GMAT Math : Algebra

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Example Questions

Example Question #1 : Solving Linear Equations With One Unknown

Solve for :

n+2=-14-n

Possible Answers:

Correct answer:

Explanation:

n+2=-14-n

n+n=-14-2

2n=-16

n=-8

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Example Question #2 : Linear Equations, One Unknown

Solve for :

-6x-20=-2x+4(1-3x)

Possible Answers:

Correct answer:

Explanation:

-6x-20=-2x+4(1-3x)

-6x-20=-2x+4-12x

-6x-20=-14x+4

-6x+14x=4+20

8x=24

x=3

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Example Question #3 : Linear Equations, One Unknown

Solve for :

-14+6b+7-2b=1+5b

Possible Answers:

Correct answer:

Explanation:

-14+6b+7-2b=1+5b

-7+4b=1+5b

4b-5b=1+7

-b=8

b=-8

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Example Question #1 : Solving Linear Equations With One Unknown

What is the midpoint coordinate of (1,4) and (7,10) ?

Possible Answers:

(4,7)

(3,3)

(7,4)

(6,6)

Correct answer:

(4,7)

Explanation:

Midpoint formula:

$\left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)$

$(1,4)\ and\ (7,10)$

$\left(\frac{1+7}{2},\frac{4+10}{2}\right)$

$\left(\frac{8}{2},\frac{14}{2}\right)$

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Example Question #5 : Linear Equations, One Unknown

What is the midpoint coordinate of (1,2) and (5,2) ?

Possible Answers:

(2,0)

(6,4)

(3,2)

(2,3)

Correct answer:

(3,2)

Explanation:

Midpoint formula:

$\left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)$

$(1,2)\ and\ (5,2)$

$\left(\frac{1+5}{2},\frac{2+2}{2}\right)$

$\left(\frac{6}{2},\frac{4}{2}\right)$

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Example Question #286 : Algebra

What is the midpoint coordinate of (-2,-1) and (-8,7) ?

Possible Answers:

(5,3)

(3,-5)

(3,5)

(-5,3)

Correct answer:

(-5,3)

Explanation:

Midpoint formula:

$\left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)$

$(-2,-1)\ and\ (-8,7)$

$\left(\frac{-2+-8}{2},\frac{-1+7}{2}\right)$

$\left(\frac{-10}{2},\frac{6}{2}\right)$

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Example Question #4 : Solving Linear Equations With One Unknown

Solve the following equation:

$2\left | x-5\right |+16=30$ .

Possible Answers:

x = {-2 ; 12}

x = 2

x = {-7 ; 7}

x = {-9 : 16}

x = {0 ; 10}

Correct answer:

x = {-2 ; 12}

Explanation:

We start by isolating the absolute value expression:

This gives us two cases when we remove the absolute value:

x - 5 = 7 and x - 5 = -7

Then we solve for each case:

$x - 5 = 7 \Rightarrow x = 7 + 5 \Rightarrow x= 12$

$x - 5 = -7 \Rightarrow x = -7 + 5\Rightarrow x= -2$

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Example Question #287 : Algebra

Solve for :

5 (N - 6)- 2(N + 4) = 7 (N + 5)

Possible Answers:

$N = -18 \frac{1}{4}$

$N = -\frac{3}{4}$

$N = -1\frac{1}{4}$

$N = -25 \frac{3}{4}$

The equation has no solution.

Correct answer:

$N = -18 \frac{1}{4}$

Explanation:

5 (N - 6)- 2(N + 4) = 7 (N + 5)

$\left (5 N - 30 \right ) - \left (2 N + 8 \right )= 7N + 35$

5 N - 30 - 2 N - 8= 7N + 35

3 N - 3 8= 7N + 35

3 N - 3 8+ 38 = 7N + 35 + 38

3 N = 7N + 73

3 N-7N = 7N + 73 -7N

-4N = 73

$N = \frac{73}{-4}= - 18 \frac{1}{4}$

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Example Question #288 : Algebra

Solve for :

$\left (4x + 7 \right ) + 2 (x + 15 ) = 3(x-17)$

Possible Answers:

$x = -4\frac{2}{3}$

$x = -9\frac{7}{9}$

$x = -29 \frac{1}{3}$

$x = -24\frac{1}{3}$

The equation has no solution.

Correct answer:

$x = -29 \frac{1}{3}$

Explanation:

$\left (4x + 7 \right ) + 2 (x + 15 ) = 3(x-17)$

$\left (4x + 7 \right ) + (2x+30) = 3x-51$

6x+37 = 3x-51

6x+37 - 3x = 3x-51 - 3x

3x+37 = -51

3x = -51 - 37

3x = -88

$x = \frac{-88}{3}= -29 \frac{1}{3}$

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Example Question #1371 : Gmat Quantitative Reasoning

Which of the following equations has the set of all real numbers as its solution set?

Possible Answers:

6(N+4) + 2N = 8 (N+3)

4(N+4) +4 N = 8 (N+3)

3(N+4) + 5N = 8 (N+3)

2(N+4) + 6N = 8 (N+3)

5(N+4) + 3N = 8 (N+3)

Correct answer:

6(N+4) + 2N = 8 (N+3)

Explanation:

The right side of each equation is 8 (N+3) , which simplifies by way of distribution to

$8 (N+3) = 8 \cdot N + 8\cdot 3 = 8N + 24$

If the left side of the equation simplifies to an identical expression, the equation has all real numbers as its solutions.

We test the left side of each equation:

2(N+4) + 6N = 8 (N+3)

$2(N+4) + 6N = 2 \cdot N + 2 \cdot 4 + 6N = 2N + 8 + 6N = 8N + 8$

3(N+4) + 5N = 8 (N+3)

$3(N+4) + 5N= 3 \cdot N + 3 \cdot 4 + 5N = 3N + 12 + 5N = 8N +12$

4(N+4) +4 N = 8 (N+3) \

$4(N+4) +4 N = 4 \cdot N + 4 \cdot 4 + 4N = 4N + 16 + 4N = 8N + 16$

5(N+4) + 3N = 8 (N+3)

$5(N+4) + 3N = 5 \cdot N + 5 \cdot 4 + 3N = 5N + 20 + 3N = 8N + 20$

6(N+4) + 2N = 8 (N+3)

$6(N+4) + 2N = 6 \cdot N + 6 \cdot 4 + 2N = 6N + 24 + 2N = 8N + 24$

Of the given choices,

6(N+4) + 2N = 8 (N+3)

can be rewritten as

8N + 24= 8N + 24 ,

which is an identity and has the set of all real numbers as its solution set.

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GMAT Math : Algebra

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1 : Solving Linear Equations With One Unknown

Example Question #2 : Linear Equations, One Unknown

Example Question #3 : Linear Equations, One Unknown

Example Question #1 : Solving Linear Equations With One Unknown

Example Question #5 : Linear Equations, One Unknown

Example Question #286 : Algebra

Example Question #4 : Solving Linear Equations With One Unknown

Example Question #287 : Algebra

Example Question #288 : Algebra

Example Question #1371 : Gmat Quantitative Reasoning

All GMAT Math Resources

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