Understanding absolute value - GMAT Quantitative

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Question

Solve $\left | 3x - 7 \right |=8$ .

Answer

$\left | 3x - 7 \right |=8$ really consists of two equations: $3x - 7 = \pm 8$

We must solve them both to find two possible solutions.

$3x - 7 = 8 \Rightarrow 3x = 15\Rightarrow x = 5$

$3x - 7 = - 8 \Rightarrow 3x = -1\Rightarrow x = -1/3$

So or $\dpi{100} -\frac{1}{3}$ .

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Question

Solve $\left | 2x - 5 \right |\geq 3$ .

Answer

It's actually easier to solve for the complement first. Let's solve $\left | 2x-5 \right |<3$ . That gives -3 < 2x - 5 < 3. Add 5 to get 2 < 2x < 8, and divide by 2 to get 1 < x < 4. To find the real solution then, we take the opposites of the two inequality signs. Then our answer becomes $x\leq 1 \textsc{ or } x\geq 4$ .

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Question

If , which of the following has the greatest absolute value?

Answer

Since , we know the following:

;

$-1<\frac{n}{2}<0$ ;

$\frac{1}{n^2}>1$ ;

$\frac{1}{n}<-1$ ;

.

Also, we need to compare absolute values, so the greatest one must be either $\left | \frac{1}{n^2}\right |$ or $\left | \frac{1}{n}\right |$ .

We also know that $\left | n^2\right |<\left | n\right |$ when .

Thus, we know for sure that $\left | \frac{1}{n^2}\right |>\left | \frac{1}{n}\right |$ .

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Question

Give the -intercept(s), if any, of the graph of the function $f (x) = \left | 4x + B \right | - 8$ in terms of .

Answer

Set and solve for :

$\left | 4x + B \right | - 8 = 0$

$\left | 4x + B \right | = 8$

Rewrite as a compound equation and solve each part separately:

$4x + B = -8 \textrm{ or } 4x + B = 8$

$4x \div 4 =\left ( -B -8 \right ) \div 4$

$x = \frac{-B -8 }{4}$

$4x \div 4 =\left ( -B +8 \right ) \div 4$

$x = \frac{-B +8 }{4}$

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Question

A number is ten less than its own absolute value. What is this number?

Answer

We can rewrite this as an equation, where is the number in question:

A nonnegative number is equal to its own absolute value, so if this number exists, it must be negative.

In thsi case, , and we can rewrite that equation as

$2N \div 2 = -10\div 2$

This is the only number that fits the criterion.

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Question

Give all numbers that are twenty less than twice their own absolute value.

Answer

We can rewrite this as an equation, where is the number in question:

$N = 2\cdot |N| - 20$

If is nonnegative, then , and we can rewrite this as

Solve:

If is negative, then , and we can rewrite this as

$N = 2\left (-N \right ) - 20$

$3N \div 3=- 20\div 3$

$N=- 6\frac{2}{3}$

The numbers $- 6\frac{2}{3}, 20$ have the given characteristics.

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Question

Solve for in the absolute value equation

$1-\frac{5}{2}\left |x-5 \right | = 15$

Answer

The correct answer is that there is no .

We start by adding to both sides giving

$-\frac{5}{2} \left|x-5\right|=14$

Then multiply both sides by .

$-5\left|x-5|=28$

Then divide both sides by

$|x-5|=-\frac{28}{5}$

Now it is impossible to go any further. The absolute value of any quantity is always positive (or sometimes ). Here we have the absolute value of something equaling a negative number. That's never possible, hence there is no that makes this a true equation.

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Question

Solve the following equation:

$32-\left | x-23\right |=13$

Answer

We start by isolating the expression with the absolute value:

$32-\left | x-23\right |=13$ becomes $-\left | x-23\right |=13-32=-19$

So: or

We then solve the two equations above, which gives us 42 and 4 respectively.

So the solution is

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Question

$\left | 2m-\frac{1}{2}\right |<4$

Which of the following could be a value of ?

Answer

To solve an inequality we need to remember what the absolute value sign says about our expression. In this case it says that

$\left | 2m-\frac{1}{2}\right |<4$

can be written as

$2m-\frac{1}{2}<4$ Of $2m-\frac{1}{2}>-4$ .

Rewriting this in one inequality we get:

$-4<2m-\frac{1}{2}<4$

From here we add one half to both sides .

$-\frac{7}{2}<2m<\frac{9}{2}$

Finally, we divide by two to isolate and solve for m.

$-\frac{7}{4}<m<\frac{9}{4}$

Only $\frac{7}{4}$ is between -1.75 and 2.25

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Question

Solve the absolute value equation for .

Answer

We proceed as follows

(Start)

(Subtract 3 from both sides)

or (Quantity inside the absolute value can be positive or negative)

or (add five to both sides)

$x = \frac{7}{2}$ or $x = \frac{3}{2}$

Another way to say this is $x = \frac{7}{2},\frac{3}{2}$

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Question

Find the possible values of :

$\left | 2k-15\right |=5$

Answer

$\left | 2k-15\right |=5$

There are two ways to solve the absolute value portion of this problem:

or

From here, you can solve each of these equations independently to arrive at the correct answer:

or

or

The solution is .

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Question

The absolute value of negative seventeen is multiplied by a number that is three fewer than twelve. The resulting number is subtracted fromnegative six. What number is yielded at the end of this sequence of operations?

Answer

This is a problem where we need to use our translating skills. We are given a word problem and asked to solve it. To do so, we need to rewrite our word problem as an equation and then use arithmetic to find the answer. In these types of problems, the hardest step is usually translating correctly, so make sure to be meticulous and work step-by-step!

1)"The absolute value of negative seventeen": Recall that absolute value means that we will just change the sign to positive. Missing that will end up giving you the trap answer .

$\left | -17\right |=17$

2)"is multiplied by a number which is three fewer than twelve." We need a number that is three fewer than twelve, so we need to subtract. Follow it up with multiplication and you get:

3)"The resulting number is subtracted fromnegative six." The key word here is "from"—make sure you aren't computing , which would result in another one of the trap answers!

The correct answer is .

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Question

Solve $\left | 5x+20\right |=55$ .

Answer

Since we are solving an absolute value equation, $\left | 5x+20\right |=55$ , we must solve for both potential values of the equation:

1.)

2.)

Solving Equation 1:

Solving Equation 2:

Therefore, for $\left | 5x+20\right |=55$ , or .

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Question

Solve $\left | 8x-16\right |=24$ .

Answer

Since we are solving an absolute value equation, $\left | 8x-16\right |=24$ , we must solve for both potential values of the equation:

1.)

2.)

Solving Equation 1:

Solving Equation 2:

Therefore, for $\left | 8x-16\right |=24$ , or .

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Question

Define an operation $\ast$ as follows:

For all real numbers ,

$a \ast b = \left | \frac{1}{4}a - \frac{1}{7}b \right |$

Evaluate: $7 \ast 4$

Answer

$a \ast b = \left | \frac{1}{4}a - \frac{1}{7}b \right |$

$7 \ast 4= \left | \frac{1}{4} \cdot 7 - \frac{1}{7} \cdot 4 \right |$

$= \left | \frac{7}{4} - \frac{4}{7} \right |$

$= \left | \frac{49}{28} - \frac{16}{28} \right |$

$= \left | \frac{33}{28} \right |$

$= \left |1 \frac{5}{28} \right |$

$= 1 \frac{5}{28}$

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Question

Define an operation $\ast$ as follows:

For all real numbers ,

$a \ast b = |a+b| + |a-b|$

Evaluate $\left (-7 \right ) \ast \left (-5 \right )$

Answer

$a \ast b = |a+b| + |a-b|$

$\left (-7 \right ) \ast \left (-5 \right )= |-7+\left (-5 \right )| + |-7-\left (-5 \right )|$

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Question

Define an operation $\ast$ as follows:

For all real numbers ,

$a \ast b = |a+b| - |a-b|$

Evaluate $\left (-7 \right ) \ast \left (-9 \right )$

Answer

$a \ast b = |a+b| - |a-b|$

$\left (-7 \right ) \ast \left (-9 \right )= |\left (-7 \right )+\left (-9 \right )| - |\left (-7 \right )-\left (-9 \right )|$

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Question

Define a function to be

Give the range of the function.

Answer

An absolute value of a number must always assume a nonnegative value, so

$|3x- 12 | \ge 0$ , and

$|3x- 12 | + 5\ge 0 +5$

Therefore,

$f(x)=|3x- 12 | + 5\ge 5$

and the range of is the set $[5, \infty)$ .

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Question

Give the range of the function

Answer

The key to answering this question is to note that this equation can be rewritten in piecewise fashion.

If $x \ge 10$ , since both and are nonnegative, we can rewrite as

, or

.

On $[10, \infty)$ , this has as its graph a line with positive slope, so it is an increasing function. The range of this part of the function is $[f(10), \infty)$ , or, since

,

$[4, \infty)$ .

If , since is negative and is positive, we can rewrite as

, or

is a constant function on this interval and its range is .

If $x \le 6$ , since both and are nonpositive, we can rewrite as

, or

.

On $[- \infty, 6)$ , this has as its graph a line with negative slope, so it is a decreasing function. The range of this part of the function is $[f(6), \infty)$ , or, since

, $[4, \infty)$ .

The union of the ranges is the range of the function - $[4, \infty)$ .

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Question

Give the range of the function

Answer

The key to answering this question is to note that this equation can be rewritten in piecewise fashion.

If , since both and are positive, we can rewrite as

, or

,

a constant function with range $\left { -4\right }$ .

If $6 \le x \le 10$ , since is negative and is positive, we can rewrite as

, or

This is decreasing, as its graph is a line with negative slope. The range is ,

or, since

and

,

.

If , since both and are negative, we can rewrite as

, or

,

a constant function with range $\left { 4\right }$ .

The union of the ranges is the range of the function - - which is not among the choices.

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