GMAT Math : Problem-Solving Questions

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Example Questions

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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Example Question #1372 : Problem Solving Questions

Consider the incomplete equation

$10(N - 18) + 5N = \square (N-12 )$

What number replaces the box in order to form an equation with no solution?

Possible Answers:

None of the other choices yields a correct answer.

Correct answer:

None of the other choices yields a correct answer.

Explanation:

Set to be the number that replaces the box.

Simplify first:

10(N - 18) + 5N =A (N-12 )

10N - 180+ 5N =A (N-12 )

15N - 180 =A N-12A

15N - 180+180-AN=A N-12A +180 - AN

15N -AN= 180 - 12A

Now solve for in terms of :

$\left (15 -A \right )N= 180 - 12A$

$N=\frac{ 180 - 12A}{15 -A }$

The only possible value of that might preclude the existence of a solution is A = 15 , since it makes the denominator 0. However, let us test this value in the original equation:

10(N - 18) + 5N =15 (N-12 )

15N - 180 = 15N - 180

As it turns out, replacing the box with 15 yields an identity, not a contradiction, so the solution set is the set of all real numbers. There is no number that fits the description.

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Example Question #11 : Linear Equations, One Unknown

Consider the incomplete equation

$4x + 17 = 6\left ( \square x - 16 \right ) - 4x$

Which of the following numbers can replace the box to form an equation whose one and only solution is 2?

Possible Answers:

$- \frac{79}{12}$

None of the other responses gives a correct answer.

$\frac{4}{3}$

$\frac{129}{12}$

$\frac{113}{4}$

Correct answer:

$\frac{129}{12}$

Explanation:

Rewrite this equation as

$4x + 17 = 6\left ( A x - 16 \right ) - 4x$

If 2 is a solution of the equation, then we can substitute 2 for to make a true arithmetic equation. Replace with 2 and solve for :

$4 \cdot 2 + 17 = 6\left ( A \cdot 2- 16 \right ) - 4 \cdot 2$

$8+ 17 = 6\left (2 A - 16 \right ) - 8$

$25 = 12 A - 96 \right ) - 8$

25 = 12 A - 104

129 = 12 A

$A = \frac{129}{12}$

This number replaces the box in order to form the equation

$4x + 17 = 6\left ( \frac{129}{12}x - 16 \right ) - 4x$

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Example Question #11 : Solving Linear Equations With One Unknown

Solve the following equation for

$\frac{6x}{4} +7 = 18 +x$

Possible Answers:

x=11

x=33

x=22

x=7

x=0

Correct answer:

x=22

Explanation:

We proceed as follows

$\frac{6x}{4} +7 = 18 +x$ (Start)

6x +28 = 72 +4x (Multiply both sides by 4. Remember to distribute the 4 to both summands on both sides.)

6x = 44 +4x (Subtract 28 from both sides)

2x = 44 (Subtract 4x from both sides)

x =22 (Divide both sides by 2)

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Example Question #291 : Algebra

Define a function as follows:

f(x) = 8x - 35

If f(N)= 47 , evaluate .

Possible Answers:

N = 10.25

N = 341

N = 96

N = 1.5

N = 656

Correct answer:

N = 10.25

Explanation:

Since f(N)= 47 , we can plug N in for x and 47 in for f(N) to get the following equation,

8N - 35 = 47

From here, we want to solve for N therefore we must isolate N on one side of the equation and all other numbers on the other side.

8N = 47 + 35

8N = 82

$N = 82 \div 8$

N = 10.25

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Example Question #11 : Linear Equations, One Unknown

Lisa went to a bargain bookstore where books were sold for dollars and magazines for (x-2) dollars. After buying six books and four magazines, she only spent $30.00. How much did the books and magazines cost?

Possible Answers:

Books: $4.20; Magazines: $2.20

Books: $1.80; Magazines: $3.80

Books: $3.80; Magazines: $5.80

Books: $3.80; Magazines: $1.80

Correct answer:

Books: $3.80; Magazines: $1.80

Explanation:

We must first write out the equation to this problem.

If we set to be the cost of books then we can set to represent the cost of magazines.
She purchased books which means she spent on books.
She also bought magazines which means she spent on magazines.
The total spent was so when these two values are added together they must equal :

6x+4(x-2)=30

We can now solve for :

6x+4x-8=30

10x-8=30

10x=38

x=3.80

Remember: the cost of magazines was x-2 so we must plug in the value to find our answer.

3.80-2=1.80

So, our answer is that books cost $ 3.80 and magazines cost $ 1.80 .

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Example Question #1376 : Problem Solving Questions

Students at a local college decide to make and sell t-shirts in order to raise money for their activities. They will pay the manufacturer $\$6.00$ per t-shirt made and a fixed fee of $\$200.00$ . If they sell each t-shirt for $\$16.00$ , how many t-shirts would they have to make and sell to raise $\$1600.00$ ?

Possible Answers:

180

200

160

140

120

Correct answer:

180

Explanation:

We set up the following equation:

16x-6x-200=1600

10x=1800

x=180

The students will need to make and sell 180 t-shirts in order to raise $\$1600.00$ .

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Example Question #12 : Solving Linear Equations With One Unknown

Solve for : $\frac{1}{2}x+6=34$

Possible Answers:

x=40

x=-56

x=56

x=52

x=68

Correct answer:

x=56

Explanation:

In order to solve for , we must isolate on one side of the equation:

$\frac{1}{2}x+6=34$

$\frac{1}{2}x=28$

x=56

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GMAT Math : Problem-Solving Questions

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #287 : Algebra

Example Question #288 : Algebra

Example Question #1371 : Gmat Quantitative Reasoning

Example Question #1372 : Problem Solving Questions

Example Question #11 : Linear Equations, One Unknown

Example Question #11 : Solving Linear Equations With One Unknown

Example Question #291 : Algebra

Example Question #11 : Linear Equations, One Unknown

Example Question #1376 : Problem Solving Questions

Example Question #12 : Solving Linear Equations With One Unknown

All GMAT Math Resources

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