Tribunal Constitucional “sin ideología” de partidos altruista

Discrete Quantitative Questions

1. If a < b < 0, then which of the following must be true?

A ab < 0 B _{a + b > 0} C a b < 0 D _{b – a > 0} E a – b > 0

2. If xy > 0 and yz < 0, then which of the following must be negative?

A _xyz B xy2z C _x2_y2_z D x2y2z2 E xy z

3. If ab2 _{> 0 and ac < 0, then which of the following must be true?}

A ab >0 B b > 0 C a c < 0 D b2c < 0 E a(c2) > 0

4. If 0 > x > y, then which of the following must be true? (Indicate all that apply.) A x2 – y2 < 0 B _{y – x < 0} C 1 x2 < 1 D x + y x > 0 E y2 – x2 < 0

5. If _z2 _{+ 1} > 0, then which of the following must be true? A x > 0 B x < a C x > a D xa > 0 E x + a > 0

6. If xy > 0 and x + y > 0, then which of the following must be true?

A _{x < 0} B |x| > |y| C _{x > 0} D x y > 0 E _{y > 0}

Quantitative Comparison Questions

Each of the following questions consists of two quantities, Quantity A and Quantity B. You are to compare the two quantities. You may use additional information centered above the two quantities if additional information is given. Choose

A if Quantity A is greater

B if Quantity B is greater

C if the two quantities are equal

D if the relationship between the two quantities cannot be determined

0 > a > b QUANTIT Y A QUANTIT Y B 1. a2 _b4 _{A B C D} |x| > |y| QUANTIT Y A QUANTIT Y B 2. x2 _y2 _{A B C D}

a and b do not equal zero

QUANTIT Y A QUANTIT Y B

a b = 4 QUANTIT Y A QUANTIT Y B 4. a b A B C D a b > 0 QUANTIT Y A QUANTIT Y B 5. ab 0 A B C D p2_q3_>0 p3_q2 _{> 0} QUANTIT Y A QUANTIT Y B 6. pq 0 A B C D a2_b3_c5 _>0 a3_b4_c5 _{< 0} QUANTIT Y A QUANTIT Y B 7. ab 0 A B C D

Exercise Answers

Discrete Quantitative Questions

1. D Since a and b are negative, their product and quotient must be positive.

Thus eliminate A and C. The sum of two negatives is a negative. Thus eliminate B. For D and E, choose values. Let a = –4 and b = –3. Evaluate the expression in D: –3 – (–4) = –3 + 4 = 1. 1 > 0, so D is true. Evaluate the expression in E: –4 – (–3) = –4 + 3 = –1. –1 is not greater than zero. Eliminate E.

2. B If xy is positive, then x and y must have the same sign. If yz is negative, then

y and z must have different signs. If x and y have the same sign and y and z have different signs, then x and z must have different signs, meaning xz < 0. Use this information in each choice. For A, you know that xz < 0, but you do not know the sign of y, so you do not know whether the product is negative. In B, you know that y2 _{> 0 (because of the even exponent) and that xz < 0.}

(–)(+) = (–), so Choice B is negative. In C, you know that x2_y2 _{must be positive}

because of the even exponents, but you do not know the sign of z. In D, all the variables are raised to even exponents, so the result must be positive. In E, you know that xy > 0, but you do not know the sign of the denominator. The only choice that must be negative is B.

3. C, D, and E In the first inequality, you know that a must be positive since b2

is positive. In the second inequality, you know that a and c must have different signs. If a > 0, and a and c have different signs, then c < 0.

4. A, B, and D Since this is a “must be true” question, it is helpful to plug in

values for x and y. First, plug in integers, and then fractions.

Case 1: x = –0.5 and y = –0.75. Plug these values into the choices: A: –0.52 _{– (–0.75)}2 < 0 → True

B: –0.75 – (–0.5) < 0 → True C: _–0.51 2 < 1 → False

D: –0.5 + –0.75_–0.5 > 0 → True E: –0.752 _{– (–0.5)}2 < 0 → False

Now plug in integers for the choices that yielded true in Case 1. Let x = –2 and y = –3.

A: –22 _{– (–3)}2 < 0 → True

B: –3 – (–2) < 0 → True E: –2 + –3_–2 > 0 → True

Since 1, 2, and 4 remain true for both conditions, those are the answers.

5. C For a fraction to be positive, the numerator and denominator must have the

same sign. Since z is raised to an even exponent, you know that z2 _{must equal}

at least zero. Therefore, z2 _{+ 1 must be positive. If the denominator of this}

fraction is positive, then the numerator is also positive. Thus x – a > 0. Add a to both sides: x > a.

6. C, D, and E If xy > 0, then x and y are both positive or both negative. If x and

y are both negative, then x + y < 0. The condition that x and y are both negative does not satisfy the given information. Thus to satisfy the given information, x and y must both be positive. If x and y are both positive, then choices C, D, and E are true.

Quantitative Comparison Questions

1. B From the stem, you know that a and b are both negative, and that b is more

negative than a. Since the exponents in the columns are even, the signs of a and b will not affect the value of the outcome. Instead, what is relevant is the absolute values of a and b. Since b is more negative than a, its absolute value is greater. When you raise b to the 4th power, the result will therefore be greater than when you raise a to the 2nd power.

2. A Plugging in numbers is a good strategy here. Since the columns have even

exponents, the signs will not matter. So choose fractions and integers.

Case 1: x = 1₂ and y = 1₃. In this case, the value of column A is 1₄ and the value of column B is 1₉. Quantity A is greater, so the answer is A or D.

Case 2: Plug in new values to prove D: x = 3 and y = 2. In this case, the

value of column A is 9, and the value of column B is 4. Quantity A is still greater.

3. B Though it would appear that you do not have sufficient information about a

and b, keep in mind that the even exponents and the signs will help you make inferences. In Quantity A, a2 _{and b}4 _{must be positive (because of the even}

exponents). Thus their product is positive. Multiply this product by –1, and the result is negative. In Quantity B, –a2 _{and –b}4 _{are both positive. Thus their}

product is positive. The value in Column B is greater.

4. D Since a_b equals a positive number, a and b must have the same sign. However, you do not know what the signs are. If a = 8, then b = 2, and Quantity A is greater. But if a = –8, then b = –2, and Quantity B is greater. There is more than one relationship, so the answer is D.

5. A If a_b > 0, then a and b must have the same sign. If a and b have the same sign, then their product must be positive. Thus ab > 0, and the correct answer is A.

6. A Your first step should be to make inferences from the stem. Inequality 1:

you know that p2_q3 _{> 0. Because of the even exponent, p}2 _{must be positive.}

Thus q3 _{must be positive. If q}3 _{is positive, then q > 0 (remember, odd exponents}

preserve the sign of the base). Inequality 2: you know that p3_q2 _{> 0. Because}

of the even exponent, q2 _{must be positive. Thus p}3 _{must be positive. If p}3 _is

positive, then p > 0. Positive × positive = positive, so pq > 0.

7. B Your first step should be to make inferences from the stem. Inequality 1:

because of the even exponent, you know that a2 _{> 0. Thus b}3_c5 _{> 0. Because of}

the odd exponents, b3 _{and c}5 _{will have the same signs as b and c, respectively.}

Thus bc > 0. Inequality 2: because of the even exponent, you know that b4 _{> 0.}

Thus a3_c5 _{< 0. Because of the odd exponents, a}3 _{and c}5 _{will have the same signs}

If bc > 0, then b and c must have the same sign. If ac < 0, then a and c must have different signs. Therefore, a and b must have different signs, and their product must be negative.