GMAT Math : GMAT Quantitative Reasoning

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Example Questions

Example Question #1481 : Gmat Quantitative Reasoning

For what value(s) of does this quadratic equation have exactly one solution?

$x^{2} + Bx + 50 = 0$

Possible Answers:

$B = - 5 \sqrt{ 2 }\textrm{ or }B =5 \sqrt{ 2 }$

$B = - 10 \sqrt{ 2 }\textrm{ or }B = 10 \sqrt{ 2 }$

$B = - 5 \sqrt{ 5 }\textrm{ or }B =5 \sqrt{ 5 }$

$B = - 10 \textrm{ or }B = 10$

$B = - 5 \textrm{ or }B = 5$

Correct answer:

$B = - 10 \sqrt{ 2 }\textrm{ or }B = 10 \sqrt{ 2 }$

Explanation:

For the equation $A x^{2}+B x + C = 0$ to have exactly one solution, it must hold that discriminant $B^{2} -4AC = 0$ , so we solve for in this equation, setting A = 1,C= 50 :

$B^{2} -4AC = 0$

$B^{2} -4 \cdot 1 \cdot 50 = 0$

$B^{2} -200 = 0$

$B^{2} = 200$

$B =\sqrt{ 200} = \sqrt{ 100} \cdot \sqrt{ 2 } = 10 \sqrt{ 2 }$

$B =-\sqrt{ 200} = - 10 \sqrt{ 2 }$

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Example Question #1482 : Gmat Quantitative Reasoning

For what value of does this equation have exactly one solution?

$Ax^{2} - 4x + 8 = 0$

Possible Answers:

$A = \frac{1}{2}$

A = 1

A = 2

$A = \frac{1}{4}$

$A = \frac{1}{8}$

Correct answer:

$A = \frac{1}{2}$

Explanation:

For the equation $A x^{2}+B x + C = 0$ to have exactly one solution, it must hold that discriminant $B^{2} -4AC = 0$ .

Solve for , setting B = -4, C = 8 :

$(-4)^{2} -4 \cdot A \cdot 8 = 0$

16 -32 A = 0

16 -32 A -16 = 0-16

-32 A = -16

$-32 A\div (-32) = -16 \div (-32)$

$A = \frac{1}{2}$

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Example Question #1483 : Gmat Quantitative Reasoning

For what value(s) of does this equation have exactly one solution?

$2x^{2} - x+ C = 0$

Possible Answers:

$C = \frac{1}{8}$

$C = \frac{1}{16}$

$C = \frac{1}{2}$

C=1

$C = \frac{1}{4}$

Correct answer:

$C = \frac{1}{8}$

Explanation:

The discriminant of a quadratic expression $A x^{2}+B x + C$ is $B^{2} -4AC$ .

For the equation $A x^{2}+B x + C = 0$ to have exactly one solution, it must hold that discriminant $B^{2} -4AC = 0$ .

Solve for , setting A = 2, B = -1 :

$\left ( -1\right ) ^{2} -4 \cdot 2 \cdot C = 0$

1 -8 C = 0

1 -8 C -1 = 0 -1

-8 C = -1

$-8 C \div (-8) = -1 \div (-8)$

$C = \frac{1}{8}$

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Example Question #11 : Solving Quadratic Equations

Solve the following for x:

x^2 -x-6 =0

Possible Answers:

x= -6 or x=1

x= -2 or x=3

x= 2 or x=-3

x= 2 or x=3

x= 1 or x=-6

Correct answer:

x= -2 or x=3

Explanation:

There are 2 ways to solve this - factoring or using the quadratic formula.

First, for factoring we wish to find a way to factor this into an expression that looks like (cx+m)(dx+n)= 0. This way, we know the possible values for x will be -m/c and -n/d. (You can check by plugging in and getting the result algebraically.)

To do this, we need to look at the terms of the quadratic. Quadratics are expressed as ax^2+bx+c . In this case, a = 1, b = -1, and c = -6.

When solving for factors, the easiest starting point is to find the factors of a and c. Since a= 1, the only factor is 1, Therefore we know our c and d value will be 1. So our factored expression will look like (1x+m)(1x+n).

If we FOIL (x+m)(x+n) we see that we get x^2+mx+nx+mn = x^2+ x(m+n) +mn

So to find m and n, we look at our factors for c. Since c=-6, our possible factors are 1,2,3, and 6. Since we also know that m+n must equal b, we can try and find a fitting m and n so that the sum is -1. However, we know that c is negative; therefore we know that one of our factors will be positive and one will be negative. Therefore, we are looking to find a DIFFERENCE of 1 between our factors, which we can find between 2 and 3.

So we know that -2*3 = -6, and 2*-3 = -6. We also know that m+n =b, so -2+3 =1 (NOT WHAT WE WANT!) and 2+-3 = -1 (SCORE!). Since we have all the pieces to the puzzle, we see that this factors into (x+2)(x-3). Thus, solving our equation of (x+2)(x-3) = 0, we solve x+2=0 AND x-3=0. Thus our answer: x is -2 or 3.

ALTERNATE SOLUTIONS

If you are having trouble with the factoring, or if you find a quadratic that is not factorable, you can always use the quadratic equation. This says that when ax^2+bx+c = 0 , x = $\frac{-b\pm \sqrt{b^{2}- 4 a c}}{2a}$

Plugging in we get x= $\frac{1\pm \sqrt{1- 4(1)(-6)}}{2(1)} = \frac{1\pm \sqrt{25}}{2} = \frac{1\pm5}{2}$

So x = -2 or x=3.

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Example Question #12 : Solving Quadratic Equations

Solve the following expression for :

$3x^{2}+7x+7 = 5$

Possible Answers:

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

x = -5

x = {-7 ; 7}

x = 2

$x={-2 ; -\frac{1}{3}}$

Correct answer:

$x={-2 ; -\frac{1}{3}}$

Explanation:

Let us start by simplifying the equation:

$3x^{2}+7x+7 = 5 \Leftrightarrow 3x^{2}+7x+2 = 0 \Leftrightarrow (3x+1)(x+2)=0$

Now that we have the equation simplified into two products, we solve for each case:

3x + 1 = 0 or x + 2 = 0

x = -1/3 or x = -2

So the answer is the set {-2 ; -1/3}

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Example Question #402 : Algebra

How many real number solutions are there to the equation 4x^2 + 2x= 5 when solving for ?

Possible Answers:

Correct answer:

Explanation:

Since the problem didn't ask for what actually is, we don't have to solve for it. Instead we check the discriminant of the quadratic equation; $\sqrt{b^2-4ac}$

First we put our equation in standard form, 4x^2+2x-5 =0 and we then see that a=4, b=2, c=-5 . Substituting these into the discriminant gives us $\sqrt{2^2-4\times4\times-5} = \sqrt{84}$ . Since the number inside the square root is greater than , our original equation has real roots.

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Example Question #2 : How To Use The Quadratic Function

The length of a rectangular piece of land is two feet more than three times its width. If the area of the land is $125\:ft^2$ , what is the width of that piece of land?

Possible Answers:

$\frac{1-2\sqrt{94}}{3}\:ft^2$

$\frac{-1-2\sqrt{94}}{3}\:ft^2$

$\frac{2-4\sqrt{94}}{6}\:ft^2$

$\frac{-1+4\sqrt{94}}{3}\:ft^2$

$\frac{-1+2\sqrt{94}}{3}\:ft^2$

Correct answer:

$\frac{-1+2\sqrt{94}}{3}\:ft^2$

Explanation:

The area of a rectangle is the product of its length by its width, which we know to be equal to $125\:ft^2$ in our problem. We also know that the length is equal to 2+3w , where represents the width of the land. Therefore, we can write the following equation:

w(2+3w)=125

Distributing the outside the parentheses, we get:

$3w^{2}+2w=125$

Subtracting 125 from each side of the equation, we get:

$3w^{2}+2w-125=0$

We get a quadratic equation, and since there is no factor of and that adds up to , we use the quadratic formula to solve this equation.

a=3 b=2 c=-125

$w=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

We can first calculate the discriminant (i.e. the part under the square root)

$b^{2}-4ac= 2^{2}-4\times3 \times(-125)$

$4-4\times3\times(-125)$

$4-12\times(-125)$

4-(-1500)=1504

We replace that value in the quadratic formula, solving both the positive version of the formula (on the left) and the negative version of the formula (on the right):

$w=\frac{-2+\sqrt{1504}}{6}$ $w=\frac{-2+\sqrt{1504}}{6}$

Breaking down the square root:

$w=\frac{-2+\sqrt{2\times2\times2\times2\times2\times 47}}{6}$ $w=\frac{-2-\sqrt{2\times2\times2\times2\times2\times 47}}{6}$

We can pull two of the twos out of the square root and place a outside of it:

$w=\frac{-2+4\sqrt{2\times 47}}{6}$ $w=\frac{-2-4\sqrt{2\times 47}}{6}$

We can then multiply the and the :

$w=\frac{-2+4\sqrt{94}}{6}$ $w=\frac{-2-4\sqrt{94}}{6}$

At this point, we can reduce the equations, since each of the component parts of their right sides has a factor of :

$w=\frac{-1+2\sqrt{94}}{3}$ $w=\frac{-1+2\sqrt{94}}{3}$

Since width is a positive value, the answer is:

$w=\frac{-1+2\sqrt{94}}{3}$

$w=\frac{-1+2\times9.695}{3}$

$w=\frac{-1+19.39}{3}$

$w=\frac{18.39}{3}=6.13$

The width of the piece of land is approximately $6.13\:ft^2$ .

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Example Question #13 : Solving Quadratic Equations

How many real solutions and how many imaginary solutions are there to the following equation?

$x^{3} + 18x = 9x^{2}$

Possible Answers:

Three real solutions, no imaginary solutions

No real solutions, two imaginary solutions

One real solution, two imaginary solutions

No real solutions, three imaginary solutions

Two real solutions, no imaginary solutions

Correct answer:

Three real solutions, no imaginary solutions

Explanation:

Write the equation in standard form:

$x^{3} + 18x = 9x^{2}$

$x^{3} - 9x^{2} + 18x = 9x^{2} - 9x^{2}$

$x^{3} - 9x^{2} + 18x = 0$

Factor out the greatest common factor of :

$x \left ( x^{2} - 9x + 18 \right ) = 0$

Factor the trinomial by writing (x+?)(x+?) , replacing the question marks with two integers with product and sum . These integers are -6,-3 , so the above becomes

x( x-6) (x-3)= 0

Since the expression factors out to three distinct linear expressions, the expression has three real zeros,and the correct choice is three real solutions.

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Example Question #1488 : Gmat Quantitative Reasoning

Solve for :

$\frac{z^3 -15z^2 +20z -300}{z^2+20}$

Possible Answers:

2z -315

z^3 - 300

z-15

$\frac{1}{z^2 + 20}$

z^2 +20

Correct answer:

z-15

Explanation:

We start by splitting the numerator into two expressions, and factoring them:

$\frac{z^3 -15z^2 +20z -300}{z^2+20} = \frac{z^2(z -15) + 20(z-15)}{z^2 + 20}$

From there, we can pull (z-15) out of the numerator:

$\frac{z^2(z -15) + 20(z-15)}{z^2 + 20} = \frac{(z^2 + 20)(z-15)}{z^2 + 20}$

z^2 + 20/z^2 +20 = 1

so we are left with z-15.

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Example Question #1 : Understanding Factoring

Factor $4y^{2}+4y-15$ .

Possible Answers:

$\dpi{100} \small (2y-3)(2y+5)$

$\dpi{100} \small (2y+3)(2y-5)$

$\dpi{100} \small (2y-5)(2y+5)$

$\dpi{100} \small (2y-3)(2y+3)$

$\dpi{100} \small (y-3)(y+5)$

Correct answer:

$\dpi{100} \small (2y-3)(2y+5)$

Explanation:

To factor this, we need two numbers that multiply to $\dpi{100} \small 4\times -15=60$ and sum to $\dpi{100} \small 4$ . The numbers $\dpi{100} \small -6$ and $\dpi{100} \small 10$ work.

$4y^{2}+4y-15 = 4y^{2} - 6y + 10y - 15 = 2y(2y-3) + 5(2y-3) = (2y-3)(2y+5)$

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GMAT Math : GMAT Quantitative Reasoning

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1481 : Gmat Quantitative Reasoning

Example Question #1482 : Gmat Quantitative Reasoning

Example Question #1483 : Gmat Quantitative Reasoning

Example Question #11 : Solving Quadratic Equations

Example Question #12 : Solving Quadratic Equations

Example Question #402 : Algebra

Example Question #2 : How To Use The Quadratic Function

Example Question #13 : Solving Quadratic Equations

Example Question #1488 : Gmat Quantitative Reasoning

Example Question #1 : Understanding Factoring

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