GRE Math : GRE Quantitative Reasoning

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Example Questions

Example Question #43 : Algebra

Solve for $\displaystyle x$:

$\displaystyle 4x + 9x + 13 = 0$

Possible Answers:

$\frac{12}{13}$ $\displaystyle \frac{12}{13}$

$\frac{13}{9}$ $\displaystyle \frac{13}{9}$

$\displaystyle -1$

$\frac{13}{4}$ $\displaystyle \frac{13}{4}$

$\displaystyle 1$

Correct answer:

$\displaystyle -1$

Explanation:

First combine like terms. In this case, 4x and 9x can be added together:

13x + 13 = 0

Subtract 13 from both sides:

13x = -13

Divide both sides by 13 to isolate x:

x = -13/13

x = -1

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Example Question #53 : Equations / Inequalities

Ben is walking three dogs that weigh an average of 75 pounds each. Ben begins to walk a fourth dog, and the average weight of the dogs decreases to 70 pounds. What is the weight in pounds of the fourth dog?

Possible Answers:

$55$ $\displaystyle 55$

$45$ $\displaystyle 45$

$50$ $\displaystyle 50$

$60$ $\displaystyle 60$

$40$ $\displaystyle 40$

Correct answer:

$55$ $\displaystyle 55$

Explanation:

The total weight of the first three dogs is 225 pounds. This amount, plus the weight of the fourth dog, divided by total number of dogs, is the new average weight:

$\frac{d + 225}{4} = 70$ $\displaystyle \frac{d + 225}{4} = 70$

$d + 225 = 280$ $\displaystyle d + 225 = 280$

$d = 55 lbs$ $\displaystyle d = 55 lbs$

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Example Question #43 : Algebra

Pets Plus makes bird houses. Their monthly fixed expenses are $750. The cost for each bird house is $15. The bird houses sell for $40.

What is the monthly break-even point at Pets Plus?

Possible Answers:

$25$ $\displaystyle 25$

$50$ $\displaystyle 50$

$30$ $\displaystyle 30$

$40$ $\displaystyle 40$

$35$ $\displaystyle 35$

Correct answer:

$30$ $\displaystyle 30$

Explanation:

Let $x=$ $\displaystyle x=$ the number of bird houses sold each month.

$Revenue = 40x$ $\displaystyle Revenue = 40x$

$Costs=15x+750$ $\displaystyle Costs=15x+750$

The break-even point is where the revenue is the same as the costs:

$Revenue=Costs$ $\displaystyle Revenue=Costs$

$40x=15x+750$ $\displaystyle 40x=15x+750$

Solve for $x$ $\displaystyle x$:

$x=30$ $\displaystyle x=30$

Therefore, Pets Plus must sell 30 bird houses to break-even.

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Example Question #111 : Linear / Rational / Variable Equations

Pets Plus makes bird houses. Their monthly fixed expenses are $750. The cost for each bird house is $15. The bird houses sell for $40.

If Pets Plus sells 50 bird houses, what is the profit?

Possible Answers:

$\$500$ $\displaystyle \$500$

$\$750$ $\displaystyle \$750$

$\$625$ $\displaystyle \$625$

$\$300$ $\displaystyle \$300$

$\$250$ $\displaystyle \$250$

Correct answer:

$\$500$ $\displaystyle \$500$

Explanation:

Let $x$ $\displaystyle x$ = the number of birdhouses sold each month.

$Revenue=40x$ $\displaystyle Revenue=40x$

$Costs=15x+750$ $\displaystyle Costs=15x+750$

$Profit = Revenue-Costs$ $\displaystyle Profit = Revenue-Costs$

$=40x-15x-750$ $\displaystyle =40x-15x-750$

$=25x-750$ $\displaystyle =25x-750$

Substituting in 50 for $x$ $\displaystyle x$ gives an answer of 500, so the profit on 50 birdhouses is $500.

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Example Question #111 : Equations / Inequalities

George is three times older than Joey. The sum of their ages is 16. What is the product of their ages?

Possible Answers:

$\displaystyle 54$

$\displaystyle 27$

$\displaystyle 48$

$\displaystyle 64$

$\displaystyle 36$

Correct answer:

$\displaystyle 48$

Explanation:

Let $\displaystyle x$ = Joey's age and $\displaystyle 3x$ = George's age.

Then the equation to solve becomes $\displaystyle x + 3x = 16$.

4x=16 $\displaystyle 4x=16$

x=4 $\displaystyle x=4$

Therefore, Joey is 4 years old and George is 12 years old, so the product of their ages is 48.

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Example Question #61 : Linear / Rational / Variable Equations

Three consecutive even numbers add to 42. What is the middle number?

Possible Answers:

$18$ $\displaystyle 18$

$16$ $\displaystyle 16$

$14$ $\displaystyle 14$

$12$ $\displaystyle 12$

$10$ $\displaystyle 10$

Correct answer:

$14$ $\displaystyle 14$

Explanation:

Let $x$ $\displaystyle x$ = 1st even number, $x+2$ $\displaystyle x+2$ = 2nd even number, and $x+4$ $\displaystyle x+4$ = 3rd even number.

Then the equation to solve becomes $x+(x+2)+(x+4)=42$ $\displaystyle x+(x+2)+(x+4)=42$.

$3x+6=42$ $\displaystyle 3x+6=42$

Thus $x=12,x+2=14,\ and\ x+4=16$ $\displaystyle x=12,x+2=14,\ and\ x+4=16$, so the middle number is 14.

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Example Question #61 : Equations / Inequalities

Consider the following equation:

$\left | 2x-9 \right | = x-3$ $\displaystyle \left | 2x-9 \right | = x-3$

Which of the following must be true?

$I) \quad x=6$ $\displaystyle I) \quad x=6$

$II) \quad x \mbox{ is positive }$ $\displaystyle II) \quad x \mbox{ is positive }$

$III) \quad x \mbox{ is even }$ $\displaystyle III) \quad x \mbox{ is even }$

Possible Answers:

$I, II \mbox{ and } III$ $\displaystyle I, II \mbox{ and } III$

$II \mbox{ and } III \mbox{ only }$ $\displaystyle II \mbox{ and } III \mbox{ only }$

$II \mbox{ only }$ $\displaystyle II \mbox{ only }$

$\mbox{ none of these }$ $\displaystyle \mbox{ none of these }$

$III \mbox{ only }$ $\displaystyle III \mbox{ only }$

Correct answer:

$II \mbox{ and } III \mbox{ only }$ $\displaystyle II \mbox{ and } III \mbox{ only }$

Explanation:

The quantity inside the absolute value brackets must equal either x-3 $\displaystyle x-3$ or -(x-3) $\displaystyle -(x-3)$, depending on whether the quantity inside the brackets is positive or negative. We therefore have two seperate equations:

$\displaystyle 2x-9 = x-3$

$\displaystyle 2x-9 =-(x-3)$

To solve the first equation, add 9 to both sides:

$\displaystyle 2x = x+6$

Subtract $\displaystyle x$ from both sides:

x=6 $\displaystyle x=6$

This is the first solution. Now let's look at the second equation. The distributive law gives us:

$\displaystyle 2x-9 =-x+3$

Add 9 to both sides:

$\displaystyle 2x =-x+12$

Add $\displaystyle x$ to both sides:

3x = 12 $\displaystyle 3x = 12$

Divide both sides by 3:

x=4 $\displaystyle x=4$

Therefore, $\displaystyle x$ is either 4 or 6.

Statement $\displaystyle I$ does NOT have to be true because $\displaystyle x$ can also equal 4.

Statement $\displaystyle II$ must be true because both 4 and 6 are positive .

Finally, statement III $\displaystyle III$ always holds because 4 and 6 are both even.

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Example Question #112 : Equations / Inequalities

$5\textup{,}100 = 300(2v + 7)$ $\displaystyle 5\textup{,}100 = 300(2v + 7)$,

then v = $\displaystyle v =$

Possible Answers:

$\displaystyle 1$

$\frac{1}{5}$ $\displaystyle \frac{1}{5}$

$\displaystyle 5$

$\displaystyle 50$

500 $\displaystyle 500$

Correct answer:

$\displaystyle 5$

Explanation:

Divide both sides by 300 to get $\displaystyle 17 = 2v + 7$. Subtract 7 and divide by two to get v = 5 $\displaystyle v = 5$.

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Example Question #91 : How To Find The Solution To An Equation

If b – 3 = a, then (a – b)² =

Possible Answers:

The answer cannot be determined from the given information.

–9

–3

Correct answer:

Explanation:

The quantity can be regrouped to be –3 = a – b. Thus, (a – b)² = (–3)² = 9.

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Example Question #1 : Quadratic Equation

Solve for x: x² + 4x = 5

Possible Answers:

-1 or 5

-5

None of the other answers

-5 or 1

-1

Correct answer:

-5 or 1

Explanation:

Solve by factoring. First get everything into the form Ax² + Bx + C = 0:

x² + 4x - 5 = 0

Then factor: (x + 5) (x - 1) = 0

Solve each multiple separately for 0:

X + 5 = 0; x = -5

x - 1 = 0; x = 1

Therefore, x is either -5 or 1

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GRE Math : GRE Quantitative Reasoning

Study concepts, example questions & explanations for GRE Math

All GRE Math Resources

Example Questions

Example Question #43 : Algebra

Example Question #53 : Equations / Inequalities

Example Question #43 : Algebra

Example Question #111 : Linear / Rational / Variable Equations

Example Question #111 : Equations / Inequalities

Example Question #61 : Linear / Rational / Variable Equations

Example Question #61 : Equations / Inequalities

Example Question #112 : Equations / Inequalities

Example Question #91 : How To Find The Solution To An Equation

Example Question #1 : Quadratic Equation

All GRE Math Resources

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