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ISEE Upper Level Quantitative : 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 #241 : Algebraic Concepts

Simplify:

$\left (4x \right )^{4} +\left ( 2y \right )^{2}$

Possible Answers:

$1,024x^{4} y ^{2}$

$64x^{4} y ^{2}$

$16x^{4} +4y ^{2}$

$4x^{4} +2y ^{2}$

$256x^{4} +4y ^{2}$

Correct answer:

$256x^{4} +4y ^{2}$

Explanation:

$\left (4x \right )^{4} +\left ( 2y \right )^{2} =4^{4}x^{4} + 2^{2}y^{2}=256x^{4} +4y ^{2}$

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Example Question #244 : Algebraic Concepts

Expand: $(x-5)^{3}$

Possible Answers:

$x^{3}-25x^{2}+50x-125$

$x^{3}-125$

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

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

$x^{3}+25x^{2}-50x-125$

Correct answer:

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

Explanation:

A binomial can be cubed using the pattern:

$(A-B)^{3} = A^{3} - 3A^{2}B+ 3AB^{2} - B^{3}$

Set A = x, B = 5

$(x-5)^{3}$

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

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

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Example Question #245 : Algebraic Concepts

Factor completely:

$3x^{2} - x -14$

Possible Answers:

$\left ( x-7 \right ) \left (3x+2 \right )$

$\left ( x-2 \right ) \left (3x+ 7 \right )$

$\left ( x+2 \right ) \left (3x- 7 \right )$

$\left ( x-2 \right ) \left (3x- 7 \right )$

$\left ( x+7 \right ) \left (3x-2 \right )$

Correct answer:

$\left ( x+2 \right ) \left (3x- 7 \right )$

Explanation:

A trinomial whose leading term has a coefficent other than 1 can be factored using the -method. We split the middle term using two numbers whose product is 3 (-14) = -42 and whose sum is . These numbers are -7,6 , so:

$3x^{2} - x -14$

$= 3x^{2} - 7x +6x -14$

$= \left (3x^{2} - 7x \right ) +\left ( 6x -14 \right )$

$=x \left (3x- 7 \right ) +2 \left (3x- 7 \right )$

$=\left ( x+2 \right ) \left (3x- 7 \right )$

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Example Question #246 : Algebraic Concepts

Multiply:

$\left ( x + y + 3\right ) \left ( x + y - 3\right )$

Possible Answers:

$x^{2} - 9 xy + y^{2}$

$x^{2} - 2xy + y^{2} - 9$

$x^{2} + 2xy + y^{2} - 9$

$x^{2} + y^{2} - 9$

$x^{2} + xy + y^{2} - 9$

Correct answer:

$x^{2} + 2xy + y^{2} - 9$

Explanation:

This can be achieved by using the pattern of difference of squares:

$\left ( x + y + 3 \right ) \left ( x + y - 3\right )$

$=\left [\left ( x + y \right ) + 3 \right ] \left [\left ( x + y \right ) - 3 \right ]$

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

$= ( x + y ) ^{2} - 9$

Applying the binomial square pattern:

$= x^{2} + 2xy + y^{2} - 9$

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Example Question #251 : Algebraic Concepts

Simplify:

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

Possible Answers:

$x ^{3 } + 3 x^{2} +x +\frac{1}{27}$

$x ^{3 } + x^{2} + \frac{1}{3} x +\frac{1}{27}$

$x ^{3 } + \frac{1}{3} x^{2} + \frac{1}{9} x +\frac{1}{27}$

$x ^{3 } + 2 x^{2} + \frac{2}{3} x +\frac{1}{27}$

$x ^{3 } + \frac{2}{3} x^{2} + \frac{2}{9} x +\frac{1}{27}$

Correct answer:

$x ^{3 } + x^{2} + \frac{1}{3} x +\frac{1}{27}$

Explanation:

The cube of a sum pattern can be applied here:

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

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

$= x ^{3 } + x^{2} +3\cdot x \cdot \frac{1}{9} +\frac{1}{27}$

$= x ^{3 } + x^{2} + \frac{1}{3} x +\frac{1}{27}$

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Example Question #252 : Algebraic Concepts

Fill in the box to form a perfect square trinomial:

$x ^{2} - 13x + \square$

Possible Answers:

$\frac{169}{4}$

169

$\frac{169}{2}$

Correct answer:

$\frac{169}{4}$

Explanation:

To obtain the constant term of a perfect square trinomial, divide the linear coefficient, which here is , by 2, and square the quotient. The result is

$\left (\frac{ -13}{2} \right) ^{2} = \frac{(-13)^{2}}{2^{2}}= \frac{169}{4}$

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Example Question #1 : How To Multiply Exponential Variables

Fill in the box to form a perfect square trinomial:

$x^{2} - 18x + \square$

Possible Answers:

144

324

162

Correct answer:

Explanation:

To obtain the constant term of a perfect square trinomial, divide the linear coefficient, which here is , by 2, and square the quotient. The result is

$\left (\frac{-18}{2} \right )^{2} =\left ( -9\right ) ^{2} = 81$

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Example Question #8 : How To Multiply Exponential Variables

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

Which is the greater quantity?

(a) The coefficient of $x^{4}$

(b) The coefficient of $x^{6}$

Possible Answers:

(b) is greater.

(a) is greater.

It is impossible to tell from the information given.

(a) and (b) are equal.

Correct answer:

(a) is greater.

Explanation:

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 n = 10 for both r = 4 and r = 6 :

(a) $\frac{10!}{(10-4)!4!} \cdot 2 ^{10-4} = \frac{10!}{6!4!} \cdot 2 ^{6}$

(b) $\frac{10!}{(10-6)!6!} \cdot 2 ^{10-6} = \frac{10!}{4!6!} \cdot 2 ^{4} = \frac{10!}{6!4!} \cdot 2 ^{4}$

(a) is the greater quantity.

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Example Question #9 : How To Multiply Exponential Variables

x > 3

Which is the greater quantity?

(a) $16x^{2} - 72x + 81$

(b) 8

Possible Answers:

(a) and (b) are equal

It is impossible to tell from the information given

(b) is greater

(a) is greater

Correct answer:

(a) is greater

Explanation:

$16x^{2} - 72x + 81$

$= 4^{2} x^{2} -2\cdot 4x \cdot 9 + 9^{2}$

$=\left ( 4 x \right ) ^{2} - 2 \cdot 4x \cdot 9 + 9^{2}$

$= \left ( 4x -9 \right )^{2}$

Since x > 3 ,

$4x - 9 > 4 \cdot3 - 9 = 3$ , so

$16x^{2} - 72x + 81= \left ( 4x -9 \right )^{2} > 3^{2} = 9 >8$

making (a) greater.

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Example Question #1 : How To Multiply Exponential Variables

Which is the greater quantity?

(a) $y ^{2} + 4y + 4$

(b) $y^{2} + 6y + 9$

Possible Answers:

(a) and (b) are equal.

(b) is greater.

It is impossible to tell from the information given.

(a) is greater.

Correct answer:

It is impossible to tell from the information given.

Explanation:

We show that either polynomial can be greater by giving two cases:

Case 1: y = 0

$y ^{2} + 4y + 4 = 0 ^{2} + 4 \cdot 0 + 4 = 4$

$y^{2} + 6y + 9 = 0^{2} + 6 \cdot 0 + 9 = 9$

$y^{2} + 6y + 9 > y ^{2} + 4y + 4$

Case 2: y = -3

$y ^{2} + 4y + 4 = (-3) ^{2} + 4 \cdot (-3) + 4 = 9-12 + 4 = 1$

$y^{2} + 6y + 9 = (-3)^{2} + 6 \cdot (-3) + 9 = 9 - 18 + 9 = 0$

$y ^{2} + 4y + 4 < y^{2} + 6y + 9$

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ISEE Upper Level Quantitative : Variables

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

All ISEE Upper Level Quantitative Resources

Example Questions

Example Question #241 : Algebraic Concepts

Example Question #244 : Algebraic Concepts

Example Question #245 : Algebraic Concepts

Example Question #246 : Algebraic Concepts

Example Question #251 : Algebraic Concepts

Example Question #252 : Algebraic Concepts

Example Question #1 : How To Multiply Exponential Variables

Example Question #8 : How To Multiply Exponential Variables

Example Question #9 : How To Multiply Exponential Variables

Example Question #1 : How To Multiply Exponential Variables

All ISEE Upper Level Quantitative Resources

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