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ISEE Upper Level Math : Variables

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

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Example Questions

Example Question #6 : Variables And Exponents

Simplify:

(2x-y)^2-(x-2y)^2

Possible Answers:

3x^2+3y^2-4xy

3x^2+3y^2

x^2+y^2

3x^2-3y^2

3x^2+3y^2+4xy

Correct answer:

3x^2-3y^2

Explanation:

Expand each term by using FOIL:

(2x-y)^2-(x-2y)^2 = [(2x)^2-(2) (2x)(y)+(y)^2]-[(x)^2+(2) (x)(2y)-(2y)^2]

=4x^2-4xy+y^2-x^2+4xy-4y^2

Rearrange to group like-terms together.

(4x^2-x^2)+(-4xy+4xy)+(y^2-4y^2)

Simplify by combining like-terms.

3x^2-3y^2

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Example Question #7 : Variables And Exponents

Simplify:

4x^3y-x^2y^2-3x^3y+2x^2y^2

Possible Answers:

-x^3y+x^2y^2

-x^3y-x^2y^2

x^3y-x^2y^2

x^3y+x^2y^2

The expression can not be simplified further

Correct answer:

x^3y+x^2y^2

Explanation:

Start by reordering the expression to group like-terms together.

4x^3y-x^2y^2-3x^3y+2x^2y^2

(4x^3y-3x^3y)+(-x^2y^2+2x^2y^2)

Combine like-terms to simplify.

(4x^3y-3x^3y)+(-x^2y^2+2x^2y^2)

x^3y+x^2y^2

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Example Question #11 : Variables And Exponents

Simplify:

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

Possible Answers:

$-18x^{15}$

$-\frac{x^{2}}{216}$

$\frac{x^{15}}{216}$

$216x^{2}$

$-18x^{2}$

Correct answer:

$\frac{x^{15}}{216}$

Explanation:

Apply the power of a product property:

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

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

$=\frac{1}{216}\cdot x ^{-5 \cdot (-3) }$

$=\frac{1}{216}\cdot x ^{15 }$

$=\frac{x ^{15 }}{216}$

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Example Question #12 : Variables And Exponents

What is the coefficient of $x^{5}$ in the expansion of $(x+1)^{12}$ .

Possible Answers:

7,920

9,504

792

Correct answer:

792

Explanation:

By the Binomial Theorem, if $(x + 1)^{n}$ is expanded, the coefficient of $x^{r}$ is

$_{n}\textrm{C} _{r} = \frac{n!}{(n-r)!r!}$ .

Substitute n = 12, r = 5 : The coefficient of $x^{5}$ is:

$_{12}\textrm{C} _{5} = \frac{12!}{(12-5)!5!} = \frac{12!}{7!5!}$

$= \frac{12 \cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}$

$= \frac{12 \cdot 11\cdot 10\cdot 9\cdot 8}{ 5\cdot 4\cdot 3\cdot 2\cdot 1} = \frac{11\cdot 9\cdot 8}{ 1} = 792$

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Example Question #11 : Variables And Exponents

Simplify the expression: $\left [\left ( x ^{3} \right )^{3} \right ]^{3 }$

Possible Answers:

$x^ {18}$

$x^{27}$

$x ^{9}$

$27x^{3}$

729x

Correct answer:

$x^{27}$

Explanation:

Apply the power of a power property twice:

$\left [\left ( x ^{3} \right )^{3} \right ]^{3 } = ( x ^{3\; \cdot \; 3 } )^{3 } = ( x ^{9 } )^{3 } = x ^{9\cdot 3} =x ^{27}$

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Example Question #12 : Variables And Exponents

What is the coefficient of $x^{4}$ in the expansion of $(2x + 3)^{6}$ ?

Possible Answers:

240

2,160

135

5,320

Correct answer:

2,160

Explanation:

By the Binomial Theorem, the $x^{r}$ term of $(ax + b)^{n}$ is

$_{n}\textrm{C} _{r} b^{n-r}\left ( ax \right )^{r}= \frac{n!}{(n-r)!r!} a^{r} b^{n-r} x^{r}$ ,

making the coefficient of $x^{r}$

$\frac{n!}{(n-r)!r!} a^{r} b^{n-r}$ .

We can set a = 2,b=3,n=6,r=4 in this expression:

$\frac{6!}{(6-4)!4!} \cdot 2^{4} \cdot 3^{6-4}$

$=\frac{6!}{2!4!}\cdot 2^{4}\cdot 3^{2}$

$=\frac{720}{2\cdot 24}\cdot 16\cdot 9$

=2,160

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Example Question #2 : How To Find The Exponent Of Variables

What is the coefficient of $x^{8}$ in the expansion of $(x + 3)^{10}$ ?

Possible Answers:

405

3,240

135

Correct answer:

405

Explanation:

By the Binomial Theorem, the $x^{r}$ term of $(x + b)^{n}$ is

$_{n}\textrm{C} _{r} b^{n-r} x^{r}= \frac{n!}{(n-r)!r!} b^{n-r} x^{r}$ .

Substitute b = 3, n = 10, r= 8 and this becomes

$\frac{10!}{(10-8)!8!} 3^{10-8} x^{8}$ .

The coefficient is

$\frac{10!}{(10-8)!8!} \cdot 3^{10-8} = \frac{10!}{2!8!}\cdot 3^{2} = \frac{9 \cdot 10}{2}\cdot 9 = 405$ .

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Example Question #13 : Variables And Exponents

Evaluate:

$\left [ (x^4)^4 \right ]^5$

Possible Answers:

$x^{40}$

$x^{80}$

$x^{60}$

$x^{12}$

$x^{16}$

Correct answer:

$x^{80}$

Explanation:

We need to apply the power of power rule twice:

$\left [ (x^4)^4 \right ]^5=(x^{4\times 4})^5=(x^{16})^5=x^{16\times 5}=x^{80}$

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Example Question #14 : Variables And Exponents

$(x^{-4})^{a}=x^{8}$

Solve for .

Possible Answers:

Correct answer:

Explanation:

Based on the power of a product rule we have:

$(x^{-4})^{a}=x^{8}\Rightarrow x^{-4\times a}=x^8$

The bases are the same, so we can write:

$-4a=8\Rightarrow a=\frac{8}{-4}=-2$

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Example Question #5 : How To Find The Exponent Of Variables

Simplify:

$(\frac{2x^5}{3y^2})^{-4}$

Possible Answers:

$\frac{y^8}{16x^{20}}$

$\frac{81y^8}{x^{20}}$

$\frac{81y^8}{16x^{16}}$

$\frac{81y^9}{16x^{20}}$

$\frac{81y^8}{16x^{20}}$

Correct answer:

$\frac{81y^8}{16x^{20}}$

Explanation:

First, recognize that raising the fraction to a negative power is the same as raising the inverted fraction to a positive power.

$(\frac{2x^5}{3y^2})^{-4}=(\frac{3y^2}{2x^5})^4$

Apply the exponent within the parentheses and simplify.

$\frac{(3y^2)^4}{(2x^5)^4}$

$\frac{3^4(y^2)^4}{2^4(x^5)^4}$

$\frac{81y^{(2\times 4)}}{16x^{(5\times 4)}}$

$\frac{81y^8}{16x^{20}}$

This fraction cannot be simplified further.

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All ISEE Upper Level Math Resources

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ISEE Upper Level Math : Variables

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

All ISEE Upper Level Math Resources

Example Questions

Example Question #6 : Variables And Exponents

Example Question #7 : Variables And Exponents

Example Question #11 : Variables And Exponents

Example Question #12 : Variables And Exponents

Example Question #11 : Variables And Exponents

Example Question #12 : Variables And Exponents

Example Question #2 : How To Find The Exponent Of Variables

Example Question #13 : Variables And Exponents

Example Question #14 : Variables And Exponents

Example Question #5 : How To Find The Exponent Of Variables

All ISEE Upper Level Math Resources

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