GMAT Math : GMAT Quantitative Reasoning

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Example Questions

Example Question #353 : Algebra

Solve $\left | 2x+12\right |=20$ .

Possible Answers:

x=8 or x=-16

x=4 or x=-16

x=-16

x=4

x=8

Correct answer:

x=4 or x=-16

Explanation:

Since we are solving an absolute value equation, $\left | 2x+12\right |=20$ , we must solve for both potential values of the equation:

1.) 2x+12=20

2.) 2x+12=-20

Solving Equation 1:

2x+12=20

2x=8

x=4

Solving Equation 2:

2x+12=-20

2x=-32

x=-16

Therefore, for $\left | 2x+12\right |=20$ , x=4 or x=-16 .

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Example Question #42 : Equations

Solve for : 3y-4x=7x+33

Possible Answers:

$y=-\frac{3}{11}x+11$

$y=-\frac{11}{3}x+11$

$y=\frac{3}{11}x+11$

Not enough information provided

$y=\frac{11}{3}x+11$

Correct answer:

$y=\frac{11}{3}x+11$

Explanation:

In order to solve 3y-4x=7x+33 for , we need to isolate on one side of the equation:

3y-4x=7x+33

3y=11x+33

$y=\frac{11}{3}x+11$

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Example Question #354 : Algebra

Which of the following is a solution to the equation 7x+3y=22 ?

Possible Answers:

x=1, y=5

Two of the other answers are correct.

x=5, y=1

x=4, y=2

x=2, y=3

Correct answer:

x=1, y=5

Explanation:

In order to find values of and for which 7x+3y=22 , we need to plug the values into the equation:

1.) x=1, y=5 : 7(1)+3(5)=7+15=22 Correct

2.) x=5, y=1 : 7(5)+3(1)=35+3=38 Incorrect

3.) x=2, y=3 : 7(2)+3(3)=14+9=23 Incorrect

4.) x=4, y=2 : 7(4)+3(2)=28+6=34 Incorrect

Therefore, the only correct answer is x=1, y=5 .

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Example Question #361 : Algebra

Solve for : $\frac{1}{3}x+\frac{2}{3}y=\frac{7}{9}$

Possible Answers:

$y=\frac{7}{3}-2x$

$x=\frac{7}{3}-2y$

$x=\frac{2}{3}-7y$

$x=\frac{3}{7}-2y$

Not enough information provided

Correct answer:

$x=\frac{7}{3}-2y$

Explanation:

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

$\frac{1}{3}x+\frac{2}{3}y=\frac{7}{9}$

$\frac{1}{3}x=\frac{7}{9}-\frac{2}{3}y$

$x=\frac{21}{9}-2y$

Reducing the fraction,

$x=\frac{7}{3}-2y$

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Example Question #361 : Algebra

Solve for in the equation

3d + ad + y = 10

Possible Answers:

$d= \frac{10 - y }{3+a}$

$d= \frac{10 + y }{3+a}$

$d= \frac{10 - y+a }{3 }$

$d= \frac{10 - y-a }{3 }$

$d= \frac{y-10 }{3+a}$

Correct answer:

$d= \frac{10 - y }{3+a}$

Explanation:

3d + ad + y = 10

3d + ad + y - y= 10 - y

3d + ad = 10 - y

$\left (3 + a \right )d = 10 - y$

$\frac{\left (3 + a \right )d }{3+a}= \frac{10 - y }{3+a}$

$d= \frac{10 - y }{3+a}$

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Example Question #363 : Algebra

Solve for in the equation:

$v = 1+ \sqrt[3]{w-7}$

Possible Answers:

$w = v^{3} -3v^{2}+3v+6$

$w = v^{3} +6$

$w = v^{3} +3v^{2}+3v-6$

$w = v^{3} +3v^{2}+3v+8$

$w = v^{3} -3v^{2}+3v-8$

Correct answer:

$w = v^{3} -3v^{2}+3v+6$

Explanation:

$v = 1+ \sqrt[3]{w-7}$

$v- 1 = 1+ \sqrt[3]{w-7} - 1$

$\sqrt[3]{w-7} = v- 1$

$\left (\sqrt[3]{w-7}\right )^{3} = \left (v- 1 \right )^{3}$

$w-7= v^{3} -3v^{2}+3v-1$

$w-7+ 7= v^{3} -3v^{2}+3v-1 + 7$

$w = v^{3} -3v^{2}+3v+6$

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Example Question #51 : Equations

Solve for in the equation

$a^{3}- y + 1= c^{2}$

Possible Answers:

$a=\sqrt[3]{ c^{2}} + y - 1$

$a=\sqrt[3]{ c^{2} + y - 1}$ or $a= -\sqrt[3]{ c^{2} + y - 1}$

$a=\sqrt[3]{ c^{2}} + y - 1$ or $a=-\sqrt[3]{ c^{2}} + y - 1$

$a=\sqrt[3]{ c^{2}} + y - 1$ or $a=-\sqrt[3]{ c^{2}} - y + 1$

$a=\sqrt[3]{ c^{2} + y - 1}$

Correct answer:

$a=\sqrt[3]{ c^{2} + y - 1}$

Explanation:

$a^{3}- y + 1= c^{2}$

$a^{3}- y + 1+ y - 1= c^{2} + y - 1$

$a^{3} = c^{2} + y - 1$

$\sqrt[3]{ a^{3} } =\sqrt[3]{ c^{2} + y - 1}$

$a=\sqrt[3]{ c^{2} + y - 1}$

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Example Question #52 : Equations

Solve for in the equation:

$v^{2}+ 4vt + 4t^{2} = h+30$

Possible Answers:

$t=\frac{- v + h + 60 }{2}$ or $t=\frac{- v - h - 60 }{2}$

t=- 2v + h + 30 or t=- 2v - h - 30

$t=- 2v + \sqrt{ h + 30 }$ or $t=- 2v - \sqrt{ h + 30 }$

$t=\frac{- v + h + 30 }{2}$ or $t=\frac{- v - h - 30 }{2}$

$t=\frac{- v + \sqrt{ h + 30 }}{2}$ or $t=\frac{- v - \sqrt{ h + 30 } }{2}$ .

Correct answer:

$t=\frac{- v + \sqrt{ h + 30 }}{2}$ or $t=\frac{- v - \sqrt{ h + 30 } }{2}$ .

Explanation:

$v^{2}+ 4vt + 4t^{2}$ is a perfect square trinomial:

$v^{2}+ 4vt + 4t^{2} = v^{2}+ 2 \cdot 2 t + (2t)^{2} = (v+2t)^{2}$

The equation can be rewritten as

$(v+2t)^{2} = h + 30$

By the square-root property, since no assumption was made about the sign of any variable,

$v+2t = \pm\sqrt{ h + 30 }$

$v+2t -v = \pm \sqrt{ h + 30 }-v$

$2t= -v\pm \sqrt{ h + 30 }$

$\frac{2t}{2}=\frac{- v\pm \sqrt{ h + 30 } }{2}$

$t=\frac{- v\pm \sqrt{ h + 30 }}{2}$

Therefore,

$t=\frac{- v + \sqrt{ h + 30 }}{2}$ or $t=\frac{- v - \sqrt{ h + 30 } }{2}$ .

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Example Question #53 : Equations

Solve for in the equation

$y^{2} + 6y + Q = 0$

Possible Answers:

$y = -3 - \sqrt{9+Q}$ or $y = -3 \p+\sqrt{9+Q}$

$y = -3 - \sqrt{9-Q}$ or $y = -3 \p+\sqrt{9-Q}$

$y = 3 + \sqrt{9+Q}$ or $y = -3 +\sqrt{9+Q}$

$y = 3 - \sqrt{9-Q}$ or $y = 3 \p+\sqrt{9-Q}$

$y = 3 - \sqrt{9-Q}$ or $y = -3 -\sqrt{9-Q}$

Correct answer:

$y = -3 - \sqrt{9-Q}$ or $y = -3 \p+\sqrt{9-Q}$

Explanation:

The statement is a quadratic equation in , so it can be solved using the quadratic formula,

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

where a = 1,b = 6, c = Q

$y = \frac{-6 \pm \sqrt{6^{2}-4(1)(Q)}}{2 (1)}$

$y = \frac{-6 \pm \sqrt{36-4Q}}{2}$

$y = \frac{-6 \pm \sqrt{4(9-Q)}}{2}$

$y = \frac{-6 \pm 2 \sqrt{9-Q}}{2}$

$y = \frac{2\left (-3 \pm \sqrt{9-Q} \right )}{2}$

$y = -3 \pm \sqrt{9-Q}$

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Example Question #54 : Equations

How many distinct solutions are there to the following equation?

$\small 3x^2-27=0$

Possible Answers:

Infinitely Many

Correct answer:

Explanation:

We are given a classic quadratic equation, but we aren't asked for the solutions, just how many distinct solutions there are. Remember, distinct solutions are different solutions. If we get two solutions that are the same numbers, they do not count.

The quickest way to solve this involves some factoring.

Start by pulling out a 3

$\small 3(x^2-9)=0$

Now, within our parentheses, we have a classic difference of squares. The interior factors further to look like this.

$\small \small 3{(x+3)(x-3)}=0$

From here we can either solve the equation and count our solutions, or we can recognize that the two factors are different and therefore will give different solutions. Let's solve it by using the Zero Product Property

Solution 1

$\small x+3=0$

$\small x=-3$

Solution 2

$\small x-3=0$

$\small x=3$

Thus, we have two distinct solutions!

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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 #353 : Algebra

Example Question #42 : Equations

Example Question #354 : Algebra

Example Question #361 : Algebra

Example Question #361 : Algebra

Example Question #363 : Algebra

Example Question #51 : Equations

Example Question #52 : Equations

Example Question #53 : Equations

Example Question #54 : Equations

All GMAT Math Resources

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