1. If xy>1 and z<0, which of the following statements must be true?

I. x>z
II.xyz<-1
III.xy/z<1/z

A. I. only
B. II. only
C. III. only
D. II. and III.
E. None

2.

ab>0
1/a<1/b

Which is greater, a or b?

3.

p/4<q/12

Which is greater, (p-1)/4 or (q-9)/12?

4.

a and b are positive integers.
7<ab<21

Which is greater, a+b or ab?

5.

-1<=x-3<=6
5<=y+1<=9

Which is greater, the largest possible value of x/y or 2?

6.

a+b=5
b<7/3

Which is greater, a or 8/3?

These require some thought, and it serves you not for someone to do that for you.

example 1.
xy>1 implies x,y are both positive, or both negative.
z<0 means z is negative. Not think out the choices
I. x>z cannot be true always, x could be -100 and z could be -88
II.xyz<-1 xy is positive, z is negative, so sign is right. but the product of xyz could be -1/2 (example xy=2, z=-.5
III
III.xy/z<1/z divide both sides by xy.
1/z<1/xyz . now consider z=-1/2, and xy 10
1/-.5<1/-5
-2<-.2 true the right side is always greater than the left,

A. I. only
B. II. only
C. III. only
D. II. and III.
E. None

1. To determine which statements must be true, we can analyze each option individually.

Statement I: x > z
Given xy > 1 and z < 0, we do not have enough information to determine whether x is greater than z. It is possible for x to be less than z, so Statement I is not necessarily true.

Statement II: xyz < -1
Given xy > 1, we can multiply both sides of the inequality by z (-z) without changing the inequality's direction: xyz > -z. Because we know z < 0, we can flip the direction of the inequality: xyz < -z. Hence, Statement II is true.

Statement III: xy/z < 1/z
From xy > 1 and z < 0, we can divide both sides of the first inequality by z: xy/z > 1/z. Also, since z < 0, dividing by z flips the direction of the inequality: xy/z < 1/z. Thus, Statement III is true.

Based on the analysis, Statements II and III must be true. Therefore, the correct answer is D. II. and III.

2. To determine which variable, a or b, is greater, we can compare the given inequalities directly.

Given ab > 0, we know that a and b have the same sign (either both positive or both negative).

Given 1/a < 1/b, we can cross-multiply the inequalities as follows: ab < ba. Since a and b have the same sign, we can divide both sides of the inequality by ab (assuming ab ≠ 0) to get 1 < b/a.

From the two inequalities, 1 < b/a, and ab > 0, we can conclude that b/a > 1.

Therefore, b is greater than a.

3. To compare (p-1)/4 and (q-9)/12, we need to simplify and analyze the expressions.

Given p/4 < q/12, we can multiply both sides of the inequality by 3 to eliminate the denominators: 3(p/4) < 3(q/12).

This simplifies to (3p)/4 < (3q)/12, which further simplifies to (p/4) < (q/4).

Now, we can add -1/4 to both sides of the inequality: (p/4) - (1/4) < (q/4) - (1/4).

Simplifying, (p-1)/4 < (q-1)/4.

Comparing this with the original expressions, we can conclude that (p-1)/4 is less than (q-9)/12.

Therefore, (q-9)/12 is greater than (p-1)/4.

4. To determine whether a+b or ab is greater, we can analyze the given inequality.

Given 7 < ab < 21, we know that ab is bounded between 7 and 21.

To compare a+b and ab, we need to consider different scenarios.

Scenario 1: a = 1 and b = 7 (minimum values within the given range).
In this case, a+b = 1+7 = 8, and ab = 1 x 7 = 7. Hence, a+b is greater than ab.

Scenario 2: a = 4 and b = 5 (maximum values within the given range).
In this case, a+b = 4+5 = 9, and ab = 4 x 5 = 20. Hence, a+b is less than ab.

Since the question does not specify whether a or b should be chosen to maximize the sum or the product, both options are possible. Therefore, we cannot determine whether a+b or ab is greater.

5. To compare the largest possible value of x/y with 2, we need to analyze the given inequalities.

Given -1 ≤ x-3 ≤ 6 and 5 ≤ y+1 ≤ 9, we can rearrange the inequalities as follows: 2 ≤ x ≤ 9 and 4 ≤ y ≤ 8.

To find the largest possible value of x/y, we need to consider the worst-case scenario, where x is at its maximum value (9) and y is at its minimum value (4).

In this scenario, the largest possible value of x/y = 9/4 = 2.25.

Comparing 2.25 with 2, we can conclude that the largest possible value of x/y is greater than 2.

Therefore, the largest possible value of x/y is greater than 2.

6. To compare the values of a and 8/3, we can analyze the given equations.

Given a+b = 5 and b < 7/3, we can rearrange the first equation as follows: b = 5 - a.

Substituting this value into the second equation, we get 5 - a < 7/3.

To make the comparison easier, we can multiply both sides of the inequality by 3: 15 - 3a < 7.

Further simplifying, we have -3a < -8. Dividing both sides of the inequality by -3 (remembering to flip the direction of the inequality), we get a > 8/3.

Therefore, a is greater than 8/3.