GMAT Math : Algebra

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Example Questions

Example Question #24 : Solving Equations

To convert Fahrenheit temperature to the equivalent in Celsius , use the formula

$C = \frac{5}{9}(F-32)$

To the nearest tenth of a degree, convert $70 ^{\circ } F$ to degrees Celsius.

Possible Answers:

$21.1 ^{\circ }C$

$70.9^{\circ }C$

$6.9^{\circ }C$

$158.0 ^{\circ }C$

$87.8 ^{\circ }C$

Correct answer:

$21.1 ^{\circ }C$

Explanation:

$C = \frac{5}{9}(F-32) = C = \frac{5}{9}(70-32) = \frac{5}{9}\cdot 38 \approx 21.1 ^{\circ }C$

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Example Question #1422 : Problem Solving Questions

Solve for :

| 0.25 - 0.4 x | = 0.1

Possible Answers:

$x = -0.875\textrm{ or } x = 0.875$

$x = 0.375\textrm{ or }x = 0.875$

The equation has no solution.

$x = -0.875\textrm{ or } x = -0.375$

$x =- 0.375\textrm{ or }x = 0.375$

Correct answer:

$x = 0.375\textrm{ or }x = 0.875$

Explanation:

Rewrite as a compound equation, then solve each equation individually:

| 0.25 - 0.4 x | = 0.1

$0.25 - 0.4 x= - 0.1 \textrm{ or }0.25 - 0.4 x= 0.1$

0.25 - 0.4 x= - 0.1

0.25 - 0.4 x- 0.25 = - 0.1 - 0.25

- 0.4 x = - 0.35

$- 0.4 x \div \left ( - 0.4 \right ) = - 0.35 \div \left ( - 0.4 \right )$

x = 0.875

0.25 - 0.4 x= 0.1

0.25 - 0.4 x- 0.25 = 0.1 - 0.25

- 0.4 x = - 0.15

$- 0.4 x \div \left ( - 0.4 \right ) = - 0.15 \div \left ( - 0.4 \right )$

x = 0.375

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Example Question #21 : Solving Equations

Solve for :

$\frac{8 }{x^{2}} - \frac{13}{x} = 6$

Possible Answers:

The equation has no solution.

$x = -2 \textrm{ or } x= \frac{3}{8 }$

$x= -\frac{8 }{3}\textrm{ or }x = \frac{1}{2}$

$x= -\frac{3}{8 }\textrm{ or }x = 2$

$x = - \frac{1}{2} \textrm{ or } x= \frac{8 }{3}$

Correct answer:

$x= -\frac{8 }{3}\textrm{ or }x = \frac{1}{2}$

Explanation:

Eliminate the denominators by multiplying by $x^{2}$ , then solve the resulting equation:

$\frac{8 }{x^{2}} - \frac{13}{x} = 6$

$\left (\frac{8 }{x^{2}} - \frac{13}{x} \right )\cdot x ^{2} = 6 \cdot x ^{2}$

$\frac{8 }{x^{2}} \cdot\frac{ x ^{2}}{1} - \frac{13}{x} \cdot\frac{ x ^{2}}{1} = 6 x ^{2}$

$8 - 13x= 6 x ^{2}$

$8 - 13x+ 13 x - 8 = 6 x ^{2} + 13 x - 8$

$6 x ^{2} + 13 x - 8 = 0$

Solve using the method:

$6 x ^{2} -3x + 16 x - 8 = 0$

$3x \left ( 2 x - 1 \right ) + 8 \left ( 2 x - 1 \right )= 0$

$\left ( 3x+ 8 \right ) \left ( 2 x - 1 \right )= 0$

$3x+ 8 = 0 \textrm{ or }2x-1 = 0$

3x+8 = 0

3x+8 -8 = 0 -8

3x = -8

$3x\div 3 = -8 \div 3$

$x= -\frac{8 }{3}$

2x-1 = 0

2x-1+1 = 0+1

2x = 1

$2x \div 2 = 1 \div 2$

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

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Example Question #1424 : Problem Solving Questions

Solve for . Give all real solutions:

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

Possible Answers:

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

The equation has no real solution.

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

x = 25

x = -16

Correct answer:

x = 25

Explanation:

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

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

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

(t-5)(t+4)= 0

Set each binomial to 0 and solve separately:

t-5 = 0

t = 5

$\sqrt{x}= 5$

x = 25

t+4= 0

t=-4

$\sqrt{x}=-4$

Since no real number has as its principal square root, this yields no solution.

The only solution is x = 25 .

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Example Question #31 : Solving Equations

Solve for , giving all solutions, real and imaginary:

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Factor the expression:

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

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

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

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

Rewrite:

x- 4 = 0

x=4

$x^{2} +3 =0$

$x^{2} = -3$

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

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Example Question #1424 : Problem Solving Questions

Solve for :

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

Possible Answers:

The equation has no solution.

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

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

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

x = 4

Correct answer:

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

Explanation:

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

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

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

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

Set the exponents equal to each other:

2x+5 = -2 ( x-3 )

2x+5 = -2 x+6

2x+5 +2x-5 = -2 x+6+2x-5

4x = 1

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

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Example Question #341 : Algebra

Daniel has 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:

Correct answer:

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:

$A = 3 + 2 \cdot D$

and

D = 7

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

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Example Question #32 : Solving Equations

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

What is the value of ?

Possible Answers:

1.8

1.6

1.7

1.5

1.9

Correct answer:

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.

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

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

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

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

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

Anything raised to the zero just becomes one.

2=x+0.5

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

x=2-0.5

x=1.5

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Example Question #31 : Solving Equations

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

By definition:

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

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

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Example Question #342 : Algebra

$125^{N} = x$

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

Possible Answers:

$\frac{2}{3}N$

$N \ln 5$

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

$\frac{3}{2}N$

$N \sqrt{N }$

Correct answer:

$\frac{3}{2}N$

Explanation:

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

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

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

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GMAT Math : Algebra

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #24 : Solving Equations

Example Question #1422 : Problem Solving Questions

Example Question #21 : Solving Equations

Example Question #1424 : Problem Solving Questions

Example Question #31 : Solving Equations

Example Question #1424 : Problem Solving Questions

Example Question #341 : Algebra

Example Question #32 : Solving Equations

Example Question #31 : Solving Equations

Example Question #342 : Algebra

All GMAT Math Resources

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