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ISEE Upper Level Quantitative : How to add exponential variables

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

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All ISEE Upper Level Quantitative Resources

8 Diagnostic Tests 234 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #211 : Algebraic Concepts

Simplify:

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

Possible Answers:

The expression cannot be simplified further

$11x^{2}y^{2}+ 11 xy +7$ $\displaystyle 11x^{2}y^{2}+ 11 xy +7$

$22x^{3}y^{3}+7$ $\displaystyle 22x^{3}y^{3}+7$

$29x^{3}y^{3}$ $\displaystyle 29x^{3}y^{3}$

$30x^{2}y^{2}+ 11 xy +7$ $\displaystyle 30x^{2}y^{2}+ 11 xy +7$

Correct answer:

$11x^{2}y^{2}+ 11 xy +7$ $\displaystyle 11x^{2}y^{2}+ 11 xy +7$

Explanation:

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

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

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

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

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

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Example Question #51 : Variables

$\displaystyle x > 0, y < 0$

Which is the greater quantity?

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

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

Possible Answers:

It is impossible to tell from the information given.

(b) is greater.

(a) is greater.

(a) and (b) are equal.

Correct answer:

(b) is greater.

Explanation:

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

Since $\displaystyle x$ and $\displaystyle y$ have different signs,

xy<0 $\displaystyle xy< 0$, and, subsequently,

2xy < 0 $\displaystyle 2xy < 0$

Therefore,

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

This makes (b) the greater quantity.

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Example Question #1 : How To Add Exponential Variables

Assume that $\displaystyle x$ and $\displaystyle y$ are not both zero. Which is the greater quantity?

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

(b) 4xy $\displaystyle 4xy$

Possible Answers:

(a) and (b) are equal.

It is impossible to tell from the information given.

(a) is greater.

(b) is greater.

Correct answer:

It is impossible to tell from the information given.

Explanation:

Simplify the expression in (a):

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

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

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

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

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

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Example Question #1 : Variables And Exponents

x > 0 $\displaystyle x > 0$

Which is the greater quantity?

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

(b) $\displaystyle 4x$

Possible Answers:

It is impossible to tell from the information given

(b) is greater

(a) and (b) are equal

(a) is greater

Correct answer:

It is impossible to tell from the information given

Explanation:

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

Case 1: x = 2 $\displaystyle x = 2$

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

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

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

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

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

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

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Example Question #1 : Variables And Exponents

Assume all variables to be nonzero.

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

Possible Answers:

$15x^{10}y ^{8}z^{6}$ $\displaystyle 15x^{10}y ^{8}z^{6}$

None of the answer choices are correct.

$36x^{10}y ^{8}z^{6}$ $\displaystyle 36x^{10}y ^{8}z^{6}$

$36x^{5}y ^{4}z^{3}$ $\displaystyle 36x^{5}y ^{4}z^{3}$

$15x^{5}y ^{4}z^{3}$ $\displaystyle 15x^{5}y ^{4}z^{3}$

Correct answer:

None of the answer choices are correct.

Explanation:

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

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

None of the given expressions are correct.

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All ISEE Upper Level Quantitative Resources

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ISEE Upper Level Quantitative : How to add exponential variables

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

All ISEE Upper Level Quantitative Resources

Example Questions

Example Question #211 : Algebraic Concepts

Example Question #51 : Variables

Example Question #1 : How To Add Exponential Variables

Example Question #1 : Variables And Exponents

Example Question #1 : Variables And Exponents

All ISEE Upper Level Quantitative Resources

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