Algebra Practice Questions — GRE General Test — Page 11

Question 201Question →

In the $xy$ -plane, line $m$ is defined by the equation $ax + by = c$ , where $a$ , $b$ , and $c$ are non-zero real numbers such that $ab < 0$ and $ac > 0$ . Line $k$ is perpendicular to line $m$ and intersects line $m$ at its $x$ -intercept. Which of the following statements must be true? Select all such statements.

Select all that apply

Line

m

has a positive slope.

Line

k

passes through Quadrant II.

The

y

-intercept of line

k

is negative.

The point of intersection of line

m

and line

k

lies in Quadrant I.

Line

m

passes through Quadrant II.

Show answer & explanation

Answer: Line $m$ has a positive slope.; Line $k$ passes through Quadrant II.

Answer

The statements asserting that line

m

has a positive slope and that line

k

passes through Quadrant II must be true.

The statement regarding line

m

having a positive slope is correct because

ab < 0

implies

a

and

b

have opposite signs, making

-\frac{a}{b} > 0

. The statement asserting line

k

passes through Quadrant II is correct because line

k

possesses a negative slope

\frac{b}{a} < 0

and a positive

y

-intercept

-\frac{bc}{a^2} > 0

, ensuring it enters Quadrant II.

Step-by-Step Solution

1

Determine the slope and intercepts of line

m

.

Line

m

:

y = -\frac{a}{b}x + \frac{c}{b}

. Slope is

-\frac{a}{b} > 0

because

ab < 0

.

x

-intercept is

\left(\frac{c}{a}, 0\right)

where

\frac{c}{a} > 0

because

ac > 0

.

y

-intercept is

\left(0, \frac{c}{b}\right)

where

\frac{c}{b} < 0

because

b

and

c

have opposite signs.

Converting standard line equations to slope-intercept form exposes the signs of slopes and intercepts based on coefficient products.

2

Determine the slope, equation, and properties of line

k

.

Since line

k

is perpendicular to line

m

, its slope is the negative reciprocal of

-\frac{a}{b}

, which is

\frac{b}{a} < 0

. Line

k

passes through

\left(\frac{c}{a}, 0\right)

, giving equation

y = \frac{b}{a}\left(x - \frac{c}{a}\right) = \frac{b}{a}x - \frac{bc}{a^2}

.

Perpendicular lines have slopes whose product is

-1

.

3

Analyze quadrant coverage for both lines and verify statements.

Line

m

has positive slope and negative

y

-intercept

\implies

passes through Quadrants I, III, IV. Line

k

has negative slope and positive

y

-intercept

-\frac{bc}{a^2} > 0 \implies

passes through Quadrants I, II, IV. Intersection is at

\left(\frac{c}{a}, 0\right)

on the positive

x

-axis.

Systematic sign analysis determines quadrant trajectory and exact axis locations.

Key Concept

Properties of lines, perpendicular slopes, and sign analysis of intercepts in coordinate geometry.

Question 202Question →

Courier A departs from Warehouse X heading toward Warehouse Y at 8:00 AM traveling at a constant speed of $40$ miles per hour. Courier B departs from Warehouse Y heading toward Warehouse X along the same straight route at 9:00 AM traveling at a constant speed of $60$ miles per hour. If the total distance between Warehouse X and Warehouse Y is $190$ miles, at what time will the two couriers meet?

9:54 AM

10:00 AM

10:30 AM

11:00 AM

11:15 AM

Show answer & explanation

Answer: 10:30 AM

Answer

10:30 AM

The correct answer is 10:30 AM. Between 8:00 AM and 9:00 AM, Courier A travels 40 miles alone. At 9:00 AM, the distance remaining between them is 150 miles. Because they travel toward each other, their speeds combine to 100 mph (40 + 60). Dividing 150 miles by 100 mph gives 1.5 hours (1 hour and 30 minutes). Adding 1 hour and 30 minutes to 9:00 AM gives a meeting time of 10:30 AM.

Step-by-Step Solution

1

Calculate the distance traveled by Courier A before Courier B starts moving.

From 8:00 AM to 9:00 AM (1 hour), Courier A travels

40 \text{ mph} \times 1 \text{ hour} = 40 \text{ miles}

.

Courier A has a 1-hour head start.

2

Determine the remaining distance to be covered between the two couriers at 9:00 AM.

Remaining distance

= 190 - 40 = 150 \text{ miles}

.

Subtract Courier A's distance from the total distance of 190 miles.

3

Calculate the combined rate of both couriers and solve for elapsed time after 9:00 AM.

Combined rate

= 40 + 60 = 100 \text{ mph}

. Elapsed time

t = \frac{150}{100} = 1.5 \text{ hours} = 1 \text{ hour } 30 \text{ minutes}

.

Since they move toward each other, their speeds add up to close the gap.

4

Add the elapsed combined time to 9:00 AM to find the meeting time.

9:00 AM

+ 1 \text{ hour } 30 \text{ minutes} = 10:30 \text{ AM}

.

The combined movement began at 9:00 AM when Courier B started traveling.

Key Concept

Distance-rate-time relationship with staggered start times

Estimated Time:1m 30s

Question 203Question →

Two automated assembly robots, Robot P and Robot Q, produce identical components. Robot P operates at a constant rate of $p$ components per hour, and Robot Q operates at a constant rate of $q$ components per hour, where $p > q > 0$ . During a shift, Robot P worked for $4$ hours and Robot Q worked for $6$ hours to produce a combined total of $360$ components. If $T$ represents the total number of components produced when Robot P works for $7$ hours and Robot Q works for $3$ hours, which of the following statements must be true? Select all such statements.

Select all that apply

The rate of Robot P,

p

, must be greater than

36

components per hour.

The rate of Robot Q,

q

, must be less than

36

components per hour.

The total number of components

T

must be greater than

360

and less than

630

.

The rate of Robot P,

p

, must be less than

60

components per hour.

The total number of components

T

could be equal to

320

.

Show answer & explanation

Answer: The rate of Robot P, $p$ , must be greater than $36$ components per hour.; The rate of Robot Q, $q$ , must be less than $36$ components per hour.; The total number of components $T$ must be greater than $360$ and less than $630$ .

Answer

The correct statements are: the rate of Robot P,

p

, must be greater than

36

components per hour; the rate of Robot Q,

q

, must be less than

36

components per hour; and the total number of components

T

must be greater than

360

and less than

630

.

The system of equations

2p + 3q = 180

combined with

p > q > 0

strictly bounds

p

between

36

and

90

, and

q

between

0

and

36

. Substituting these boundary conditions into the total expression

T = 5p + 180

yields the strict range

360 < T < 630

. Consequently, the three statements asserting

p > 36

,

q < 36

, and

360 < T < 630

are all mathematically required.

Step-by-Step Solution

1

Set up the linear equation from the initial production shift and simplify.

4p + 6q = 360 \implies 2p + 3q = 180 \implies q = 60 - \frac{2}{3}p

This establishes the exact relationship between the production rates

p

and

q

.

2

Apply the given constraints

p > q > 0

to determine the domain bounds for

p

and

q

.

p > 60 - \frac{2}{3}p \implies \frac{5}{3}p > 60 \implies p > 36

. Also,

q > 0 \implies 60 - \frac{2}{3}p > 0 \implies p < 90

. Thus,

36 < p < 90

and

0 < q < 36

.

Determining extreme bounds for

p

automatically constrains both individual rates.

3

Formulate

T

in terms of

p

and evaluate its numerical boundaries.

T = 7p + 3q = 7p + (180 - 2p) = 5p + 180

. Substituting

36 < p < 90

gives

360 < T < 630

.

Substituting

3q = 180 - 2p

simplifies

T

into a single-variable linear modeling equation.

Key Concept

Linear word problem modeling, variable elimination, and system inequality constraint analysis

Estimated Time:1m 45s

Question 204Question →

If $x$ is a real number that satisfies the inequality $|3x - 3| \le 9$ , which of the following statements must be true? Select all that apply.

Select all that apply

-4 \le 2x \le 8

|x - 1| \le 3

x^2 \le 16

x \ge 0

1 - x \le 2

Show answer & explanation

Answer: $-4 \le 2x \le 8$ ; $|x - 1| \le 3$ ; $x^2 \le 16$

Answer

The statements that must be true are

-4 \le 2x \le 8

,

|x - 1| \le 3

, and

x^2 \le 16

.

Solving

|3x - 3| \le 9

gives

-9 \le 3x - 3 \le 9

, which simplifies to

-2 \le x \le 4

. Multiplying this range by 2 yields

-4 \le 2x \le 8

. Subtracting 1 gives

-3 \le x - 1 \le 3

, which is

|x - 1| \le 3

. Squaring values in

[-2, 4]

yields non-negative numbers up to 16, so

x^2 \le 16

is also true.

Step-by-Step Solution

1

Unfold the absolute value inequality into a compound inequality.

-9 \le 3x - 3 \le 9

By definition,

|u| \le k

(where

k \ge 0

) is equivalent to

-k \le u \le k

.

2

Add 3 to all parts of the compound inequality.

-6 \le 3x \le 12

Isolating the variable term

3x

.

3

Divide all parts by 3 to solve for

x

.

-2 \le x \le 4

Dividing by a positive constant preserves the direction of the inequality signs.

4

Test each proposed statement against the interval

[-2, 4]

.

-4 \le 2x \le 8

is true;

|x - 1| \le 3

is true;

x^2 \le 16

is true;

x \ge 0

fails for

x = -1

;

1 - x \le 2

fails for

x = -2

.

Determining which properties hold for every real number in the solution set.

Key Concept

Linear Inequalities and Absolute Value

Estimated Time:1m 30s

Question 205Question →

In the $xy$ -plane, line $k$ passes through the point $(2, -1)$ and is perpendicular to the line $3x - 2y = 6$ . Line $m$ is parallel to line $k$ . If the distance between line $k$ and line $m$ is $\sqrt{13}$ units and line $m$ has a positive $y$ -intercept, what is the $y$ -intercept of line $m$ ?

\frac{14}{3}

\frac{17}{2}

\frac{5}{2}

4

7

Show answer & explanation

Answer: $\frac{14}{3}$

Answer

\frac{14}{3}

First, find the slope of the given line

3x - 2y = 6

, which is

\frac{3}{2}

. Because line

k

is perpendicular, its slope is

-\frac{2}{3}

. Using the point

(2, -1)

, line

k

has equation

2x + 3y - 1 = 0

. Line

m

is parallel, so it has equation

2x + 3y + C = 0

. The distance between two parallel lines

Ax + By + C_1 = 0

and

Ax + By + C_2 = 0

is given by

d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

. Setting

\frac{|C - (-1)|}{\sqrt{2^2 + 3^2}} = \sqrt{13}

yields

|C + 1| = 13

, which gives

C = 12

or

C = -14

. The

y

-intercept of line

m

is

-\frac{C}{3}

. Since the

y

-intercept must be positive,

C

must be negative, so

C = -14

. Therefore, the

y

-intercept is

-\frac{-14}{3} = \frac{14}{3}

.

Step-by-Step Solution

1

Determine the slope of line

k

.

The line

3x - 2y = 6

has slope

m_1 = \frac{3}{2}

. Because line

k

is perpendicular to it, the slope of line

k

is

m_k = -\frac{2}{3}

.

Perpendicular lines have negative reciprocal slopes.

2

Write the standard form equation of line

k

.

Using point-slope form with

(2, -1)

:

y - (-1) = -\frac{2}{3}(x - 2) \implies 2x + 3y - 1 = 0

.

Standard form

Ax + By + C_1 = 0

is required to apply the distance formula between parallel lines.

3

Set up the equation for line

m

and use the distance formula between parallel lines.

Since line

m

is parallel to line

k

, its equation is

2x + 3y + C = 0

. The distance between line

k

and line

m

is

d = \frac{|C - (-1)|}{\sqrt{2^2 + 3^2}} = \frac{|C + 1|}{\sqrt{13}} = \sqrt{13}

.

The distance between parallel lines

Ax + By + C_1 = 0

and

Ax + By + C_2 = 0

is given by

\frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

.

4

Solve for constant

C

and calculate the

y

-intercept of line

m

.

|C + 1| = 13 \implies C + 1 = 13

or

C + 1 = -13

, giving

C = 12

or

C = -14

. The

y

-intercept of

2x + 3y + C = 0

is

-\frac{C}{3}

. For a positive

y

-intercept,

C

must be negative, so

C = -14

. Thus, the

y

-intercept is

-\frac{-14}{3} = \frac{14}{3}

.

The problem states that line

m

has a positive

y

-intercept.

Key Concept

Distance between parallel lines and perpendicular slope relationships in coordinate geometry

Estimated Time:3m 0s

Question 206Question →

Let $P(x) = x^2 - mx + n$ be a quadratic polynomial with real coefficients $m$ and $n$ , having two distinct real roots $\alpha$ and \beta. If the roots satisfy the system of equations $\alpha^3 + \beta^3 = m(n + 7)$ and $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{10}{n^2}$ , which of the following statements MUST be true? Select all such statements.

Select all that apply

The product of the roots,

n

, is equal to

1.5

.

The sum of the squares of the roots,

\alpha^2 + \beta^2

, is equal to

10

.

The discriminant of the polynomial

P(x)

is equal to

7

.

The absolute value of the sum of the roots,

|\alpha + \beta|

, is equal to

5

.

The roots

\alpha

and

\beta

have opposite signs.

Show answer & explanation

Answer: The product of the roots, $n$ , is equal to $1.5$ .; The sum of the squares of the roots, $\alpha^2 + \beta^2$ , is equal to $10$ .; The discriminant of the polynomial $P(x)$ is equal to $7$ .

Answer

The correct statements are those asserting that the product of the roots is

1.5

, the sum of the squares of the roots is

10

, and the discriminant of

P(x)

is

7

.

Using Vieta's formulas and algebraic identity expansions for

\alpha^3 + \beta^3

and

\frac{1}{\alpha^2} + \frac{1}{\beta^2}

establishes the system of equations

m^2 = 4n + 7

and

m^2 = 2n + 10

. Solving this system gives

n = 1.5

,

m^2 = 13

, and a discriminant

\Delta = m^2 - 4n = 7

. Thus, the product of roots is

1.5

, the sum of squares

\alpha^2 + \beta^2 = m^2 - 2n = 10

, and the discriminant is

7

.

Step-by-Step Solution

1

Apply Vieta's formulas to express sum and product of roots.

\alpha + \beta = m \quad \text{and} \quad \alpha\beta = n

Vieta's relations link polynomial coefficients directly to symmetrical root expressions.

2

Expand

\alpha^3 + \beta^3

in terms of

m

and

n

.

\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = m^3 - 3mn = m(m^2 - 3n)

Using algebraic identities converts root powers into functions of

m

and

n

.

3

Equate the expression from Step 2 to

m(n + 7)

to find an equation for

m^2

.

m(m^2 - 3n) = m(n + 7) \implies m^2 - 3n = n + 7 \implies m^2 = 4n + 7

Since the roots are distinct,

m \neq 0

, allowing division by

m

.

4

Simplify the second given equation

\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{10}{n^2}

.

\frac{\alpha^2 + \beta^2}{\alpha^2\beta^2} = \frac{m^2 - 2n}{n^2} = \frac{10}{n^2} \implies m^2 - 2n = 10 \implies m^2 = 2n + 10

Combining fractions over a common denominator

(\alpha\beta)^2 = n^2

isolates

m^2 - 2n

.

5

Solve for

n

,

m^2

, and the discriminant

\Delta

.

4n + 7 = 2n + 10 \implies 2n = 3 \implies n = 1.5; \quad m^2 = 13; \quad \Delta = m^2 - 4n = 13 - 6 = 7

Equating the two expressions for

m^2

yields unique values for

n

,

m^2

, and

\Delta

.

Key Concept

Quadratic Equations, Vieta's Formulas, and Symmetric Polynomial Expressions

Question 207Question →

Two cyclists, Clara and Dan, start simultaneously from opposite ends of a $120$ -mile trail and ride directly toward each other along the same path. Clara rides at a constant speed of $c$ miles per hour, and Dan rides at a constant speed of $d$ miles per hour. If they maintain their initial speeds, they will meet in $3$ hours. However, if Clara increases her speed by $20\%$ and Dan increases his speed by $33\frac{1}{3}\%$ , they will meet $36$ minutes earlier. What is Clara's original speed $c$ , in miles per hour?

15

20

22.5

25

30

Show answer & explanation

Answer: 25

Answer

Clara's original speed

c

is 25 miles per hour.

The correct answer is 25 miles per hour. Initial combined speed is

c + d = \frac{120}{3} = 40

mph. Meeting 36 minutes earlier means a new duration of 2.4 hours, giving a new combined speed of

\frac{120}{2.4} = 50

mph. Setting up the equation

\frac{6}{5}c + \frac{4}{3}(40 - c) = 50

yields

c = 25

.

Step-by-Step Solution

1

Determine the initial combined rate of Clara and Dan.

Since they cover

120

miles in

3

hours, their combined speed is

c + d = \frac{120}{3} = 40

miles per hour.

Distance equals rate multiplied by time (

D = r \cdot t

). When moving toward each other, combined rate is

c + d

.

2

Calculate the new travel time and the new combined rate.

The new travel time is

3\text{ hours} - 36\text{ minutes} = 3 - 0.6 = 2.4

hours (or

\frac{12}{5}

hours). The new combined rate is

\frac{120}{2.4} = 50

miles per hour.

36 minutes is equivalent to

\frac{36}{60} = 0.6

hours.

3

Express the modified individual rates algebraically.

Clara's new speed is

1.2c = \frac{6}{5}c

. Dan's new speed is

\left(1 + \frac{1}{3}\right)d = \frac{4}{3}d

.

Increasing by

20\%

multiplies rate by

1.2

, and increasing by

33\frac{1}{3}\%

multiplies rate by

\frac{4}{3}

.

4

Formulate and solve the system of linear equations.

We have the system:
1)

c + d = 40 \implies d = 40 - c

2)

\frac{6}{5}c + \frac{4}{3}d = 50

Substituting

d = 40 - c

into equation 2:

\frac{6}{5}c + \frac{4}{3}(40 - c) = 50

Multiplying the entire equation by

15

to eliminate denominators:

18c + 20(40 - c) = 750

18c + 800 - 20c = 750

-2c = -50 \implies c = 25

Solving for

c

yields Clara's original speed in miles per hour.

Key Concept

Distance-Rate-Time modeling using systems of linear equations with percentage changes in individual rates

Estimated Time:2m 30s

Question 208Question →

For all real numbers

x

and

y

, the custom operation

\diamondsuit

is defined by

x \diamondsuit y = x^2 - y^2 - 2x + 2y

The function

f

is defined for all real numbers

t

by

f(t) = t + 1

. Which of the following statements must be true for all real numbers

a

and

b

? Select all such statements.

Select all that apply

f(a + b) = f(a) + f(b)

If

a \neq b

, then

a \diamondsuit b = 0

if and only if

a + b = 2

.

f(a) \diamondsuit f(b) = a^2 - b^2

a \diamondsuit (-a) = 0

(a + b) \diamondsuit 0 = a^2 + b^2 - 2a - 2b

Show answer & explanation

Answer: If $a \neq b$ , then $a \diamondsuit b = 0$ if and only if $a + b = 2$ .; $f(a) \diamondsuit f(b) = a^2 - b^2$

Answer

The statements 'If

a \neq b

, then

a \diamondsuit b = 0

if and only if

a + b = 2

' and '

f(a) \diamondsuit f(b) = a^2 - b^2

' must be true.

The statement claiming that if

a \neq b

,

a \diamondsuit b = 0

if and only if

a + b = 2

is correct because factoring

a \diamondsuit b

gives

(a - b)(a + b - 2) = 0

, which requires

a + b = 2

when

a - b \neq 0

. The statement claiming

f(a) \diamondsuit f(b) = a^2 - b^2

is also correct because substituting

f(a) = a + 1

and

f(b) = b + 1

into the operation yields

[(a + 1) - (b + 1)][(a + 1) + (b + 1) - 2] = (a - b)(a + b) = a^2 - b^2

.

Step-by-Step Solution

1

Factor the custom operation definition

x \diamondsuit y = (x^2 - y^2) - 2(x - y) = (x - y)(x + y) - 2(x - y) = (x - y)(x + y - 2)

Factoring groups terms into products that simplify statement testing.

2

Analyze the condition for

a \diamondsuit b = 0

when

a \neq b

(a - b)(a + b - 2) = 0 \implies a + b - 2 = 0 \implies a + b = 2

Since

a \neq b

,

a - b \neq 0

, allowing division by

(a - b)

to isolate

a + b = 2

.

3

Evaluate

f(a) \diamondsuit f(b)

using

f(t) = t + 1

f(a) = a + 1, f(b) = b + 1 \implies f(a) \diamondsuit f(b) = [(a + 1) - (b + 1)][(a + 1) + (b + 1) - 2] = (a - b)(a + b) = a^2 - b^2

Direct substitution into the factored form simplifies the evaluation.

4

Evaluate remaining statements to identify counterexamples

f(a + b) = a + b + 1 \neq a + b + 2 = f(a) + f(b)

;

a \diamondsuit (-a) = -4a \neq 0

;

(a + b) \diamondsuit 0 = a^2 + 2ab + b^2 - 2a - 2b \neq a^2 + b^2 - 2a - 2b

Confirming false statements ensures accurate multi-selection.

Key Concept

Algebraic manipulation of custom binary operations and function evaluation

Question 209Question →

An electronics manufacturer models the monthly demand $N$ for a specific model of tablet as $N(P) = 1,200 - 3P$ , where $P$ is the selling price per tablet in dollars. The monthly total cost of producing these tablets consists of a fixed overhead cost of $\$15,000$ plus a variable cost of $\$100$ per tablet produced. If the manufacturer earned a monthly net profit of $\$45,000$ , what is the smaller of the two possible selling prices, in dollars, that could yield this profit?

Show answer & explanation

Answer: 200

Answer

The smaller of the two possible selling prices is 200 dollars.

Total revenue is given by

R(P) = P \cdot N(P) = P(1,200 - 3P) = 1,200P - 3P^2

, and total cost is given by

C(P) = 15,000 + 100(1,200 - 3P) = 135,000 - 300P

. The profit equation

\text{Profit}(P) = R(P) - C(P) = 45,000

yields

-3P^2 + 1,500P - 135,000 = 45,000

. Rearranging gives

-3P^2 + 1,500P - 180,000 = 0

, which simplifies to

P^2 - 500P + 60,000 = 0

. Factoring gives

(P - 200)(P - 300) = 0

, leading to prices of

\$200

and

\$300

. The smaller price is

\$200

.

Step-by-Step Solution

1

Formulate algebraic functions for monthly revenue and monthly total cost in terms of price

P

.

Revenue

R(P) = P(1,200 - 3P) = 1,200P - 3P^2

and Total Cost

C(P) = 15,000 + 100(1,200 - 3P) = 135,000 - 300P

.

Total revenue equals unit price multiplied by demand quantity, while total cost equals fixed overhead plus variable cost per unit multiplied by demand quantity.

2

Formulate the net profit function

\text{Profit}(P) = R(P) - C(P)

and equate it to the target profit of

\$45,000

.

(1,200P - 3P^2) - (135,000 - 300P) = 45,000

, which simplifies to

-3P^2 + 1,500P - 180,000 = 0$.

Net profit is defined as total revenue minus total cost.

3

Divide the resulting quadratic equation by

-3

to normalize coefficients and solve for

P

.

P^2 - 500P + 60,000 = 0 \implies (P - 200)(P - 300) = 0

, giving roots

P = 200

and

P = 300$.

Simplifying the quadratic equation reveals two factored linear roots.

4

Identify the smaller of the two valid price solutions.

P = 200 dollars.

The question specifically requests the smaller of the two possible selling prices.

Key Concept

Algebraic Modeling of Revenue, Cost, and Profit Functions using Quadratic Equations

Question 210Question →

If $x$ is a real number that satisfies the inequality $|4 - 3x| \le 11$ , what is the maximum possible value of the expression $\frac{2 - x}{-3}$ ?

1

-\frac{13}{9}

-1

\frac{5}{3}

3

Show answer & explanation

Answer: $1$

Answer

1

Expanding the absolute value inequality

|4 - 3x| \le 11

gives

-11 \le 4 - 3x \le 11

. Subtracting

4

from all sides yields

-15 \le -3x \le 7

. Dividing by

-3

and reversing the inequality signs gives

-\frac{7}{3} \le x \le 5

. Simplifying the target expression gives

\frac{2 - x}{-3} = \frac{x - 2}{3}

. Because

\frac{x - 2}{3}

increases with

x

, substituting the maximum allowed value

x = 5

gives

\frac{5 - 2}{3} = 1

.

Step-by-Step Solution

1

Unfold the absolute value inequality into a compound inequality

-11 \le 4 - 3x \le 11

For any real number

k \ge 0

,

|A| \le k

is equivalent to

-k \le A \le k

.

2

Isolate the variable term by subtracting 4 from all parts

-15 \le -3x \le 7

Subtracting a constant maintains the direction of the inequality.

3

Divide by -3 and flip the inequality signs

5 \ge x \ge -\frac{7}{3}

, or equivalently

-\frac{7}{3} \le x \le 5

Dividing an inequality by a negative number reverses the inequality direction.

4

Simplify the target expression and maximize it over the valid domain of x

\frac{2 - x}{-3} = \frac{x - 2}{3}

. Maximum occurs at

x = 5

:

\frac{5 - 2}{3} = 1

Since

\frac{x - 2}{3}

increases as

x

increases, its maximum occurs at the largest possible value of

x

.

Key Concept

Linear Inequalities and Absolute Value Bounds

Question 211Question →

Consider the following system of linear equations in variables

x

,

y

, and

z

, where

a

,

b

, and

c

are real constants:

\begin{aligned} 2x - y + 3z &= a \\ x + 2y - z &= b \\ 7x + 4y + 3z &= c \end{aligned}

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

Select all that apply

If

c = 2a + 3b

, the system has infinitely many solutions.

There exist no real values of

a

,

b

, and

c

for which the system has a unique solution.

If

a = 0

,

b = 0

, and

c = 0

, the system has

(0, 0, 0)

as its only solution.

If

c \neq 2a + 3b

, the system has infinitely many solutions.

If

a = 1

,

b = 2

, and

c = 7

, the system has at least one solution.

Show answer & explanation

Answer: If $c = 2a + 3b$ , the system has infinitely many solutions.; There exist no real values of $a$ , $b$ , and $c$ for which the system has a unique solution.

Answer

The correct statements are: 'If

c = 2a + 3b

, the system has infinitely many solutions' and 'There exist no real values of

a

,

b

, and

c

for which the system has a unique solution.'

The correct options accurately reflect the structural properties of the system. First, scaling the first equation by 2 and the second by 3 yields

7x + 4y + 3z = 2a + 3b

. Comparing this with the third equation,

7x + 4y + 3z = c

, shows that when

c = 2a + 3b

, the third equation provides no new constraints, leaving 2 independent equations in 3 variables and thus producing infinitely many solutions. Second, because the coefficient matrix has linearly dependent rows, its rank is 2 (less than the 3 variables), making a unique solution impossible regardless of the constants

a

,

b

, and

c

.

Step-by-Step Solution

1

Analyze the linear dependence of the left-hand sides of the equations.

Observe that

2(2x - y + 3z) + 3(x + 2y - z) = (4x + 3x) + (-2y + 6y) + (6z - 3z) = 7x + 4y + 3z

.

Finding a linear combination of the first two equations that produces the left-hand side of the third equation allows us to analyze system consistency.

2

Determine the condition for consistency.

The system is consistent if and only if

2a + 3b = c

.

If

c = 2a + 3b

, the third equation is a linear combination of the first two, resulting in a system of 2 independent equations in 3 variables, which yields infinitely many solutions.

3

Evaluate the possibility of a unique solution.

The rank of the coefficient matrix is 2, which is strictly less than the number of variables (3).

A system of linear equations has a unique solution if and only if the rank of the coefficient matrix equals the number of variables. Thus, no choice of

a, b, c

can produce a unique solution.

4

Verify specific numerical options.

For

a=0, b=0, c=0

,

c = 2(0)+3(0)=0

, giving infinitely many solutions. For

a=1, b=2, c=7

,

2(1)+3(2)=8 \neq 7

, giving zero solutions.

Testing specific constant values confirms consistency or inconsistency based on whether

c = 2a + 3b

is satisfied.

Key Concept

Consistency and Number of Solutions in 3x3 Systems of Linear Equations

Question 212Question →

If $x$ is an integer that satisfies both $|7 - 2x| \le 9$ and $\frac{1 - x}{-2} > 2$ , what is the sum of all possible values of $x$ ?

Show answer & explanation

Answer: 21

Answer

The sum of all possible integer values of

x

is 21.

To solve the system of inequalities, analyze each part separately. For

|7 - 2x| \le 9

, express it as

-9 \le 7 - 2x \le 9

. Subtracting

7

yields

-16 \le -2x \le 2

. Dividing all parts by

-2

requires reversing the inequality symbols, giving

8 \ge x \ge -1

, or

-1 \le x \le 8

. For

\frac{1 - x}{-2} > 2

, multiplying by

-2

reverses the inequality symbol to

1 - x < -4

. Subtracting

1

gives

-x < -5

, and dividing by

-1

reverses the symbol again to

x > 5

. The integer values satisfying both

-1 \le x \le 8

and

x > 5

are

6, 7,

and

8

. Summing these integer values yields

6 + 7 + 8 = 21

.

Step-by-Step Solution

1

Solve the absolute value inequality

|7 - 2x| \le 9

-1 \le x \le 8

Unfolding the absolute value gives

-9 \le 7 - 2x \le 9

. Subtracting 7 yields

-16 \le -2x \le 2

, and dividing by

-2

reverses the inequalities to give

-1 \le x \le 8

.

2

Solve the linear inequality \frac{1 - x}{-2} > 2

x > 5

Multiplying by

-2

flips the inequality sign to yield

1 - x < -4

. Subtracting 1 gives

-x < -5

, and dividing by

-1

flips the inequality sign again to yield

x > 5

.

3

Find the overlapping integer values for x

x \in \{6, 7, 8\}

The integers that are greater than 5 and less than or equal to 8 are 6, 7, and 8.

4

Sum the integer solutions

21

Calculating 6 + 7 + 8 = 21.

Key Concept

Solving absolute value inequalities and compound linear inequalities with sign reversal when dividing by negative numbers.

Question 213Question →

The quadratic equation $x^2 - (k - 2)x + (k - 5) = 0$ has two distinct real roots, $\alpha$ and $\beta$ . If $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 1$ , what is the value of the constant $k$ ?

\frac{9}{4}

\frac{11}{4}

\frac{7}{2}

\frac{31}{8}

\frac{31}{4}

Show answer & explanation

Answer: $\frac{11}{4}$

Answer

The value of the constant

k

is

\frac{11}{4}

.

Using Vieta's formulas for the quadratic equation

x^2 - (k - 2)x + (k - 5) = 0

, we have

\alpha + \beta = k - 2

and

\alpha\beta = k - 5

. The given condition

\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 1

simplifies to

\frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2} = 1

. Substituting the expressions in terms of

k

yields

\frac{(k - 2)^2 - 2(k - 5)}{(k - 5)^2} = 1

, which expands to

k^2 - 6k + 14 = k^2 - 10k + 25

. Subtracting

k^2

from both sides gives

4k = 11

, so

k = \frac{11}{4}

.

Step-by-Step Solution

1

Apply Vieta's formulas to express root sum and product in terms of

k

\alpha + \beta = k - 2

and

\alpha\beta = k - 5

For any standard quadratic

ax^2 + bx + c = 0

, the sum of roots is

-\frac{b}{a}

and the product of roots is

\frac{c}{a}

.

2

Rewrite the given sum of reciprocal squares using algebraic identities

\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}

Combining fractions over a common denominator allows substitution of the known sum

\alpha+\beta

and product

\alpha\beta

.

3

Substitute the expressions for sum and product into the equation

\frac{(k - 2)^2 - 2(k - 5)}{(k - 5)^2} = 1

Set the algebraic expression equal to the given target value of 1.

4

Expand both numerator and denominator and solve for

k

\frac{k^2 - 4k + 4 - 2k + 10}{k^2 - 10k + 25} = 1 \implies k^2 - 6k + 14 = k^2 - 10k + 25 \implies 4k = 11 \implies k = \frac{11}{4}

Equating numerator and denominator eliminates the quadratic

k^2

terms, yielding a linear equation in

k

.

5

Verify that

k = \frac{11}{4}

yields real, distinct, non-zero roots

Discriminant

D = (k-2)^2 - 4(k-5) = 0.5625 + 9 = 9.5625 > 0

, and

\alpha\beta = -2.25 \neq 0

Ensures the quadratic has two distinct real roots as required by the problem prompt.

Key Concept

Vieta's Formulas and Symmetric Polynomial Transformations of Quadratic Roots

Question 214Question →

In the $xy$ -plane, line $\ell_1$ passes through the point $(3, 7)$ and is perpendicular to line $\ell_2$ , which is defined by the equation $2x - 5y = 10$ . Line $\ell_3$ is parallel to line $\ell_1$ and has a $y$ -intercept that is $6$ units greater than the $y$ -intercept of line $\ell_1$ . If line $\ell_3$ intersects the $x$ -axis at the point $(a, 0)$ , what is the value of $a$ ?

Show answer & explanation

Answer: 8.2

Answer

The value of

a

is 8.2 (or

\frac{41}{5}

).

Converting

2x - 5y = 10

to slope-intercept form yields

y = \frac{2}{5}x - 2

, so the slope of line

\ell_2

is

\frac{2}{5}

. Line

\ell_1

is perpendicular to

\ell_2

, giving it a slope of

m_1 = -\frac{5}{2}

. Using the point

(3, 7)

, the line equation for

\ell_1

is

y - 7 = -2.5(x - 3)

, which simplifies to

y = -2.5x + 14.5

. Line

\ell_3

is parallel to

\ell_1

, so

m_3 = -2.5

, and its

y

-intercept is

14.5 + 6 = 20.5

. Writing the equation for line

\ell_3

as

y = -2.5x + 20.5

and setting

y = 0

gives

0 = -2.5a + 20.5 \implies 2.5a = 20.5 \implies a = 8.2

.

Step-by-Step Solution

1

Determine the slope of line

\ell_2

The slope of line

\ell_2

is

\frac{2}{5}

.

Convert

2x - 5y = 10

into slope-intercept form

y = \frac{2}{5}x - 2

.

2

Determine the slope of line

\ell_1

The slope of line

\ell_1

is

-\frac{5}{2}

.

Perpendicular lines have slopes that are negative reciprocals of each other.

3

Find the

y

-intercept of line

\ell_1

The

y

-intercept of

\ell_1

is

\frac{29}{2} = 14.5

.

Apply point-slope form with point

(3, 7)

:

y - 7 = -\frac{5}{2}(x - 3) \implies y = -\frac{5}{2}x + \frac{29}{2}

.

4

Construct the equation for line

\ell_3

The equation of

\ell_3

is

y = -\frac{5}{2}x + \frac{41}{2}

.

Line

\ell_3

has the same slope as

\ell_1

(

-\frac{5}{2}

) and its

y

-intercept is

14.5 + 6 = 20.5 = \frac{41}{2}

.

5

Calculate the

x

-intercept coordinate

a

of line

\ell_3

a = 8.2

.

Substitute

y = 0

into the equation for

\ell_3

:

0 = -\frac{5}{2}a + \frac{41}{2} \implies 5a = 41 \implies a = 8.2

.

Key Concept

Perpendicular and parallel slopes, point-slope equation derivation, line transformations, and intercept determination.

Estimated Time:2m 30s

Question 215Question →

For all real numbers

x

, the function

f

is defined by

f(x) = x^2 - 2x

. The custom binary operation

\otimes

is defined for all real numbers

a

and

b

by

a \otimes b = f(a + b) - f(a - b)

If

k

is a constant such that

(k \otimes 3) \otimes 2 = 88

, what is the value of

k

?

\frac{5}{12}

\frac{11}{12}

2

\frac{17}{6}

6

Show answer & explanation

Answer: 2

Answer

The value of

k

is 2.

Expanding the definition

a \otimes b = f(a+b) - f(a-b)

using

f(x) = x^2 - 2x

gives

[(a+b)^2 - 2(a+b)] - [(a-b)^2 - 2(a-b)] = (a^2 + 2ab + b^2 - 2a - 2b) - (a^2 - 2ab + b^2 - 2a + 2b) = 4ab - 4b = 4b(a-1)

. Evaluating

k \otimes 3

yields

12k - 12

. Substituting this into

(12k - 12) \otimes 2

yields

4(2)(12k - 12 - 1) = 96k - 104

. Setting

96k - 104 = 88

gives

96k = 192

, so

k = 2

.

Step-by-Step Solution

1

Express the custom operation

a \otimes b

in simplified algebraic terms.

a \otimes b = f(a+b) - f(a-b) = [(a+b)^2 - 2(a+b)] - [(a-b)^2 - 2(a-b)] = 4ab - 4b = 4b(a - 1)

.

Expanding and canceling common terms simplifies the binary operation definition.

2

Evaluate the inner operation

k \otimes 3

.

k \otimes 3 = 4(3)(k - 1) = 12k - 12

.

Substitute

a = k

and

b = 3

into the simplified operation formula

4b(a - 1)

.

3

Evaluate the outer operation

(12k - 12) \otimes 2

.

(12k - 12) \otimes 2 = 4(2)[(12k - 12) - 1] = 8(12k - 13) = 96k - 104

.

Substitute

a = 12k - 12

and

b = 2

into

4b(a - 1)

.

4

Set the resulting expression equal to 88 and solve for

k

.

96k - 104 = 88 \implies 96k = 192 \implies k = 2

.

Linear algebraic equation solving yields the value of

k

.

Key Concept

Custom Binary Operations and Nested Function Evaluation

Question 216Question →

A solar power facility operates two types of solar panel arrays: Array Alpha and Array Beta. When operational, Array Beta produces electricity at a constant hourly rate that is $25\%$ greater than the constant hourly rate of Array Alpha. On a clear day, Array Alpha operated for $8$ hours and Array Beta operated for $6$ hours, together generating a total of $3,100$ kilowatt-hours (kWh) of electricity. What was the hourly production rate of Array Alpha, in kWh per hour?

160

193.75

200

248

250

Show answer & explanation

Answer: 200

Answer

200 kWh per hour

Let the hourly rate of Array Alpha be

r

kWh per hour. Since Array Beta produces at a rate

25\%

greater, its hourly rate is

1.25r

kWh per hour. Multiply each rate by the respective number of hours operated: Array Alpha produced

8r

kWh and Array Beta produced

6 \times 1.25r = 7.5r

kWh. Combining these gives

8r + 7.5r = 15.5r = 3,100

. Solving for

r

yields

r = 200

kWh per hour.

Step-by-Step Solution

1

Define variables for the hourly rates of Array Alpha and Array Beta.

Let

r

be the hourly rate of Array Alpha in kWh per hour. Since Array Beta's rate is

25\%

greater, Array Beta's rate is

r + 0.25r = 1.25r

kWh per hour.

Establishing the linear relationship between the two unknown rates.

2

Set up the total electricity output equation using rate times time for each array.

\text{Total Energy} = (8 \text{ hours} \times r) + (6 \text{ hours} \times 1.25r) = 3,100

Total production is the sum of production from Array Alpha and Array Beta.

3

Simplify the algebraic equation and solve for

r

.

8r + 7.5r = 3,100 \implies 15.5r = 3,100 \implies r = \frac{3,100}{15.5} = 200

Isolating

r

gives the hourly rate of Array Alpha.

Key Concept

Linear Algebraic Modeling of Combined Rates and Percentages

Estimated Time:1m 30s

Question 217Question →

An investor allocates a total of $\$24,000$ among three accounts: Account A, which earns $3\%$ annual simple interest; Account B, which earns $5\%$ annual simple interest; and Account C, which earns $7\%$ annual simple interest. The total annual interest earned from all three accounts combined at the end of one year is $\$1,260$ . If the amount invested in Account C is $\$2,000$ more than twice the amount invested in Account A, what is the amount invested in Account B?

\$1,000

\$4,000

\$11,000

\$15,000

\$19,000

Show answer & explanation

Answer: $\$19,000$

Answer

The amount invested in Account B is

\$19,000

.

The system of linear equations representing the total investment, interest earned, and relative account values yields

A = \$1,000

,

C = \$4,000

, and

B = \$19,000

. Thus, the amount invested in Account B is

\$19,000

.

Step-by-Step Solution

1

Define variables and write the system of three linear equations based on total investment, total annual interest, and account relationship.

Let

A

,

B

, and

C

represent the dollars invested in Accounts A, B, and C respectively.
Equation (1):

A + B + C = 24,000

Equation (2):

0.03A + 0.05B + 0.07C = 1,260

, which simplifies to

3A + 5B + 7C = 126,000

Equation (3):

C = 2A + 2,000

, or

2A - C = -2,000

Translate the verbal conditions into an explicit

3 \times 3

system of linear equations.

2

Substitute

C = 2A + 2,000

into Equations (1) and (2) to reduce the system to two variables (

A

and

B

).

From Equation (1):

A + B + (2A + 2,000) = 24,000 \implies 3A + B = 22,000 \implies B = 22,000 - 3A

.
From Equation (2):

3A + 5B + 7(2A + 2,000) = 126,000 \implies 17A + 5B + 14,000 = 126,000 \implies 17A + 5B = 112,000

.

Eliminate variable

C

to simplify solving the linear system.

3

Substitute

B = 22,000 - 3A

into

17A + 5B = 112,000

to solve for

A

.

17A + 5(22,000 - 3A) = 112,000 \implies 17A + 110,000 - 15A = 112,000 \implies 2A = 2,000 \implies A = 1,000

.

Solve for the single variable

A

.

4

Determine the values of

C

and

B

.

C = 2(1,000) + 2,000 = 4,000

.

B = 22,000 - 3(1,000) = 19,000

.

Substitute

A = 1,000

back into the expressions for

C

and

B

to find the targeted investment amount.

Key Concept

Setting up and solving a system of three linear equations in three variables by substitution and elimination.

Estimated Time:2m 0s

Question 218Question →

Two industrial pumps, Pump A and Pump B, are used to fill a storage tank. Working alone at its constant standard operating rate, Pump A can fill the empty tank in $12$ hours. Working alone at its constant standard operating rate, Pump B can fill the empty tank in $18$ hours.

To fill the tank, both pumps begin operating simultaneously at their standard rates. After $4$ hours of joint operation, Pump A undergoes maintenance that reduces its operating rate by $25\%$ , while Pump B continues operating at its standard rate. Exactly $2$ hours after Pump A's rate is reduced, Pump B's operating rate is increased by $50\%$ above its standard rate due to a valve adjustment. If both pumps continue operating at these adjusted rates until the tank is full, how many total hours from the initial start does it take to completely fill the tank?

6\frac{2}{3}

hours

7\frac{1}{7}

hours

7\frac{3}{7}

hours

8\frac{1}{7}

hours

8\frac{2}{3}

hours

Show answer & explanation

Answer: $7\frac{3}{7}$ hours

Answer

The total time required to completely fill the tank from the initial start is

7\frac{3}{7}

hours.

The solution proceeds in three stages. In the first 4 hours, both pumps at standard rates completed 5/9 of the job. In the next 2 hours, Pump A operated at 1/16 tank/hr and Pump B at 1/18 tank/hr, completing an additional 17/72 of the job, bringing total completed work to 19/24 of the tank. For the final 5/24 of the tank, Pump A (1/16 tank/hr) and Pump B (1/12 tank/hr) worked at a combined rate of 7/48 tank/hr, taking 10/7 (or 1 3/7) hours. The total time is 4 + 2 + 1 3/7 = 7 3/7 hours.

Step-by-Step Solution

1

Determine individual standard rates and combined initial rate.

Pump A rate

r_A = \frac{1}{12}

tank/hr; Pump B rate

r_B = \frac{1}{18}

tank/hr. Initial combined rate

r_1 = \frac{1}{12} + \frac{1}{18} = \frac{5}{36}

tank/hr.

Work rate is the reciprocal of the total time taken to complete one full job.

2

Calculate work completed in Phase 1 (first 4 hours).

Work completed

= 4 \times \frac{5}{36} = \frac{20}{36} = \frac{5}{9}

of the tank. Remaining work

= 1 - \frac{5}{9} = \frac{4}{9}

of the tank.

Both pumps work at standard combined rate for 4 hours.

3

Adjust rates for Phase 2 (hours 4 to 6) and calculate work done.

Pump A reduced rate

= 0.75 \times \frac{1}{12} = \frac{1}{16}

tank/hr. Combined rate

r_2 = \frac{1}{16} + \frac{1}{18} = \frac{17}{144}

tank/hr. Work done in 2 hours

= 2 \times \frac{17}{144} = \frac{17}{72}

. Cumulative work

= \frac{5}{9} + \frac{17}{72} = \frac{57}{72} = \frac{19}{24}

of the tank.

Phase 2 lasts 2 hours with Pump A operating at

75\%

efficiency and Pump B at standard rate.

4

Adjust rates for Phase 3 (from hour 6 onwards) and calculate remaining time.

Pump B increased rate

= 1.50 \times \frac{1}{18} = \frac{1}{12}

tank/hr. Combined rate

r_3 = \frac{1}{16} + \frac{1}{12} = \frac{7}{48}

tank/hr. Remaining work

= 1 - \frac{19}{24} = \frac{5}{24}

. Additional time required

= \frac{5/24}{7/48} = \frac{5}{24} \times \frac{48}{7} = \frac{10}{7} = 1\frac{3}{7}

hours.

Divide remaining fractional work by the new combined rate.

5

Calculate total elapsed time.

Total time

= 4 + 2 + 1\frac{3}{7} = 7\frac{3}{7}

hours.

Sum the durations of all three phases.

Key Concept

Multi-stage work-rate problems with variable individual rates and fractional job completion tracking.

Alternative Method

Define the tank capacity as 144 units (the LCM of 12, 18, 16, 48). Pump A standard rate = 12 units/hr; Pump B standard rate = 8 units/hr. Phase 1 (4 hrs): combined rate = 20 units/hr, work done = 80 units. Phase 2 (2 hrs): Pump A rate = 9 units/hr, Pump B rate = 8 units/hr, combined rate = 17 units/hr, work done = 34 units. Total work done in 6 hrs = 114 units. Remaining work = 30 units. Phase 3: Pump A rate = 9 units/hr, Pump B rate = 12 units/hr, combined rate = 21 units/hr. Additional time = 30/21 = 10/7 hrs. Total time = 6 + 10/7 = 7 3/7 hrs.

Estimated Time:2m 30s

Question 219Question →

For any real number $x$ , the custom unary operation $\triangle$ is defined by $\triangle(x) = 2x - x^2$ . For all real numbers $u$ and $v$ , the custom binary operation $\odot$ is defined by $u \odot v = \triangle(u) + \triangle(v) + uv$ . The function $f$ is defined for all real numbers $x$ by $f(x) = x \odot (3 - x)$ . What is the maximum value of $f(x)$ ?

Show answer & explanation

Answer: 3.75

Answer

The maximum value of

f(x)

is

3.75

(or

\frac{15}{4}

).

Expanding

f(x) = x \odot (3 - x)

yields

\triangle(x) + \triangle(3 - x) + x(3 - x)

. Substituting

\triangle(t) = 2t - t^2

gives

(2x - x^2) + (6 - 2x - (9 - 6x + x^2)) + (3x - x^2) = -3x^2 + 9x - 3

. The vertex of this quadratic function occurs at

x = \frac{9}{6} = 1.5

, where

f(1.5) = -3(2.25) + 13.5 - 3 = 3.75

.

Step-by-Step Solution

1

Evaluate the custom unary operation

\triangle

for each argument

\triangle(x) = 2x - x^2

and

\triangle(3-x) = 2(3-x) - (3-x)^2 = -x^2 + 4x - 3

Substitute

x

and

3-x

into the definition

\triangle(t) = 2t - t^2

and expand carefully.

2

Compute the product term

uv

x(3 - x) = 3x - x^2

The binary definition

u \odot v

includes an additive product term

uv

.

3

Sum all components to construct the explicit quadratic expression for

f(x)

f(x) = (2x - x^2) + (-x^2 + 4x - 3) + (3x - x^2) = -3x^2 + 9x - 3

Combine like terms for

x^2

,

x

, and the constant.

4

Determine the vertex of the downward-opening parabola

f(x) = -3x^2 + 9x - 3

x = \frac{3}{2} = 1.5

, yielding

f(1.5) = 3.75$

Since the coefficient of

x^2

is negative (

-3 < 0

), the maximum occurs at

x = -\frac{b}{2a} = \frac{9}{6} = 1.5

.

Key Concept

Custom operations combined with quadratic function optimization

Estimated Time:2m 0s

Question 220Question →

In the $xy$ -plane, line $\ell_1$ passes through the points $(2, k)$ and $(k, 14)$ , where $k$ is a constant. Line $\ell_2$ passes through the point $(k, 14)$ and has a $y$ -intercept at $(0, 22)$ . If line $\ell_1$ is perpendicular to line $\ell_2$ , what is the sum of all possible values of $k$ ?

-22

-14

-6

6

8

Show answer & explanation

Answer: -6

Answer

The sum of all possible values of

k

is -6.

The correct answer is -6 because the slope of line

\ell_1

is

m_1 = \frac{14-k}{k-2}

and the slope of line

\ell_2

is

m_2 = -\frac{8}{k}

. Since the lines are perpendicular, their slopes multiply to

-1

, yielding

\frac{8(14-k)}{k(k-2)} = 1

. Solving the resulting quadratic equation

k^2 + 6k - 112 = 0

yields

k = 8

and

k = -14

. The sum of these values is

8 + (-14) = -6

.

Step-by-Step Solution

1

Calculate the slope of line

\ell_1

in terms of

k

.

m_1 = \frac{14 - k}{k - 2}

for

k \neq 2

.

The slope of a line passing through

(x_1, y_1)

and

(x_2, y_2)

is given by

\frac{y_2 - y_1}{x_2 - x_1}

.

2

Calculate the slope of line

\ell_2

using

(k, 14)

and the

y

-intercept

(0, 22)

.

m_2 = \frac{14 - 22}{k - 0} = -\frac{8}{k}

for

k \neq 0

.

The

y

-intercept

(0, 22)

provides a second point on line

\ell_2

to find its slope.

3

Set up the perpendicularity condition

m_1 \cdot m_2 = -1

.

\left(\frac{14 - k}{k - 2}\right)\left(-\frac{8}{k}\right) = -1 \implies \frac{8(14 - k)}{k(k - 2)} = 1$.

Two non-vertical lines are perpendicular if and only if the product of their slopes equals

-1

.

4

Expand and rearrange the equation into standard quadratic form.

112 - 8k = k^2 - 2k \implies k^2 + 6k - 112 = 0.

Multiplying both sides by

k(k - 2)

clears the denominator to form a quadratic equation.

5

Factor the quadratic equation to find all possible values of

k

.

(k + 14)(k - 8) = 0 \implies k = -14 \text{ or } k = 8.

The quadratic expression factors neatly, giving two valid non-zero values for

k

.

6

Calculate the sum of all possible values of

k

.

(-14) + 8 = -6.

Summing the two solutions gives the required final numerical value.

Key Concept

Perpendicular Slopes and Quadratic Line Equations

Estimated Time:2m 30s

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