Absolute Value Inequalities

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GRE Quantitative Reasoning › Absolute Value Inequalities

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The weight of the bowling balls manufactured at the factory must be lbs. with a tolerance of lbs. Which of the following absolute value inequalities can be used to assess which bowling balls are tolerable?

$\left | w-8\right | \geq 3$

CORRECT

$\left | w-8\right | \leq 3$

$\left | w-3 \right |\geq 8$

$\left | w-3 \right | \leq 8$

Explanation

The following absolute value inequality can be used to assess the bowling balls that are tolerable:

$\left | w-8\right |\geq3$

$\left | \frac{x-6}{2} \right |\leq 3$

$x\geq 0$ and $x\leq 12$

CORRECT

$x\leq 0$ and $x\geq12$

$x\geq 0$ and $x\leq 12$

There is no solution.

Explanation

$\left | \frac{x-6}{2} \right |\leq3$

$\frac{2}{1} \left | \frac{x-6}{2} \right | \leq 2\times3$

$x-6 \leq 6$

$x\leq 12$

$\frac{2}{1} \left | \frac{x-6}{2} \right | \geq 2 \times -3$

$x-6 \geq -6$

$x\geq 0$

The correct answer is $x\leq12$ and $x\geq 0.$

$5\left | x-7\right |+14<39$

CORRECT

Explanation

The first thing we must do is get the absolute value alone: $5\left | x-7\right |+14<39$

$5\left | x-7\right |<25$

$\left | x-7\right |<5$

When we're working with absolute values, we are actually solving two equations:

and

Fortunately, these can be written as one equation:

If you feel more comfortable solving the equations separately then go ahead and do so.

To get alone, we added on both sides of the inequality sign

$4\left |(x+0.3)\right | - 3 < 11$

and

CORRECT

and

There is no solution.

and

Explanation

$4\left |(x+0.3) \right | - 3< 11$

$\frac{4x}{4} < \frac{12.8}{4}$

$\frac{4x}{4}>\frac{-9.2}{4}$

The correct answer is and

A type of cell phone must be less than 9 ounces with a tolerance of 0.4 ounces. Which of the following inequalities can be used to assess which cell phones are tolerable? (w refers to the weight).

$\left | w-9\right | \leq 0.4$

CORRECT

$\left | w-0.4\right | \leq 9$

$\left | w-9\right | \geq 0.4$

$\left | w-0.4\right | \geq 9$

Explanation

The Absolute Value Inequality that can assess which cell phones are tolerable is:

$\left | w-0.4\right | \leq 9$

$\left | 6+x\right |-4 < 0$

and

CORRECT

and

Explanation

$\left | 6+x\right | -4 < 0$

The correct answer is and

$\left | -4x-5 \right |< 11$

and $x< \frac{3}{2}$

CORRECT

$x< \frac{4}{5}$ or

Explanation

Since the absolute value with x in it is alone on one side of the inequality, you set the expression inside the absolute value equal to both the positive and negative value of the other side, 11 and -11 in this case. For the negative value -11, you must also flip the inequality from less than to a greater than. You should have two inequalities looking like this.

and

Add 5 to both sides in each inequality.

and

Divide by -4 to both sides of the inequality. Remember, dividing by a negative will flip both inequality symbols and you should have this.

and $x< \frac{3}{2}$

$\frac{2}{3}\left | 9x+6\right | -2\geq 1$

$x\geq -\frac{1}{6}$ or $x\leq -\frac{7}{6}$

CORRECT

$x\leq-\frac{1}{6}$ and $x\geq- \frac{1}{2}$

There is no solution.

$x\geq \frac{1}{6}$ and $x\leq \frac{1}{2}$

Explanation

$\frac{2}{3}\left | 9x+6\right | -2\geq 1$

$\frac{2}{3}\left | 9x+6 \right |\geq 3$

$\left | 9x+6 \right |\geq \frac{9}{2}$

At this point, you've isolated the absolute value and can solve this problems for both cases, $9x+6\geq \frac{9}{2}$ and $9x+6\leq -\frac{9}{2}$ . Beginning with the first case:

$9x+6\geq \frac{9}{2}$

$9x+\frac{12}{2}\geq\frac{9}{2}$

$9x\geq -\frac{3}{2}$

$x\geq -\frac{1}{6}$

Then for the second case:

$9x+6\leq -\frac{9}{2}$

$9x+\frac{12}{2}\leq -\frac{9}{2}$

$9x\leq -\frac{21}{2}$

$x\leq -\frac{7}{6}$

Which of the following expresses the entire solution set of $\left | x-10\right |+ 3 < 12$ ?

and

CORRECT

Explanation

Before expanding the quantity within absolute value brackets, it is best to simplify the "actual values" in the problem. Thus $\left | x-10 \right |+3<12$ becomes:

$\left | x-10 \right |<9$

From there, note that the absolute value means that one of two things is true: or . You can therefore solve for each possibility to get all possible solutions. Beginning with the first: