How to find absolute value - SSAT Upper Level Quantitative

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Question

Evaluate for :

$\left | 4x - 1.4 \right | + \left | $x^{2}$ - 1 \right |$

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Answer

Substitute 0.6 for :

$\left | 4x - 1.4 \right | + \left | $x^{2}$ - 1 \right |$

$= \left | 4 \cdot 0.6 - 1.4 \right | + \left | $0.6^{2}$ - 1 \right |$

$= \left | 2.4 - 1.4 \right | + \left | 0.36 - 1 \right |$

$= \left | 1 \right | + \left | -0.64 \right |$

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Question

Given: are distinct integers such that:

Which of the following could be the least of the three?

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Answer

$0 \le |a|< b < |c|$ , which means that must be positive.

If is nonnegative, then . If is negative, then it follows that . Either way, . Therefore, cannot be the least.

We now show that we cannot eliminate or as the least.

For example, if , then is the least; we test both statements:

, which is true.

, which is also true.

If , then is the least; we test both statements:

, which is true.

, which is also true.

Therefore, the correct response is or only.

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Question

Evaluate: $\begin{vmatrix}; ; \left | 5- 3 \cdot 4\right |-8 \right ;\end{vmatrix}$

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Answer

$\begin{vmatrix}; ; \left | 5- 3 \cdot 4\right |-8 \right ;\end{vmatrix}=\begin{vmatrix}; ; \left | 5- 12\right |-8 \right ;\end{vmatrix}=\begin{vmatrix}; ; \left | -7\right |-8 \right ;\end{vmatrix}=| 7 - 8| = | -1| = 1$

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Question

Evaluate the expression if and .

$\left | 4x+3y \right | - \left | 4x-3y \right |$

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Answer

$\left | 4x+3y \right | - \left | 4x-3y \right |$

To solve, we replace each variable with the given value.

$\left | 4 (5) +3 ( 7) \right | - \left | 4 (5) -3 ( 7) \right |\ .$

$\left | 20 +21 \right | - \left | 20 -21 \right |$

Simplify. Remember that terms inside of the absolute value are always positive.

$\left | 41 \right | - \left | -1 \right |=41-1 = 40$

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Question

Evaluate for :

$\left | 2x - 18 \right | + \left | 3x - 7 \right |$

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Answer

$\left | 2x -18 \right | + \left | 3x - 7 \right |$

$= \left | 2 \cdot 2 - 18 \right | + \left | 3 \cdot 2 - 7 \right |$

$= \left | 4 - 18 \right | + \left | 6 - 7 \right |$

$= \left | -14 \right | + \left | -1 \right |$

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Question

Evaluate for :

$\left |0.5 x - 0.7 \right | - \left |0.6 x - 0.4 \right |$

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Answer

Substitute .

$\left |0.5 x - 0.7 \right | - \left |0.6 x - 0.4 \right |$

$= \left |0.5 \cdot 0.6 - 0.7 \right | - \left |0.6 \cdot 0.6 - 0.4 \right |$

$= \left |0.3- 0.7 \right | - \left |0.36 - 0.4 \right |$

$= \left |-0.4 \right | - \left | - 0.04 \right |$

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Question

Which of the following sentences is represented by the equation

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Answer

is the absolute value of , which in turn is the sum of a number and seven and a number. Therefore, can be written as "the absolute value of the sum of a number and seven". Since it is equal to , it is three less than the number, so the equation that corresponds to the sentence is

"The absolute value of the sum of a number and seven is three less than the number."

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Question

Define an operation $\blacklozenge$ as follows:

For all real numbers ,

$a \blacklozenge b = \left | 2a- 2b + 5 \right |$

Evaluate $(-4) \blacklozenge (-7)$

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Answer

$a \blacklozenge b = \left | 2a- 2b + 5 \right |$

$(-4) \blacklozenge (-7) = \left | 2 (-4)- 2(-7) + 5 \right |$

$= \left |-8- (-14) + 5 \right |$

$= \left |-8+14 + 5 \right |$

$= \left |11 \right |$

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Question

Define an operation $\triangleright$ as follows:

For all real numbers ,

$a \triangleright b = \left | a- $\frac{1}{2}$b\right | + \left | $\frac{1}{2}$ a+b\right |$

Evaluate $$\frac{1}{3}$ \triangleright 4$ .

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Answer

$a \triangleright b = \left | a- $\frac{1}{2}$b\right | + \left | $\frac{1}{2}$ a+b\right |$

$$\frac{1}{3}$ \triangleright 4 = \left | $\frac{1}{3}$ - $\frac{1}{2}$ \cdot 4\right | + \left | $\frac{1}{2}$ \cdot $\frac{1}{3}$ +4\right |$

$= \left | $\frac{1}{3}$ - 2\right | + \left | $\frac{1}{6}$ +4\right |$

$= \left | -1 $\frac{2}{3}$ \right | + \left | 4 $\frac{1}{6}$ \right |$

$= 1 $\frac{2}{3}$ + 4 $\frac{1}{6}$$

$= 5 $\frac{5}{6}$$

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Question

Define an operation $\blacktriangledown$ as follows:

For all real numbers ,

$a \blacktriangledown b= $\frac{a+1}{\left| a\right|+ \left| b\right|}$$

Evaluate: $$\frac{4}{5}$ \blacktriangledown \left (-$\frac{4}{5}$ \right )$ .

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Answer

$a \blacktriangledown b= $\frac{a+1}{\left| a\right|+ \left| b\right|}$$ , or, equivalently,

$a \blacktriangledown b=\left ( a+1 \right ) \div ( | a |+ | b | )$

$$\frac{4}{5}$ \blacktriangledown \left (-$\frac{4}{5}$ \right )=\left ( $\frac{4}{5}$+1 \right ) \div \left (\left| $\frac{4}{5}$ \right |+ \left |- $\frac{4}{5}\right| \right)$

$= $\frac{9}{5}$ \div \left( $\frac{4}{5}$ +\frac{4}{5}$ \right )$

$= $\frac{9}{5}$ \div $\frac{8}{5}$$

$= $\frac{9}{5}$ \times $\frac{5}{8}$$

$= $\frac{9}{8}$$

$=1 $\frac{1}{8}$$

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Question

Define $f(x) = |3x - $|x^{2}$- 7|; |$

Evaluate .

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Answer

$f(x) = |3x - $|x^{2}$- 7|; |$

$f(2) = |3 \cdot 2 - $|2^{2}$- 7|; |$

$= |3 \cdot 2 - |4- 7|; |$

$= |3 \cdot 2 - |-3|; |$

$= |3 \cdot 2 -3 |$

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Question

Define $g(x)= \left | \left | 1,000- $\sqrt{x}$\right | - $x^{3}$ \right |$ .

Evaluate .

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Answer

$g(x)= \left | \left | 1,000- $\sqrt{x}$\right | - $x^{3}$ \right |$

$g(16)= \left | \left | 1,000- $\sqrt{16}\right| - $16^{3}$ \right|$

$= \left| \left| 1,000- 4 \right| - $16^{3}$ \right|$

$= \left| \left| 996 \right| - $16^{3}$ \right|$

$= \left| 996 - $16^{3}$ \right |$

$= \left | 996 - 4,096 \right |$

$= \left | -3,100 \right |$

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Question

Define $p(x) = $\frac{\left|x+2 \right|-1}{\left|x+1 \right|-2}$$ .

Evaluate $p \left (-1 $\frac{1}{5}$ \right )$ .

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Answer

$p(x) = $\frac{\left|x+2 \right|-1}{\left|x+1 \right|-2}$$ , or, equivalently,

$p(x) =\left ( \left |x+2 \right |-1 \right ) \div \left ( \left |x+1 \right |-2 \right )$

$p\left (-1 $\frac{1}{5}$ \right )=\left ( \left |-1 $\frac{1}{5}$+2 \right |-1 \right ) \div \left ( \left |-1 $\frac{1}{5}$+1 \right |-2 \right )$

$=\left ( \left | $\frac{4}{5}$ \right |-1 \right ) \div \left ( \left |- $\frac{1}{5}$ \right |-2 \right )$

$=\left ( $\frac{4}{5}$ -1 \right ) \div \left ( $\frac{1}{5}$ -2 \right )$

$= - $\frac{1}{5}$ \div \left ( - $\frac{9}{5}$ \right )$

$= $\frac{1}{5}$ \div $\frac{9}{5}$$

$= $\frac{1}{5}$ \times $\frac{5}{9}$$

$= $\frac{1}$ {9}$

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Question

Given: are distinct integers such that:

Which of the following could be the least of the three?

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Answer

$0 \le |m|< n < |p|$ , which means that must be positive.

If is nonnegative, then . If is negative, then it follows that . Either way, . Therefore, cannot be the least.

Now examine the statemtn . If , then - but we are given that and are distinct. Therefore, is nonzero, , and

and

.

cannot be the least either.

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Question

, , and are distinct integers. and . Which of the following could be the least of the three?

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Answer

$0 \le |a|<b$ , so must be positive. Therefore, since $0 \le |b| < c$ , it follows that , so must be positive, and

If is negative or zero, it is the least of the three. If is positive, then the statement becomes

,

and is still the least of the three. Therefore, must be the least of the three, and the correct choice is "None of the other responses is correct."

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Question

, , and are distinct integers. and . Which of the following could be the greatest of the three?

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Answer

$0 \le |a|<b$ , so must be positive. Therefore, since $0 \le |b| < c$ , equivalently, , so must be positive, and

If is negative or zero, it is the least of the three. If is positive, then the statement becomes

,

and is still the least of the three. Therefore, must be the greatest of the three.

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Question

Give the solution set:

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Answer

When dealing with absolute value bars, it is important to understand that whatever is inside of the absolute value bars can be negative or positive. This means that an inequality can be made.

In this particular case if , then, equivalently,

From here, isolate the variable by adding seven to each side.

In interval notation, this is .

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Question

Give the solution set:

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Answer

If , then either or . Solve separately:

or

The solution set, in interval notation, is $(-\infty, -11) \cup (23, \infty)$ .

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Question

Define an operation $\bigstar$ on the real numbers as follows:

If , then $a \bigstar b = |a + b-1|$

If , then $a \bigstar b = 4$

If , then $a \bigstar b = |a - b+1|$

If $p = 0 \bigstar 5$ , $q = 0 \bigstar 0$ , and $r=0 \bigstar( -5)$

then which of the following is a true statement?

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Answer

$p = 0 \bigstar 5$

Since , evaluate

$a \bigstar b = |a - b+1|$ , setting :

$0 \bigstar 5 = |0 - 5+1|$

$0 \bigstar 5 = |-4| = 4$

$q = 0 \bigstar 0$

Since , then select the pattern

$0 \bigstar 0 = 4$

$r=0 \bigstar( -5)$

Since , evaluate

$a \bigstar b = |a + b-1|$ , setting :

$0 \bigstar (-5) = |0 + (-5)-1|$

$0 \bigstar (-5) = |-6| = 6$

, so the correct choice is that .

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Question

Solve the following expression for when .

$\left |4- $x^{2}\right |$

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Answer

First you plug in for and squre it.

This gives the expression which is equal to .

Since the equation is within the absolute value lines, you must make it the absolute value which is the amount of places the number is from zero.

This makes your answer .

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