Variables and Exponents - ISEE Upper Level Quantitative Reasoning

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Question

Half of one hundred divided by five and multiplied by one-tenth is .

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Answer

Let's take this step by step. "Half of one hundred" is 100/2 = 50. Then "half of one hundred divided by five" is 50/5 = 10. "Multiplied by one-tenth" really is the same as dividing by ten, so the last step gives us 10/10 = 1.

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Question

Simplify: $\frac{8x^{-6}}{2x^{-3}}$

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Answer

$\frac{8x^{-6}}{2x^{-3}} = \frac{8}{2} \cdot \frac{x^{-6}}{x^{-3}} = 4 \cdot x ^{-6 - (-3)} = 4 \cdot x ^{-3} =4 \cdot \frac{1}{x^{3}} = \frac{4}{x^{3}}$

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Question

Simplify:

$\frac{10 x ^{-6} y ^{-4}}{20x ^{-8}y^{-2}}$

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Answer

Break the fraction up and apply the quotient of powers rule:

$\frac{10 x ^{-6} y ^{-4}}{20x ^{-8}y^{-2}}$

$= \frac{10 }{20} \cdot \frac{ x ^{-6} }{x ^{-8}} \cdot \frac{y ^{-4}}{y^{-2}}$

$= \frac{1 }{2} \cdot x ^{-6- (-8)} \cdot y ^{-4- (-2)}$

$= \frac{1 }{2} \cdot x ^{2} \cdot y ^{-2}$

$= \frac{1 }{2} \cdot x ^{2} \cdot \frac{1}{y ^{2}}$

$= \frac{x ^{2} }{2y ^{2} }$

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Question

Simplify: $\frac{16x^4}{12x^2}$

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Answer

To simplify this expression, look at the like terms separately. First, simplify $\frac{16}{12}$ . This becomes $\frac{4}{3}$ . Then, deal with the $\frac{x^4}{x^2}$ . Since the bases are the same and you're dividing, you can subtract exponents. This gives you Since the exponent is positive, you put in the numerator. This gives you a final answer of $\frac{4x^2}{3}$ .

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Question

is a negative number.

Which is the greater quantity?

(a) The reciprocal of $x^{7}$

(b) The reciprocal of $x^{8}$

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Answer

A negative number raised to an odd power is negative; a negative number raised to an even power is positive. It follows that $x^{7}$ is negative and $x^{8}$ is positive. Also, the reciprocal of a nonzero number assumes the same sign as the number itself, so the reciprocal of $x^{8}$ is positive and that of $x^{7}$ is negative. It follows that the reciprocal of $x^{8}$ is the greater of the two.

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Question

Simplify:

$\frac{10 x ^{-6} y ^{-4}}{20x ^{-8}y^{-2}}$

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Answer

Break the fraction up and apply the quotient of powers rule:

$\frac{10 x ^{-6} y ^{-4}}{20x ^{-8}y^{-2}}$

$= \frac{10 }{20} \cdot \frac{ x ^{-6} }{x ^{-8}} \cdot \frac{y ^{-4}}{y^{-2}}$

$= \frac{1 }{2} \cdot x ^{-6- (-8)} \cdot y ^{-4- (-2)}$

$= \frac{1 }{2} \cdot x ^{2} \cdot y ^{-2}$

$= \frac{1 }{2} \cdot x ^{2} \cdot \frac{1}{y ^{2}}$

$= \frac{x ^{2} }{2y ^{2} }$

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Question

Simplify:

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

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Answer

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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Question

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

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Answer

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 : 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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Question

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

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Answer

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 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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Question

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

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Answer

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 in this expression:

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

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

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

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Question

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

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Answer

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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Question

Evaluate:

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

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Answer

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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Question

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

Solve for .

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Answer

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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Question

Simplify:

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

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Answer

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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Question

Simplify:

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

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Answer

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

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

Apply the exponent within the parentheses and simplify.

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

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

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

$\frac{4z^6}{x^4y^2}$

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Question

Simplify if $x\neq 0,-1$ and $y\neq 0$ .

$\frac{x^3y+2x^2y+xy}{xy(x+1)}$

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Answer

Begin by factoring the numerator and denominator. can be factored out of each term.

$\frac{x^3y+2x^2y+xy}{xy(x+1)}=\frac{xy(x^2+2x+1)}{xy(x+1)}$

can be canceled, since it appears in both the numerator and denomintor.

$\frac{x^2+2x+1}{x+1}$

Next, factor the numerator.

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

Simplify.

$\frac{(x+1)(x+1)}{x+1}=x+1$

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Question

$\frac{1}{3}+ \sqrt{y}= \frac{2}{5}$

Evaluate .

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Answer

$\frac{1}{3}+ \sqrt{y}= \frac{2}{5}$

$\frac{1}{3}+ \sqrt{y}- \frac{1}{3}= \frac{2}{5} - \frac{1}{3}$

$\sqrt{y} = \frac{2}{5} - \frac{1}{3}$

$\sqrt{y} = \frac{3 \cdot 2}{3 \cdot 5} - \frac{1 \cdot 5}{3 \cdot 5}$

$\sqrt{y} = \frac{6}{1 5} - \frac{ 5}{15}$

$\sqrt{y} = \frac{6- 5}{15}$

$\sqrt{y} = \frac{1}{15}$

$y = (\sqrt{y}) ^{2}= \left (\frac{1}{15} \right )^{2} = \frac{1^{2}}{15 ^{2}} = \frac{1}{225 }$

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Question

$\sqrt{y}-\frac{1}{3} = \frac{2}{5}$

Evaluate .

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Answer

To solve for the variable isolate it on one side of the equation with all of constants on the other side.

$\sqrt{y}-\frac{1}{3} = \frac{2}{5}$

First add one third to both sides.

$\sqrt{y}-\frac{1}{3} + \frac{1}{3} = \frac{2}{5} + \frac{1}{3}$

$\sqrt{y} = \frac{2}{5} + \frac{1}{3}$

Calculate a common denominator to add the two fractions.

$\sqrt{y} = \frac{3 \cdot 2}{3 \cdot 5} + \frac{1 \cdot 5}{3 \cdot 5}$

$\sqrt{y} = \frac{6}{1 5} + \frac{ 5}{15}$

$\sqrt{y} = \frac{6+ 5}{15}$

$\sqrt{y} = \frac{11}{15}$

Square both sides to solve for y.

$y = (\sqrt{y}) ^{2}= \left (\frac{11}{15} \right )^{2} = \frac{11^{2}}{15 ^{2}} = \frac{121}{225 }$

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Question

$t ^{3} = -8$

$u^{9} = -12$

Evaluate $(tu) ^{9}$ .

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Answer

By the Power of a Product Principle,

$(tu) ^{9} = t ^{9} \cdot u ^{9}$

Also, by the Power of a Power Principle,

$(t^{3} ) ^{3} =t ^{3 \cdot 3} = t^{9}$

Combining these ideas, then substituting:

$(tu) ^{9} = (t^{3} ) ^{3} \cdot u ^{9}$

$= (-8) ^{3} \cdot (-12)$

$= (-8^{3} ) \cdot (-12)$

$= (-512 ) \cdot (-12)$

$= +(512 \cdot 12)$

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Question

$a^{3} = -10$

Evaluate $(2a) ^{6}$ .

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Answer

By the Power of a Power Principle,

$(a^{3} ) ^{2} =a ^{3 \cdot 2} = a ^{6}$

So

$a ^{6} = (a^{3} ) ^{2} = (-10) ^{2} =+ (10 ^{2} )= 100$

Also, by the Power of a Product Principle,

$(2a) ^{6} = 2^{6} \cdot a ^{6}$

$2 ^{6} = 64$ , so, substituting,

$(2a) ^{6} = 64 ( 100 ) = 6,400$ .

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