GMAT Math : Algebra

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Example Questions

Example Question #1301 : Problem Solving Questions

Give the third term of a sequence $a_{n}$ .

Statement 1: The first and second terms are 10 and 20, respectively.

Statement 2: The fourth and fifth terms are 40 and 50, respectively.

Possible Answers:

BOTH STATEMENTS TOGETHER do NOT provide sufficient information to answer the question.

BOTH STATEMENTS TOGETHER provide sufficient information to answer the question, but NEITHER STATEMENT ALONE provides sufficient information to answer the question.

STATEMENT 2 ALONE provides sufficient information to answer the question, but STATEMENT 1 ALONE does NOT provide sufficient information to answer the question.

STATEMENT 1 ALONE provides sufficient information to answer the question, but STATEMENT 2 ALONE does NOT provide sufficient information to answer the question.

EITHER STATEMENT ALONE provides sufficient information to answer the question.

Correct answer:

BOTH STATEMENTS TOGETHER do NOT provide sufficient information to answer the question.

Explanation:

The two statements are insuffcient to determine the third term. Between them, only four of the terms are given. While the sequence seems to be arithmetic with common difference 10, this is not explicitly stated; no specific rule is given for the sequence.

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Example Question #80 : Understanding Functions

Piecewise

Let be the piecewise-defined function graphed above. Define function

g(x) = 2x-7 .

Give the domain of the function $f \circ g$ .

Possible Answers:

$\left [-1, \frac{1}{2}\right ] \cup \left [1\frac{1}{2}, 6 \right )$

$(-\infty, \infty)$

$\left [ -\frac{1}{2}, 7 \right ]$

$\left [ -25, -19\right ] \cup \left [-11, 3 \right )$

$\left [ -23, 7\right ]$

Correct answer:

$\left [ -\frac{1}{2}, 7 \right ]$

Explanation:

is a polynomial function, so its own domain is the set of all real numbers. This does not restrict the domain of $f \circ g$ . However, since $(f \circ g)(x) = f[g(x)]$ , it follows that g(x) must fall within the domain of , which is [-8, 7] .

Therefore,

$-8 \le g(x) \le 7$

$-8 \le 2x- 7 \le 7$

$-8 + 7 \le 2x- 7+ 7 \le 7 + 7$

$-1 \le 2x \le 14$

$-1\div 2 \le 2x \div 2 \le 14 \div 2$

$-\frac{1}{2} \le x \le 7$

The domain of $f \circ g$ is the set $\left [ -\frac{1}{2}, 7 \right ]$ .

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Example Question #221 : Algebra

Give the common difference of an arithmetic sequence $a_{n}$ .

Statement 1: $a_{5} - a_{1} = 100$

Statement 2: $a_{1} + 200 = a_{9}$

Possible Answers:

STATEMENT 2 ALONE provides sufficient information to answer the question, but STATEMENT 1 ALONE does NOT provide sufficient information to answer the question.

BOTH STATEMENTS TOGETHER provide sufficient information to answer the question, but NEITHER STATEMENT ALONE provides sufficient information to answer the question.

EITHER STATEMENT ALONE provides sufficient information to answer the question.

STATEMENT 1 ALONE provides sufficient information to answer the question, but STATEMENT 2 ALONE does NOT provide sufficient information to answer the question.

BOTH STATEMENTS TOGETHER do NOT provide sufficient information to answer the question.

Correct answer:

EITHER STATEMENT ALONE provides sufficient information to answer the question.

Explanation:

The common difference of an arithmetic sequence is the difference of two consecutive terms, $a_{n} - a_{n-1}$ . It can be calculated from the difference of any two terms $a_{n}$ and $a_{m}$ from the formula

$d = \frac{a_{n}-a_{m}}{n-m }$

By Statement 1 alone, setting n = 5 and m = 1 ,

$d = \frac{a_{n}-a_{m}}{n-m } = \frac{a_{5}-a_{1}}{5-1 } = \frac{100}{4} = 25$ .

By Statement 2 alone, setting n = 9 and m = 1 ,

$d = \frac{a_{n}-a_{m}}{n-m } = \frac{a_{9}-a_{1}}{9-1 } = \frac{200}{8} = 25$ .

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Example Question #81 : Understanding Functions

f(x) = |8-x|

g(x) = |x-8|

How many values of make

(f + g )(x) = 0

a true statement?

Possible Answers:

Two

Infinitely many

One

Four

None

Correct answer:

One

Explanation:

(f + g )(x) = f(x)+g(x) = |8 - x| + |x-8| , so the problem is equivalent to solving

|8 - x| + |x-8| = 0 .

It can be seen that -(x-8)= -x+8 = 8 - x ; since the expressions within the two sets of absolute value bars are opposites, it follows that

|8 - x| = |x-8| ,

and the equation is equivalent to

2 |x-8| = 0

$2 |x-8| \div 2 = 0 \div 2$

|x-8| = 0

x-8 = 0

x = 8

This is the only solution, so the correct choice is one.

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Example Question #82 : Functions/Series

f(x) = |17-x|

g(x) = |x-17|

How many values of make

(f - g )(x) = 0

a true statement?

Possible Answers:

None

One

Four

Infinitely many

Two

Correct answer:

Infinitely many

Explanation:

(f + g )(x) = f(x)+g(x) = |17 - x| + |x-17| , so the problem is equivalent to solving

|17 - x| - |x-17| = 0 .

It can be seen that -(x-17)= -x+17 = 17 - x ; since the expressions within the two sets of absolute value bars are opposites, it follows that

|17 - x| = |x-17| ,

and the equation is equivalent to

|x-17| - |x-17| = 0

0 = 0

Therefore, the statement (f - g )(x) = 0 is true for all real values of , and the correct response is infinitely many.

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Example Question #83 : Functions/Series

The domain and the range of a function are both the set $\left \{ 1, 2, 3, 4 \right \}$ . Also, $f^{-1}$ exists.

Which of the following tables cannot show the values of for each of the given domain elements?

Possible Answers:

None of the other responses is correct.

$\begin{matrix} \underline{x }&\underline{f(x)} \\ 1& 4\\ 2&3\\ 3&2 \\ 4& 1 \end{matrix}$

$\begin{matrix} \underline{x }&\underline{f(x)} \\ 1& 1\\ 2&4\\ 3&3 \\ 4& 2 \end{matrix}$

$\begin{matrix} \underline{x }&\underline{f(x)} \\ 1& 2\\ 2&3 \\ 3&4 \\ 4& 1 \end{matrix}$

$\begin{matrix} \underline{x }&\underline{f(x)} \\ 1& 1\\ 2&2 \\ 3&3 \\ 4& 4 \end{matrix}$

Correct answer:

None of the other responses is correct.

Explanation:

A function has an inverse if and only if, if a ,b are in the domain of , then f(a) = f(b) implies that a =b , or, contrapositively, if $a \ne b$ , then $f(a) \ne f(b)$ . In each of the four choices, no two values of are matched with the same value of f(x) , so $a \ne b$ implies that $f(a) \ne f(b)$ in all four choices. Since the entire domain is given to be $\left \{ 1, 2, 3, 4 \right \}$ , all of the given functions have inverses.

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Example Question #85 : Functions/Series

Which of the following is an example of a relation which is not a function?

Pairing each positive integer with...

Possible Answers:

...with the word "green" if is a prime integer, the word "red" if is a composite integer, and the word "white" if N = 1 .

...the word "yellow" if has four or more factors, and the word "black" if has fewer than four factors.

...the word "orange" if is odd, and the word "turquoise" if is divisible by 2.

...the word "brown" if and 20 are relatively prime, and the word "gray" if is divisible by either 4, 5, or both.

...the word "blue" if there exists a positive number with absolute value , and the word "purple" if there exists a negative number with absolute value .

Correct answer:

...the word "blue" if there exists a positive number with absolute value , and the word "purple" if there exists a negative number with absolute value .

Explanation:

A relation is a function if and only if, for each value in the domain, there is one and only one value in the range that can be paired with . Let us examine all of the relations.

Pairing each positive integer with the word "orange" if is odd, and the word "turquoise" if is is divisible by 2:

Every positive integer is either odd (not divisible by 2) or even (divisible by 2); no integer is both.

Pairing each positive integer with the word "brown" if and 20 are relatively prime, and the word "gray" if divisible by either 2, 5, or both:

If is relatively prime to 20, then it shares only one factor with 20, which is 1; otherwise, shares at least one prime factor with 20 - 2, 5, or both.

Pairing each positive integer with the word "yellow" if has four or more factors, and the word "black" if has fewer than four factors:

This is simply saying that either does fall in a set or does not fall in the same set.

Pairing each positive integer with with the word "green" if is a prime integer, the word "red" if is a composite integer, and the word "white" if .

Every positive integer except 1 is either prime or composite, and no number is both; 1 is considered neither.

In each of the above cases, each positive integer belongs in exactly one of the described sets, so is paired with only one word. Each of these relations is a function.

Now consider this relation:

Pairing each positive integer with the word "blue" if there exists a positive number with absolute value , and the word "purple" if there exists a negative number with absolute value .

For each positive integer , there exists one positive integer and one negative integer with absolute value ; for example, if N = 7 , then

|7| = |-7| = 7 .

Therefore, each positive integer is paired with two values, "blue" and "purple", and the relation is not a function.

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Example Question #81 : Understanding Functions

For any values , , define the operation $\Gamma$ as follows:

$a\; \Gamma\; b = 2ab - a - b$

Which of the following expressions is equal to $2x\; \Gamma\; 2y$ ?

Possible Answers:

4xy +2x-2y

8xy -2x-2y

8xy

4xy -2x-2y

Correct answer:

8xy -2x-2y

Explanation:

Substitute and for and in the expression for $a\; \Gamma\; b$ :

$2x\; \Gamma\; 2y = 2(2x)(2y) - 2x - 2y = 8xy -2x-2y$

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Example Question #1 : Simplifying Algebraic Expressions

Factor $\frac{x^{2}+6x+5}{x^{2}+10x+25}.$

Possible Answers:

$\dpi{100} \small 1$

$\frac{1}{(x-5)^{2}}$

$\frac{1}{(x+5)^{2}}$

$\frac{x+1}{x+5}$

$\frac{x+1}{x-5}$

Correct answer:

$\frac{x+1}{x+5}$

Explanation:

Let's first look at the numerator and denominator separately.

$x^{2}+6x+5$ : We need two numbers that multiply to 5 and add to 6. The numbers 1 and 5 work. So, $x^{2}+6x+5 = (x+5)(x+1)$

$x^{2}+10x + 25$ : We need two numbers that multiply to 25 and add to 10. The numbers 5 and 5 work. So, $x^{2}+10x + 25 = (x+5)(x+5)$

Putting this together, $\frac{x^{2}+6x+5}{x^{2}+10x+25} = \frac{(x+5)(x+1)}{(x+5)(x+5)} = \frac{x+1}{x+5}$

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Example Question #222 : Algebra

Find the solutions to the equation x+3y+x-3=2y+2x+y .

Possible Answers:

x=3y

all real numbers

x=3, y is any real number

no solution

x=2,y=0

Correct answer:

no solution

Explanation:

Let's combine like terms.

2x+3y-3=2x+3y

-3=0 , so the equation has no solution.

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GMAT Math : Algebra

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1301 : Problem Solving Questions

Example Question #80 : Understanding Functions

Example Question #221 : Algebra

Example Question #81 : Understanding Functions

Example Question #82 : Functions/Series

Example Question #83 : Functions/Series

Example Question #85 : Functions/Series

Example Question #81 : Understanding Functions

Example Question #1 : Simplifying Algebraic Expressions

Example Question #222 : Algebra

All GMAT Math Resources

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