GMAT Math : Equations

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Example Questions

Example Question #361 : Algebra

Solve for $\displaystyle a$ in the equation

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

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Example Question #52 : Equations

Solve for $\displaystyle t$ in the equation:

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

Possible Answers:

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

$\displaystyle t=- 2v + h + 30$ or $\displaystyle t=- 2v - h - 30$

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

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

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

Correct answer:

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

Explanation:

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

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

The equation can be rewritten as

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

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

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

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

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

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

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

Therefore,

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

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Example Question #53 : Equations

Solve for $\displaystyle y$ in the equation

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

where $\displaystyle a = 1,b = 6, c = Q$

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

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

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

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

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

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

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Example Question #54 : Equations

How many distinct solutions are there to the following equation?

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

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

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

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

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

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

$\small x+3=0$ $\displaystyle \small x+3=0$

$\small x=-3$ $\displaystyle \small x=-3$

Solution 2

$\small x-3=0$ $\displaystyle \small x-3=0$

$\small x=3$ $\displaystyle \small x=3$

Thus, we have two distinct solutions!

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Example Question #55 : Solving Equations

Solve for $\displaystyle g$ in the equation

$h = f+ \sqrt{3g+ 4}$ $\displaystyle h = f+ \sqrt{3g+ 4}$

Possible Answers:

$y =\frac{ \sqrt{h - f} - 4}{3}$ $\displaystyle y =\frac{ \sqrt{h - f} - 4}{3}$

$g= \frac{h^{2}-2hf+f^{2} - 12}{3}$ $\displaystyle g= \frac{h^{2}-2hf+f^{2} - 12}{3}$

$g= \frac{h^{2}-2hf+f^{2} - 4}{3}$ $\displaystyle g= \frac{h^{2}-2hf+f^{2} - 4}{3}$

$g= \frac{h^{2} +f^{2} - 4}{3}$ $\displaystyle g= \frac{h^{2} +f^{2} - 4}{3}$

$y =\frac{ \sqrt{h - f} -12}{3}$ $\displaystyle y =\frac{ \sqrt{h - f} -12}{3}$

Correct answer:

$g= \frac{h^{2}-2hf+f^{2} - 4}{3}$ $\displaystyle g= \frac{h^{2}-2hf+f^{2} - 4}{3}$

Explanation:

$h = f+ \sqrt{3g+ 4}$ $\displaystyle h = f+ \sqrt{3g+ 4}$

$h- f = f+ \sqrt{3g+ 4} - f$ $\displaystyle h- f = f+ \sqrt{3g+ 4} - f$

$\sqrt{3g+ 4} = h- f$ $\displaystyle \sqrt{3g+ 4} = h- f$

$\left (\sqrt{3g+ 4} \right )^{2}= \left (h- f \right )^{2}$ $\displaystyle \left (\sqrt{3g+ 4} \right )^{2}= \left (h- f \right )^{2}$

$3g+4= h^{2}-2hf+f^{2}$ $\displaystyle 3g+4= h^{2}-2hf+f^{2}$

$3g+4- 4= h^{2}-2hf+f^{2} - 4$ $\displaystyle 3g+4- 4= h^{2}-2hf+f^{2} - 4$

$3g = h^{2}-2hf+f^{2} - 4$ $\displaystyle 3g = h^{2}-2hf+f^{2} - 4$

$\frac{3g }{3}= \frac{h^{2}-2hf+f^{2} - 4}{3}$ $\displaystyle \frac{3g }{3}= \frac{h^{2}-2hf+f^{2} - 4}{3}$

$g= \frac{h^{2}-2hf+f^{2} - 4}{3}$ $\displaystyle g= \frac{h^{2}-2hf+f^{2} - 4}{3}$

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Example Question #56 : Solving Equations

$\displaystyle K$ is 44% of $\displaystyle P$.

2K+3P $\displaystyle 2K+3P$ is what percent of $\displaystyle P$?

Possible Answers:

$121 \frac{1}{3} \%$ $\displaystyle 121 \frac{1}{3} \%$

$388 \%$ $\displaystyle 388 \%$

$135 \%$ $\displaystyle 135 \%$

$332\%$ $\displaystyle 332\%$

$314\frac{2}{3} \%$ $\displaystyle 314\frac{2}{3} \%$

Correct answer:

$388 \%$ $\displaystyle 388 \%$

Explanation:

$\displaystyle K$ is 44% of $\displaystyle P$, so $\displaystyle 2K$ is $2 \times 44 \% = 88 \%$ $\displaystyle 2 \times 44 \% = 88 \%$ of $\displaystyle P$.

Also, $\displaystyle 3P$ is 300% of $\displaystyle P$.

Add these:

2K+3P $\displaystyle 2K+3P$ is $88 \% + 300 \% = 388 \%$ $\displaystyle 88 \% + 300 \% = 388 \%$ of $\displaystyle P$

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Example Question #361 : Algebra

Solve for $\displaystyle x$:

$\displaystyle -3x+18=7x-2$

Possible Answers:

x=-5 $\displaystyle x=-5$

x=2 $\displaystyle x=2$

x=-4 $\displaystyle x=-4$

x=-2 $\displaystyle x=-2$

x=5 $\displaystyle x=5$

Correct answer:

x=2 $\displaystyle x=2$

Explanation:

To solve the equation, we first group the $\displaystyle x$ terms on one side and the constants on the other side:

$\displaystyle -3x+18=7x-2$

$\displaystyle -10x=-20$

Now we can simply divide both sides by $\displaystyle -10$ to solve for $\displaystyle x$:

$\displaystyle -10x=-20$

x=2 $\displaystyle x=2$

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Example Question #371 : Algebra

Solve the following equation:

$\displaystyle 2x-5+3x=18-6x+10$

Possible Answers:

x=5 $\displaystyle x=5$

x=3 $\displaystyle x=3$

x=6 $\displaystyle x=6$

x=11 $\displaystyle x=11$

x=1 $\displaystyle x=1$

Correct answer:

x=3 $\displaystyle x=3$

Explanation:

To solve the equation, we must simplify it by adding together the like terms. We can group the $\displaystyle x$ terms on the left side of the equation and the constants on the right side of the equation:

$\displaystyle 2x-5+3x=18-6x+10$

$\displaystyle 2x+3x+6x=18+10+5$

11x=33 $\displaystyle 11x=33$

$x=\frac{33}{11}$ $\displaystyle x=\frac{33}{11}$

x=3 $\displaystyle x=3$

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Example Question #372 : Algebra

Solve the following equation for $\displaystyle x$:

3x+5=26 $\displaystyle 3x+5=26$

Possible Answers:

$x=\frac{31}{3}$ $\displaystyle x=\frac{31}{3}$

x=6 $\displaystyle x=6$

x=7 $\displaystyle x=7$

x=0 $\displaystyle x=0$

x=26 $\displaystyle x=26$

Correct answer:

x=7 $\displaystyle x=7$

Explanation:

To solve, we must isolate $\displaystyle x$. First, subtract $\displaystyle 5$ from both sides and then divide both sides by $\displaystyle 3$.

$\displaystyle 3x+5-5=26-5$

3x=21 $\displaystyle 3x=21$

$\frac{3x}{3}=\frac{21}{3}$ $\displaystyle \frac{3x}{3}=\frac{21}{3}$

x=7 $\displaystyle x=7$

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Example Question #51 : Solving Equations

Solve for $\displaystyle x$:

2x+5=23 $\displaystyle 2x+5=23$

Possible Answers:

x=14 $\displaystyle x=14$

x=9 $\displaystyle x=9$

x=2 $\displaystyle x=2$

x=16 $\displaystyle x=16$

Correct answer:

x=9 $\displaystyle x=9$

Explanation:

To solve, isolate $\displaystyle x$.

2x+5=23 $\displaystyle 2x+5=23$

2x=18 $\displaystyle 2x=18$

x=9 $\displaystyle x=9$

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GMAT Math : Equations

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #361 : Algebra

Example Question #52 : Equations

Example Question #53 : Equations

Example Question #54 : Equations

Example Question #55 : Solving Equations

Example Question #56 : Solving Equations

Example Question #361 : Algebra

Example Question #371 : Algebra

Example Question #372 : Algebra

Example Question #51 : Solving Equations

All GMAT Math Resources

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