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SAT Math : Arithmetic

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Example Questions

Example Question #881 : Arithmetic

Which of the following sentences is represented by the equation

| x + 7 | = x - 3

Possible Answers:

The sum of three and the absolute value of the sum of a number is three greater than the number.

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

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

None of the other responses are correct.

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

Correct answer:

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

Explanation:

| x + 7 | is the absolute value of x+ 7 , 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 x - 3 , 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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Example Question #174 : Integers

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

Evaluate f(2) .

Possible Answers:

None of the other responses is correct.

Correct answer:

Explanation:

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

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

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

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

$= |3 \cdot 2 -3 |$

= |6 -3 |

= | 3 |

= 3

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Example Question #242 : New Sat

Define an operation $\blacktriangledown$ as follows:

For all real numbers a,b ,

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

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

Possible Answers:

$1 \frac{1}{8}$

$\frac{5}{8}$

None of the other responses is correct.

The expression is undefined.

Correct answer:

$1 \frac{1}{8}$

Explanation:

$a \blacktriangledown b= \frac{a+1}{\left | a\right |+ \left | b\right |}$ , 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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Example Question #3 : Absolute Value

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

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

Possible Answers:

$-\frac{1} {9}$

$-\frac{1}{11}$

$\frac{1} {9}$

$\frac{1}{11}$

Correct answer:

$\frac{1} {9}$

Explanation:

$p(x) = \frac{\left |x+2 \right |-1}{\left |x+1 \right |-2}$ , 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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Example Question #1 : Absolute Value

Define an operation $\triangleright$ as follows:

For all real numbers a,b ,

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

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

Possible Answers:

$5 \frac{5}{6}$

$2\frac{1}{2}$

$1\frac{1}{2}$

$\text{None of the other responses are correct.}$

$6\frac{1}{2}$

Correct answer:

$5 \frac{5}{6}$

Explanation:

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

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

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

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

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

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

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Example Question #881 : Arithmetic

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

Evaluate g(16) .

Possible Answers:

$\textup{The expression is undefined.}$

$5\textup{,}100$

$3\textup{,}092$

$5\textup{,}092$

$3\textup{,}100$

Correct answer:

$3\textup{,}100$

Explanation:

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

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

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

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

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

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

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

=3,100

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Example Question #243 : Integers

Solve

$\left | x -7\right | = 3$

Possible Answers:

x=10, 4

x=-4

No solution

x=10

x=0

Correct answer:

x=10, 4

Explanation:

Since this is an absolute value equation, we must set the left hand side equal to both the positive and negative versions of the right side. Therefore,

x-7 = 3

x-7 = -3

Solving the first equation, we see that x=10

Solving the second, we see that x=4

When we substitute each value back into the original equation $\left | x -7\right | = 3$ , we see that they both check.

$\left | 10 -7\right | = \left | 3 \right | =3$

$\left | 4 -7\right | = \left | -3 \right | =3$

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Example Question #12 : How To Find Absolute Value

Solve:

|4x -2 | = 14

Possible Answers:

x=-3

x=4

None of the given answers.

x=-3, 4

x=-4, 3

Correct answer:

x=-3, 4

Explanation:

To solve this equation, we want to set 4x-2 equal to both and and solve for .

Therefore:

4x-2 = 14

4x = 16

x=4

and

4x-2=-14

4x = -12

x=-3

We can check both of these answers by plugging them back into the inequality to see if the statement is true.

|4(4) -2| = |16 -2| = |14| = 14

and

|4(-3)-2|=|-14| = 14

Both answers check, so our final answer is x=-3, 4

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Example Question #884 : Arithmetic

Solve:

$| \frac{x}{2}+\pi | =4\pi$

Possible Answers:

$6\pi, -10\pi$

6, 10

$6\pi, 10\pi$

Correct answer:

$6\pi, -10\pi$

Explanation:

To solve this problem, we want to set what's inside the absolute value signs equal to the positive and negative value on the right side of the equation. That's because the value inside the absolute value symbols could be equivalent to $4\pi$ or $-4\pi$ , and the equation would still hold true.

So let's set $\frac{x}{2}+\pi$ equal to $4\pi$ and $-4\pi$ separately and solve for our unknown.

First:

$\frac{x}{2}+\pi = 4\pi$

$\frac{x}{2}=3\pi$

$x=6\pi$

Second:

$\frac{x}{2}+\pi = -4\pi$

$\frac{x}{2} = -5\pi$

$x=-10\pi$

Therefore, our answers are $6\pi$ and $-10\pi$ .

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Example Question #882 : Arithmetic

Evaluate the expression if x=5 and y=7 .

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

Possible Answers:

Correct answer:

Explanation:

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

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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SAT Math : Arithmetic

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #881 : Arithmetic

Example Question #174 : Integers

Example Question #242 : New Sat

Example Question #3 : Absolute Value

Example Question #1 : Absolute Value

Example Question #881 : Arithmetic

Example Question #243 : Integers

Example Question #12 : How To Find Absolute Value

Example Question #884 : Arithmetic

Example Question #882 : Arithmetic

All SAT Math Resources

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