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ISEE Upper Level Math : Variables and Exponents

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All ISEE Upper Level Math Resources

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Example Questions

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Example Question #81 : Variables

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

Evaluate .

Possible Answers:

$\frac{121}{225 }$

$\frac{1}{4}$

$\frac{1}{225 }$

$\frac{9}{64}$

Correct answer:

$\frac{1}{225 }$

Explanation:

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

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

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

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

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

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

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

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

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Example Question #82 : Variables

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

Evaluate .

Possible Answers:

$\frac{1}{4}$

$\frac{9}{64}$

$\frac{1}{225 }$

$\frac{121}{225 }$

Correct answer:

$\frac{121}{225 }$

Explanation:

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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Example Question #83 : Variables

Simplify the following:

(7t^6h^4b^9)^3

Possible Answers:

$343t^{18}h^{12}b^{27}$

$21t^{9}h^{7}b^{12}$

$343t^{9}h^{7}b^{12}$

$11t^{9}h^{7}b^{12}$

Correct answer:

$343t^{18}h^{12}b^{27}$

Explanation:

Simplify the following:

(7t^6h^4b^9)^3

Let's recall the rules for distributing exponents.

We treat coefficients (like the 7) like regular numbers and raise them to the new exponent.

We deal with variables (like the t, h, and b) by multiplying their current exponent by the new exponent.

Doing so yields:

$7^3t^{6*3}h^{3*4}b^{3*9}$

Simplify to get:

$343t^{18}h^{12}b^{27}$

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Example Question #81 : Variables

$a^{3} = -10$

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

Possible Answers:

-6,400

6,400

200

Correct answer:

6,400

Explanation:

By the Power of a Power Principle,

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

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

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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Example Question #85 : Variables

$t ^{3} = -8$

$u^{9} = -12$

Evaluate $(tu) ^{9}$ .

Possible Answers:

6,144

288

-6,144

Correct answer:

6,144

Explanation:

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:

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

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

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

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

$= +(512 \cdot 12)$

=6,144

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Example Question #1 : How To Subtract Exponential Variables

Simplify the expression:

$7x^{3} - 5x^{2} + 9x^{3}$

Possible Answers:

$16 x^{3} - 5x^{2}$

$2x^{2} + 9x^{3}$

$-2x^{3} - 5x^{2}$

The expression cannot be simplified further.

$11x^{5}$

Correct answer:

$16 x^{3} - 5x^{2}$

Explanation:

Group, then collect like terms:

$7x^{3} - 5x^{2} + 9x^{3}$

$=7x^{3} + 9x^{3} - 5x^{2}$

$=\left ( 7 + 9 \right )x^{3} - 5x^{2}$

$=16x^{3} - 5x^{2}$

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Example Question #2 : How To Subtract Exponential Variables

Simplify:

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

Possible Answers:

-8xy

8xy

$\frac{2x^{2}+8y^{2}}{xy}$

Correct answer:

Explanation:

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

$=\frac{(x^{2}+4xy+4y^{2})-(x^{2}-4xy+4y^{2})}{xy}$

$=\frac{x^{2}-x^{2}+4xy+4xy+4y^{2}-4y^{2}}{xy}$

$=\frac{8xy}{xy} = 8$

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Example Question #3 : How To Subtract Exponential Variables

Assume that $x \neq 0$ . Which of the following expressions is equivalent to:

$\frac{\left (x+1 \right ) ^{3} - \left ( x^{3 } + 1\right ) }{x^{2}}$ ?

Possible Answers:

$\frac{ 3 }{x} + \frac{ 3 }{x^{2}}$

$3 + \frac{ 3 }{x}$

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

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

Correct answer:

$3 + \frac{ 3 }{x}$

Explanation:

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

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

$= \frac{x^{3} -x^{3} +3x^{2} + 3x + 1-1}{x^{2}}$

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

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

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

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Example Question #1 : How To Subtract Exponential Variables

If $x\neq 0$ , simplify:

$\frac{(x-1)^2-(x^2-1)}{x-1}$

Possible Answers:

x-1

-2(x-1)

2(x-1)

Correct answer:

Explanation:

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

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

$=\frac{2(1-x)}{x-1}$

$=\frac{2(1-x)}{-1(1-x)}$

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

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Example Question #1 : How To Subtract Exponential Variables

If $x,y\neq 0$ , simplify:

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

Possible Answers:

4xy

Correct answer:

Explanation:

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

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

$=\frac{4xy}{xy}=4$

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ISEE Upper Level Math : Variables and Exponents

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

All ISEE Upper Level Math Resources

Example Questions

Example Question #81 : Variables

Example Question #82 : Variables

Example Question #83 : Variables

Example Question #81 : Variables

Example Question #85 : Variables

Example Question #1 : How To Subtract Exponential Variables

Example Question #2 : How To Subtract Exponential Variables

Example Question #3 : How To Subtract Exponential Variables

Example Question #1 : How To Subtract Exponential Variables

Example Question #1 : How To Subtract Exponential Variables

All ISEE Upper Level Math Resources

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