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ISEE Upper Level Math : Equations

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Example Questions

Example Question #171 : How To Find The Solution To An Equation

Define $f (x) = x^{2}$ .

If (fg)(2) = 64 , which of the following cannot be a valid definition of ?

Possible Answers:

g(x) = 8x

$g(x) = x^{2} + 12$

$g(x) = \frac{32}{x}$

$g(x) = x ^{3}$

Correct answer:

$g(x) = x ^{3}$

Explanation:

$f (x) = x^{2}$ , so, substituting 2 for , we see that

$f (2) = 2^{2} =4$ .

By definition,

$(fg)(2) = f(2) \cdot g(2)$ , so

(fg)(2) = 64

Substituting and solving for g(2) :

$f(2) \cdot g(2) = 64$

$4\cdot g(2) = 64$

$4\cdot g(2) \div 4 = 64 \div 4$

g(2) = 16

Examine all four alternatives by, again, substituting 2 for in each and finding for which one $g(2) \ne 16$ :

$g(x) = x^{2} + 12$

$g(2) = 2^{2} + 12 = 4 + 12 = 16$

$g(x) = \frac{32}{x}$

$g(2) = \frac{32}{2} = 16$

g(x) = 8x

$g(2) = 8 \cdot 2 = 16$

$g(x) = x ^{3}$

$g(2) = 2 ^{3} = 8$

Therefore, of the four choices, only $g(x) = x ^{3}$ is not a valid definition, since it does not match the conditions.

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Example Question #171 : Equations

Define $f(x) = x^{5}$ .

If $\frac{f}{g} (5) = 50$ , then which of the following could be the definition of g(x) ?

Possible Answers:

$g(x) =\frac{ x^{3}}{2}$

$g(x) = 2x^{3}$

$g(x) =\frac{2}{ x^{3}}$

$g(x) =\frac{1}{ 2x^{3}}$

Correct answer:

$g(x) =\frac{ x^{3}}{2}$

Explanation:

By definition,

$\frac{f}{g} (5) = \frac{f(5)}{g(5)}$ .

Also, by substitution:

$f(x) = x^{5}$

$f(5) = 5^{5} = 3,125$

Therefore, the question is equivalent to asking for which definition of g(x) it holds that

$\frac{3,125}{g(5)} = 50$

Each definition in the given choices can be evaluated for x = 5 by substitution, with each value of g(5) tested in turn:

$g(x) =\frac{1}{ 2x^{3}}$

$g(5) =\frac{1}{ 2 \cdot 5^{3}} = \frac{1}{ 2 \cdot 125} = \frac{1}{250} = 0.004$

$\frac{f}{g} (5) = \frac{3,125}{0.008} = 3,125 \div 0.008 = 390,625$

$g(x) =\frac{2}{ x^{3}}$

$g(x) =\frac{2}{ 5^{3}}= \frac{2}{125} = 0.016$

$\frac{f}{g} (5) = \frac{3,125}{0.016} = 3,125 \div 0.016 =195, 312 .5$

$g(x) = 2x^{3}$

$g(5) = 2 \cdot 5^{3} = 2 \cdot 125 = 250$

$\frac{f}{g} (5) = \frac{3,125}{250} = 12.5$

$g(x) =\frac{ x^{3}}{2}$

$g(x) =\frac{ 5^{3}}{2} = \frac{ 125}{2} = 62.5$

$\frac{f}{g} (5) = \frac{3,125}{62.5} = 50$

This makes $g(x) =\frac{ x^{3}}{2}$ the correct choice.

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Example Question #172 : Equations

Define f (x) = 7x - 23 .

If (f-g) (8) = 34 , then which of the following could be a valid definition of the function ?

Possible Answers:

g(x) = x - 6

g(x) = x - 7

g(x) = x - 9

g(x) = x- 8

Correct answer:

g(x) = x - 9

Explanation:

f (x) = 7x - 23 , so, substituting 8 for , we see that

$f (8) = 7 \cdot 8 - 23= 56 - 23 = 33$

By definition,

(f-g) (8) = f(8) - g (8) , so

(f-g) (8) = 34

f(8) - g (8) = 34

33 - g (8) = 34

33 - g (8) - 33 = 34 - 33

-g(8)= 1

g(8) = -1

Examine all four alternatives by, again, substituting 8 for , and find the one for which g(8) = -1 .

g(x) = x - 6

g(x) = 8 - 6 = 2

g(x) = x - 7

g(x) = 8 - 7 = 1

g(x) = x- 8

g(x) = 8- 8 = 0

g(x) = x - 9

g(x) = 8 - 9 = -1

Of the four choices, g(x) = x - 9 is the definition such that g(8) = -1 .

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Example Question #171 : Algebraic Concepts

Solve the following equation for t, when d is 6.

5t^2=6d+34

Possible Answers:

t=70

$t=\sqrt{14}$

t=14

$t=\pm\sqrt{14}$

Correct answer:

$t=\pm\sqrt{14}$

Explanation:

Solve the following equation for t, when d is 6.

5t^2=6d+34

Let's begin by plugging 6 in for d and then, using algebra, we will find t.

5t^2=6(6)+34

5t^2=36+34

5t^2=70

Divide by 5:

$t^2=\frac{70}{5}=14$

Square root both sides to finish up

$t=\sqrt{14}$

Almost there, but because we are square rooting, we need plus or minus the square root of 14

$t=\pm\sqrt{14}$

The reason for this is that positive or negative square root of 14 will get us positive 14 when we square it. Therefore, we technically have two answers.

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Example Question #172 : Algebraic Concepts

Which solution makes this equation true:

5x-8=12

Possible Answers:

x=3

x=7

x=4

x=15

x=5

Correct answer:

x=4

Explanation:

To solve for x, we want x to stand alone. So, we get

5x-8=12

5x-8+8=12+8

5x-0=20

5x=20

$\frac{5x}{5} = \frac{20}{5}$

x=4

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Example Question #173 : Algebraic Concepts

Solve the following equation when y is equal to 12.

$\sqrt{y^2}-4=c^3$

Possible Answers:

Not enough information provided.

c=-2

c=2

c=12

Correct answer:

c=2

Explanation:

Solve the following equation when y is equal to 12.

$\sqrt{y^2}-4=c^3$

To begin, let's realize that

$\sqrt{y^2}=y$

So, let's update our original equation.

y-4=c^3

Now, let's plug in 12 for y

12-4=c^3

8=c^3

So, what number cubed equals 8?

$c=\sqrt[3]{8}=2$

So our answer is 2

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Example Question #174 : Algebraic Concepts

Solve the following equation for q.

3q^3-147=12

Possible Answers:

$q=\sqrt{53}$

$q=\sqrt[3]{45}$

$q=\sqrt[3]{59}$

$q=\sqrt[3]{53}$

Correct answer:

$q=\sqrt[3]{53}$

Explanation:

Solve the following equation for q.

3q^3-147=12

First, let's divide everything by three. This will work because 147 and 12 are both divisible by 3

$(3q^3-147=12)\div3\rightarrow q^3-49=4$

Now, add 49 to both sides to get the variable by itself

q^3=53

$q=\sqrt[3]{53}$

Now, this doesn't come out to a neat decimal, so we will just leave it as is.

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Example Question #175 : Algebraic Concepts

Which of the following makes this equation true:

7y-13=29

Possible Answers:

y=6

y=7

y=8

y=9

y=2

Correct answer:

y=6

Explanation:

To answer the question, we will need to solve for y. We get

7y-13=29

7y-13+13=29+13

7y-0=42

7y=42

$\frac{7y}{7} = \frac{42}{7}$

y=6

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Example Question #176 : Algebraic Concepts

Solve the following equation for j.

14j-12=58

Possible Answers:

j=2

No real solutions

j=-5

j=5

Correct answer:

j=5

Explanation:

Solve the following equation for j.

14j-12=58

We can solve this with basic algebra.

First, add 12 to both sides:

14j-12(+12)=58(+12)

14j=70

Next, divide both sides by 14

$14j\div14=70 \div 14$

$j=\frac{70}{14}=5$

And we end up with...

j=5

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Example Question #177 : Algebraic Concepts

Which of the following makes the equation true:

$\frac{x}{5} +9 = 12$

Possible Answers:

x=7

x=12

x=15

x=5

x=3

Correct answer:

x=15

Explanation:

To find the answer, we will solve for x. So, we get

$\frac{x}{5} +9 = 12$

$\frac{x}{5} +9 -9 = 12-9$

$\frac{x}{5} + 0 = 3$

$\frac{x}{5} = 3$

$\frac{x}{5} \cdot 5 = 3 \cdot 5$

x=15

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ISEE Upper Level Math : Equations

Study concepts, example questions & explanations for ISEE Upper Level Math

All ISEE Upper Level Math Resources

Example Questions

Example Question #171 : How To Find The Solution To An Equation

Example Question #171 : Equations

Example Question #172 : Equations

Example Question #171 : Algebraic Concepts

Example Question #172 : Algebraic Concepts

Example Question #173 : Algebraic Concepts

Example Question #174 : Algebraic Concepts

Example Question #175 : Algebraic Concepts

Example Question #176 : Algebraic Concepts

Example Question #177 : Algebraic Concepts

All ISEE Upper Level Math Resources

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