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GMAT-QUANT-TOPIC-3-Inequalities-Absolute-Value-Solutions

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GMAT QUANT TOPIC 3
(INEQUALITIES + ABSOLUTE VALUE)
SOLUTIONS
1.
There are two characteristics of x that dictate its exponential behavior.
First of all, it is a decimal with an absolute value of less than 1. Secondly,
it is a negative number.

I. True. x3 will always be negative (negative × negative × negative =
negative), and x2 will always be positive (negative × negative = positive),
so x3 will always be less than x2.

II. True. x5 will always be negative, and since x is negative, 1 – x will
always be positive because the double negative will essentially turn 1 –
x into 1 + |x|. Therefore, x5 will always be less than 1 – x.

III. True. One useful method for evaluating this inequality is to plug in a
number for x. If x = - 0.5,

x4 = (-0.5)4 = 0.0625

x2 = (-0.5)2 = 0.25

To understand why this works, it helps to think of the negative aspect of
x and the decimal aspect of x separately.

Because x is being taken to an even exponent in both instances, we can
essentially ignore the negative aspect because we know the both results
will be positive.

The rule with decimals between 0 and 1 is that the number gets smaller
and smaller in absolute value as the exponent gets bigger and bigger.
Therefore, x4 must be smaller in absolute value than x2.

The correct answer is E.

,2.
(1) INSUFFICIENT: We can solve this absolute value inequality by
considering both the positive and negative scenarios for the absolute
value expression |x + 3|.
If x > -3, making (x + 3) positive, we can rewrite |x + 3| as x + 3:
x+3<4
x<1
If x < -3, making (x + 3) negative, we can rewrite |x + 3| as -(x + 3):
-(x + 3) < 4
x + 3 > -4
x > -7
If we combine these two solutions we get -7 < x < 1, which means we
can’t tell whether x is positive.

(2) INSUFFICIENT: We can solve this absolute value inequality by
considering both the positive and negative scenarios for the absolute
value expression |x – 3|.
If x > 3, making (x – 3) positive, we can rewrite |x – 3| as x – 3:
x–3<4
x<7
If x < 3, making (x – 3) negative, we can rewrite |x – 3| as -(x – 3) OR 3 –
x
3–x<4
x > -1
If we combine these two solutions we get -1 < x < 7, which means we
can’t tell whether x is positive.

(1) AND (2) INSUFFICIENT: If we combine the solutions from statements
(1) and (2) we get an overlapping range of -1 < x < 1. We still can’t
tell whether x is positive.

The correct answer is E.

3.

The question asks: is x + n < 0?
(1) INSUFFICIENT: This statement can be rewritten as x + n < 2n – 4. This
rephrased statement is consistent with x + n being either negative or
non-negative. (For example if 2n – 4 = 1,000, then x + n could be any
integer, negative or not, that is less than 1,000.) Statement (1) is

,insufficient because it answers our question by saying "maybe yes,
maybe no".
(2) SUFFICIENT: We can divide both sides of this equation by -2 to get x <
-n (remember that the inequality sign flips when we multiply or divide by
a negative number). After adding n to both sides of resulting inequality,
we are left with x + n < 0.
The correct answer is B.

4.
This is a multiple variable inequality problem, so you must solve it by
doing algebraic manipulations on the inequalities.

(1) INSUFFICIENT: Statement (1) relates b to d, while giving us no
knowledge about a and c. Therefore statement (1) is insufficient.

(2) INSUFFICIENT: Statement (2) does give a relationship between a and
c, but it still depends on the values of b and d. One way to see this
clearly is by realizing that only the right side of the equation contains the
variable d. Perhaps ab2 – b is greater than b2c – d simply because of the
magnitude of d. Therefore there is no way to draw any conclusions
about the relationship between a and c.

(1) AND (2) SUFFICIENT: By adding the two inequalities from statements
(1) and (2) together, we can come to the conclusion that a > c. Two
inequalities can always be added together as long as the direction of the
inequality signs is the same:

ab2 – b > b2c – d
(+) b > d
-----------------------
ab2 > b2c

Now divide both sides by b2. Since b2 is always positive, you don't have
to worry about reversing the direction of the inequality. The final result:
a > c.

The correct answer is C.


5.

, Since this question is presented in a straightforward way, we can
proceed right to the analysis of each statement. On any question that
involves inequalities, make sure to simplify each inequality as much as
possible before arriving at the final conclusion.
(1) INSUFFICIENT: Let’s simplify the inequality to rephrase this
statement:
-5x > -3x + 10
5x – 3x < -10 (don't forget: switch the sign when multiplying or dividing
by a negative)
2x < -10
x < -5

Since this statement provides us only with a range of values for x, it is
insufficient.

(2) INSUFFICIENT: Once again, simplify the inequality to rephrase the
statement:
-11x – 10 < 67
-11x < 77
x > -7

Since this statement provides us only with a range of values for x, it is
insufficient.

(1) AND (2) SUFFICIENT: If we combine the two statements together, it
must be that
-7 < x < -5. Since x is an integer, x = -6.

The correct answer is C.

6. We can start by solving the given inequality for x:

8x > 4 + 6x
2x > 4
x>2

So, the rephrased question is: "If the integer x is greater than 2, what is
the value of x?"

(1) SUFFICIENT: Let's solve this inequality for x as well:

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