GMAT Math : Algebra

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Example Questions

Example Question #8 : Solving Equations

Which is NOT a solution to the equation 2x-4y+z=8

Possible Answers:

$(\frac{7}{2},0,1)$

$(4+2a-\frac{b}{2},a,b)$

(2,4,6)

$(\frac{11}{2},1,1)$

(6,1,0)

Correct answer:

(2,4,6)

Explanation:

To solve this, we need to plug the answer choices into the equation and see which choice does NOT work. Let's go through the answer choices.

$\dpi{100} (2,4,6):2(2)-4(4)+6=-6$

This is the correct answer. If this had been a solution to the equation, the equation would have produced 8.

$\dpi{100} (\frac{11}{2},1,1):2(\frac{11}{2}) -4(1)+1)=8$

(6,1,0):2(6)-4(1)+0=8

$(\frac{7}{2},0,1):2(\frac{7}{2})-4(0)+1=8$

$(4+2a-\frac{b}{2}):$

Let's let a=1 and b=0 . Then this ordered pair becomes (6,1,0) .

2(6)-4(1)+0=8 . Any other answer choices of this form will also work.

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Example Question #9 : Solving Equations

Given the equation $x\cdot(x+k)=-7$ , if can be any integer, how many different sets of integer solutions are there for ?

Possible Answers:

Infinitely many

Correct answer:

Explanation:

$x\cdot(x+k)=-7$ is the same as x^2 + kx + 7 = 0 .

For solutions sets of integers, we are able to divide it as: $(x+\_\_)(x+\_\_) = 0$

When we multiply this, we notice that the constants must return 7. Since 7 is a prime number, we now know that these constants are 1 and 7. The signs of these can still switch however. If both of the numbers are negative, it will also return a product of positive 7. These are the only ways to make the constant term work. If they are both positive numbers, then we get k=8 . If they're both negative numbers then k= -8 .

For the values of , we simply notice the only way for the overall product to be 0 is for one (or both) of the pieces to be zero. This is done by having equal the additive inverse of the constant piece.

Thus we have 2 integer solution sets: $\left\{1,7\right\}$ and $\left\{-1,-7\right\}$

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Example Question #10 : Solving Equations

The volume of a fixed mass of gas varies inversely as the atmospheric pressure, as measured in millibars, acting on it, and directly as the temperature, as measured in kelvins, acting on it.

A balloon is filled to capacity at a point in time at which the atmospheric pressure is 104 millibars and the temperature is 295 kelvins. Six hours later, the temperature has increased to 305 kelvins, but the volume of the gas has not changed at all. What is the current atmospheric pressure?

Possible Answers:

$104 \textrm{ mbar}$

$108 \textrm{ mbar}$

$112 \textrm{ mbar}$

$101 \textrm{ mbar}$

The information provided is insufficient to answer this question.

Correct answer:

$108 \textrm{ mbar}$

Explanation:

The following variation equation can be set up:

$\frac{V_{1} P _{1}}{T_{1}} = \frac{V_{2} P _{2}}{T_{2}}$

But since the initial volume and the current volume are equal, or, equivalently, $V_{1} = V_{2}$ ,

$\frac{V_{1} P _{1}}{T_{1}} \div V_{1} = \frac{V_{2} P _{2}}{T_{2}}\div V_{2}$

$\frac{P _{1}}{T_{1}}= \frac{P _{2}}{T_{2}}$

We substitute $P _{1}= 104, T_{1}=295 ,T_{2}=305$ , and solve for $P _{2}$ :

$\frac{104}{295}= \frac{P _{2}}{305}$

$\frac{104}{295}\cdot 305 = \frac{P _{2}}{305} \cdot 305$

$P_{2} = \frac{104}{295}\cdot 305 \approx 108 \textrm{ mbar}$

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Example Question #321 : Algebra

Solve for :

4 (x +8) + 13 = 6 (x + 9) - x

Possible Answers:

$x = 4 \frac{1}{2}$

The equation has no solution.

x=9

$x = -4 \frac{1}{2}$

x = -9

Correct answer:

x = -9

Explanation:

4 (x +8) + 13 = 6 (x + 9) - x

$4 \cdot x +4 \cdot 8 + 13 = 6 \cdot x + 6 \cdot 9 - x$

4 x +32 + 13 = 6x + 54 - x

4 x +45 = 5x + 54

4 x +45 -4x -54 = 5x + 54 -4x -54

-9 = x

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Example Question #322 : Algebra

Give all real solutions of the following equation:

$x^{4} - 10x^{2} + 9 = 0$

Possible Answers:

$\left \{ 1, 3\right \}$

$\left \{ -9, -1, 1, 9\right \}$

$\left \{1, 9\right \}$

The equation has no solution.

$\left \{ -3, -1, 1, 3\right \}$

Correct answer:

$\left \{ -3, -1, 1, 3\right \}$

Explanation:

By substituting $u = x^{2}$ and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ , this equation be rewritten as a quadratic equation, and solved as such:

$x^{4} - 10x^{2} + 9 = 0$

$u^{2} - 10u + 9 = 0$

We are looking to factor the quadratic expression as (u+?)(u+?) , replacing the two question marks with integers with a product of 9 and a sum of ; these integers are -9,-1 .

(u-9)(u-1) = 0

Substitute back:

$(x^{2}-9)(x^{2}-1) = 0$

These factors can themselves be factored as the difference of squares:

(x+3)(x-3)(x+1)(x-1) = 0

Set each factor to zero and solve:

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

$x-1 = 0 \Rightarrow x = 1$

$x+1 = 0 \Rightarrow x = - 1$

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

The solution set is $\left \{ -3, -1, 1, 3\right \}$ .

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Example Question #323 : Algebra

Solve for :

$x - 7 \sqrt{x} + 12 = 0$

Possible Answers:

The equation has no solution.

$x = 9 \textrm{ or } x = -9$

$x = \sqrt{3} \textrm{ or } x = 2$

$x = 3 \textrm{ or } x = 4$

$x = 9 \textrm{ or } x = 16$

Correct answer:

$x = 9 \textrm{ or } x = 16$

Explanation:

Substitute $u = \sqrt{x}$ , and, subsequently, $u^{2} = x$ , to rewrite this equation as quadratic, then solve by factoring.

$x - 7 \sqrt{x} + 12 = 0$

$u^{2} - 7 u + 12 = 0$

We can rewrite the quadratic expression as (x + ?) (x + ?) , where the question marks are replaced with integers whose product is 12 and whose sum is ; these integers are $-3\ and -4$ .

(u-3) (u-4)= 0

Set each factor to zero and solve for ; then substitute back and solve for :

u-3 = 0

u=3

$\sqrt{x} =3$

x = 9

u-4 = 0

u=4

$\sqrt{x} =4$

x = 16

The solution set, which can be confirmed by substitution, is $\left \{ 9,16\right \}$ .

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Example Question #324 : Algebra

Find all real solutions to the following equation:

$\frac{1}{x^{2}} - \frac{8}{x} +12 = 0$

Possible Answers:

$x = \frac{1}{6} \textrm{ or } x = \frac{1}{2}$

$x = 2 \textrm{ or } x = 6$

$x = -6 \textrm{ or } x = -2$

$x = - \frac{1}{2} \textrm{ or } x = - \frac{1}{6}$

The equation has no solution.

Correct answer:

$x = \frac{1}{6} \textrm{ or } x = \frac{1}{2}$

Explanation:

This can be best solved by substituting $u = \frac{1}{x}$ , and, subsequently, $u ^{2}=\left ( \frac{1}{x} \right ) ^{2} =\frac{1}{x ^{2}}$ , then solving the resulting quadratic equation.

$\frac{1}{x^{2}} - \frac{8}{x} + 12 = 0$

$\frac{1}{x^{2}} - 8 \cdot \frac{1}{x} + 12 = 0$

$u^{2} - 8u + 12 = 0$

Factor the expression on the left by finding two integers whose product is 12 and whose sum is :

(u-6)(u-2) = 0

Set each linear binomial factor to 0, solve separately for , and substitute back:

u-6 = 0

u=6

$\frac{1}{x} = 6$

$x = \frac{1}{6}$

u-2 = 0

u=2

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

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

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Example Question #325 : Algebra

The period of a pendulum - that is, the time it takes for the pendulum to swing once and back - varies directly as the square root of its length.

The pendulum of a giant clock is 18 meters long and has period 8.5 seconds. If the pendulum is lengthened to 21 meters, what will its period be, to the nearest tenth of a second?

Possible Answers:

$11.2 \textrm{ sec}$

$7.9 \textrm{ sec}$

$10.4 \textrm{ sec}$

$11.6 \textrm{ sec}$

$9.2 \textrm{ sec}$

Correct answer:

$9.2 \textrm{ sec}$

Explanation:

The variation equation for this situation is

$\frac{P_{1}}{\sqrt{L_{1}}} = \frac{P_{2}}{\sqrt{L}_{2}}$

Set $P_{1} = 8.5 , L_{1} = 18, {L}_{2}= 21$ , and solve for $P_{2}$ ;

$\frac{8.5}{\sqrt{18}} = \frac{P_{2}}{\sqrt{21}}$

$\frac{8.5}{\sqrt{18}} \cdot{\sqrt{21}} = \frac{P_{2}}{\sqrt{21}} \cdot{\sqrt{21}}$

$P_{2} = \frac{8.5}{\sqrt{18}} \cdot{\sqrt{21}} \approx 9.2 \textrm{ sec}$

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Example Question #326 : Algebra

$x = 8^{N}$

Which of these expressions is equal to $\log_{32} x$ ?

Possible Answers:

$\sqrt[5]{N^{3}}$

$\ln \frac{5}{3}N$

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

$\frac{3}{5} N$

$\frac{5}{3} N$

Correct answer:

$\frac{3}{5} N$

Explanation:

$2^{5} = 32$

$2 = \sqrt[5]{32}$

$8 = 2^{3} = \left ( \sqrt[5]{32} \right ) ^{3} = 32^{ \frac{3}{5} }$

$x = 8^{N}= \left (32^{ \frac{3}{5} } \right \)^ {N } =32^{ \frac{3}{5} N }$

$\log_{32} x =\log_{32}32^{ \frac{3}{5} N } = \frac{3}{5} N$

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Example Question #327 : Algebra

Solve for :

| 4x - 18 | - 18 = 57

Possible Answers:

$x= -14.25 \textrm{ or }x=23.25$

x=23.25

$x= -14.25 \textrm{ or }x=14.25$

x= -14.25

$x= -23.25 \textrm{ or }x=23.25$

Correct answer:

$x= -14.25 \textrm{ or }x=23.25$

Explanation:

First, isolate the absolute value expression on one side:

| 4x - 18 | - 18 = 57

| 4x - 18 | -18+ 18 = 57+18

| 4x - 18 | = 75

Rewrite as a compound sentence:

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

Solve each separately:

4x - 18 = - 75

4x - 18+18 = -75+18

4x= - 57

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

x= - 14.25

4x - 18 = 75

4x - 18+18 = 75+18

4x= 93

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

x = 23.25

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

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GMAT Math : Algebra

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #8 : Solving Equations

Example Question #9 : Solving Equations

Example Question #10 : Solving Equations

Example Question #321 : Algebra

Example Question #322 : Algebra

Example Question #323 : Algebra

Example Question #324 : Algebra

Example Question #325 : Algebra

Example Question #326 : Algebra

Example Question #327 : Algebra

All GMAT Math Resources

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