GRE Math : GRE Quantitative Reasoning

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Example Questions

Example Question #41 : Evaluating Expressions

$\displaystyle x^2-7x=-6$

Quantity A: $\displaystyle 3x$

Quantity B: $\displaystyle 2$

Possible Answers:

The relationship cannot be determined.

Quantity B is greater.

Quantity A is greater.

The two quantities are equal.

Correct answer:

Quantity A is greater.

Explanation:

Since there is an x^2 $\displaystyle x^2$ term in the equation

$\displaystyle x^2-7x=-6$

We must acknowledge the possibility of multiple values that satisfy the equation.

Rewrite it so that it is set equal to zero:

$\displaystyle x^2-7x+6$

For an equation of the form

$\displaystyle ax^2+bx+c=0$

The solutions for $\displaystyle x$ can be found using the quadratic formula:

$x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $\displaystyle x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x_{1,2}=\frac{-(-7) \pm \sqrt{(-7)^2-4(1)(6)}}{2(1)}=\frac{7 \pm \sqrt{25}}{2(1)}=1,6$ $\displaystyle x_{1,2}=\frac{-(-7) \pm \sqrt{(-7)^2-4(1)(6)}}{2(1)}=\frac{7 \pm \sqrt{25}}{2(1)}=1,6$

Quantity A:

Since this is $\displaystyle 3x$, it can have the values of $\displaystyle 3$ or $\displaystyle 18$. Both are bigger than $\displaystyle 2.$

Quantity A is greater.

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Example Question #361 : Gre Quantitative Reasoning

$\displaystyle x^2+3x=10$

Quantity A: x^2 $\displaystyle x^2$

Quantity B: $\displaystyle 5$

Possible Answers:

The relationship cannot be determined.

Quantity B is greater.

Quantity A is greater.

The two quantities are equal.

Correct answer:

The relationship cannot be determined.

Explanation:

Since there is an x^2 $\displaystyle x^2$ term in the equation

$\displaystyle x^2+3x=10$

We must acknowledge the possibility of multiple values that satisfy the equation.

Rewrite it so that it is set equal to zero:

$\displaystyle x^2+3x-10=0$

For an equation of the form

$\displaystyle ax^2+bx+c=0$

The solutions for $\displaystyle x$ can be found using the quadratic formula:

$x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $\displaystyle x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x_{1,2}=\frac{-3 \pm \sqrt{3^2-4(1)(-10)}}{2(1)}=\frac{-3 \pm \sqrt49}{2}=2,-5$ $\displaystyle x_{1,2}=\frac{-3 \pm \sqrt{3^2-4(1)(-10)}}{2(1)}=\frac{-3 \pm \sqrt49}{2}=2,-5$

Quantity A:

The two possible values are $2^2,(-5)^2\rightarrow 4,25$ $\displaystyle 2^2,(-5)^2\rightarrow 4,25$. Only one of them is greater than five.

The relationship cannot be determined.

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Example Question #41 : How To Evaluate Algebraic Expressions

$\displaystyle x^2+5x+6=0$

Quantity A: $\displaystyle x$

Quantity B: 1.5 $\displaystyle 1.5$

Possible Answers:

The relationship cannot be determined.

Quantity B is greater.

Quantity A is greater.

The two quantities are equal.

Correct answer:

Quantity B is greater.

Explanation:

Since there is an x^2 $\displaystyle x^2$ term in the equation

$\displaystyle x^2+5x+6=0$

We must acknowledge the possibility of multiple values that satisfy the equation.

For an equation of the form

$\displaystyle ax^2+bx+c=0$

The solutions for $\displaystyle x$ can be found using the quadratic formula:

$x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $\displaystyle x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x_{1,2}=\frac{-5 \pm \sqrt{5^2-4(1)(6)}}{2(1)}=\frac{-5\pm \sqrt1}{2}=-2,-3$ $\displaystyle x_{1,2}=\frac{-5 \pm \sqrt{5^2-4(1)(6)}}{2(1)}=\frac{-5\pm \sqrt1}{2}=-2,-3$

Both roots are negative.

Quantity B is greater.

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Example Question #361 : Algebra

At a store, pasta is sold in three sizes. A large box costs the same as four medium boxes or eight small boxes. If James buys an equal amount of large and medium boxes of pasta for the price needed to buy one hundred small boxes, how many medium boxes of pasta does he buy?

Possible Answers:

$\displaystyle 10$

$\displaystyle 18$

$\displaystyle 20$

$\displaystyle 12$

$\displaystyle 14$

Correct answer:

$\displaystyle 10$

Explanation:

To approach this problem, assign variables. Since one large box equals four medium boxes or eight small boxes in price

L=4M $\displaystyle L=4M$

L=8S $\displaystyle L=8S$

From this we can say that

4M=8S $\displaystyle 4M=8S$

or that

M=2S $\displaystyle M=2S$

We're told that James buys an equal number of large and medium boxes, and that the total price is equal to that of 100 small boxes:

$\displaystyle xL+xM=100S$

$\displaystyle x(L+M)=100S$

Rewrite this equation in terms of just the price of small boxes:

$\displaystyle x(8S+2S)=100S$

$\displaystyle x(10S)=100S$

x=10 $\displaystyle x=10$

James buys ten medium and ten large boxes of large pasta.

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Example Question #42 : How To Evaluate Algebraic Expressions

When the integer $\displaystyle a$ is multiplied by $\displaystyle 8$, the result is $\displaystyle 5$ more than $\displaystyle 24$ times the integer $\displaystyle b$. What is a-3b $\displaystyle a-3b$ ?

Possible Answers:

$\frac{5}{8}$ $\displaystyle \frac{5}{8}$

$-\frac{5}{8}$ $\displaystyle -\frac{5}{8}$

$\frac{8}{5}$ $\displaystyle \frac{8}{5}$

$\displaystyle 40$

$-\frac{8}{5}$ $\displaystyle -\frac{8}{5}$

Correct answer:

$\frac{5}{8}$ $\displaystyle \frac{5}{8}$

Explanation:

The key is to write the problem into mathematical terms.

We're told

"When the integer $\displaystyle a$ is multiplied by $\displaystyle 8$, the result is $\displaystyle 5$ more than $\displaystyle 24$ times the integer $\displaystyle b$."

Breaking this down piece by piece:

When the integer $\displaystyle a$ is multiplied by $\displaystyle 8$:

$\displaystyle 8a$

When the integer $\displaystyle a$ is multiplied by $\displaystyle 8$, the result is :

8a= $\displaystyle 8a=$

When the integer $\displaystyle a$ is multiplied by $\displaystyle 8$, the result is $\displaystyle 5$ more than:

8a=5+ $\displaystyle 8a=5+$

When the integer $\displaystyle a$ is multiplied by $\displaystyle 8$, the result is $\displaystyle 5$ more than $\displaystyle 24$ times the integer $\displaystyle b$:

$\displaystyle 8a=5+24b$

Now, solve this for $\displaystyle a:$

$a=3b+\frac{5}{8}$ $\displaystyle a=3b+\frac{5}{8}$

$a-3b=(3b+\frac{5}{8})-3b=\frac{5}{8}$ $\displaystyle a-3b=(3b+\frac{5}{8})-3b=\frac{5}{8}$

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Example Question #41 : Evaluating Expressions

$\displaystyle x^2-7x=-12$

Quantity A: $\displaystyle x$

Quantity B: $\displaystyle 3$

Possible Answers:

Quantity B is greater.

Quantity A is greater.

The relationship cannot be determined.

The two quantities are equal.

Correct answer:

The relationship cannot be determined.

Explanation:

Since there is an x^2 $\displaystyle x^2$ term in the equation

$\displaystyle x^2-7x=-12$

We must acknowledge the possibility of multiple values that satisfy the equation.

Rewrite it so that it is set equal to zero:

$\displaystyle x^2-7x+12=0$

For an equation of the form

$\displaystyle ax^2+bx+c=0$

The solutions for $\displaystyle x$ can be found using the quadratic formula:

$x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $\displaystyle x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x_{1,2}=\frac{-(-7) \pm \sqrt{(-7)^2-4(1)(12)}}{2(1)}=\frac{7\pm \sqrt{1}}{2}=3,4$ $\displaystyle x_{1,2}=\frac{-(-7) \pm \sqrt{(-7)^2-4(1)(12)}}{2(1)}=\frac{7\pm \sqrt{1}}{2}=3,4$

Quantity A can be either three or four. Since one of these is equal to Quantity B while the other is greater, the relationship cannot be determined.

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Example Question #81 : Expressions

When the integer $\displaystyle a$ is multiplied by $\displaystyle 2$, the result is thrice the difference of $\displaystyle 4$ and $\displaystyle 6$ times the integer $\displaystyle b$. What is the value of $\frac{1}{3}a+3b$ $\displaystyle \frac{1}{3}a+3b$ ?

Possible Answers:

$\displaystyle 4$

$\displaystyle 6$

$\displaystyle 2$

$\displaystyle -2$

$\displaystyle -6$

Correct answer:

$\displaystyle 2$

Explanation:

To approach this problem, write out the problem statement in mathematical terms.

We're told

"When the integer $\displaystyle a$ is multiplied by $\displaystyle 2$, the result is thrice the difference of $\displaystyle 4$ and $\displaystyle 6$ times the integer $\displaystyle b$."

Write this out step by step.

When the integer $\displaystyle a$ is multiplied by $\displaystyle 2$:

$\displaystyle 2a$

When the integer $\displaystyle a$ is multiplied by $\displaystyle 2$, the result is:

2a= $\displaystyle 2a=$

When the integer $\displaystyle a$ is multiplied by $\displaystyle 2$, the result is thrice:

2a=3() $\displaystyle 2a=3()$

When the integer $\displaystyle a$ is multiplied by $\displaystyle 2$, the result is thrice the difference:

2a=3(-) $\displaystyle 2a=3(-)$

When the integer $\displaystyle a$ is multiplied by $\displaystyle 2$, the result is thrice the difference of $\displaystyle 4$:

$\displaystyle 2a=3(4-)$

When the integer $\displaystyle a$ is multiplied by $\displaystyle 2$, the result is thrice the difference of $\displaystyle 4$ and $\displaystyle 6$ times the integer $\displaystyle b$:

$\displaystyle 2a=3(4-6b)$

Now, solve for $\displaystyle a:$

a=6-9b $\displaystyle a=6-9b$

$\frac{1}{3}a+3b=\frac{1}{3}(6-9b)+3b=2-3b+3b=2$ $\displaystyle \frac{1}{3}a+3b=\frac{1}{3}(6-9b)+3b=2-3b+3b=2$

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Example Question #47 : How To Evaluate Algebraic Expressions

$\displaystyle x^2-8x=-12$

Quantity A: $\displaystyle x$

Quantity B: $\displaystyle 1$

Possible Answers:

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined.

Correct answer:

Quantity A is greater.

Explanation:

Since there is an x^2 $\displaystyle x^2$ term in the equation

$\displaystyle x^2-8x=-12$

We must acknowledge the possibility of multiple values that satisfy the equation.

Rewrite it so that it is set equal to zero:

$\displaystyle x^2-8x+12=0$

For an equation of the form

$\displaystyle ax^2+bx+c=0$

The solutions for $\displaystyle x$ can be found using the quadratic formula:

$x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $\displaystyle x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x_{1,2}=\frac{-(-8) \pm \sqrt{(-8)^2-4(1)(12)}}{2(1)}=\frac{8\pm \sqrt{16}}{2}=2,6$ $\displaystyle x_{1,2}=\frac{-(-8) \pm \sqrt{(-8)^2-4(1)(12)}}{2(1)}=\frac{8\pm \sqrt{16}}{2}=2,6$

Both possible values for Quantity A are greater than $\displaystyle 1.$

Quantity A is greater.

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Example Question #361 : Algebra

$x \neq 0, -1$ $\displaystyle x \neq 0, -1$

Take the expression x@y $\displaystyle x@y$ to be equivalent to $\frac{x}{y}+x$ $\displaystyle \frac{x}{y}+x$. What is the expression $\displaystyle x^3@x^2@x$ equivalent to?

Possible Answers:

$\frac{x^2+x +1}{x^3+1}$ $\displaystyle \frac{x^2+x +1}{x^3+1}$

$\frac{x^4+x^3+x^2}{x^3+1}$ $\displaystyle \frac{x^4+x^3+x^2}{x^3+1}$

$\frac{x^3+x^2+x}{x+1}$ $\displaystyle \frac{x^3+x^2+x}{x+1}$

$\frac{x^2+x+1}{x+1}$ $\displaystyle \frac{x^2+x+1}{x+1}$

$\frac{x^4+x^3+x^2}{x+1}$ $\displaystyle \frac{x^4+x^3+x^2}{x+1}$

Correct answer:

$\frac{x^4+x^3+x^2}{x+1}$ $\displaystyle \frac{x^4+x^3+x^2}{x+1}$

Explanation:

When considering $\displaystyle x^3@x^2@x$ , begin at the right of the equation. Since

x@y $\displaystyle x@y$ is equivalent to $\frac{x}{y}+x$ $\displaystyle \frac{x}{y}+x$, x^2@x $\displaystyle x^2@x$ must be equivalent to

$\frac{x^2}{x}+x^2=x+x^2$ $\displaystyle \frac{x^2}{x}+x^2=x+x^2$

Now we need only consider $\displaystyle x^3@(x+x^2)$

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

$\frac{x^4+x^3+x^2}{x+1}$ $\displaystyle \frac{x^4+x^3+x^2}{x+1}$

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Example Question #41 : Evaluating Expressions

$\displaystyle x^2-11x=-10$

Quantity A: |x-5| $\displaystyle |x-5|$

Quantity B: $\displaystyle 3$

Possible Answers:

Quantity B is greater.

The relationship cannot be determined.

The two quantities are equal.

Quantity A is greater.

Correct answer:

Quantity A is greater.

Explanation:

Since there is an x^2 $\displaystyle x^2$ term in the equation

$\displaystyle x^2-11x=-10$

We must acknowledge the possibility of multiple values that satisfy the equation.

Rewrite it so that it is set equal to zero:

$\displaystyle x^2-11x+10=0$

For an equation of the form

$\displaystyle ax^2+bx+c=0$

The solutions for $\displaystyle x$ can be found using the quadratic formula:

$x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $\displaystyle x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x_{1,2}=\frac{-(-11) \pm \sqrt{(-11)^2-4(1)(10)}}{2(1)}=\frac{11 \pm \sqrt{81}}{2}=1,10$ $\displaystyle x_{1,2}=\frac{-(-11) \pm \sqrt{(-11)^2-4(1)(10)}}{2(1)}=\frac{11 \pm \sqrt{81}}{2}=1,10$

Since there are two roots for x, Quantity A has two possible values:

$\displaystyle |1-5|=|-4|=4;|10-5|=|5|=5$

Both of these are greater than $\displaystyle 3$.

Quantity A is greater.

Report an Error

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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 #41 : Evaluating Expressions

Example Question #361 : Gre Quantitative Reasoning

Example Question #41 : How To Evaluate Algebraic Expressions

Example Question #361 : Algebra

Example Question #42 : How To Evaluate Algebraic Expressions

Example Question #41 : Evaluating Expressions

Example Question #81 : Expressions

Example Question #47 : How To Evaluate Algebraic Expressions

Example Question #361 : Algebra

Example Question #41 : Evaluating Expressions

All GRE Math Resources

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