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SSAT Upper Level Math : Algebra

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Example Questions

Example Question #31 : Algebra

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:

The expression is undefined.

$\frac{5}{8}$

None of the other responses is correct.

$1 \frac{1}{8}$

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 #11 : How To Find 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:

$-\frac{1} {9}$

$\frac{1} {9}$

$\frac{1}{11}$

$-\frac{1}{11}$

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 #11 : 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:

$5 \frac{5}{6}$

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

$6\frac{1}{2}$

$2\frac{1}{2}$

$1\frac{1}{2}$

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 #1 : How To Find Absolute Value

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

Evaluate g(16) .

Possible Answers:

$\textup{The expression is undefined.}$

$3\textup{,}100$

$3\textup{,}092$

$5\textup{,}092$

$5\textup{,}100$

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 #4 : Absolute Value

Define an operation $\blacklozenge$ as follows:

For all real numbers a,b ,

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

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

Possible Answers:

Both and

Correct answer:

Explanation:

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

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

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

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

$= \left |11 \right |$

= 11

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Example Question #31 : Algebra

Given: a,b,c are distinct integers such that:

|a|< b < |c|

a+|b| = |c|

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

Possible Answers:

or only

, , or

or only

only

Correct answer:

or only

Explanation:

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

If is nonnegative, then a = |a| < b . If is negative, then it follows that a < 0 < b . Either way, a < b . Therefore, cannot be the least.

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

For example, if a = 4, b= 6, c = 10 , then is the least; we test both statements:

|a|< b < |c|

|4|< 6 < |10|

4 < 6< 10 , which is true.

a+|b| = |c|

4+|6| = |10|

4+6 =10 , which is also true.

If a = 4, b= 6, c = - 10 , then is the least; we test both statements:

|a|< b < |c|

|4|< 6 < |-10|

4 < 6< 10 , which is true.

a+|b| = |c|

4+|6| = |-10|

4+6 =10 , which is also true.

Therefore, the correct response is or only.

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Example Question #33 : Algebra

, , and are distinct integers. |a|<b and |b| < c . Which of the following could be the greatest of the three?

Possible Answers:

only

, , or

only

None of the other responses is correct.

Correct answer:

only

Explanation:

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

|a|<b < c

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

a < b < c ,

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

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Example Question #11 : How To Find Absolute Value

Give the solution set:

|x- 6| > 17

Possible Answers:

$(-\infty, -11) \cup (23, \infty)$

$(-\infty, -11) \cup (11, \infty)$

$(-\infty, -23) \cup (23, \infty)$

$(-\infty,-23 ) \cup (11, \infty)$

$(-\infty,-23 ) \cup (-11, \infty)$

Correct answer:

$(-\infty, -11) \cup (23, \infty)$

Explanation:

If |x- 6| > 17 , then either x -6 > 17 or x- 6 < -17 . Solve separately:

x -6 > 17

x -6 + 6 > 17 + 6

x> 23

x- 6 < -17

x- 6 + 6< -17 + 6

x< -11

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

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Example Question #31 : Algebra

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

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

If a = b , then $a \bigstar b = 4$

If a< b , 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?

Possible Answers:

p < q = r

r < p = q

p = q = r

q = r< p

p = q < r

Correct answer:

p = q < r

Explanation:

$p = 0 \bigstar 5$

Since 0 < 5 , evaluate

$a \bigstar b = |a - b+1|$ , setting a = 0, b=5 :

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

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

$q = 0 \bigstar 0$

Since 0 = 0 , then select the pattern

$0 \bigstar 0 = 4$

$r=0 \bigstar( -5)$

Since 0 > -5 , evaluate

$a \bigstar b = |a + b-1|$ , setting a = 0, b=-5 :

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

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

p = q= 4, r = 6 , so the correct choice is that p = q < r .

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Example Question #36 : Algebra

Given: m,n,p are distinct integers such that:

|m|< n < |p|

|m|+n = p

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

Possible Answers:

only

, , or

only

or only

only

Correct answer:

only

Explanation:

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

If is nonnegative, then m = |m| < n . If is negative, then it follows that m < 0 < n . Either way, m < n . Therefore, cannot be the least.

Now examine the statemtn |m|+n = p . If m = 0 , then n = p - but we are given that and are distinct. Therefore, is nonzero, |m| > 0 , and

|m| +n > 0 + n

and

p > n .

cannot be the least either.

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SSAT Upper Level Math : Algebra

Study concepts, example questions & explanations for SSAT Upper Level Math

All SSAT Upper Level Math Resources

Example Questions

Example Question #31 : Algebra

Example Question #11 : How To Find Absolute Value

Example Question #11 : Absolute Value

Example Question #1 : How To Find Absolute Value

Example Question #4 : Absolute Value

Example Question #31 : Algebra

Example Question #33 : Algebra

Example Question #11 : How To Find Absolute Value

Example Question #31 : Algebra

Example Question #36 : Algebra

All SSAT Upper Level Math Resources

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