?

Indicate all such statements.

• a = b
• a > b
• a is positive

35. a is directly proportional to b. If a = 8 when b = 2, what is a when b = 4?

(A)10

(B)16

(C)32

(D)64

(E)128

 

Algebra Answers

1. 676. Distribute, group like terms, and solve for x:

4(–3x – 8)	=	8(–x + 9)
–12x – 32	=	–8x + 72
–32	=	4x + 72
–104	=	4x
–26	=	x

Then, multiply 26 by 26 in the calculator (or –26 by –26, although the negatives will cancel each other out) to get x², which is 676.

2. –6.

2x(4 – 6)	=	–2x + 12
2x(–2)	=	–2x + 12
–4x	=	–2x + 12
–2x	=	12
x	=	–6

3. –2. = –6

Multiply both sides by 2x, distribute the left side, combine like terms, and solve:

3(6 – x)	=	–6(2x)
18 – 3x	=	–12x
18	=	–9x
–2	=	x

4. –6. = 17

Multiply both sides by the expression 2 – x, distribute both sides, combine like terms, and solve:

8 – 2(–4 + 10x)	=	17(2 – x)
8 + 8 – 20x	=	34 – 17x
16 – 20x	=	34 – 17x
16	=	34 + 3x
–18	=	3x
–6	=	x

5. (A). Translate the question stem into an equation and solve for z:

–5	=	–z + 7
–12	=	–z
12	=	z

Because z = 12 > –12, Quantity A is greater.

6. (E). Begin by square-rooting both sides of the equation, but remember that 225 could be the square of either 15 or –15. (The calculator will not remind you of this! It’s your job to keep this in mind). So:

x + 3	=	15
x	=	12
so, x – 1	=	11

OR

x + 3	=	–15
x	=	–18
so, x – 1	=	–19

Only –19 appears in the choices.

7. (B). To evaluate the expression in Quantity A, replace x with 2.

x2 – 4x + 3	=
(2)2 – 4(2) + 3	=
4 – 8 + 3	=	–1 < 1

Therefore, Quantity B is greater.

8. (A). To find the value of p, first replace c with 100 to find the value for Quantity A:

p	=	300c2 – c
p	=	300(100)2 – 100
p	=	300(10,000) – 100
p	=	3,000,000 – 100 = 2,999,900

Since c = 100, the value for Quantity B is 29,000(100) = 2,900,000. Quantity A is greater.

9. (A). First, solve for x:

–(x)3	=	64
(x)3	=	–64

The GRE calculator will not do a cube root. As a result, cube roots on the GRE tend to be quite small and easy to puzzle out. What number times itself three times equals –64? The answer is x = –4.

Since x is negative, Quantity A is positive (a negative number times itself four times is positive) and Quantity B is negative (a negative number times itself five times is negative). No further calculations are needed to conclude that Quantity A is greater. Notice that solving for the value of x here was not strictly necessary. Knowing that the cube root of a negative number is negative gives you all the information you need to solve.

10. (C). First, solve for t:

3t3 – 7	=	74
3t3	=	81
t3	=	27
t	=	3

Now, plug t = 3 into t² – t:

     (3)² – 3 = 9 – 3 = 6

11. 30. Since each equation is already solved for y, set the right side of each equation equal to the other.

4x + 10	=	7x – 5
10	=	3x – 5
15	=	3x
5	=	x

Substitute 5 for x in the first equation and solve for y.

y	=	4(5) + 10
y	=	30

x = 5 and y = 30. Be sure to answer for y, not x.

12. 3. Notice that the first equation has the term –y while the second equation has the term +y. While it is possible to use the substitution method, summing the equations together will make –y and y cancel, so this is the easiest way to solve for x.

	x – y̸	=	4
	2x + y̸	=	5
	3x	=	9
	x	=	3

13. 13. This question contains only two equations, but three variables. To isolate y, both x and z must be eliminated. Notice that the coefficients of x and z are the same in both equations. Subtract the second equation from the first to eliminate x and z.

	4x + y + 3z	=	34
	–(4x       + 3z)	=	21)
	y	=	13

14. (C). FOIL the terms in Quantity A:

     (x + 2)(x – 3) = x² – 3x + 2x – 6 = x² – x – 6

The two quantities are equal.

15. (B). FOIL the terms in Quantity A:

     (2x – y)(x + 4y) = 2x² + 8xy – xy – 4y² = 2x² + 7xy – 4y²

Since 2x² and –4y² appear in both quantities, eliminate them. Quantity A is now equal to 7xy and Quantity B is now equal to 8xy. Because xy > 0, Quantity B is greater. (Don’t assume! If xy were zero, the two quantities would have been equal. If xy were negative, Quantity A would have been greater.)

16. (D). Factor x² – 2x = 0:

     x² – 2x = 0
     x(x – 2) = 0
     x = 0 OR (x – 2) = 0

x = 0 or 2.

Thus, Quantity A could be less than or equal to Quantity B. The relationship cannot be determined from the information given.

(Note that you cannot simply divide both sides of the original equation by x. It is illegal to divide by a variable unless it is certain that the variable does not equal zero.)

17. (D). In Quantity A, multiply d by every term in the parentheses:

     d(d² – 2d + 1) =
     (d × d²) – (d × 2d) + (d × 1) =
     d³ – 2d² + d

In Quantity B, multiply d by the two terms in the parentheses:

     d(d² – 2d) + 1 =
     (d × d²) – (d × 2d) + 1 =
     d³ – 2d² + 1

Because d³ – 2d² is common to both quantities, it can be ignored. The comparison is really between d and 1. Without more information about d, the relationship cannot be determined from the information given.

18. (C). In Quantity A, the term xy²z on the outside of the parentheses must be multiplied by each of the three terms inside the parentheses. Then simplify the expression as much as possible.

Taking one term at a time, the first is xy²z × x²z = x³y²z², because there are three factors of x, two factors of y, and two factors of z. Similarly, the second term is xy²z × yz² = xy³z³ and the third is xy²z × (–xy²) = –x²y⁴z. Adding these three terms together gives the distributed form of Quantity A: x³y²z² + xy³z³ – x²y⁴z.

This is identical to Quantity B, so the two quantities are equal.

19. (D). Since a is common to both quantities, it can be ignored. The comparison is really between b and c. Because 2b = 4c, it is true that b = 2c, so the comparison is really between 2c and c. Watch out for negatives. If the variables are positive, Quantity A is greater, but if the variables are negative, Quantity B is greater.

20. (D). If the variables are positive, Quantity A is greater. However, all three variables could equal zero, in which case the two quantities are equal. Watch out for the word “non-negative,” which means “positive or zero.”

21. (C). The following relationships are given: a = , b = , and d = 3c. Pick one variable and put everything in terms of that variable. For instance, variable a:

     b = 2a
     c = 3b = 3(2a) = 6a
     d = 3c = 3(6a) = 18a

Substitute into the quantities and simplify.

Quantity A:

Quantity B:

The two quantities are equal.

22. |x| = |y| and . Since x² – y² = 0, add y² to both sides to get x² = y². It might look as though x = y, but this is not necessarily the case. For example, x could be 2 and y could be –2. Algebraically, taking the square root of both sides of x² = y² does not yield x = y, but rather |x| = |y|. Thus, the 1st statement is not necessarily true and the 2nd statement is true. The 3rd statement is also true and can be generated algebraically:

x2 – y2	=	0
x2	=	y2
	=	1

23. (C). This question may at first look difficult, as there are three variables and only two equations. However, notice that the top equation can be divided by 3, yielding x + 2y = 9. This can be plugged into the second equation:

(x + 2y) + z	=	11
(9) + z	=	11
z	=	2

Quantity A is thus 2 + 5 = 7. For Quantity B, remember that x + 2y = 9. Thus, Quantity B is 9 – 2 = 7.

The two quantities are equal.

24. (B). The factored form of the Difference of Squares (one of the “special products” you need to memorize for the exam) is comprised of the terms given in this problem:

     x² – y² = (x + y)(x – y)

Substitute the values and in place of (x – y) and (x + y), respectively:

     x² – y² = ×

Combine 12 and 3 under the same root sign and solve:

x2 – y2	=
x2 – y2	=
x2 – y2	=	6

25. (B). Plug in any two unequal values for a and b, and Quantity A will always be equal to –1. This is because a negative sign can be factored out of the top or bottom of the fraction to show that the top and bottom are the same, except for their signs:

26. (D). To compare a and c, put c in terms of a. Multiply the first equation by 2 to find that b = 2a. Substitute into the second equation: c = 3b = 3(2a) = 6a. If all three variables are positive, then 6a > a. If all three variables are negative, then a > 6a. Finally, all three variables could equal zero, making the two quantities equal.

27. (C). The Difference of Squares (one of the “special products” you need to memorize for the exam) is x² – y² = (x + y)(x – y). This pattern works for any perfect square minus another perfect square. Thus, x³⁶ – y³⁶ will factor according to this pattern. Note that , or x³⁶ = (x¹⁸)². First, factor x³⁶ – y³⁶ in the numerator, then cancel x¹⁸ + y¹⁸ with the x¹⁸ + y¹⁸ on the bottom:

The x¹⁸ – y¹⁸ in the numerator will also factor according to this pattern. Then cancel x⁹ + y⁹ with the x⁹ + y⁹ on the bottom:

28. (B). First, recognize that x² + 2xy + y² = (x + y)². This is one of the “special products” you need to memorize for the exam. Factor the top, then cancel:

29. (D). It is possible to simplify first and then plug in examples, or to just plug in examples without simplifying. For instance, if x = 2 and y = 1:

Quantity A:
Quantity B:

In this case, Quantity A is greater. Next, try negatives. If x = –1 and y = –2 (remember, x must be greater than y):

Quantity A:
Quantity B:

Quantity A is still greater. However, before assuming that Quantity A is always greater, make sure you have tried every category of possibilities for x and y. What if x is positive and y is negative? For instance, x = 2 and y = –2:

Quantity A:
Quantity B:

In this case, Quantity B is greater. The relationship cannot be determined from the information given.

30. –3. One of the “special products” you need to memorize for the GRE is x² + 2xy + y² = (x + y)². Write this pattern on your paper, plug in the given values, and simplify, solving for 2xy:

x2 + 2xy + y2	=	(x + y)2
(x2 + y2) + 2xy	=	(x + y)2
(12) + 2xy	=	(–3)2
12 + 2xy	=	9
2xy	=	–3

31. 6. The Difference of Squares (one of the “special products” you need to memorize for the exam) is x² – y² = (x + y)(x – y). Write this pattern on your paper and plug in the given values, solving for x + y:

x2 – y2	=	(x + y)(x – y)
3	=	(x + y)(1/2)
6	=	x + y

32. 4. One of the “special products” you need to memorize for the exam is x² – 2xy + y² = (x – y)². Write this pattern on your paper and plug in the given values:

x2 – 2xy + y2	=	(x – y)2
84 + y2	=	(–10)2
84 + y2	=	100
y2	=	16
y	=	4 or –4, so |y| = 4.

33. (B). First, multiply out (remember FOIL = First, Outer, Inner, Last) each of the terms in parentheses:

     (x² – 2x – 2x + 4) + (x² – 1x – 1x + 1) + (x²) + (x² + 1x + 1x + 1) + (x² + 2x + 2x + 4)

Note that some of the terms will cancel each other out (e.g., –x and x, –2x and 2x):

     (x² + 4) + (x² + 1) + (x²) + (x² + 1) + (x² + 4)

Finally, combine:

     5x² + 10

34. a > b and a is positive. Distribute for a: a = (x + y)² = x² + 2xy + y². Since b = x² + y², a and b are the same except for the “extra” 2xy in a. Since xy is positive, a is greater than b. The 1st statement is false and the 2nd statement is true.

Each term in the sum for a is positive: xy is given as positive, and x² and y² are definitely positive, as they are squared and not equal to zero. Therefore, a = x² + 2xy + y² is positive. The 3rd statement is true.

35. (B). To answer this question, it is important to understand what is meant by the phrase “directly proportional.” It means that a = kb, where k is a constant. In alternative form: = k, where k is a constant.

So, because they both equal the constant, . Plugging in values: . Cross-multiply and solve:

     32 = 2a_new
     a_new = 16