GMAT Math : Algebra

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Example Questions

Example Question #68 : Solving Equations

AB = 100,000

AC = 25

BD = 40

Which of the following is equal to ?

Possible Answers:

160,000

0.01

0.00000625

100

62,500

Correct answer:

0.01

Explanation:

AC = 25 and BD = 40 , so

$AC \cdot BD = 25 \cdot 40 = 1,000$

$AB \cdot CD = 1,000$

Since AB= 100,000 , it follows that

$100,000 \cdot CD = 1,000$

$CD = 1,000 \div 100,000 = 0.01$

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Example Question #1461 : Problem Solving Questions

True or false: N = 5 .

Statement 1: $(N+2)^{2} = 2N + 19$

Statement 2: $1 + \frac{10}{N-5} = 0$

Possible Answers:

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

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.x

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

Correct answer:

EITHER STATEMENT ALONE provides sufficient information to answer the question.

Explanation:

Substitute 5 for in both statements, and it becomes immediately apparent that it is a solution of neither statement:

$(N+2)^{2} = 2N + 19$

$(5+2)^{2} = 2 \cdot 5 + 19$

$7^{2} = 10+19$

49= 29 , a false statement.

$1 + \frac{10}{N-5} = 0$

$1 + \frac{10}{5-5} = 0$

$1 + \frac{10}{0} = 0$ , a false statement since the left quantity, having a zero denominator, is undefined.

Therefore, it follows from either statement alone that N = 5 is false.

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Example Question #70 : Solving Equations

is a rational number. True or false: x = 5

Statement 1: $x^{2} - 3x - 10 = 0$

Statement 2: $1 - \frac{7}{x+2} = 0$

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.

EITHER STATEMENT ALONE provides sufficient information to answer the question.

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

Correct answer:

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

Explanation:

Assume Statement 1 alone and solve for :

$x^{2} - 3x - 10 = 0$

(x-5)(x+2) = 0

Either:

x - 5 = 0 , in which case x = 5 ,

x+2 = 0 , in which case x = -2 .

Therefore, Statement 1 alone is inconclusive.

Assume Statement 2 alone and solve for :

$1 - \frac{7}{x+2} = 0$

$1 =\frac{7}{x+2}$

$1 \cdot (x+ 2) =\frac{7}{x+2} \cdot (x+ 2)$

x+ 2 = 7

x = 5

The only solution to this equation is x = 5 , so Statement 2 alone answers the question.

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Example Question #381 : Algebra

D = E - 12

$E = \sqrt{F}$

F = 2 G

Express in terms of .

Possible Answers:

$G = \sqrt{2D}- 12$

$G = 4D^{2} - 96D + 576$

$G = \sqrt{2D-24}$

$G = \frac{1}{2} D^{2} -12D +72$

$G = \frac{1}{2} D^{2} +12D +72$

Correct answer:

$G = \frac{1}{2} D^{2} +12D +72$

Explanation:

F = 2 G

$F\div 2 = 2 G \div 2$

$\frac{1}{2} F = G$

$E = \sqrt{F}$

$E^{2} = (\sqrt{F})^{2}$

$E^{2} =F$

D = E - 12

D + 12 = E - 12 + 12

D + 12 = E

By two substitutions:

$G = \frac{1}{2} F = \frac{1}{2} E^{2} = \frac{1}{2} (D+12)^{2}$

$G = \frac{1}{2} (D^{2} +2 \cdot 12 \cdot D +12^{2})$

$G = \frac{1}{2} (D^{2} +24 D +144)$

$G = \frac{1}{2} D^{2} +12D +72$

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Example Question #382 : Algebra

Which of the following is the solution set of the equation:

$x-3 = \sqrt{2x+9}$

Possible Answers:

$\left \{ 0,8\right \}$

$\left \{ 8\right \}$

$\left \{ -8\right \}$

$\left \{ 0\right \}$

$\left \{ -8, 0\right \}$

Correct answer:

$\left \{ 8\right \}$

Explanation:

Square both sides of the equation, then solve the resulting quadratic equation by factoring:

$x-3 = \sqrt{2x+9}$

$\left (x-3 \right ) ^{2}=\left ( \sqrt{2x+9 \right )}^{2}$

$x^{2}-6x+9=2x+9$

$x^{2}-8x=0$

x(x-8)=0

Either x=0

x-8=0

x=8

Checking both solutions, however:

$0-3 = \sqrt{2\cdot 0+9}$

-3 = 3

This eliminates 0 as a solution.

$8-3 = \sqrt{2\cdot 8+9}$

5=5

8 turns out to be the only solution.

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Example Question #381 : Algebra

Solve for .

x^2+4x+9=15

Possible Answers:

$x=-4\pm2\sqrt{10}$

$x=\pm 3$

$x=4\pm4\sqrt{10}$

$x=-2\pm \sqrt{10}$

Correct answer:

$x=-2\pm \sqrt{10}$

Explanation:

Solve using the quadratic formula:

$ax^2+bx+c=0;\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

x^2+4x+9=15

$x=\frac{-4\pm\sqrt{4^2-(4)(1)(-6)}}{(2)(1)}$

$x=\frac{-4\pm\sqrt{16+24}}{2}$

$x=\frac{-4\pm\sqrt{40}}{2}$

$x=\frac{-4\pm2\sqrt10}{2}$

$x=-2\pm\sqrt{10}$

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Example Question #2 : Solving Quadratic Equations

Multiply: $(x^{2} - 5x + 3) (x - 2)$

Possible Answers:

$x^{3} - 7 x^{2}+7x - 6$

$x^{3} +3 x^{2}+7x - 6$

$x^{3} +7 x^{2}+10x - 6$

$x^{3} - 7 x^{2}+13x - 6$

$x^{3} +3 x^{2}+13x - 6$

Correct answer:

$x^{3} - 7 x^{2}+13x - 6$

Explanation:

Distribute:

$(x^{2} - 5x + 3) (x - 2)$

$=x^{2}(x - 2) - 5x(x - 2) + 3(x - 2)$

$=x^{2}\cdot x - x^{2}\cdot 2- 5x \cdot x +5x \cdot 2 + 3 \cdot x - 3 \cdot 2$

$=x^{3} - 2 x^{2}- 5 x^{2} +10x + 3 x - 6$

$=x^{3} - 7 x^{2}+13x - 6$

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Example Question #1 : Solving Quadratic Equations

Consider the equation $x^{2} -Bx + B = 0$ . For what value(s) of does the equation have two real solutions?

Possible Answers:

B< -4 or B >4

B< -4

B< 0 or B >4

B >4

B< -4 or B >0

Correct answer:

B< 0 or B >4

Explanation:

The discriminant of the expression $x^{2} -Bx + B$ is $\left (-B \right )^{2} - 4 \cdot1\cdot B = B^{2} -4B$ . For the equation $x^{2} -Bx + B = 0$ to have two real solutions, this discriminant must be positive, so:

$B^{2} -4B > 0$

$B \left (B -4 \right ) > 0$

One of two things happens:

Case 1:

B> 0 and B -4 >0

B> 0 and B >4

But this is the same as simply saying B >4

Case 2:

B< 0 and B -4 <0

B< 0 and B <4

But this is the same as simply saying B< 0

Therefore, the equation has two real solutions if and only if either B< 0 or B >4

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Example Question #2 : Solving Quadratic Equations

Find all real solutions for :

$x^{4} + 9x ^{2}- 10 = 0$

Possible Answers:

$x \in \left \{-\sqrt{10},-1,1, \sqrt{10} \right \}$

$x \in \left \{ -1,1 \right \}$

$x \in \left \{1, \sqrt{10} \right \}$

The equation has no real solution.

$x \in \left \{-\sqrt{10}, \sqrt{10} \right \}$

Correct answer:

$x \in \left \{ -1,1 \right \}$

Explanation:

Substitute $u = x^{2}$ , and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ , and solve the resulting quadratic equation.

$x^{4} + 9x ^{2}- 10 = 0$

$u^{2} + 9u- 10 = 0$

We can rewrite the quadratic expression as (x + ?) (x + ?) , where the question marks are replaced with integers whose product is and whose sum is 9; the integers are -1,10 :

(u-1)(u+10)= 0

Set each factor to zero and solve for ; then substitute back and solve for :

u - 1 = 0

u = 1

$x^{2}= 1$

$x = -1 \textrm{ or } x = 1$

u + 10 = 0

u = -10

$x^{2}= -10$ , which has no real solution.

Therefore, the solution set is $\left \{ -1,1\right \}$

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Example Question #3 : Solving Quadratic Equations

What is the minimum value of the function $f(x) = x^{2} - 7x + 7$ for all real values of ?

Possible Answers:

$-3\frac{1}{2}$

$-19 \frac{1}{4}$

f(x) does not have a minimum value.

$-5 \frac{1}{4}$

Correct answer:

$-5 \frac{1}{4}$

Explanation:

We find the -coordinate of the vertex of the parabola for f(x) . First, we find its -coordinate using the formula

$x = -\frac{b}{2a}$ , setting a = 1, b = -7 .

$x = -\frac{b}{2a} = -\frac{-7}{2 \cdot 1} = \frac{7}{2}$

$f \left ( \frac{7}{2} \right)$ is the -coordindate of the vertex, and, subsequently, the minimum value of f(x) :

$f(x) = x^{2} - 7x + 7$

$f\left (\frac{7}{2} \right ) = \left (\frac{7}{2} \right ) ^{2} - 7\cdot \left (\frac{7}{2} \right )+ 7$

$f\left (\frac{7}{2} \right ) = \frac{49}{4} - \frac{49}{2} + 7$

$f\left (\frac{7}{2} \right ) = \frac{49}{4} - \frac{98}{4} + \frac{28}{4}$

$f\left (\frac{7}{2} \right ) = -\frac{21}{4} = -5 \frac{1}{4}$

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GMAT Math : Algebra

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #68 : Solving Equations

Example Question #1461 : Problem Solving Questions

Example Question #70 : Solving Equations

Example Question #381 : Algebra

Example Question #382 : Algebra

Example Question #381 : Algebra

Example Question #2 : Solving Quadratic Equations

Example Question #1 : Solving Quadratic Equations

Example Question #2 : Solving Quadratic Equations

Example Question #3 : Solving Quadratic Equations

All GMAT Math Resources

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