GMAT Math : Algebra

Study concepts, example questions & explanations for GMAT Math

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Example Questions

Example Question #12 : Inequalities

What value of \(\displaystyle m\) satisfies both of the following inequalities?

\(\displaystyle 9-3m\leq5\)

\(\displaystyle 2m+7\leq13-m\)

Possible Answers:

\(\displaystyle \frac{8}{3}\)

\(\displaystyle \frac{3}{2}\)

\(\displaystyle \frac{7}{3}\)

\(\displaystyle \frac{5}{4}\)

\(\displaystyle \frac{2}{3}\)

Correct answer:

\(\displaystyle \frac{3}{2}\)

Explanation:

Solve for \(\displaystyle m\) in both inequalities:

(1)

\(\displaystyle 9-3m\leq5\)

Subtract 9 from each side of the inequality:

\(\displaystyle -3m\leq-4\)

Then, divide by \(\displaystyle -3\). Remember to switch the direction of the sign from "less than or equal to" to "greater than or equal to."

\(\displaystyle m\geq\frac{4}{3}\)

(2)

\(\displaystyle 2m+7\leq13-m\)

Add \(\displaystyle m\) to both sides of the inequality:

\(\displaystyle 3m+7\leq13\)

Subtract 7 from each side of the inequality:

\(\displaystyle 3m\leq6\)

Divide each side of the inequality by 3:

\(\displaystyle m\leq2\)

 

From solving both inequalities, we find \(\displaystyle m\) such that:

\(\displaystyle \frac{4}{3}\leq m\leq 2\)

\(\displaystyle \frac{4}{3}\approx 1.33\)

Only \(\displaystyle \frac{3}{2}\)  is in that interval \(\displaystyle (\frac{3}{2}=1.5)\).

\(\displaystyle \frac{5}{4}\) and \(\displaystyle \frac{2}{3}\) are less than 1.33, so they are not in the interval.

\(\displaystyle \frac{7}{3}\) and \(\displaystyle \frac{8}{3}\) are more than 2, so they are not in the interval.

Example Question #2 : How To Find The Solution To An Inequality With Division

Which of the following could be a value of \(\displaystyle x\), given the following inequality?

\(\displaystyle 8-7x\leq11x+4\)

Possible Answers:

\(\displaystyle -\frac{2}{9}\)

\(\displaystyle -\frac{2}{3}\)

\(\displaystyle -\frac{1}{3}\)

\(\displaystyle \frac{1}{9}\)

\(\displaystyle \frac{4}{9}\)

Correct answer:

\(\displaystyle \frac{4}{9}\)

Explanation:

The inequality that is presented in the problem is:

\(\displaystyle 8-7x\leq11x+4\)

Start by moving your variables to one side of the inequality and all other numbers to the other side:

\(\displaystyle -7x-11x\leq4-8\)

\(\displaystyle -18x\leq-4\)

Divide both sides of the equation by \(\displaystyle -18\). Remember to flip the direction of the inequality's sign since you are dividing by a negative number!

\(\displaystyle x\geq\frac{-4}{-18}\)

Reduce:

\(\displaystyle x\geq\frac{2}{9}\)

The only answer choice with a value greater than \(\displaystyle \frac{2}{9}\) is \(\displaystyle \frac{4}{9}\).

Example Question #123 : Algebra

\(\displaystyle 6m-7>11\)

\(\displaystyle n< 2n-9\)

If \(\displaystyle m\) and \(\displaystyle n\) are two integers, which of the following inequalities would be true?

Possible Answers:

\(\displaystyle m*n< 3\)

\(\displaystyle -m*n>0\)

\(\displaystyle m+n>12\)

\(\displaystyle m+n< -12\)

\(\displaystyle \frac{m}{n}< 0\)

Correct answer:

\(\displaystyle m+n>12\)

Explanation:

First let's solve each of the inequalities:

 \(\displaystyle 6m-7>11\)

\(\displaystyle 6m>18\)

\(\displaystyle m>3\)

 

\(\displaystyle n< 2n-9\)

\(\displaystyle -n< -9\)

Don't forget to flip the direction of the sign when dividing by a negative number:

\(\displaystyle n>9\)

\(\displaystyle m+n>12\) is the correct answer. \(\displaystyle m\) is an integer greater than 3 and \(\displaystyle n\) is greater than 9. Therefore, the sum of \(\displaystyle m\) and \(\displaystyle n\) is greater than 12.

\(\displaystyle m+n< -12\) is not true. \(\displaystyle m\) and \(\displaystyle n\) are two positive integers as \(\displaystyle m\) is greater than 3 and \(\displaystyle n\) is greater than 9. The sum of two positive integers cannot be a negative number.

\(\displaystyle \frac{m}{n}< 0\) is not true. \(\displaystyle m\) and \(\displaystyle n\) are two positive integers as \(\displaystyle m\) is greater than 3 and \(\displaystyle n\) is greater than 9. The division of two positive numbers is positive and therefore cannot be less than 0.

\(\displaystyle m*n< 3\) is not true. \(\displaystyle m\) is greater than 3 and \(\displaystyle n\) is greater than 9. The product of \(\displaystyle m\) and \(\displaystyle n\) cannot be less than 3.

\(\displaystyle -m*n>0\) is not true. \(\displaystyle m\) and \(\displaystyle n\) are positive. Therefore, the product of \(\displaystyle -m\) and \(\displaystyle n\) is negative and cannot be greater than 0.

Example Question #1201 : Gmat Quantitative Reasoning

\(\displaystyle 3-x>8-2x\)

\(\displaystyle y-2>9\)

If an integer \(\displaystyle m\) satisfies both of the above inequalities, which of the following is true about \(\displaystyle m\)?

Possible Answers:

\(\displaystyle m< 11\)

\(\displaystyle m>11\)

\(\displaystyle 5< m< 11\)

\(\displaystyle m=5\)

\(\displaystyle m< 5\)

Correct answer:

\(\displaystyle m>11\)

Explanation:

First, we solve both inequalities:

\(\displaystyle 3-x>8-2x\)

\(\displaystyle 3-x+2x>8\)

\(\displaystyle x>5\)

 

\(\displaystyle y-2>9\)

\(\displaystyle y>11\)

If \(\displaystyle m\) satisfies both inequalities, then \(\displaystyle m\) is greater than 5 AND \(\displaystyle m\) is greater than 11. Therefore \(\displaystyle m\) is greater than 11.

\(\displaystyle m>11\) is the correct answer.

Example Question #1205 : Gmat Quantitative Reasoning

Solve the following inequality:

\(\displaystyle 7x-4< 13x+26\)

Possible Answers:

\(\displaystyle x< 3\)

\(\displaystyle x>-5\)

\(\displaystyle x< -5\)

\(\displaystyle x< -3\)

\(\displaystyle x>3\)

Correct answer:

\(\displaystyle x>-5\)

Explanation:

We start by simplifying our inequality like any other equation:

\(\displaystyle 7x-4< 13x+26\)

\(\displaystyle -6x< 30\)

Now we must remember that when we divide by a negative, the inequality is flipped, so we obtain:

\(\displaystyle x>-5\)

Example Question #11 : Inequalities

Solve the following inequality:

\(\displaystyle \small 4x-7\geq13x+11\)

Possible Answers:

\(\displaystyle \small \small \small \small \small x\geq -4\)

\(\displaystyle \small \small \small \small \small x\geq -2\)

\(\displaystyle \small \small \small \small x\geq 2\)

\(\displaystyle \small \small x\leq-2\)

\(\displaystyle \small \small \small x\leq2\)

Correct answer:

\(\displaystyle \small \small x\leq-2\)

Explanation:

Solving inequalities is very similar to solving equations, but we need to remember an important rule:

If we multiply or divide by a negative number, we must switch the direction of the inequality. So a "greater than" sign will become a "less than" sign and vice versa.

We are given

\(\displaystyle 4x-7\geq13x+11\)

Start by moving the \(\displaystyle 7\) and the \(\displaystyle 13x\) over:

\(\displaystyle 4x-7({\color{Blue} -13x})({\color{Red} +7})\geq13x+11({\color{Red} +7})({\color{Blue} -13x})\)

Simplify to get the following:

\(\displaystyle -9x\geq18\)

Then, we will divide both sides of the equation by \(\displaystyle -9\). Remember to switch the direction of the inequality sign!

\(\displaystyle \frac{-9x}{-9}\geq\frac{18}{-9}\)

\(\displaystyle \small x\leq\frac{18}{-9}\)

So,

\(\displaystyle \small \small x\leq-2\)

Example Question #13 : Inequalities

\(\displaystyle x = 2y- 6\) and \(\displaystyle x = 10 - 5z\). If \(\displaystyle x\) the greatest number of the three, then what are the possible values of \(\displaystyle x\) ?

Possible Answers:

\(\displaystyle x< 1 \frac{2}{3}\)

\(\displaystyle x>6\)

\(\displaystyle x< 6\)

\(\displaystyle x>1 \frac{2}{3}\)

\(\displaystyle 1\frac{2}{3} < x< 6\)

Correct answer:

\(\displaystyle x>6\)

Explanation:

The question is equivalent to asking when \(\displaystyle x > y\) and \(\displaystyle x > z\) are both true statements. 

We can solve for \(\displaystyle y\) in the first equation:

\(\displaystyle x = 2y- 6\)

\(\displaystyle 2y = x + 6\)

\(\displaystyle y = \frac{1}{2}(x+6)\)

\(\displaystyle y = \frac{1}{2}x+3\)

Now use substitution to solve the inequality

\(\displaystyle x > y\)

\(\displaystyle x > \frac{1}{2}x+3\)

\(\displaystyle x - \frac{1}{2}x > \frac{1}{2}x+3 - \frac{1}{2}x\)

\(\displaystyle \frac{1}{2}x > 3\)

\(\displaystyle 2 \cdot \frac{1}{2}x >2 \cdot 3\)

\(\displaystyle x > 6\)

Therefore, \(\displaystyle x > y\) if and only if \(\displaystyle x > 6\).

 

Similarly, we can solve for \(\displaystyle z\) in the first equation:

\(\displaystyle x = 10 - 5z\)

\(\displaystyle x + 5z - x = 10 - 5z + 5z - x\)

\(\displaystyle 5z = 10-x\)

\(\displaystyle z = \frac{1}{5}(10-x)\)

\(\displaystyle z = 2 -\frac{1}{5}x\)

Now use substitution to solve the inequality

\(\displaystyle x > z\)

\(\displaystyle x >2 -\frac{1}{5}x\)

\(\displaystyle x+ \frac{1}{5}x >2 -\frac{1}{5}x + \frac{1}{5}x\)

\(\displaystyle \frac{6}{5}x >2\)

\(\displaystyle \frac{5}{6} \cdot \frac{6}{5}x >\frac{5}{6} \cdot 2\)

\(\displaystyle x > \frac{5}{3} = 1\frac{2}{3}\)

Therefore, \(\displaystyle x > z\) if and only if \(\displaystyle x > 1\frac{2}{3}\).

 

We combine these results, and conclude that both \(\displaystyle x > y\) and \(\displaystyle x > z\) hold - that is, \(\displaystyle x\) is the greatest of the three - if and only if \(\displaystyle x > 6\).

Example Question #21 : Solving Inequalities

Solve for \(\displaystyle x\):

\(\displaystyle 8< -5x+18< 103\)

Possible Answers:

\(\displaystyle 5.2< x< 24.2\)

\(\displaystyle x\textup{ is undefined}\)

\(\displaystyle -17< x< 2\)

\(\displaystyle x< 5.2\textup{ or }x>24.2\)

\(\displaystyle x< -17\textup{ or }x>2\)

Correct answer:

\(\displaystyle -17< x< 2\)

Explanation:

The first step to solving this inequality is to subtract 18 from all three segments of the equation:

8 (- 18)  < -5x + 18 ( - 18) < 103 ( - 18)

-10 < -5x < 85

We then divide the entire equation by -5. According to inequality rules, when we multiply or divide an inequality by a negative number, we must flip the inequality signs.

(-10/-5) > (-5x/-5) > (85/-5)

2 > x > -17 or in conventional layout -17 < x < 2

Example Question #21 : Solving Inequalities

Solve:  \(\displaystyle -45 < -6x+15< 15\)

Possible Answers:

\(\displaystyle -10 > x\) or \(\displaystyle x > 0\)

\(\displaystyle 10 > x > 0\)

\(\displaystyle 10< x\)  or  \(\displaystyle x< 0\)

\(\displaystyle -60 < x < 0\)

\(\displaystyle -10< x< 0\)

Correct answer:

\(\displaystyle 10 > x > 0\)

Explanation:

In order to solve this inequality, we must split it into two inequalities:

\(\displaystyle -45 < -6x + 15\) and \(\displaystyle -6x + 15 < 15\)

We then solve each, remembering of course that, when multiplying or dividing by a negative number, we must switch the direction of the inequality sign.

\(\displaystyle -45 - 15 < -6x + 15 - 15\)

\(\displaystyle -60 < -6x\)

\(\displaystyle \frac{-60}{-6} > \frac{-6x}{-6}\)

\(\displaystyle 10 > x\)

and

\(\displaystyle -6x + 15 < 15\)

\(\displaystyle -6x + 15 - 15 < 15 - 15\)

\(\displaystyle -6x < 0\)

\(\displaystyle \frac{-6x}{-6} > \frac{0}{-6}\)

\(\displaystyle x> 0\)

Therefore:

\(\displaystyle 0< x < 10\)

Example Question #128 : Algebra

Give the solution set of the inequality

\(\displaystyle \frac{1}{x} > \frac{2}{x-4}\)

Possible Answers:

\(\displaystyle (-\infty, -4) \cup (0, 4)\)

\(\displaystyle (-4, 0) \cup (4, \infty)\)

\(\displaystyle (-\infty, -4) \cup (4, \infty)\)

\(\displaystyle (-\infty, 0) \cup (4, \infty)\)

\(\displaystyle (-4, 4 )\)

Correct answer:

\(\displaystyle (-\infty, -4) \cup (0, 4)\)

Explanation:

To solve a rational inequality, move all expressions to the left first:

\(\displaystyle \frac{1}{x} > \frac{2}{x-4}\)

\(\displaystyle \frac{1}{x} - \frac{2}{x-4} > \frac{2}{x-4} - \frac{2}{x-4}\)

\(\displaystyle \frac{1}{x} - \frac{2}{x-4} >0\)

\(\displaystyle \frac{1(x-4 )}{x(x-4 )} - \frac{2x}{x(x-4 )} >0\)

\(\displaystyle \frac{x-4}{x(x-4 )} - \frac{2x}{x(x-4 )} >0\)

\(\displaystyle \frac{-x-4}{x(x-4 )} >0\)

\(\displaystyle -1 \cdot \frac{-x-4}{x(x-4 )} < -1 \cdot 0\)

\(\displaystyle \frac{x+4}{x(x-4 )} < 0\)

The boundary points of the solution set will be the points at which:

\(\displaystyle x+4 = 0\) - that is, \(\displaystyle x = -4\);

\(\displaystyle x = 0\);

\(\displaystyle x-4 = 0\); that is, \(\displaystyle x = 4\).

None of these values will be included in the solution set, since equality is not allowed by the inequality symbol.

Test the intervals

\(\displaystyle (-\infty, -4) , (-4, 0) , (0, 4), (4, \infty)\)

by choosing a value in each interval and testing the truth of the inequality.

\(\displaystyle (-\infty, -4)\): test \(\displaystyle x= -5\)

\(\displaystyle \frac{1}{x} > \frac{2}{x-4}\)

\(\displaystyle \frac{1}{-5} > \frac{2}{-5-4}\)

\(\displaystyle -\frac{1}{5}>-\frac{2}{9}\)

True; include the interval \(\displaystyle (-\infty, -4)\)

 

\(\displaystyle (-4, 0)\): test \(\displaystyle x = -1\)

\(\displaystyle \frac{1}{-1} > \frac{2}{-1-4}\)

\(\displaystyle \frac{1}{-1} > \frac{2}{-5}\)

\(\displaystyle -1 > -\frac{2}{5}\)

False; exclude the interval \(\displaystyle (-4, 0)\)

 

\(\displaystyle (0, 4)\): test \(\displaystyle x =1\)

\(\displaystyle \frac{1}{1} > \frac{2}{1-4}\)

\(\displaystyle \frac{1}{1} > \frac{2}{-3}\)

\(\displaystyle 1 > - \frac{2}{3}\)

True; include the interval \(\displaystyle (0, 4)\).

 

\(\displaystyle (4, \infty)\): test \(\displaystyle x = 5\)

\(\displaystyle \frac{1}{5} > \frac{2}{5-4}\)

\(\displaystyle \frac{1}{5} > \frac{2}{1}\)

\(\displaystyle \frac{1}{5} > 2\)

False; exclude the interval \(\displaystyle (4, \infty)\).

 

The solution set of the inequality is \(\displaystyle (-\infty, -4) \cup (0, 4)\).

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