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ISEE Upper Level Quantitative : Algebraic Concepts

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 #161 : How To Find The Solution To An Equation

, , and all stand for negative quantities.

$t = \frac{1}{4} u -6$

$t = \frac{1}{6} v - 4$

Which is the greater quantity?

(a)

(b)

Possible Answers:

(a) is the greater quantity

(a) and (b) are equal

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

(b) is the greater quantity

Correct answer:

(a) is the greater quantity

Explanation:

Solve the first equation for in terms of , using the properties of equality to isolate the :

$\frac{1}{4} u -6 = t$

$\frac{1}{4} u -6 + 6 = t + 6$

$\frac{1}{4} u = t+ 6$

$4 \cdot \frac{1}{4} u =4 \cdot ( t+ 6)$

$u = 4 \cdot t+4 \cdot 6$

u = 4t+24

Solve for in the second equation similarly:

$\frac{1}{6} v - 4 = t$

$\frac{1}{6} v - 4 + 4 = t + 4$

$\frac{1}{6} v = t + 4$

$6 \cdot \frac{1}{6} v =6 \cdot ( t + 4)$

$v = 6t + 6 \cdot 4$

v = 6t + 24

t < 0 , so by the properties of inequality,

4 < 6

4t> 6t

4t + 24 > 6t + 24

u > v

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Example Question #162 : How To Find The Solution To An Equation

Define f(x) = (x+ 1) (x-1) and g(x) = 2x + 1 .

(f+g) (c) = 8

Which is the greater quantity?

(a)

(b) 2

Possible Answers:

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

(a) and (b) are equal

(a) is the greater quantity

(b) is the greater quantity

Correct answer:

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

Explanation:

The definition f(x) can be rewritten by noting that this is the product of a sum and a difference of the same two terms, and that the product is the difference of their squares:

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

By definition,

$(f+g) (x) = f(x)+ g(x) = (x^{2} - 1 ) + (2x+1) = x^{2} + 2x + 1 - 1 = x^{2} + 2x$

Since (f+g) (c) = 8 , it holds that

$c^{2} + 2c = 8$ , or

$c^{2} + 2c - 8 = 0$

We can factor the trinomial using two integers whose sum is 2 and whose product is ; by a little trial and error we find 4 and , so

(c-2) (c+4) = 0 .

By the Zero Product Principle,

c- 2 = 0 , in which case c= 2 ; or,

c + 4 = 0 , in which case c= -4 < 2 .

It is therefore unclear whether is less than or equal to 2.

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Example Question #163 : How To Find The Solution To An Equation

Define $f(x) = x^{2}$ .

(f+g)(5) = 60 .

Which is the greater quantity?

(a) f(5)

(b) g(5)

Possible Answers:

(a) and (b) are equal

(a) is the greater quantity

(b) is the greater quantity

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

Correct answer:

(b) is the greater quantity

Explanation:

$f(x) = x^{2}$ , so, setting x = 5 ,

$f(5) = 5^{2} = 25$

By definition,

(f+g) (5) = f(5)+ g(5)

so, by substitution,

60 = 25+ g(5)

60 -25 = 25+ g(5)- 25

g(5) = 35

Therefore, g(5) > f(5 ) .

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Example Question #164 : How To Find The Solution To An Equation

Define $f(x) = \left ( x+ 2\right )^{2}$ and $g(x) = x^{2}$ .

(f-g) (c) = 0

Which is the greater quantity?

(a) 0

(b)

Possible Answers:

(b) is the greater quantity

(a) and (b) are equal

(a) is the greater quantity

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

Correct answer:

(b) is the greater quantity

Explanation:

$f(x) = \left ( x+ 2\right )^{2}$ can be rewritten using the square of a binomial pattern:

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

By definition,

$(f-g)(x) = f(x) - g(x) = (x^{2}+4x + 4) - x^{2} = 4x + 4$

(f-g)(c) = 4c + 4

Since

(f-g)(c) = 0 , it holds that

4c + 4 = 0

Solving for :

4c + 4 - 4 = 0 - 4

4c= -4

$4c\div 4 = -4 \div 4$

c = -1 , which is less than 0.

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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 #161 : Algebraic Concepts

Solve for :

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

Possible Answers:

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

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

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

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

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

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 : Operations

Multiply:

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

Possible Answers:

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

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

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

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

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

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 : Variables

Multiply:

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

Possible Answers:

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

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

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

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

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

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 : Operations

Simplify:

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

Possible Answers:

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

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

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

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

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

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 : Operations

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

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

(b) 0

Possible Answers:

(a) is greater

(a) and (b) are equal

(b) is greater

It is impossible to tell from the information given

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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All ISEE Upper Level Quantitative Resources

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ISEE Upper Level Quantitative : Algebraic Concepts

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

All ISEE Upper Level Quantitative Resources

Example Questions

Example Question #161 : How To Find The Solution To An Equation

Example Question #162 : How To Find The Solution To An Equation

Example Question #163 : How To Find The Solution To An Equation

Example Question #164 : How To Find The Solution To An Equation

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

Example Question #161 : Algebraic Concepts

Example Question #1 : Operations

Example Question #1 : Variables

Example Question #3 : Operations

Example Question #2 : Operations

All ISEE Upper Level Quantitative Resources

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