GMAT Math : Solving equations

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Example Questions

Example Question #31 : Equations

Solve for $\displaystyle x$. Give all real solutions:

$x - \sqrt{x} - 20 = 0$ $\displaystyle x - \sqrt{x} - 20 = 0$

Possible Answers:

x = -16 $\displaystyle x = -16$

$x = -16 \textrm{ or }x = 25$ $\displaystyle x = -16 \textrm{ or }x = 25$

x = 25 $\displaystyle x = 25$

The equation has no real solution.

$x = -25 \textrm{ or }x = 25$ $\displaystyle x = -25 \textrm{ or }x = 25$

Correct answer:

x = 25 $\displaystyle x = 25$

Explanation:

One way is to substitute $t = \sqrt{ x}$ $\displaystyle t = \sqrt{ x}$, and, subsequently, $t ^{2} = x$ $\displaystyle t ^{2} = x$

$x - \sqrt{x} - 20 = 0$ $\displaystyle x - \sqrt{x} - 20 = 0$

$t^{2}- t - 20 = 0$ $\displaystyle t^{2}- t - 20 = 0$

$\displaystyle (t-5)(t+4)= 0$

Set each binomial to 0 and solve separately:

t-5 = 0 $\displaystyle t-5 = 0$

t = 5 $\displaystyle t = 5$

$\sqrt{x}= 5$ $\displaystyle \sqrt{x}= 5$

x = 25 $\displaystyle x = 25$

t+4= 0 $\displaystyle t+4= 0$

t=-4 $\displaystyle t=-4$

$\sqrt{x}=-4$ $\displaystyle \sqrt{x}=-4$

Since no real number has $\displaystyle -4$ as its principal square root, this yields no solution.

The only solution is x = 25 $\displaystyle x = 25$.

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Example Question #32 : Equations

Solve for $\displaystyle x$, giving all solutions, real and imaginary:

$x^{3} - 4x^{2}+ 3x - 12 = 0$ $\displaystyle x^{3} - 4x^{2}+ 3x - 12 = 0$

Possible Answers:

$x \in \left \{-4,3,4\right \}$ $\displaystyle x \in \left \{-4,3,4\right \}$

$x \in \left \{ -\sqrt{3 }, \sqrt{3 },4\right \}$ $\displaystyle x \in \left \{ -\sqrt{3 }, \sqrt{3 },4\right \}$

$x \in \left \{ 4, i \sqrt{3 }, -i \sqrt{3 }\right \}$ $\displaystyle x \in \left \{ 4, i \sqrt{3 }, -i \sqrt{3 }\right \}$

$x \in \left \{ -3,3,4\right \}$ $\displaystyle x \in \left \{ -3,3,4\right \}$

$x \in \left \{ -3i,3i,4\right \}$ $\displaystyle x \in \left \{ -3i,3i,4\right \}$

Correct answer:

$x \in \left \{ 4, i \sqrt{3 }, -i \sqrt{3 }\right \}$ $\displaystyle x \in \left \{ 4, i \sqrt{3 }, -i \sqrt{3 }\right \}$

Explanation:

Factor the expression:

$x^{3} - 4x^{2}+ 3x - 12 = 0$ $\displaystyle x^{3} - 4x^{2}+ 3x - 12 = 0$

$\left ( x^{3} - 4x^{2} \right ) +\left ( 3x - 12 \right )= 0$ $\displaystyle \left ( x^{3} - 4x^{2} \right ) +\left ( 3x - 12 \right )= 0$

$x^{2} \left ( x - 4 \right ) + 3 \left ( x - 4 \right )= 0$ $\displaystyle x^{2} \left ( x - 4 \right ) + 3 \left ( x - 4 \right )= 0$

$\left (x^{2}+ 3 \right ) \left ( x - 4 \right ) = 0$ $\displaystyle \left (x^{2}+ 3 \right ) \left ( x - 4 \right ) = 0$

Rewrite:

$\displaystyle x- 4 = 0$

x=4 $\displaystyle x=4$

$x^{2} +3 =0$ $\displaystyle x^{2} +3 =0$

$x^{2} = -3$ $\displaystyle x^{2} = -3$

$x = \pm \sqrt{ -3} = \pm i \sqrt{3}$ $\displaystyle x = \pm \sqrt{ -3} = \pm i \sqrt{3}$

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Example Question #33 : Equations

Solve for $\displaystyle x$:

$6 ^{2x+5} =\left ( \frac{1}{36} \right )^{x-3}$ $\displaystyle 6 ^{2x+5} =\left ( \frac{1}{36} \right )^{x-3}$

Possible Answers:

$x = \frac{4}{11}$ $\displaystyle x = \frac{4}{11}$

$x =2 \frac{3}{4}$ $\displaystyle x =2 \frac{3}{4}$

$x = \frac{1}{4}$ $\displaystyle x = \frac{1}{4}$

x = 4 $\displaystyle x = 4$

The equation has no solution.

Correct answer:

$x = \frac{1}{4}$ $\displaystyle x = \frac{1}{4}$

Explanation:

$\frac{1}{36} = 6 ^{-2}$ $\displaystyle \frac{1}{36} = 6 ^{-2}$, so the equation can be rewritten as follows:

$6 ^{2x+5} =\left ( \frac{1}{36} \right )^{x-3}$ $\displaystyle 6 ^{2x+5} =\left ( \frac{1}{36} \right )^{x-3}$

$6 ^{2x+5} =\left ( 6 ^{-2} \right )^{x-3}$ $\displaystyle 6 ^{2x+5} =\left ( 6 ^{-2} \right )^{x-3}$

$6 ^{2x+5} = 6 ^{-2\left ( x-3 \right )}$ $\displaystyle 6 ^{2x+5} = 6 ^{-2\left ( x-3 \right )}$

Set the exponents equal to each other:

$\displaystyle 2x+5 = -2 ( x-3 )$

$\displaystyle 2x+5 = -2 x+6$

$\displaystyle 2x+5 +2x-5 = -2 x+6+2x-5$

4x = 1 $\displaystyle 4x = 1$

$x = \frac{1}{4}$ $\displaystyle x = \frac{1}{4}$

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Example Question #34 : Equations

Daniel has $\displaystyle 7$ candy bars. Andy has three more candy bars than the double of Daniel's candy bars.

How many candy bars does Andy have?

Possible Answers:

$\displaystyle 12$

$\displaystyle 17$

$\displaystyle 23$

$\displaystyle 15$

$\displaystyle 13$

Correct answer:

$\displaystyle 17$

Explanation:

Let A be the number of Andy's candy bars and D be the number of Daniel's candy bars.

We start by setting up the equation:

$A = 3 + 2 \cdot D$ $\displaystyle A = 3 + 2 \cdot D$

and

D = 7 $\displaystyle D = 7$

$A = 3 + 2 \cdot 7 = 3 + 14 = 17$ $\displaystyle A = 3 + 2 \cdot 7 = 3 + 14 = 17$

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Example Question #35 : Equations

$\frac{2}{(5^{-1})x+0.1}=5$ $\displaystyle \frac{2}{(5^{-1})x+0.1}=5$

What is the value of $\displaystyle x$?

Possible Answers:

1.7 $\displaystyle 1.7$

1.6 $\displaystyle 1.6$

1.9 $\displaystyle 1.9$

1.8 $\displaystyle 1.8$

1.5 $\displaystyle 1.5$

Correct answer:

1.5 $\displaystyle 1.5$

Explanation:

To find the value of x we need to isolate x on one side of the equation and the rest of the numbers on the other side.

$\frac{2}{(5^{-1})x+0.1}=5$ $\displaystyle \frac{2}{(5^{-1})x+0.1}=5$

First, we multiply what is in the denominator on the left had side by the numerators on both sides.

$2=5\times((5^{-1})x +0.1)$ $\displaystyle 2=5\times((5^{-1})x +0.1)$

Then we distribute the 5 to both terms in the binomial. Doing this we get a zero in the exponent.

$2=(5^{0})x+0.5$ $\displaystyle 2=(5^{0})x+0.5$

Anything raised to the zero just becomes one.

2=x+0.5 $\displaystyle 2=x+0.5$

From here we subtract 0.5 from each side to solve for x.

x=2-0.5 $\displaystyle x=2-0.5$

x=1.5 $\displaystyle x=1.5$

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Example Question #36 : Equations

Define $f (x) = x \cdot |x |$ $\displaystyle f (x) = x \cdot |x |$ . Which of the following would be a valid alternative way of expressing the definition of $\displaystyle f$?

Possible Answers:

$f (x) = \left\{\begin{matrix} -x^{2} & \textrm{ if } x < 0\\ x^{2} & \textrm{ if } x \geq 0 \end{matrix}\right.$ $\displaystyle f (x) = \left\{\begin{matrix} -x^{2} & \textrm{ if } x < 0\\ x^{2} & \textrm{ if } x \geq 0 \end{matrix}\right.$

$f (x) = \left\{\begin{matrix}0 & \textrm{ if } x < 0\\ x^{2} & \textrm{ if } x \geq 0 \end{matrix}\right.$ $\displaystyle f (x) = \left\{\begin{matrix}0 & \textrm{ if } x < 0\\ x^{2} & \textrm{ if } x \geq 0 \end{matrix}\right.$

$f (x) = \left\{\begin{matrix} x^{2} & \textrm{ if } x < 0\\ -x^{2} & \textrm{ if } x \geq 0 \end{matrix}\right.$ $\displaystyle f (x) = \left\{\begin{matrix} x^{2} & \textrm{ if } x < 0\\ -x^{2} & \textrm{ if } x \geq 0 \end{matrix}\right.$

$f (x) = -x^{2}$ $\displaystyle f (x) = -x^{2}$

$f (x) = \left\{\begin{matrix} 0 & \textrm{ if } x < 0\\ -x^{2} & \textrm{ if } x \geq 0 \end{matrix}\right.$ $\displaystyle f (x) = \left\{\begin{matrix} 0 & \textrm{ if } x < 0\\ -x^{2} & \textrm{ if } x \geq 0 \end{matrix}\right.$

Correct answer:

Explanation:

By definition:

If $x \geq 0$ $\displaystyle x \geq 0$, then |x| = x $\displaystyle |x| = x$ ,and subsequently, $f (x) = x \cdot |x | = x \cdot x= x^{2}$ $\displaystyle f (x) = x \cdot |x | = x \cdot x= x^{2}$

If x < 0 $\displaystyle x < 0$, then $\displaystyle |x| = -x$ ,and subsequently, $f (x) = x \cdot |x | = x \cdot \left (- x \right )= - x^{2}$ $\displaystyle f (x) = x \cdot |x | = x \cdot \left (- x \right )= - x^{2}$

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Example Question #37 : Equations

$125^{N} = x$ $\displaystyle 125^{N} = x$

Which of the following expressions is equal to $\log_{25}x$ $\displaystyle \log_{25}x$ ?

Possible Answers:

$N \ln 5$ $\displaystyle N \ln 5$

$\sqrt[3]{N^2}$ $\displaystyle \sqrt[3]{N^2}$

$N \sqrt{N }$ $\displaystyle N \sqrt{N }$

$\frac{2}{3}N$ $\displaystyle \frac{2}{3}N$

$\frac{3}{2}N$ $\displaystyle \frac{3}{2}N$

Correct answer:

$\frac{3}{2}N$ $\displaystyle \frac{3}{2}N$

Explanation:

$5 ^{2} = 25$ $\displaystyle 5 ^{2} = 25$, so $5 = \sqrt{25}$ $\displaystyle 5 = \sqrt{25}$, and $125 = 5 ^{3}= \left (\sqrt{25} \right )^{3} = 25^\frac{3}{2}$ $\displaystyle 125 = 5 ^{3}= \left (\sqrt{25} \right )^{3} = 25^\frac{3}{2}$.

$x= 125^{N} = \left ( 25^\frac{3}{2} \right )^{N} =25^{\frac{3}{2}N} \right$ $\displaystyle x= 125^{N} = \left ( 25^\frac{3}{2} \right )^{N} =25^{\frac{3}{2}N} \right$

$\log_{25} x = \frac{3}{2}N$ $\displaystyle \log_{25} x = \frac{3}{2}N$

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Example Question #351 : Algebra

A factory makes barrels of the same shape but different sizes; the amount of water they hold varies directly as the cube of their height. The four-foot-high barrel holds 20 gallons of water; how much water would the six-foot-high barrel hold?

Possible Answers:

$64\textrm{ gal}$ $\displaystyle 64\textrm{ gal}$

$80\textrm{ gal}$ $\displaystyle 80\textrm{ gal}$

$67 \frac{1}{2}\textrm{ gal}$ $\displaystyle 67 \frac{1}{2}\textrm{ gal}$

$30\textrm{ gal}$ $\displaystyle 30\textrm{ gal}$

$45\textrm{ gal}$ $\displaystyle 45\textrm{ gal}$

Correct answer:

$67 \frac{1}{2}\textrm{ gal}$ $\displaystyle 67 \frac{1}{2}\textrm{ gal}$

Explanation:

Let $\displaystyle h$ be the height of a barrel and $\displaystyle V$ be its volume. Since $\displaystyle V$ varies directly as the cube of $\displaystyle h$, the variation equation is

$V = k h^{3}$ $\displaystyle V = k h^{3}$

for some constant of variation $\displaystyle k$.

We find $\displaystyle k$ by substituting $\displaystyle V = 20, h= 4$ from the smaller barrels:

$20 = k \cdot 4^{3}$ $\displaystyle 20 = k \cdot 4^{3}$

$20 = k \cdot 64$ $\displaystyle 20 = k \cdot 64$

$k = \frac{20 }{64}= \frac{5 }{16}$ $\displaystyle k = \frac{20 }{64}= \frac{5 }{16}$

Then the variation equation is:

$V = \frac{5 }{16} h^{3}$ $\displaystyle V = \frac{5 }{16} h^{3}$

Now we can substitute h = 6 $\displaystyle h = 6$ to find the volume of the larger barrel:

$V = \frac{5 }{16} \cdot 6^{3} = \frac{5 }{16} \cdot 216 = \frac{5 }{16} \cdot \frac{216}{1} = \frac{5 }{2} \cdot \frac{27}{1} = \frac{135 }{2}= 67 \frac{1 }{2}$ $\displaystyle V = \frac{5 }{16} \cdot 6^{3} = \frac{5 }{16} \cdot 216 = \frac{5 }{16} \cdot \frac{216}{1} = \frac{5 }{2} \cdot \frac{27}{1} = \frac{135 }{2}= 67 \frac{1 }{2}$

The larger barrel holds $67 \frac{1 }{2}$ $\displaystyle 67 \frac{1 }{2}$ gallons.

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Example Question #31 : Solving Equations

Solve for $\displaystyle x$.

$\displaystyle -x-3xy+4>0$

Possible Answers:

$x<\frac{-4}{1+3y}$ $\displaystyle x< \frac{-4}{1+3y}$

$x<\frac{4}{1+3y}$ $\displaystyle x< \frac{4}{1+3y}$

$x>\frac{4}{1+3y}$ $\displaystyle x>\frac{4}{1+3y}$

$x>\frac{-4}{1+3y}$ $\displaystyle x>\frac{-4}{1+3y}$

Correct answer:

$x<\frac{4}{1+3y}$ $\displaystyle x< \frac{4}{1+3y}$

Explanation:

$-x-3xy+4>0\Rightarrow 4>x+3xy$ $\displaystyle -x-3xy+4>0\Rightarrow 4>x+3xy$

$4>x(1+3y) \Rightarrow \frac{4}{1+3y}>x$ $\displaystyle 4>x(1+3y) \Rightarrow \frac{4}{1+3y}>x$

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Example Question #353 : Algebra

To convert Celsius temperature $\displaystyle C$ to the equivalent in Fahrenheit temperature $\displaystyle F$, use the formula

$F = \frac{9}{5}C + 32$ $\displaystyle F = \frac{9}{5}C + 32$

To the nearest tenth of a degree, convert $70^{\circ } C$ $\displaystyle 70^{\circ } C$ to degrees Fahrenheit.

Possible Answers:

$68.4 ^{\circ } F$ $\displaystyle 68.4 ^{\circ } F$

$158.0 ^{\circ } F$ $\displaystyle 158.0 ^{\circ } F$

$21.1 ^{\circ } F$ $\displaystyle 21.1 ^{\circ } F$

$94.0 ^{\circ } F$ $\displaystyle 94.0 ^{\circ } F$

$183.6 ^{\circ } F$ $\displaystyle 183.6 ^{\circ } F$

Correct answer:

$158.0 ^{\circ } F$ $\displaystyle 158.0 ^{\circ } F$

Explanation:

$F = \frac{9}{5}C + 32 = \frac{9}{5} \cdot 70 + 32 = 126 + 32 = 158$ $\displaystyle F = \frac{9}{5}C + 32 = \frac{9}{5} \cdot 70 + 32 = 126 + 32 = 158$

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GMAT Math : Solving equations

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #31 : Equations

Example Question #32 : Equations

Example Question #33 : Equations

Example Question #34 : Equations

Example Question #35 : Equations

Example Question #36 : Equations

Example Question #37 : Equations

Example Question #351 : Algebra

Example Question #31 : Solving Equations

Example Question #353 : Algebra

All GMAT Math Resources

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