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ISEE Upper Level Quantitative : ISEE Upper Level (grades 9-12) Quantitative Reasoning

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 #165 : How To Find The Solution To An Equation

Define f(x) = 6x .

$(g \circ f) \left ( \frac{1}{2} \right ) = 6$

Which is the greater quantity?

(a) f(3)

(b) g(3)

Possible Answers:

It is impossible to determine which is greater from the information given

(b) is the greater quantity

(a) and (b) are equal

(a) is the greater quantity

Correct answer:

(a) is the greater quantity

Explanation:

f(x) = 6x , so, by substitution,

$f \left ( \frac{1}{2} \right ) = 6 \cdot \frac{1}{2} = 3$ .

By way of the definition of a composition of functions,

$(g \circ f) \left ( \frac{1}{2} \right ) = g \left [ f \left ( \frac{1}{2} \right ) \right ] = g(3 )$ .

Since $(g \circ f) \left ( \frac{1}{2} \right ) =6$ , it follows that g(3) = 6 .

Also, by substitution,

$f(3 )= 6 \cdot 3 = 18$ .

Therefore, f(3 )> g(3) .

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Example Question #166 : How To Find The Solution To An Equation

Solve for :

$x\left (3x+7 \right ) =20$

Possible Answers:

$x=-5 \textrm{ or }x = 1\frac{1}{3}$

$x=-4 \textrm{ or }x = 1\frac{2}{3}$

$x=-1\frac{2}{3} \textrm{ or }x = 4$

$x=4 \frac{1}{3} \textrm{ or } x = 20$

$x=-1\frac{1}{3} \textrm{ or }x = 5$

Correct answer:

$x=-4 \textrm{ or }x = 1\frac{2}{3}$

Explanation:

First, rewrite the quadratic equation in standard form by distributing the through the product on the left, then collecting all of the terms on the left side:

$x\left (3x+7 \right ) =20$

$x\left \cdot \; 3x+x\left \cdot\; 7 =20$

$3x ^{2}+7x =20$

$3x ^{2}+7x -20=20 -20$

$3x ^{2}+7x -20=0$

Use the method to factor the quadratic expression $3x ^{2}+7x -20$ ; we are looking to split the linear term by finding two integers whose sum is 7 and whose product is 3 (-20) = -60 . These integers are -5,12 , so:

$\left (3x ^{2}-5x \right ) + \left ( 12x -20 \right ) =0$

$x \left (3x-5 \right ) +4 \left ( 3x -5 \right ) =0$

$\left (x +4 \right )\left (3x-5 \right )=0$

Set each expression equal to 0 and solve:

x+4 = 0

x=-4

3x-5=0

3x=5

$x = \frac{5}{3} = 1\frac{2}{3}$

The solution set is $\left \{ -4, 1\frac{2}{3}\right \}$ .

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Example Question #1 : Variables

Multiply:

(0.6x-0.4y)(0.6x+0.4y)

Possible Answers:

$0.36x^{2}-0.48xy- 0.16y^{2}$

$0.36x^{2}-0.24xy+ 0.16y^{2}$

$0.36x^{2}- 0.16y^{2}$

$0.36x^{2}+0.16y^{2}$

$0.36x^{2}-0.48xy+ 0.16y^{2}$

Correct answer:

$0.36x^{2}- 0.16y^{2}$

Explanation:

This can be solved using the pattern for the sum times difference of two terms:

(0.6x-0.4y)(0.6x+0.4y)

$=\left ( 0.6x \right )^{2} -\left ( 0.4y \right ) ^{2}$

$=0.6^{2}x^{2}- 0.4^{2}y^{2}$

$=0.36x^{2}- 0.16y^{2}$

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Example Question #1 : Operations

Multiply:

$\left (\frac{4}{3}x + \frac{5}{3} \right )\left (\frac{4}{3}x - \frac{5}{3} \right )$

Possible Answers:

$\frac{16}{9} x ^{2} - \frac{40}{9} x- \frac{25}{9}$

$\frac{16}{9} x ^{2} - \frac{40}{9} x+ \frac{25}{9}$

$\frac{16}{9} x ^{2} - \frac{20}{9} x- \frac{25}{9}$

$\frac{16}{9} x ^{2}- \frac{25}{9}$

$\frac{16}{9} x ^{2}+ \frac{25}{9}$

Correct answer:

$\frac{16}{9} x ^{2}- \frac{25}{9}$

Explanation:

This can be solved using the pattern for the sum times difference of two terms:

$\left (\frac{4}{3}x + \frac{5}{3} \right )\left (\frac{4}{3}x - \frac{5}{3} \right )$

$=\left (\frac{4}{3}x \right ) ^{2}- \left (\frac{5}{3} \right )^{2}$

$=\frac{4 ^{2}}{3 ^{2}}\cdot x ^{2}- \frac{5^{2}}{3^{2}}$

$=\frac{16}{9} x ^{2}- \frac{25}{9}$

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Example Question #3 : Variables

Simplify:

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

Possible Answers:

$\frac{16}{9}x^{2} - \frac{20}{9}x +\frac{25}{9}$

$\frac{16}{9}x^{2} - \frac{40}{9}x -\frac{25}{9}$

$\frac{16}{9}x^{2} - \frac{20}{9}x -\frac{25}{9}$

$\frac{16}{9}x^{2} - \frac{25}{9}$

$\frac{16}{9}x^{2} - \frac{40}{9}x +\frac{25}{9}$

Correct answer:

$\frac{16}{9}x^{2} - \frac{40}{9}x +\frac{25}{9}$

Explanation:

This can be solved using the pattern for the square of a binomial:

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

$=\left (\frac{4}{3}x \right ) ^{2} -2 \cdot \frac{4}{3}x \cdot \frac{5}{3} + \left ( \frac{5}{3} \right ) ^{2}$

$=\frac{4^{2}}{3^{2}}\cdot x^{2} -2 \cdot \frac{4}{3}\cdot \frac{5}{3} \cdot x +\frac{5^{2}}{3^{2}}$

$=\frac{16}{9}x^{2} - \frac{40}{9}x +\frac{25}{9}$

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Example Question #2 : How To Multiply Variables

N is a positive integer; N > 1 . Which is greater?

(a) $N^{2} - 2N + 1$

(b) 0

Possible Answers:

It is impossible to tell from the information given

(b) is greater

(a) and (b) are equal

(a) is greater

Correct answer:

(a) is greater

Explanation:

$N^{2} - 2N + 1 = (N - 1)^{2}$ . Since N > 1 , N - 1 > 0 , and $N^{2} - 2N + 1 = (N - 1)^{2} > 0$ . This makes (a) greater.

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Example Question #3 : How To Multiply Variables

N is a positive integer; N > 1 . Which is greater?

(a) $N^{2} - 4N + 4$

(b)

Possible Answers:

(a) and (b) are equal.

(a) is greater.

(b) is greater.

It is impossible to tell from the information given.

Correct answer:

It is impossible to tell from the information given.

Explanation:

If N = 2 , then $N^{2} - 4N + 4 = 2 ^{2} - 4 \cdot 2 + 4= 4 - 8 + 4 = 0$ .

If N = 3 , then $N^{2} - 4N + 4 = 3 ^{2} - 4 \cdot 3 + 4= 9 - 12 + 4 = 1$ .

Therefore, at least two possibilities can be demonstrated.

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Example Question #2 : Operations

Which is the greater quantity?

(a) $-8 (x^{2}-x-6 )$

(b) $-8 x^{2}+8x-48$

Possible Answers:

It is impossible to tell from the information given.

(a) is greater.

(b) is greater.

(a) and (b) are equal.

Correct answer:

(a) is greater.

Explanation:

$-8 (x^{2}-x-6 )$

$=( -8 ) x^{2}-( -8 )x-( -8 )6$

$=-8 x^{2}-( -8x )-( -48 )$

$=-8 x^{2}+ 8x + 48$

Since 48 > -48 ,

$-8 x^{2}+ 8x + 48 > -8 x^{2}+ 8x - 48$ .

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Example Question #5 : Operations

Assume is nonzero. Which expression is not equivalent to $16 y^{4}$ ?

Possible Answers:

$8 y^{2} \cdot 2y^{2}$

$\frac{32y^{8}}{2y^{4}}$

All four expressions given in the other choices are equivalent to $16 y^{4}$ .

$\left ( 4y^{4}\right )^{4}$

$4y^{4} + 4y^{4} +4y^{4} + 4y^{4}$

Correct answer:

$\left ( 4y^{4}\right )^{4}$

Explanation:

Each expression can be simplified using the properties of exponents, among others:

$\frac{32y^{8}}{2y^{4}} = \frac{32 }{2 } \cdot \frac{ y^{8}}{ y^{4}} = 16 y^{8-4} =16y^{4}$

$8 y^{2} \cdot 2y^{2} = 8 \cdot 2 \cdot y^{2} \cdot y^{2} = 16 y ^{2+2}= 16 y ^{4}$

$4y^{4} + 4y^{4} +4y^{4} + 4y^{4} = (4+4+4+4)y^{4} = 16y^{4}$

$\left ( 4y^{4}\right )^{4} = \left ( 4 \right )^{4} \cdot \left ( y^{4}\right )^{4} = 256y^{4 \cdot 4} = 256y^{16}$

The correct choice is $\left ( 4y^{4}\right )^{4}$ , as it is the only choice not equivalent to $16 y^{4}$ .

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Example Question #5 : How To Multiply Variables

Let be negative. Which of the following is the greater quantity?

(A) -4y

(B) -5y

Possible Answers:

(A) and (B) are equal

(B) is greater

It is impossible to determine which is greater from the information given

(A) is greater

Correct answer:

(B) is greater

Explanation:

-4 > -5 ,

So by the multiplication property of inequality, when each is multiplied by the negative number ,

-4y < -5y

and (B) is greater.

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All ISEE Upper Level Quantitative Resources

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ISEE Upper Level Quantitative : ISEE Upper Level (grades 9-12) Quantitative Reasoning

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

All ISEE Upper Level Quantitative Resources

Example Questions

Example Question #165 : How To Find The Solution To An Equation

Example Question #166 : How To Find The Solution To An Equation

Example Question #1 : Variables

Example Question #1 : Operations

Example Question #3 : Variables

Example Question #2 : How To Multiply Variables

Example Question #3 : How To Multiply Variables

Example Question #2 : Operations

Example Question #5 : Operations

Example Question #5 : How To Multiply Variables

All ISEE Upper Level Quantitative Resources

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