Variables and Exponents - ISEE Upper Level (grades 9-12) Quantitative Reasoning

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Question

Simplify:

$5x^{2} y ^{2}+7 + 6x^{2}y^{2} + 11 xy$

Answer

Group and combine like terms $5x^{2} y ^{2},6x^{2}y^{2}$ :

$5x^{2} y ^{2}+7 + 6x^{2}y^{2} + 11 xy$

$= 5x^{2} y ^{2}+ 6x^{2}y^{2}+ 11 xy +7$

$=\left ( 5+ 6 \right )x^{2}y^{2}+ 11 xy +7$

$=11x^{2}y^{2}+ 11 xy +7$

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Question

Assume that and are not both zero. Which is the greater quantity?

(a) $\frac{(x+y)^{2} + (x - y)^{2}}{x^{2}+y^{2}}$

(b)

Answer

Simplify the expression in (a):

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

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

$=\frac{ x^{2}+x^{2} +2xy -2xy+ y^{2}+ y^{2}}{x^{2}+y^{2}}$

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

Therefore, whether (a) or (b) is greater depends on the values of and , neither of which are known.

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Question

Which is the greater quantity?

(a) $(x + y)^{2}$

(b) $x^{2}+ y^{2}$

Answer

$(x + y)^{2} = x^{2}+2xy + y^{2}$

Since and have different signs,

, and, subsequently,

Therefore,

$(x + y)^{2} = x^{2}+2xy + y^{2} < x^{2} + y^{2}$

This makes (b) the greater quantity.

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Question

Which is the greater quantity?

(a) $x^{2} + 3x$

(b)

Answer

We give at least one positive value of for which (a) is greater and at least one positive value of for which (b) is greater.

Case 1:

(a) $x^{2} + 3x = 2^{2} + 3 \cdot 2 = 4 + 6 = 10$

(b) $4x = 4 \cdot 2 = 8$

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

(a) $x^{2} + 3x = \left ( \frac{1}{2} \right ) ^{2} + 3 \cdot \frac{1}{2} = \frac{1}{4} + \frac{3}{2} = \frac{7}{4}= 1 \frac{3}{4}$

(b) $4x = 4 \cdot \frac{1}{2} = 2$

Therefore, either (a) or (b) can be greater.

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Question

Assume all variables to be nonzero.

Simplify: $\left (12x^{5}y ^{4}z^{3} \right )^{0} + \left (3x^{5}y ^{4}z^{3} \right )^{0}$

Answer

Any nonzero expression raised to the power of 0 is equal to 1. Therefore,

$\left (12x^{5}y ^{4}z^{3} \right )^{0} + \left (3x^{5}y ^{4}z^{3} \right )^{0} = 1 + 1 = 2$ .

None of the given expressions are correct.

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Question

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

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

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:

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

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

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 $x^{8}$

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:

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

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:

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

Answer

$\left (\frac{x^{3}}{2y^{4}} \right ) ^{-3} = \left (\frac{2y^{4}}{x^{3}} \right ) ^{3} = \frac{2^{3} \left (y^{4}\right ) ^{3} }{\left (x^{3} \right )^{3}}= \frac{8 y^{4 \cdot 3} }{x^{3\cdot 3} }= \frac{8 y^{12} }{x^{9} }$

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Question

Which is greater?

(a) $y ^{-2}$

(b) $y^{2}$

Answer

If , then $y ^{2}> 1^{2} = 1$ and $y ^{-2}= \frac{1}{y ^{2}} < \frac{1}{1} = 1$

$y ^{2}> 1> y ^{-2}$ , so by transitivity, $y ^{2}> y ^{-2}$ , and (b) is greater

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Question

Which is greater?

(a) $\left (-2 \right )^{-3}$

(b) $\left ( \frac{1}{2} \right ) ^{3}$

Answer

$\left (-2 \right )^{-3} = \left ( -\frac{1}{2} \right )^{3}$

A negative number to an odd power is negative, so the expression in (a) is negative. The expression in (b) is positive since the base is positive. (b) is greater.

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Question

Expand: $(x + 1)^{10}$

Which is the greater quantity?

(a) The coefficient of $x^{3}$

(b) The coefficient of $x^{7}$

Answer

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

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

(a) Substitute : The coerfficient of $x^{3}$ is

$_{10}\textrm{C} _{3} = \frac{10!}{(10-3)!3!} = \frac{10!}{7!3!}$ .

(b) Substitute : The coerfficient of $x^{7}$ is

$_{10}\textrm{C} _{7} = \frac{10!}{(10-7)!7!} = \frac{10!}{3!7!} = \frac{10!}{7!3!}$ .

The two are equal.

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Question

Expand: $(x + 2)^{10}$

Which is the greater quantity?

(a) The coefficient of

(b) The coefficient of $x^{2}$

Answer

Using the Binomial Theorem, if $(x + 2)^{n}$ is expanded, the $x^{r}$ term is

$_{n}\textrm{C} _{r} \cdot x^{r} \cdot 2 ^{n-r}$

$= \frac{n!}{(n-r)!r!} \cdot 2 ^{n-r} \cdot x^{r}$ .

This makes $\frac{n!}{(n-r)!r!} \cdot 2 ^{n-r}$ the coefficient of $x^{r}$ .

We compare the values of this expression at for both and .

(a) If and , the coefficient is

$\frac{10!}{(10-1)!1!} \cdot 2 ^{10-1} = \frac{10!}{9!} \cdot 2 ^{9} =10 \cdot 512 =5,120$ .

This is the coefficient of .

(b) If and , the coefficient is

$\frac{10!}{(10-2)!2!} \cdot 2 ^{10-2} = \frac{10!}{8!2!} \cdot 2 ^{8} = \frac{ 9 \cdot 10}{2} \cdot 256 =11,520$ .

This is the coefficient of $x^{2}$ .

(b) is the greater quantity.

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Question

Which is the greater quantity?

(a) $(x + 5)^{3}- (x - 5)^{3}$

(b)

Answer

Simplify the expression in (a):

$(x + 5)^{3}- (x - 5)^{3}$

$=(x^{3}+ 3 \cdot x^{2} \cdot 5 + 3 \cdot x \cdot 5^{2}+ 5^{3}) -(x^{3}- 3 \cdotx^{2} \cdot 5 + 3 \cdot x \cdot 5^{2}- 5^{3})$

$=(x^{3}+15 x^{2}+75 x + 125) - (x^{3}-15 x^{2}+75 x- 125)$

$=x^{3}-x^{3}+15 x^{2}+15 x^{2}+75 x-75 x + 125+125$

$=30 x^{2} + 250$

Since ,

$(x + 5)^{3}- (x - 5)^{3}=30 x^{2} + 250 > 250$ ,

making (a) greater.

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Question

Consider the expression $\left (x^{2} y \right )^{3} \left (x y ^{2} \right )^{3}$

Which is the greater quantity?

(a) The expression evaluated at

(b) The expression evaluated at

Answer

Use the properties of powers to simplify the expression:

$\left (x^{2} y \right )^{3} \left (x y ^{2} \right )^{3}$

$= \left (x^{2} \right )^{3} y ^{3} \cdot x ^{3}\left ( y ^{2} \right )^{3}$

$= \left (x^{2} \right )^{3} \cdot x ^{3}\cdot y ^{3} \cdot \left ( y ^{2} \right )^{3}$

$= \left (x^{2 \cdot 3} \right ) \cdot x ^{3}\cdot y ^{3} \cdot \left ( y ^{2\cdot 3} \right )$

$= x^{6} \cdot x ^{3}\cdot y ^{3} \cdot y ^{6}$

$= x^{6+3} \cdot y ^{3+ 6}$

$= x^{9} y ^{9} = (xy)^{9}$

(a) If , then

$\left (x^{2} y \right )^{3} \left (x y ^{2} \right )^{3}= (xy)^{9} = ( 6 \cdot 10)^{9} = 60^{9}$

(b) If , then

$\left (x^{2} y \right )^{3} \left (x y ^{2} \right )^{3}= (xy)^{9} = ( 8 \cdot 8)^{9} = 64^{9}$

(b) is greater.

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Question

Which of the following expressions is equivalent to

$\left (100 + \sqrt{x} \right )^{2}$ ?

Answer

Use the square of a binomial pattern as follows:

$\left (100 + \sqrt{x} \right )^{2}$

$= 100^{2}+ 2 \cdot 100 \cdot \sqrt {x} +(\sqrt{x})^{2}$

$= 10,000+ 2 00 \sqrt {x} +x$

This expression is not equivalent to any of the choices.

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Question

$y = (x+ 3)^{2}$

Express in terms of .

Answer

$y = (x+ 3)^{2}$

$= x^{2}+ 2 \cdot 3 \cdot x + 3^{2}$

$= x^{2}+ 6 x + 9$

, so

$16z = x^{2}+ 6 x + 9$

$\frac{16z}{16} = \frac{x^{2}+ 6 x + 9}{16}$

$z = \frac{1}{16}x^{2}+ \frac{3}{8}x+ \frac{9}{16}$

, so

$= \frac{1}{16}x^{2}+ \frac{3}{8}x+ \frac{9}{16}+6$

$= \frac{1}{16}x^{2}+ \frac{3}{8}x+ \frac{9}{16} + \frac{96}{16}$

$= \frac{1}{16}x^{2}+ \frac{3}{8}x+ \frac{105}{16}$

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