Problem-Solving Questions - GMAT Quantitative

Card 0 of 16904

Question

$125^{N} = x$

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

Answer

$5 ^{2} = 25$ , 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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Question

Solve for :

Answer

First, isolate the absolute value expression on one side:

Rewrite as a compound sentence:

$4x - 18 = - 75 \textrm{ or } 4x - 18 = 75$

Solve each separately:

$4x \div 4= - 57 \div 4$

or

$4x \div 4=93 \div 4$

The solution set is $\left { -14.25, 23.25 \right }$

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Question

Solve for :

Answer

First, isolate the absolute value expression on one side:

Rewrite as a compound sentence:

$4x - 18 = - 39 \textrm{ or } 4x - 18 = 39$

Solve each separately:

$4x \div 4= - 21 \div 4$

or

$4x \div 4=57 \div 4$

The solution set is $\left { -5.25, 14.25\right }$ .

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Question

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?

Answer

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

$V = k h^{3}$

for some constant of variation .

We find by substituting from the smaller barrels:

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

$20 = k \cdot 64$

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

Then the variation equation is:

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

Now we can substitute to find the volume of the larger barrel:

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

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

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Question

What is the y-intercept of ?

Answer

To solve for the y-intercept, you set to zero and solve for :

$y=\frac{4}{5}$

Therefore, the y-intercept is: $\left(0,\frac{4}{5}\right)$

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Question

What is the -intercept of ?

Answer

To solve for the -inter, you set to zero and solve for :

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

x-intercept:

$\left(\frac{1}{3},0\right)$

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Question

What is the y-intercept of ?

Answer

To solve for the y-intercept of , you set to zero and solve for :

y-intercept:

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Question

Give the solution set of the equation:

$\left | x ^{2} - 2x \right | = 3x -6$

Answer

Write this as a compound equation and solve each separately.

$\left | x ^{2} - 2x \right | = 3x -6$

$x ^{2} - 2x = 3x -6 \textrm{ or }x ^{2} - 2x = -\left ( 3x -6 \right )$

$x ^{2} - 2x = 3x -6$

$x ^{2} - 2x - 3x + 6 = 3x -6 - 3x + 6$

$x ^{2} - 5x + 6 =0$

$x - 3 = 0 \Rightarrow x = 3$

$x - 2 = 0 \Rightarrow x =2$

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

$x ^{2} - 2x = -3x +6$

$x ^{2} - 2x+ 3x - 6= -3x +6 + 3x - 6$

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

$x + 3 = 0 \Rightarrow x = -3$

$x - 2 = 0 \Rightarrow x =2$

This gives us three possible solutions - $\left { -3, 2, 3\right }$ . We check all three.

$\left | x ^{2} - 2x \right | = 3x -6$

$\left | \left ( -3\right ) ^{2} - 2 (-3) \right | = 3 (-3) -6$

$\left | \left9 - (-6) \right | = -9-6$

$\left |15| = -15$

This is a false statement so we can eliminate as a false "solution".

$\left | x ^{2} - 2x \right | = 3x -6$

$\left |2^{2} - 2 \cdot 2 \right | = 2 \cdot 3 -6$

$\left | \left 4-4 \right | = 6-6$

2 is a solution.

$\left | x ^{2} - 2x \right | = 3x -6$

$\left | 3 ^{2} - 2 \cdot 3 \right | = 3 \cdot 3 -6$

$\left | \left 9 - 6\right | = 9-6$

$\left |3 | = 3$

3 is a solution.

The solution set is $x \in \left { 2, 3\right }$ .

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Question

Find the roots of Separate the answers with a comma.

Answer

This can be found by factoring the equation. Doing this we get

We can solve this equation happen when or So the roots are .

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Question

Give the solution set of the equation:

$x^{2} =|x| + 6$

Answer

Case 1: is nonnegative - that is, .

The equation becomes

$x^{2} =x+ 6$

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

$x^{2} -x - 6=0$

Either , in which case ,

or , in which case , which is impossible, as we are assuming that is nonnegative.

case 2: is negative - that is,

The equation becomes

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

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

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

Either , in which case , which is impossible since we are assuming that is negative,

or , in which case .

and can both be confirmed as solutions.

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Question

Solve for :

$4 ^{5x - 7} = 8 ^{2x -1}$

Answer

Since $2 ^{2} = 4$ and $2^{3} = 8$ , replace, and use the exponent rules:

$4 ^{5x - 7} = 8 ^{2x -1}$

$\left ( 2 ^{2} \right ) ^{5x - 7} = \left ( 2^{3} \right ) ^{2x -1}$

$2 ^{2 \left ( 5x - 7 \right )}=2 ^{3 \left ( 2x - 1 \right )}$

Set the exponents equal to each other and solve for :

$2 \left ( 5x - 7 \right )=3 \left ( 2x - 1 \right )$

$4x \div 4 = 11 \div 4$

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Question

Give the solution set of the equation:

$2x + \left | x\right | = 60$

Answer

Case 1: $x \geq 0$

Then $\left | x\right | = x$ , and the equation becomes

$3x \div 3 = 60 \div 3$

Case 2:

Then $\left | x\right | = -x$ , and the equation becomes

$2x +\left ( - x \right ) = 60$

However, this conflicts with the fact that , so this is a false solution.

The only solution is therefore .

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Question

Solve for :

Answer

We need to isolate . Move all other terms to the right hand side of the equation:

Combine like terms:

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Question

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.

Answer

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

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Question

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

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

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

Answer

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

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Question

Solve for , giving all solutions, real and imaginary:

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

Answer

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:

or

$x^{2} +3 =0$

$x^{2} = -3$

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

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Question

Solve for :

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

Answer

$\frac{1}{36} = 6 ^{-2}$ , 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:

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

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Question

Solve for :

Answer

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

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

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

or

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

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Question

Solve for . Give all real solutions:

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

Answer

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

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

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

Set each binomial to 0 and solve separately:

$\sqrt{x}= 5$

or

$\sqrt{x}=-4$

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

The only solution is .

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Question

Solve for :

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

Answer

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:

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

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

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

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

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

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

or

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

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

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