Algebra Practice Questions — GRE General Test — Page 2

Question 21Question →

The quadratic equation $x^2 + bx + 45 = 0$ , where $b$ is a constant, has two negative integer roots $p$ and $q$ such that $p < q$ . If $q - p = 4$ , what is the value of $b$ ?

Show answer & explanation

Answer: 14

Answer

The value of

b

is 14.

For the quadratic equation

x^2 + bx + 45 = 0

, the roots

p

and

q

must satisfy

pq = 45

and

p + q = -b

. The negative integer factor pairs of 45 with

p < q

are

(-45, -1)

,

(-15, -3)

, and

(-9, -5)

. Calculating the difference

q - p

for each pair yields 44, 12, and 4, respectively. The condition

q - p = 4

uniquely identifies the roots as

p = -9

and

q = -5

. Summing these roots gives

p + q = -14

, so

b = -(-14) = 14

.

Step-by-Step Solution

1

Set up the relationships for the roots of the quadratic equation.

pq = 45

and

p + q = -b

.

For any quadratic equation

x^2 + bx + c = 0

, the product of the roots equals

c

and the sum of the roots equals

-b

.

2

Find all negative integer factor pairs

(p, q)

of 45 such that

p < q

.

The possible pairs are

(-45, -1)

,

(-15, -3)

, and

(-9, -5)

.

Since both roots are negative integers, their product is positive 45.

3

Determine the difference

q - p

for each pair to match the given condition

q - p = 4

.

For

(-45, -1)

,

q - p = -1 - (-45) = 44

. For

(-15, -3)

,

q - p = -3 - (-15) = 12

. For

(-9, -5)

,

q - p = -5 - (-9) = 4

.

The pair

(-9, -5)

satisfies

q - p = 4

, establishing

p = -9

and

q = -5

.

4

Calculate the coefficient

b

.

b = -(p + q) = -(-9 + (-5)) = -(-14) = 14

.

Substituting the root values into

b = -(p + q)

yields the final answer.

Key Concept

Factoring Quadratics and Relationships Between Roots and Coefficients

Question 22Question →

If $x$ and $y$ are non-zero real numbers such that $9x^2 - 12xy + 4y^2 = 0$ , what is the value of $\frac{3x + y}{2y}$ ?

\frac{3}{2}

\frac{5}{6}

3

\frac{1}{2}

\frac{5}{2}

Show answer & explanation

Answer: $\frac{3}{2}$

Answer

The value of the expression is

\frac{3}{2}

.

The given expression

9x^2 - 12xy + 4y^2

factors into the perfect square

(3x - 2y)^2 = 0

. Setting the base equal to zero gives

3x = 2y

. Substituting

2y

in place of

3x

in the target expression yields

\frac{2y + y}{2y} = \frac{3y}{2y} = \frac{3}{2}

.

Step-by-Step Solution

1

Factor the quadratic expression

9x^2 - 12xy + 4y^2 = (3x - 2y)^2 = 0

The equation is a perfect square trinomial of the form

(a - b)^2 = a^2 - 2ab + b^2

where

a = 3x

and

b = 2y

.

2

Solve for the relationship between

x

and

y

3x - 2y = 0 \implies 3x = 2y

Taking the square root of both sides gives a linear relationship between

3x

and

2y

.

3

Substitute

3x = 2y

into the targeted expression

\frac{3x + y}{2y} = \frac{2y + y}{2y} = \frac{3y}{2y} = \frac{3}{2}

Replacing

3x

with

2y

eliminates

x

and simplifies the expression directly.

Key Concept

Perfect Square Trinomial Factoring

Question 23Question →

For all real numbers $x$ and $y$ such that $4x^2 + 6xy + 9y^2 \neq 0$ and $2x + 3y \neq 0$ , which of the following expressions are equivalent to $\frac{8x^3 - 27y^3}{4x^2 + 6xy + 9y^2}$ ? Select all such expressions.

Select all that apply

2x - 3y

-(3y - 2x)

\frac{4x^2 - 9y^2}{2x + 3y}

2x + 3y

(2x - 3y)^2

Show answer & explanation

Answer: $2x - 3y$ ; $-(3y - 2x)$ ; \frac{4x^2 - 9y^2}{2x + 3y}

Answer

The equivalent expressions are

2x - 3y

,

-(3y - 2x)

, and

\frac{4x^2 - 9y^2}{2x + 3y}

.

Factoring

8x^3 - 27y^3

as

(2x - 3y)(4x^2 + 6xy + 9y^2)

allows canceling the denominator, leaving

2x - 3y

. The expression

2x - 3y

is directly correct. Rearranging terms in

-(3y - 2x)

yields

-3y + 2x = 2x - 3y

, making it correct. Factoring

\frac{4x^2 - 9y^2}{2x + 3y}

as

\frac{(2x - 3y)(2x + 3y)}{2x + 3y}

simplifies to

2x - 3y

, making it also correct.

Step-by-Step Solution

1

Factor the numerator of the rational expression using the difference of cubes formula.

8x^3 - 27y^3 = (2x)^3 - (3y)^3 = (2x - 3y)((2x)^2 + (2x)(3y) + (3y)^2) = (2x - 3y)(4x^2 + 6xy + 9y^2)

Recognizing

8x^3

as

(2x)^3

and

27y^3

as

(3y)^3

enables full polynomial factoring.

2

Simplify the fraction by canceling the common non-zero quadratic factor.

\frac{(2x - 3y)(4x^2 + 6xy + 9y^2)}{4x^2 + 6xy + 9y^2} = 2x - 3y

Since

4x^2 + 6xy + 9y^2 \neq 0

, dividing common factors reduces the expression to linear form.

3

Evaluate each provided choice for algebraic equivalence to

2x - 3y

.

The expressions

2x - 3y

,

-(3y - 2x) = 2x - 3y

, and

\frac{4x^2 - 9y^2}{2x + 3y} = \frac{(2x-3y)(2x+3y)}{2x+3y} = 2x - 3y

are all algebraically identical.

Re-expressing or factoring alternative choices proves their equivalence to the simplified expression.

Key Concept

Factoring difference of cubes

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

and difference of squares

a^2 - b^2 = (a - b)(a + b)

Question 24Question →

The quadratic equation $2x^2 + px + q = 0$ , where $p$ and $q$ are constants, has roots $r$ and $s$ . The quadratic equation $x^2 + (p - 2)x + 24 = 0$ has roots $r + 2$ and $s + 2$ . What is the value of $q$ ?

Show answer & explanation

Answer: 32

Answer

The value of

q

is 32.

Applying Vieta's formulas to

2x^2 + px + q = 0

gives

r + s = -p/2

and

rs = q/2

. For the second equation

x^2 + (p-2)x + 24 = 0

, the sum of roots is

(r+2) + (s+2) = -(p-2)

, which simplifies to

(r+s) + 4 = 2 - p

. Substituting

r+s = -p/2

yields

-p/2 + 4 = 2 - p

, solving to

p = -4

and

r+s = 2

. The product of roots for the second equation is

(r+2)(s+2) = rs + 2(r+s) + 4 = 24

. Substituting

rs = q/2

and

r+s = 2

gives

q/2 + 4 + 4 = 24

, which simplifies to

q/2 = 16

and

q = 32

.

Step-by-Step Solution

1

Express the sum and product of roots

r

and

s

in terms of

p

and

q

using Vieta's formulas.

r + s = -\frac{p}{2}

and

rs = \frac{q}{2}

.

For a quadratic equation

ax^2 + bx + c = 0

, the sum of the roots is

-\frac{b}{a}

and the product is

\frac{c}{a}

.

2

Relate the sum of the shifted roots

(r + 2)

and

(s + 2)

to the coefficients of the second quadratic equation.

(r + 2) + (s + 2) = -(p - 2) \implies (r + s) + 4 = 2 - p

.

The coefficient of

x

in

x^2 + (p - 2)x + 24 = 0

is

(p - 2)

, so the sum of its roots equals

-(p - 2)

.

3

Substitute

r + s = -\frac{p}{2}

into the sum relation to determine

p

.

-\frac{p}{2} + 4 = 2 - p \implies \frac{p}{2} = -2 \implies p = -4

.

Solving the linear equation for

p

yields

p = -4

, which means

r + s = 2

.

4

Expand the product of the shifted roots

(r + 2)(s + 2) = 24

and solve for

q

.

rs + 2(r + s) + 4 = 24 \implies \frac{q}{2} + 2(2) + 4 = 24 \implies \frac{q}{2} + 8 = 24 \implies q = 32

.

Substituting

rs = \frac{q}{2}

and

r + s = 2

isolates

q

, yielding

q = 32

.

Key Concept

Relating roots and coefficients of quadratic equations using Vieta's formulas and algebraic expansion.

Question 25Question →

For all real numbers $x$ and $y$ , the custom operation $\blacktriangle$ is defined by $x \blacktriangle y = 2x^2 - y^2 + xy$ . The function $g$ is defined for all real numbers $n$ by $g(n) = n \blacktriangle 3$ . If $k$ is a positive constant such that $g(k) = 0$ , what is the value of $k$ ?

\frac{2}{3}

1

\frac{3}{2}

3

9

Show answer & explanation

Answer: $\frac{3}{2}$

Answer

\frac{3}{2}

Evaluating

g(k) = k \blacktriangle 3

using the rule

x \blacktriangle y = 2x^2 - y^2 + xy

gives

2k^2 - 3^2 + 3k = 2k^2 + 3k - 9

. Setting this quadratic equal to zero gives

(2k - 3)(k + 3) = 0

. The roots are

k = \frac{3}{2}

and

k = -3

. Because

k

is specified to be positive, the only valid solution is

\frac{3}{2}

.

Step-by-Step Solution

1

Apply the definition of the custom operation

\blacktriangle

to evaluate

g(k)

.

g(k) = k \blacktriangle 3 = 2(k)^2 - (3)^2 + k(3) = 2k^2 - 9 + 3k = 2k^2 + 3k - 9

Substitute

x = k

and

y = 3

into the given expression

x \blacktriangle y = 2x^2 - y^2 + xy

.

2

Set the expression for

g(k)

equal to

0

and factor the quadratic equation.

2k^2 + 3k - 9 = (2k - 3)(k + 3) = 0

We need to find values of

k

for which

g(k) = 0

.

3

Solve for

k

and apply the constraint that

k

is a positive constant.

2k - 3 = 0 \implies k = \frac{3}{2}

or

k + 3 = 0 \implies k = -3

. Since

k > 0

,

k = \frac{3}{2}

.

Disregard the negative root

k = -3

because the problem specifies that

k

must be positive.

Key Concept

Combining custom symbol operations with function notation to solve quadratic equations

Estimated Time:1m 30s

Question 26Question →

For all non-zero real numbers $x$ , the function $f$ satisfies the equation $f(x) + 2f\left(\frac{1}{x}\right) = 3x$ . What is the value of $f(2)$ ?

-1

0

1

2

3

Show answer & explanation

Answer: $-1$

Answer

-1

Evaluating the functional equation at

x = 2

gives

f(2) + 2f(1/2) = 6

. Evaluating the equation at

x = 1/2

gives

f(1/2) + 2f(2) = 3/2

. Substituting

f(1/2) = 3/2 - 2f(2)

into the first equation produces

-3f(2) + 3 = 6

, which simplifies to

f(2) = -1

.

Step-by-Step Solution

1

Substitute

x = 2

into the given functional equation

f(x) + 2f\left(\frac{1}{x}\right) = 3x

.

f(2) + 2f\left(\frac{1}{2}\right) = 6

This establishes a relation between the target value

f(2)

and the auxiliary value

f\left(\frac{1}{2}\right)

.

2

Substitute

x = \frac{1}{2}

into the original equation

f(x) + 2f\left(\frac{1}{x}\right) = 3x

.

f\left(\frac{1}{2}\right) + 2f(2) = 3\left(\frac{1}{2}\right) = \frac{3}{2}

This provides a second independent linear equation relating

f(2)

and

f\left(\frac{1}{2}\right)

.

3

Isolate

f\left(\frac{1}{2}\right)

in the second equation:

f\left(\frac{1}{2}\right) = \frac{3}{2} - 2f(2)

, and substitute this expression into the first equation.

f(2) + 2\left(\frac{3}{2} - 2f(2)\right) = 6 \implies f(2) + 3 - 4f(2) = 6 \implies -3f(2) + 3 = 6

Eliminating

f\left(\frac{1}{2}\right)

converts the system into a single linear equation in terms of

f(2)

.

4

Solve

-3f(2) + 3 = 6

for

f(2)

.

-3f(2) = 3 \implies f(2) = -1

Subtracting

3

from both sides gives

-3f(2) = 3

, and dividing by

-3

yields

f(2) = -1

.

Key Concept

Solving Functional Equations via Variable Substitution

Alternative Method

Generalize

f(x)

by setting

f(x) = Ax + \frac{B}{x}

. Substituting into

f(x) + 2f(1/x) = 3x

gives

(Ax + B/x) + 2(A/x + Bx) = (A + 2B)x + (2A + B)/x = 3x

. Equating coefficients:

A + 2B = 3

and

2A + B = 0 \implies B = -2A \implies A - 4A = 3 \implies A = -1, B = 2

. Thus

f(x) = -x + 2/x

, so

f(2) = -2 + 2/2 = -1

.

Estimated Time:1m 30s

Question 27Question →

A food processing plant packages grain using two automated lines, Line A and Line B. Line A operates at a constant rate of $40$ bags per minute, and Line B operates at a constant rate of $60$ bags per minute. Line A begins filling a batch of $4,200$ bags alone. After $15$ minutes, Line A's rate decreases by $25\%$ , and Line B is turned on to work alongside Line A until the batch is complete. How many total minutes does Line A operate to complete the batch?

40 minutes

51 minutes

55 minutes

63 minutes

70 minutes

Show answer & explanation

Answer: 55 minutes

Answer

55 minutes

In the first 15 minutes, Line A fills

40 \times 15 = 600

bags, leaving

4,200 - 600 = 3,600

bags remaining. Line A's rate then drops to

40 \times 0.75 = 30

bags per minute. Working together with Line B (

60

bags per minute), their combined rate is

30 + 60 = 90

bags per minute. Filling the remaining

3,600

bags takes

\frac{3,600}{90} = 40

minutes. Thus, Line A operates for a total of

15 + 40 = 55

minutes.

Step-by-Step Solution

1

Calculate the number of bags filled by Line A during the first 15 minutes.

Line A filled

40 \text{ bags/min} \times 15 \text{ min} = 600 \text{ bags}

.

Line A worked alone for 15 minutes at its initial constant rate.

2

Determine the remaining number of bags to be filled.

Remaining bags =

4,200 - 600 = 3,600 \text{ bags}

.

Subtract the bags already completed from the total batch size.

3

Calculate Line A's reduced rate and the new combined rate of both lines.

Line A's new rate =

40 \times (1 - 0.25) = 30 \text{ bags/min}

. Combined rate =

30 + 60 = 90 \text{ bags/min}

.

Line A's filling rate drops by 25%, and Line B contributes 60 bags per minute simultaneously.

4

Calculate the time needed for both lines to finish the remaining bags.

Time for second phase =

\frac{3,600 \text{ bags}}{90 \text{ bags/min}} = 40 \text{ minutes}

.

Divide remaining work by the combined filling rate.

5

Calculate the total time Line A was operating.

Total operating time =

15 \text{ min} + 40 \text{ min} = 55 \text{ minutes}

.

Line A operated during both the initial 15-minute period and the subsequent 40-minute period.

Key Concept

Linear Modeling and Piecewise Work-Rate Scenarios

Estimated Time:2m 0s

Question 28Question →

For all real numbers $a$ and $b$ such that $b \neq 0$ and $a \neq -b$ , which of the following expressions is equivalent to $\frac{a^3b - ab^3}{a^2b + 2ab^2 + b^3}$ ?

\frac{a(a - b)}{a + b}

a - b

\frac{a(a + b)}{a - b}

\frac{a^2 + b^2}{a + b}

\frac{a - b}{a + b}

Show answer & explanation

Answer: $\frac{a(a - b)}{a + b}$

Answer

\frac{a(a - b)}{a + b}

Factoring the numerator yields

ab(a^2 - b^2) = ab(a - b)(a + b)

, and factoring the denominator yields

b(a^2 + 2ab + b^2) = b(a + b)^2

. Dividing both the numerator and the denominator by the common factor

b(a + b)

produces

\frac{a(a - b)}{a + b}

.

Step-by-Step Solution

1

Factor out the greatest common monomial factor from the numerator and denominator.

The numerator

a^3b - ab^3

becomes

ab(a^2 - b^2)

, and the denominator

a^2b + 2ab^2 + b^3

becomes

b(a^2 + 2ab + b^2)

.

Extracting common terms simplifies polynomials prior to applying polynomial identities.

2

Apply polynomial algebraic identities to factor the remaining terms.

a^2 - b^2 = (a - b)(a + b)

and

a^2 + 2ab + b^2 = (a + b)^2

. Thus, the expression becomes

\frac{ab(a - b)(a + b)}{b(a + b)^2}

.

The difference of squares and perfect square trinomial identities allow complete factorization.

3

Cancel non-zero common factors from the numerator and denominator.

Canceling

b

and one factor of

(a + b)

leaves

\frac{a(a - b)}{a + b}

.

Dividing both numerator and denominator by

b(a + b)

yields the simplified expression.

Key Concept

Simplifying rational expressions using common monomial factoring, difference of squares, and perfect square trinomial identities.

Estimated Time:1m 30s

Question 29Question →

If $x$ is a real number that satisfies the equation $(x - 3)^2 = 16$ , which of the following could be the value of $x^2 - 4x$ ? Select all such values.

Select all that apply

-5

5

21

45

77

Show answer & explanation

Answer: $5$ ; $21$

Answer

The possible values of

x^2 - 4x

are

5

and

21

.

Taking the square root of both sides of

(x - 3)^2 = 16

yields

x - 3 = \pm 4

. This gives two solutions for

x

:

x = 7

and

x = -1

. Substituting

x = 7

into

x^2 - 4x

gives

49 - 28 = 21

. Substituting

x = -1

into

x^2 - 4x

gives

(-1)^2 - 4(-1) = 1 + 4 = 5

. Thus, both

5

and

21

are valid values.

Step-by-Step Solution

1

Solve the quadratic equation for

x

x - 3 = 4 \implies x = 7

or

x - 3 = -4 \implies x = -1

Taking the square root of both sides of

(x - 3)^2 = 16

yields both positive and negative roots.

2

Substitute the first root into the expression

x^2 - 4x

7^2 - 4(7) = 49 - 28 = 21

Evaluating the target expression for

x = 7

.

3

Substitute the second root into the expression

x^2 - 4x

(-1)^2 - 4(-1) = 1 + 4 = 5

Evaluating the target expression for

x = -1

.

Key Concept

Quadratic Equations and Factoring

Question 30Question →

If $x$ is a real number such that $(x^2 - 4)^2 - 3(x^2 - 4) - 10 = 0$ , which of the following could be the value of $x$ ? Select all such values.

Select all that apply

-3

-\sqrt{2}

2

3

\sqrt{5}

Show answer & explanation

Answer: $-3$ ; $-\sqrt{2}$ ; $3$

Answer

The values that satisfy the equation are

-3

,

-\sqrt{2}

, and

3

.

The given equation is in quadratic form with respect to

(x^2 - 4)

. Letting

u = x^2 - 4

, we factor the equation as

(u - 5)(u + 2) = 0

, giving

u = 5

or

u = -2

. Setting

x^2 - 4 = 5

gives

x^2 = 9

, which yields

x = 3

or

x = -3

. Setting

x^2 - 4 = -2

gives

x^2 = 2

, which yields

x = \sqrt{2}

or

x = -\sqrt{2}

. Thus,

-3

,

-\sqrt{2}

, and

3

are all valid solutions.

Step-by-Step Solution

1

Use algebraic substitution to simplify the quadratic form.

Let

u = x^2 - 4

. The equation becomes

u^2 - 3u - 10 = 0

.

Recognizing the structure as a quadratic equation in terms of

(x^2 - 4)

simplifies factoring.

2

Factor the quadratic equation in terms of

u

.

(u - 5)(u + 2) = 0

, giving solutions

u = 5

or

u = -2

.

Finding the roots of the substituted variable

u

leads to the possible values for

x^2 - 4

.

3

Substitute back

x^2 - 4

for

u

and solve for

x

in each case.

For

u = 5

:

x^2 - 4 = 5 \implies x^2 = 9 \implies x = 3

or

x = -3

.
For

u = -2

:

x^2 - 4 = -2 \implies x^2 = 2 \implies x = \sqrt{2}

or

x = -\sqrt{2}

.

Taking both positive and negative square roots provides all valid real solutions.

Key Concept

Quadratic Form Substitution and Factoring

Estimated Time:1m 40s

Question 31Question →

If $x$ is a negative real number such that $(x + 5)^2 = 36$ , what is the value of $x^2 - 2x$ ?

-1

3

-99

143

99

Show answer & explanation

Answer: $143$

Answer

The value of

x^2 - 2x

is

143

.

Taking the square root of both sides of

(x + 5)^2 = 36

gives

x + 5 = 6

or

x + 5 = -6

. Solving for

x

yields

x = 1

or

x = -11

. Because

x

is given as a negative real number,

x = -11

. Evaluating the target expression at

x = -11

yields

(-11)^2 - 2(-11) = 121 + 22 = 143

.

Step-by-Step Solution

1

Solve the quadratic equation for

x

by applying the square root property.

x + 5 = \pm \sqrt{36} = \pm 6

Taking the square root of both sides yields both positive and negative root options.

2

Determine the two potential solutions for

x

.

x = 6 - 5 = 1

or

x = -6 - 5 = -11

Isolating

x

gives two real solutions.

3

Select the valid solution based on the problem constraints.

x = -11

The stem specifies that

x

must be a negative real number, so

x = 1

is rejected.

4

Substitute

x = -11

into the expression

x^2 - 2x

.

(-11)^2 - 2(-11) = 121 + 22 = 143

Squaring a negative number yields a positive result, and subtracting a negative quantity adds to the total.

Key Concept

Solving quadratic equations using square roots and evaluating algebraic expressions

Question 32Question →

Consider the following system of linear equations in variables

x

and

y

, where

k

is a real constant:

\begin{aligned} kx + 4y &= 8 \\ x + ky &= k + 2 \end{aligned}

Which of the following statements are true? Select all that apply.

Select all that apply

If

k = 2

, the system has infinitely many solutions.

If

k = -2

, the system has no solutions.

If

k = 0

, the unique solution to the system is

(x, y) = (2, 2)

.

If

k = 3

, the system has no solutions.

If

k = 1

, the solution to the system satisfies

x + y = 4

.

Show answer & explanation

Answer: If $k = 2$ , the system has infinitely many solutions.; If $k = -2$ , the system has no solutions.; If $k = 0$ , the unique solution to the system is $(x, y) = (2, 2)$ .

Answer

The correct statements are that setting

k = 2

results in infinitely many solutions, setting

k = -2

results in no solutions, and setting

k = 0

yields the unique solution

(2, 2)

.

The system has a coefficient matrix determinant of

k^2 - 4

. Setting

k = 2

produces identical equations (

x + 2y = 4

), giving infinitely many solutions. Setting

k = -2

produces parallel equations with different constants (

-x + 2y = 4

vs

-x + 2y = 0

), giving no solutions. Setting

k = 0

reduces the system directly to

y = 2

and

x = 2

, confirming the unique point

(2, 2)

.

Step-by-Step Solution

1

Analyze the determinant of the coefficient matrix to identify conditions for unique vs. non-unique solutions.

The coefficient matrix determinant is

\Delta = k\cdot k - 4\cdot 1 = k^2 - 4 = (k - 2)(k + 2)

.

If

\Delta \neq 0

(i.e.,

k \neq \pm 2

), the system has a unique solution. If

\Delta = 0

(i.e.,

k = 2

or

k = -2

), the lines are either identical or parallel.

2

Test

k = 2

in the system.

Equation 1 becomes

2x + 4y = 8 \implies x + 2y = 4

. Equation 2 becomes

x + 2y = 4

.

Since the equations are identical, the lines coincide, resulting in infinitely many solutions.

3

Test

k = -2

in the system.

Equation 1 becomes

-2x + 4y = 8 \implies -x + 2y = 4

. Equation 2 becomes

x - 2y = 0 \implies -x + 2y = 0

.

The slopes are equal (

1/2

) but the y-intercepts differ (

2

vs

0

), meaning the lines are parallel and distinct, yielding zero solutions.

4

Test

k = 0

and

k = 1

to evaluate remaining choices.

For

k = 0

,

4y = 8 \implies y = 2

and

x + 0 = 2 \implies x = 2

, giving

(2, 2)

. For

k = 1

, Equation 2 is directly

x + y = 3

.

This verifies that the statement for

k = 0

is correct, while the statement for

k = 1

claiming

x + y = 4

is false.

Key Concept

Parametric Linear Systems and Conditions for Solvability

Question 33Question →

A university archive uses two high-speed scanners, Machine A and Machine B, to digitize manuscript pages. Machine A operates at a constant rate of $60$ pages per hour, and Machine B operates at a constant rate of $80$ pages per hour. On Monday, Machine A starts scanning at 8:00 AM. At 9:30 AM, Machine B is brought online to work alongside Machine A. If both machines continue scanning together at their respective constant rates until a total of $510$ pages have been scanned, at what time will the scanning project be completed?

11:00 AM

11:38 AM

12:30 PM

2:45 PM

3:30 PM

Show answer & explanation

Answer: 12:30 PM

Answer

12:30 PM

The option stating 12:30 PM is correct. Machine A operates alone from 8:00 AM to 9:30 AM (1.5 hours), completing

60 \times 1.5 = 90

pages. This leaves

510 - 90 = 420

pages to be scanned. From 9:30 AM onward, both machines work together at a combined rate of

60 + 80 = 140

pages per hour. The remaining 420 pages require

\frac{420}{140} = 3

hours. Adding 3 hours to 9:30 AM gives a completion time of 12:30 PM.

Step-by-Step Solution

1

Calculate the number of pages scanned by Machine A during its head start

60 \text{ pages/hr} \times 1.5 \text{ hours} = 90 \text{ pages}

Machine A operates alone from 8:00 AM to 9:30 AM, which is

1.5

hours.

2

Determine the remaining number of pages to be scanned

510 - 90 = 420 \text{ pages}

Subtract the pages already completed from the total requirement of 510 pages.

3

Calculate the combined scanning rate of both machines

60 + 80 = 140 \text{ pages/hr}

When both machines operate simultaneously, their individual rates add together.

4

Calculate the time required to scan the remaining pages

\frac{420 \text{ pages}}{140 \text{ pages/hr}} = 3 \text{ hours}

Divide the remaining work by the combined rate.

5

Determine the final completion time

9:30 \text{ AM} + 3 \text{ hours} = 12:30 \text{ PM}

Add the combined work duration to the time when both machines began working together (9:30 AM).

Key Concept

Combined Work Rates with Staggered Start Times

Estimated Time:1m 30s

Question 34Question →

A IT consultant charges a flat setup fee of $\$150$ plus a standard hourly rate of $\$80$ for regular hours worked on a project. For any rush hours worked, the hourly rate increases by $50\%$ . On a recently completed project, the consultant worked a total of $t$ hours, of which $r$ hours were rush hours (where $0 \leq r \leq t$ ). Which of the following statements must be true regarding the total charge $C$ , in dollars, for this project? Select all such statements.

Select all that apply

The total charge, in dollars, can be expressed as

C = 150 + 80t + 40r

.

If the consultant worked a total of

20

hours and the total charge was

\$2,150

, then exactly

10

of those hours were rush hours.

The total charge, in dollars, can be expressed as

C = 150 + 120t - 40r

.

If no rush hours were worked (

r = 0

), the average charge per hour, including the setup fee, is

\frac{150}{t} + 80

dollars per hour.

If all hours worked were rush hours (

r = t

), the total charge is given by

C = 150 + 160t

.

Show answer & explanation

Answer: The total charge, in dollars, can be expressed as $C = 150 + 80t + 40r$ .; If the consultant worked a total of $20$ hours and the total charge was $\$2,150$ , then exactly $10$ of those hours were rush hours.; If no rush hours were worked ( $r = 0$ ), the average charge per hour, including the setup fee, is $\frac{150}{t} + 80$ dollars per hour.

Answer

The correct statements are those expressing the total charge as

C = 150 + 80t + 40r

, determining that

10

rush hours were worked when total cost is

\$2,150

for

20

hours, and calculating the average hourly cost as

\frac{150}{t} + 80

dollars per hour when no rush hours are worked.

The model correctly partitions total hours

t

into

(t - r)

regular hours at

\$80

/hr and

r

rush hours at

\$120

/hr, which simplifies algebraically to

C = 150 + 80t + 40r

. Solving this equation for

t = 20

and

C = 2150

yields

r = 10

. Furthermore, when

r = 0

, dividing total charge

150 + 80t

by

t

gives

\frac{150}{t} + 80

.

Step-by-Step Solution

1

Calculate the hourly rush rate from the given percentage increase.

The standard rate is

\$80

/hr. The rush rate is

\$80 \times (1 + 0.50) = \$120

/hr.

Rush hours cost

50\%

more than standard regular hours.

2

Set up the algebraic model for total charge

C

using standard hours

(t - r)

and rush hours

r

.

C = 150 + 80(t - r) + 120r = 150 + 80t - 80r + 120r = 150 + 80t + 40r

.

This combines the fixed setup fee with variable costs from regular and rush hours.

3

Evaluate the specific case where total time

t = 20

and total cost

C = 2150

.

2150 = 150 + 80(20) + 40r \implies 2150 = 1750 + 40r \implies 40r = 400 \implies r = 10

.

Plugging given values into the algebraic model allows solving for the unknown number of rush hours.

4

Calculate average hourly cost when

r = 0

.

Average cost

= \frac{150 + 80t}{t} = \frac{150}{t} + 80

.

Average hourly rate is total charge divided by total hours worked.

Key Concept

Linear algebraic modeling of rate problems with multiple rate components

Question 35Question →

A boutique perfume workshop creates a signature fragrance blend by combining two fragrance oils, Oil A and Oil B. Oil A costs $12 per ounce and contains 40% pure essential oil by volume. Oil B costs$ 20 per ounce and contains 80% pure essential oil by volume. The perfumer creates a 30-ounce batch of the signature blend at a total cost that averages exactly $15 per ounce. What is the total volume, in ounces, of pure essential oil contained in this 30-ounce blend?

Show answer & explanation

Answer: 16.5

Answer

The total volume of pure essential oil contained in the 30-ounce blend is 16.5 ounces.

To find the total essential oil content, first set up a system of equations for the volumes of Oil A (

x

) and Oil B (

y

):

x + y = 30

and

12x + 20y = 450

. Solving this system yields

x = 18.75

ounces and

y = 11.25

ounces. Multiplying each by its respective essential oil concentration gives

0.40 \times 18.75 = 7.5

ounces from Oil A and

0.80 \times 11.25 = 9

ounces from Oil B, totaling 16.5 ounces of essential oil.

Step-by-Step Solution

1

Set up linear equations representing the total volume and total cost of the mixture.

x + y = 30

and

12x + 20y = 450

, where

x

is ounces of Oil A and

y

is ounces of Oil B.

The total cost of the 30-ounce blend at

15 per ounce is

15 \times 30 = 450$ dollars.

2

Solve for the quantities of Oil A and Oil B used in the mixture.

x = 18.75

ounces of Oil A and

y = 11.25

ounces of Oil B.

Substituting

y = 30 - x

into

12x + 20(30 - x) = 450

yields

-8x = -150

, giving

x = 18.75

.

3

Compute the amount of pure essential oil contributed by each component and sum them.

0.40(18.75) + 0.80(11.25) = 7.5 + 9 = 16.5

ounces.

Oil A contains 40% essential oil by volume and Oil B contains 80% essential oil by volume.

Key Concept

Systems of Linear Equations and Mixture Modeling

Estimated Time:2m 0s

Question 36Question →

For all real numbers $x$ and $y$ such that $x \neq y$ , $x \neq -y$ , and $x^2 + y^2 \neq 0$ , consider the algebraic expression:

E(x, y) = \frac{x^4 - y^4}{x^3 - x^2y + xy^2 - y^3}

Which of the following expressions are equivalent to $E(x, y)$ for all valid values of $x$ and $y$ ? Select all such expressions.

Select all that apply

\frac{x^2 - y^2}{x - y}

\frac{x^3 + y^3}{x^2 - xy + y^2}

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

\frac{x^3 - y^3}{x^2 + xy + y^2}

\frac{x^2 - y^2}{x + y}

Show answer & explanation

Answer: $\frac{x^2 - y^2}{x - y}$ ; $\frac{x^3 + y^3}{x^2 - xy + y^2}$

Answer

The expressions equivalent to

E(x, y)

are

\frac{x^2 - y^2}{x - y}

and

\frac{x^3 + y^3}{x^2 - xy + y^2}

.

Simplifying

E(x, y)

by factoring both numerator and denominator yields

x + y

. The option with

\frac{x^2 - y^2}{x - y}

simplifies directly to

x + y

by canceling

(x - y)

. The option with

\frac{x^3 + y^3}{x^2 - xy + y^2}

uses the sum of cubes identity to factor the numerator into

(x + y)(x^2 - xy + y^2)

, which also cancels down to

x + y

. Both of these options are mathematically identical to

E(x, y)

.

Step-by-Step Solution

1

Factor the numerator of

E(x, y)

using the difference of squares identity twice.

x^4 - y^4 = (x^2 - y^2)(x^2 + y^2) = (x - y)(x + y)(x^2 + y^2)

Decomposing higher-power binomials into linear and quadratic factors enables algebraic simplification.

2

Factor the denominator of

E(x, y)

by grouping terms.

x^3 - x^2y + xy^2 - y^3 = x^2(x - y) + y^2(x - y) = (x - y)(x^2 + y^2)

Grouping adjacent terms with shared factors allows factoring out

(x - y)

.

3

Simplify the full expression

E(x, y)

by canceling common non-zero factors.

E(x, y) = \frac{(x - y)(x + y)(x^2 + y^2)}{(x - y)(x^2 + y^2)} = x + y

Since

x \neq y

and

x^2 + y^2 \neq 0

, the factors

(x - y)

and

(x^2 + y^2)

cancel completely.

4

Evaluate the given options to determine which simplify to

x + y

.

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

and

\frac{x^3 + y^3}{x^2 - xy + y^2} = x + y

, whereas

\frac{x^3 - y^3}{x^2 + xy + y^2} = x - y

and

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

.

Matching each option's fully simplified form to

x + y

identifies all valid equivalent expressions.

Key Concept

Simplifying rational expressions by polynomial factoring (grouping, difference of squares, and sum of cubes).

Question 37Question →

If $x > 0$ , which of the following is equivalent to the expression $\sqrt{9x^4 + 16x^4}$ ?

5x^2

7x^2

5x^4

7x^4

25x^2

Show answer & explanation

Answer: $5x^2$

Answer

The correct answer is

5x^2

.

First combine the terms under the square root to get

\sqrt{25x^4}

. Taking the square root yields

\sqrt{25} \cdot \sqrt{x^4} = 5x^2

. Thus,

5x^2

is the correct simplified expression.

Step-by-Step Solution

1

Combine like terms inside the square root.

9x^4 + 16x^4 = 25x^4

Since both terms have the same variable factor

x^4

, their coefficients can be added directly.

2

Apply the product rule for radicals:

\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}

.

\sqrt{25x^4} = \sqrt{25} \cdot \sqrt{x^4}

The square root of a product is equal to the product of the square roots.

3

Evaluate the square root of the coefficient and the variable term.

5 \cdot x^2 = 5x^2

\sqrt{25} = 5

and

\sqrt{x^4} = (x^4)^{1/2} = x^2

(since

x > 0$).

Key Concept

Simplifying algebraic radicals by combining like terms under the square root and applying power rules for exponents.

Question 38Question →

A logistics company calculates the shipping cost $C$ , in dollars, for a parcel of weight $w$ pounds shipped over a distance of $d$ miles using the following piecewise pricing model:
- A fixed base fee of $\$10$ .
- An additional weight surcharge of $\$1.50$ per pound for every pound (or fraction thereof) exceeding $5$ pounds.
- An additional distance surcharge of $\$0.05$ per mile for every mile exceeding $100$ miles.
- A flat combined heavy-distance surcharge of $\$25$ applied if and only if $w > 10$ AND $d > 500$ .

A customer ships two parcels, Parcel 1 and Parcel 2:
- Parcel 1 weighs $w_1$ pounds (where $w_1$ is an integer greater than 10) and is shipped a distance of $d_1$ miles (where $d_1$ is an integer greater than 500).
- Parcel 2 weighs $w_2 = w_1 - 4$ pounds and is shipped a distance of $d_2 = d_1 + 200$ miles.

Let $C_1$ be the shipping cost of Parcel 1 and $C_2$ be the shipping cost of Parcel 2.

Which of the following statements MUST be true? Select all such statements.

Select all that apply

If

w_1 = 12

, then Parcel 1 costs exactly

\$21

more to ship than Parcel 2.

If

w_1 > 14

, Parcel 2 always costs exactly

\$4

more to ship than Parcel 1.

The total combined shipping cost of both parcels,

C_1 + C_2

, is independent of the distance

d_1

.

Parcel 2 always costs more to ship than Parcel 1, regardless of the value of

w_1

.

If

w_1 = 15

, the shipping cost of Parcel 2 is

\$26.50

plus

0.05 d_1

.

Show answer & explanation

Answer: If $w_1 = 12$ , then Parcel 1 costs exactly $\$21$ more to ship than Parcel 2.; If $w_1 > 14$ , Parcel 2 always costs exactly $\$4$ more to ship than Parcel 1.

Answer

The correct statements are the ones indicating that if the weight of Parcel 1 is 12 pounds, Parcel 1 costs exactly

21 more than Parcel 2, and if the weight of Parcel 1 is greater than 14 pounds, Parcel 2 costs exactly

4 more than Parcel 1.

The cost functions must be broken down by conditional threshold. For Parcel 1, both weight (>10) and distance (>500) thresholds are met, giving C_1 = 1.5w_1 + 0.05d_1 + 22.5. For Parcel 2, d_2 = d_1 + 200 > 700 is always > 500, but w_2 = w_1 - 4 exceeds 10 if and only if w_1 > 14. If w_1 = 12, w_2 = 8, so Parcel 2 gets no

25 surcharge, resulting in C_1 - C_2 = 21. If w_1 > 14, w_2 > 10, so Parcel 2 gets the

25 surcharge, giving C_2 - C_1 = 4.

Step-by-Step Solution

1

Formulate the algebraic cost function C(w, d) for any parcel.

C(w, d) = 10 + 1.50(w - 5) + 0.05(d - 100) + S(w, d), where S(w, d) = 25 if w > 10 and d > 500, else S(w, d) = 0.

This establishes the piecewise linear equations governing shipping costs.

2

Express C_1 for Parcel 1 given w_1 > 10 and d_1 > 500.

Since w_1 > 10 and d_1 > 500, S(w_1, d_1) = 25. Thus, C_1 = 10 + 1.50(w_1 - 5) + 0.05(d_1 - 100) + 25 = 1.50 w_1 + 0.05 d_1 + 22.50.

Parcel 1 always meets both conditions for the $25 heavy-distance surcharge.

3

Analyze Parcel 2 when 10 < w_1 <= 14 (meaning 6 < w_2 <= 10).

For w_2 <= 10, S(w_2, d_2) = 0 even though d_2 > 500. C_2 = 10 + 1.50(w_1 - 9) + 0.05(d_1 + 100) + 0 = 1.50 w_1 + 0.05 d_1 + 1.50. Therefore, C_1 - C_2 = (1.50 w_1 + 0.05 d_1 + 22.50) - (1.50 w_1 + 0.05 d_1 + 1.50) = $21.00.

This confirms that for w_1 = 12, Parcel 1 costs exactly $21 more than Parcel 2.

4

Analyze Parcel 2 when w_1 > 14 (meaning w_2 > 10).

For w_2 > 10 and d_2 > 700 > 500, S(w_2, d_2) = 25. Thus, C_2 = 1.50 w_1 + 0.05 d_1 + 26.50. Subtracting C_1 gives C_2 - C_1 = (1.50 w_1 + 0.05 d_1 + 26.50) - (1.50 w_1 + 0.05 d_1 + 22.50) = $4.00.

This proves Parcel 2 costs exactly $4 more than Parcel 1 whenever w_1 > 14.

Key Concept

Piecewise linear modeling and conditional surcharge threshold evaluations

Question 39Question →

The function $f$ is defined for all real numbers $x$ by $f(x) = 2x^2 - x$ . For all real numbers $a$ and $b$ , the custom operation $\star$ is defined by $a \star b = f(a + b) - f(a - b)$ . What is the value of $2 \star (-3)$ ?

-42

42

-46

-18

78

Show answer & explanation

Answer: -42

Answer

The correct answer is -42.

Evaluating

a + b = 2 + (-3) = -1

gives

f(-1) = 2(-1)^2 - (-1) = 3

. Evaluating

a - b = 2 - (-3) = 5

gives

f(5) = 2(5)^2 - 5 = 45

. Subtracting the two function values yields

3 - 45 = -42

.

Step-by-Step Solution

1

Identify the values for variables

a

and

b

and evaluate the inputs for the function

f

.

Here

a = 2

and

b = -3

. Therefore,

a + b = 2 + (-3) = -1

and

a - b = 2 - (-3) = 5

.

The definition of the custom operation requires evaluating

f(a+b)

and

f(a-b)

.

2

Calculate

f(-1)

using the function rule

f(x) = 2x^2 - x

.

f(-1) = 2(-1)^2 - (-1) = 2(1) + 1 = 3

.

Substituting

x = -1

into the quadratic expression.

3

Calculate

f(5)

using the function rule

f(x) = 2x^2 - x

.

f(5) = 2(5)^2 - (5) = 2(25) - 5 = 45

.

Substituting

x = 5

into the quadratic expression.

4

Compute the difference

f(a+b) - f(a-b)

.

3 - 45 = -42

.

Evaluating

f(-1) - f(5)

according to the definition of

\star

.

Key Concept

Evaluating nested algebraic functions and custom symbol operations by direct substitution.

Estimated Time:1m 30s

Question 40Question →

If $|3x - 6| \le 12$ , which of the following inequality ranges represents all possible values of $x$ ?

-2 \le x \le 6

x \le 6

-6 \le x \le 2

-6 \le x \le 6

-2 \le x \le 2

Show answer & explanation

Answer: $-2 \le x \le 6$

Answer

-2 \le x \le 6

The expression

|3x - 6| \le 12

expands to the double inequality

-12 \le 3x - 6 \le 12

. Adding

6

to each part produces

-6 \le 3x \le 18

. Dividing all terms by

3

yields

-2 \le x \le 6

, which matches the correct solution range.

Step-by-Step Solution

1

Rewrite the absolute value inequality as a compound inequality.

-12 \le 3x - 6 \le 12

An absolute value inequality of the form

|u| \le k

(where

k \ge 0

) is equivalent to

-k \le u \le k

.

2

Add

6

to all three parts of the inequality.

-6 \le 3x \le 18

Isolate the term containing

x

.

3

Divide all three parts of the inequality by

3

.

-2 \le x \le 6

Since

3 > 0

, dividing by

3

isolates

x

without flipping the inequality signs.

Key Concept

Solving Linear Absolute Value Inequalities

Estimated Time:45s

Algebra

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