Algebra Practice Questions — GRE General Test — Page 10

A financial analyst models a company's weekly metrics using three variables—revenue

R

, operating cost

C

, and advertising expenditure

A

, measured in thousands of dollars. The metrics satisfy the following system of linear equations, where

k

is a real constant:

\begin{aligned} 4R - 3C + 2A &= 18 \\ R + 5C - 6A &= 14 \\ 5R + 2C - 4A &= k \end{aligned}

If this system of linear equations is consistent (has at least one solution), what is the value of

k

?

4

16

32

-32

252

Show answer & explanation

Answer: 32

Answer

32

Adding the left-hand sides of the first two equations yields

(4R - 3C + 2A) + (R + 5C - 6A) = 5R + 2C - 4A

, which is identical to the left-hand side of the third equation. For the linear system to have at least one solution (to be consistent), the right-hand side constant must satisfy the exact same linear combination:

k = 18 + 14 = 32

.

Step-by-Step Solution

1

Examine the linear combination of the left-hand sides of the first two equations

(4R - 3C + 2A) + (R + 5C - 6A) = 5R + 2C - 4A

Observing that the sum of the coefficients of the first two equations matches the left-hand side of the third equation.

2

Apply the condition for system consistency

Right-hand side of Equation 3 must equal Right-hand side of Equation 1 + Right-hand side of Equation 2

For a system with linearly dependent left-hand sides to be consistent, the same linear combination must hold for the right-hand constants.

3

Calculate the value of

k

k = 18 + 14 = 32

Adding the constants from the right-hand side of the first two equations gives the consistent value for

k

.

Key Concept

Consistency and Linear Dependence in Systems of Linear Equations

Question 182Question →

For how many real values of the constant

a

does the following system of linear equations in

x

,

y

, and

z

have no solution?

\begin{aligned} x + y - z &= 3 \\ x + (a-1)y + 3z &= 5 \\ x + 4y + (a+1)z &= a + 2 \end{aligned}

None

Exactly one

Exactly two

Exactly three

Infinitely many

Show answer & explanation

Answer: Exactly one

Answer

Exactly one

To find when the system has no solution, we first eliminate

x

by subtracting the first equation from the second and third equations. This produces a two-variable system in

y

and

z

:

(a-2)y + 4z = 2

and

3y + (a+2)z = a - 1

. The determinant of this system's coefficients is

(a-2)(a+2) - 12 = a^2 - 16

. Setting the determinant to zero yields two critical values:

a = 4

and

a = -4

. Testing

a = 4

simplifies both reduced equations to

y + 2z = 1

, which means the system is consistent with infinitely many solutions. Testing

a = -4

yields

-3y + 2z = 1

and

-3y + 2z = 5

, which is impossible (

1 = 5

), making the system inconsistent. Thus, there is exactly one real value of

a

(

a = -4

) for which the system has no solution.

Step-by-Step Solution

1

Eliminate the variable

x

from the second and third equations using the first equation.

Subtracting the first equation

x + y - z = 3

from the second equation yields:

(a-2)y + 4z = 2

Subtracting the first equation from the third equation yields:

3y + (a+2)z = a - 1

Reducing the

3 \times 3

system to a

2 \times 2

system in

y

and

z

simplifies the analysis of linear dependence and consistency.

2

Determine the values of

a

for which the reduced

2 \times 2

system lacks a unique solution by setting its coefficient determinant to zero.

The determinant of the coefficient matrix is:

D = (a-2)(a+2) - (3)(4) = a^2 - 4 - 12 = a^2 - 16

Setting

D = 0

yields

a^2 = 16

, which gives

a = 4

or

a = -4

.

A system of linear equations has either a unique solution (when the determinant is non-zero) or non-unique behavior—either no solution or infinitely many solutions—when the determinant is zero.

3

Test

a = 4

in the reduced system.

Substituting

a = 4

into the reduced equations gives:

2y + 4z = 2 \implies y + 2z = 1

3y + 6z = 3 \implies y + 2z = 1

Since both equations are identical, the system is consistent and has infinitely many solutions.

When equation ratios match completely including constant terms, the equations represent identical hyperplanes, yielding infinitely many solutions.

4

Test

a = -4

in the reduced system.

Substituting

a = -4

into the reduced equations gives:

-6y + 4z = 2 \implies -3y + 2z = 1

3y - 2z = -5 \implies -3y + 2z = 5

Comparing these gives

1 = 5

, which is a contradiction. Thus, for

a = -4

, the system has no solution.

When parallel equations have equal coefficient ratios but unequal constant ratios, the system is inconsistent.

Key Concept

Parametric Systems of Linear Equations and Consistency Conditions

Question 183Question →

If $x$ is a real number satisfying the exponential equation $9^{x+1} - 3^{2x+1} = 162$ , what is the value of $4^x$ ?

4

6

8

16

64

Show answer & explanation

Answer: 8

Answer

8

By converting

9^{x+1}

to

3^{2x+2}

and factoring out

3^{2x}

, the equation simplifies to

6 \cdot 3^{2x} = 162

. Dividing by 6 gives

3^{2x} = 27

, so

2x = 3

and

x = \frac{3}{2}

. Raising 4 to the power of

\frac{3}{2}

yields

(\sqrt{4})^3 = 8

.

Step-by-Step Solution

1

Rewrite terms with a common base of 3.

9^{x+1} = (3^2)^{x+1} = 3^{2(x+1)} = 3^{2x+2}

Since 9 is a power of 3 (

9 = 3^2

), applying exponent rules converts the equation to base 3.

2

Factor out the common exponential expression

3^{2x}

.

3^{2x+2} - 3^{2x+1} = 3^{2x} \cdot 3^2 - 3^{2x} \cdot 3^1 = 3^{2x}(9 - 3) = 6 \cdot 3^{2x}

Using product rule of exponents (

3^{a+b} = 3^a \cdot 3^b

) allows combining like terms.

3

Solve for

x

.

6 \cdot 3^{2x} = 162 \implies 3^{2x} = 27 \implies 3^{2x} = 3^3 \implies 2x = 3 \implies x = \frac{3}{2}

Dividing both sides by 6 yields

3^{2x} = 27

, and equating exponents of matching bases gives

x = \frac{3}{2}

.

4

Evaluate the target expression

4^x

.

4^{3/2} = (4^{1/2})^3 = 2^3 = 8

Substituting

x = \frac{3}{2}

into

4^x

means taking the square root of 4 and raising it to the third power.

Key Concept

Solving exponential equations using common bases and exponent properties

Question 184Question →

An executive chartered a private aircraft to complete a trip between two cities separated by a non-stop distance of $600$ miles. On the return flight along the exact same route, strong headwinds reduced the aircraft's average ground speed by $50$ miles per hour compared to its outbound speed. As a result, the return flight took $2$ hours longer than the outbound flight. What was the average speed, in miles per hour, of the aircraft on the outbound flight?

Show answer & explanation

Answer: 150

Answer

150 miles per hour

By setting the outbound speed to

v

and return speed to

v - 50

, the relationship between outbound flight time

\frac{600}{v}

and return flight time

\frac{600}{v - 50}

yields the rational equation

\frac{600}{v - 50} - \frac{600}{v} = 2

. Solving this equation gives the quadratic

v^2 - 50v - 15,000 = 0

, which factors as

(v - 150)(v + 100) = 0

. The positive root gives an outbound speed of 150 miles per hour.

Step-by-Step Solution

1

Define variables for the unknown outbound rate and express travel times for both legs.

Let

v

be the outbound speed in miles per hour. Outbound time is

\frac{600}{v}

hours and return time is

\frac{600}{v - 50}

hours.

Distance equals speed multiplied by time (

d = v \cdot t

), so time equals distance divided by speed.

2

Formulate the algebraic equation using the given difference in flight durations.

\frac{600}{v - 50} - \frac{600}{v} = 2

The return flight took

2

hours longer than the outbound flight.

3

Clear denominators and simplify into standard quadratic form.

v^2 - 50v - 15,000 = 0

Multiplying both sides by

v(v - 50)

yields

600v - 600v + 30,000 = 2(v^2 - 50v)

, which simplifies to

2v^2 - 100v - 30,000 = 0

or

v^2 - 50v - 15,000 = 0

.

4

Solve the quadratic equation for

v

.

v = 150

Factoring

(v - 150)(v + 100) = 0

gives solutions

v = 150

or

v = -100

. Physical speed must be positive.

Key Concept

Distance-Rate-Time Quadratic Algebraic Modeling

Question 185Question →

In the $xy$ -plane, line $p$ is defined by the equation $y = -\frac{3}{4}x + 3$ . Which of the following statements about line $p$ must be true? Select all that apply.

Select all that apply

Line

p

passes through the point

(4, 0)

.

Line

p

is perpendicular to any line with a slope of

\frac{4}{3}

.

The

y

-intercept of line

p

is

(0, -3)

.

Line

p

passes through Quadrant III.

The slope of line

p

is

-\frac{4}{3}

.

Show answer & explanation

Answer: Line $p$ passes through the point $(4, 0)$ .; Line $p$ is perpendicular to any line with a slope of $\frac{4}{3}$ .

Answer

The statements asserting that line

p

passes through the point

(4, 0)

and that line

p

is perpendicular to any line with a slope of

\frac{4}{3}

are correct.

Substituting

x = 4

into the line equation gives

y = 0

, showing that

(4, 0)

lies on line

p

. Additionally, the slope of line

p

is

-\frac{3}{4}

, and the negative reciprocal of

-\frac{3}{4}

is

\frac{4}{3}

, making any line with slope

\frac{4}{3}

perpendicular to line

p

.

Step-by-Step Solution

1

Identify the slope and

y

-intercept directly from the equation

y = -\frac{3}{4}x + 3

.

The slope is

m = -\frac{3}{4}

and the

y

-intercept is

(0, 3)

.

The equation is given in standard slope-intercept form

y = mx + b

.

2

Verify point inclusion and perpendicular slope relationship.

Substituting

x = 4

yields

y = -\frac{3}{4}(4) + 3 = 0

, confirming

(4, 0)

is on the line. The negative reciprocal of

-\frac{3}{4}

is

\frac{4}{3}

, confirming the perpendicular line slope.

A point lies on a line if its coordinates satisfy the equation, and perpendicular lines have slopes that multiply to

-1

.

3

Analyze quadrant coverage.

Line

p

connects

(0, 3)

on the positive

y

-axis to

(4, 0)

on the positive

x

-axis, covering Quadrant I. For

x < 0

,

y > 3

(Quadrant II). For

x > 4

,

y < 0

(Quadrant IV). It never enters Quadrant III where both coordinates are negative.

A line with a positive

y

-intercept and negative slope crosses Quadrants I, II, and IV only.

Key Concept

Line properties in coordinate geometry including slope, intercepts, perpendicularity, and quadrant passage.

Question 186Question →

In the $xy$ -plane, line $k$ is defined by the equation $y = 3x - 4$ . Line $L$ is perpendicular to line $k$ and passes through the point $(6, 2)$ . What is the $y$ -intercept of line $L$ ?

-16

-4

4

12

20

Show answer & explanation

Answer: $4$

Answer

The

y

-intercept of line

L

is

4

.

Line

k

has a slope of

3

. Since line

L

is perpendicular to line

k

, the slope of line

L

is

-\frac{1}{3}

. Substituting the point

(6, 2)

into the slope-intercept equation

y = mx + b

gives

2 = -\frac{1}{3}(6) + b

, which simplifies to

2 = -2 + b

, so

b = 4

. Therefore, the

y

-intercept is

4

.

Step-by-Step Solution

1

Find the slope of line

L

The slope of line

L

is

-\frac{1}{3}

.

Line

k

has equation

y = 3x - 4

, so its slope is

3

. Perpendicular lines have slopes that are negative reciprocals.

2

Substitute the point

(6, 2)

and slope

-\frac{1}{3}

into the slope-intercept form

2 = -\frac{1}{3}(6) + b \implies 2 = -2 + b

The equation of a line is

y = mx + b

, where

m

is the slope and

b

is the

y

-intercept.

3

Solve for

b

b = 4

Adding

2

to both sides yields the

y

-intercept.

Key Concept

Perpendicular Line Slopes and Slope-Intercept Form

Question 187Question →

A community library purchases two types of books: hardcover books for $\$24$ each and paperback books for $\$15$ each. The library spends a total of $\$360$ on these books and purchases at least one book of each type. Which of the following could be the total number of books purchased? Select all such numbers.

Select all that apply

15

18

20

21

24

Show answer & explanation

Answer: 18; 21

Answer

18 and 21

The linear modeling equation

24x + 15y = 360

simplifies to

8x + 5y = 120

. Since

x

and

y

must be positive integers,

x

must be a multiple of 5. The only valid solutions satisfying

x \ge 1

and

y \ge 1

are

(x=5, y=16)

and

(x=10, y=8)

, which yield total book counts of 21 and 18, respectively.

Step-by-Step Solution

1

Set up the linear equation from the given word problem context.

24x + 15y = 360

, where

x \ge 1

is the number of hardcover books and

y \ge 1

is the number of paperback books, with

x, y \in \mathbb{Z}^+

.

Total expenditure is the sum of cost per hardcover times number of hardcovers plus cost per paperback times number of paperbacks.

2

Simplify the equation by dividing both sides by the greatest common divisor, 3.

8x + 5y = 120

Simplifying coefficients reduces arithmetic complexity and isolates integer conditions.

3

Express

y

in terms of

x

to identify valid integer pairs

(x, y)

.

y = \frac{120 - 8x}{5} = 24 - \frac{8x}{5}

For

y

to be an integer,

8x

must be divisible by 5, meaning

x

must be a positive multiple of 5.

4

Test valid positive integer values for

x

such that

y \ge 1

.

If

x = 5

,

y = 24 - 8 = 16

, giving total books

x + y = 21

. If

x = 10

,

y = 24 - 16 = 8

, giving total books

x + y = 18

. If

x \ge 15

,

y \le 0

, which is invalid.

These are the only integer solutions satisfying

x \ge 1

and

y \ge 1

.

Key Concept

Linear Diophantine Equations in Word Problems

Estimated Time:1m 30s

Question 188Question →

Two automated assembly lines, Line A and Line B, produce components at constant individual rates. Under normal operating conditions, Line A operating for $3$ hours and Line B operating for $4$ hours together produce a combined total of $1,400$ units. Under adjusted operating conditions, Line A operates at a rate $20\%$ higher than its normal rate, while Line B operates at a rate $10\%$ lower than its normal rate. Operating together under these adjusted conditions for $5$ hours, the two lines produce a total of $2,100$ units. What is the normal rate of Line A, in units per hour?

160

200

240

250

300

Show answer & explanation

Answer: $200$

Answer

The normal rate of Line A is

200

units per hour.

Let

r_A

and

r_B

represent the normal production rates in units per hour for Line A and Line B, respectively. From the first condition,

3r_A + 4r_B = 1400

. From the second condition, operating for

5

hours at rates

1.20r_A

and

0.90r_B

yields

5(1.20r_A + 0.90r_B) = 2100

, which simplifies to

1.20r_A + 0.90r_B = 420

, or

4r_A + 3r_B = 1400

. Subtracting

3r_A + 4r_B = 1400

from

4r_A + 3r_B = 1400

gives

r_A - r_B = 0

, meaning

r_A = r_B

. Substituting

r_B = r_A

into

3r_A + 4r_A = 1400

gives

7r_A = 1400

, so

r_A = 200

units per hour.

Step-by-Step Solution

1

Define variables and set up the equation for normal operating conditions.

3r_A + 4r_B = 1400

Line A operates for

3

hours at rate

r_A

and Line B operates for

4

hours at rate

r_B

to produce

1,400

units.

2

Set up the equation for adjusted operating conditions.

5(1.20r_A + 0.90r_B) = 2100 \implies 1.20r_A + 0.90r_B = 420

Line A's rate increases by

20\%

(

1.20r_A

) and Line B's rate decreases by

10\%

(

0.90r_B

). Divided by

5

hours, their combined hourly adjusted rate is

420

units per hour.

3

Multiply the simplified adjusted equation by

10

to clear decimals.

12r_A + 9r_B = 4200 \implies 4r_A + 3r_B = 1400

Dividing all terms by

3

simplifies the linear equation for easier elimination.

4

Solve the system of equations for

r_A

.

r_A = 200

From Step 1,

4r_B = 1400 - 3r_A \implies r_B = 350 - 0.75r_A

. Substituting into

4r_A + 3(350 - 0.75r_A) = 1400

gives

4r_A + 1050 - 2.25r_A = 1400 \implies 1.75r_A = 350 \implies r_A = 200

.

Key Concept

Linear Modeling of Combined Work and Rates

Question 189Question →

For all real numbers $x \neq 0$ and $x \neq 1$ , the function $f$ is defined by $f(x) = \frac{x - 1}{x}$ . The composite function notation $f^n(x)$ is defined by $f^1(x) = f(x)$ and $f^n(x) = f\left(f^{n-1}(x)\right)$ for all integers $n \ge 2$ . What is the value of $f^{2026}(5)$ ?

-\frac{5}{4}

-\frac{1}{4}

\frac{4}{5}

\frac{5}{4}

5

Show answer & explanation

Answer: $\frac{4}{5}$

Answer

\frac{4}{5}

Evaluating successive compositions of

f(x) = \frac{x-1}{x}

starting at

x = 5

yields

f^1(5) = \frac{4}{5}

,

f^2(5) = -\frac{1}{4}

, and

f^3(5) = 5

. This produces a repeating sequence of period 3. Since

2026 \equiv 1 \pmod 3

,

f^{2026}(5)

equals

f^1(5) = \frac{4}{5}

.

Step-by-Step Solution

1

Evaluate the first iteration

f^1(5)

.

f^1(5) = f(5) = \frac{5 - 1}{5} = \frac{4}{5}

.

Substitute

x = 5

into the definition of

f(x)

.

2

Evaluate the second iteration

f^2(5) = f(f^1(5))

.

f^2(5) = f\left(\frac{4}{5}\right) = \frac{\frac{4}{5} - 1}{\frac{4}{5}} = \frac{-\frac{1}{5}}{\frac{4}{5}} = -\frac{1}{4}

.

Substitute

x = \frac{4}{5}

into the function definition.

3

Evaluate the third iteration

f^3(5) = f(f^2(5))

.

f^3(5) = f\left(-\frac{1}{4}\right) = \frac{-\frac{1}{4} - 1}{-\frac{1}{4}} = \frac{-\frac{5}{4}}{-\frac{1}{4}} = 5

.

Substitute

x = -\frac{1}{4}

into the function definition.

4

Identify the periodic pattern and evaluate

f^{2026}(5)

.

Since

f^3(5) = 5

, the function values repeat in a cycle of length 3:

(\frac{4}{5}, -\frac{1}{4}, 5)

. Dividing 2026 by 3 yields

2026 = 3 \times 675 + 1

, giving a remainder of 1. Therefore,

f^{2026}(5) = f^1(5) = \frac{4}{5}

.

The remainder determines the equivalent position in the 3-element repeating sequence.

Key Concept

Nested Function Composition and Periodicity

Question 190Question →

If $x$ is a real number that satisfies $|3 - 2x| \le 9$ , what is the maximum possible value of $|x - 5|$ ?

Show answer & explanation

Answer: 8

Answer

The maximum possible value of

|x - 5|

is 8.

Solving

|3 - 2x| \le 9

yields

-9 \le 3 - 2x \le 9

. Subtracting 3 gives

-12 \le -2x \le 6

. Dividing by

-2

and reversing inequality signs results in

-3 \le x \le 6

. To maximize

|x - 5|

, find the point in

[-3, 6]

that is farthest from 5. Evaluating at the endpoints gives

|-3 - 5| = 8

and

|6 - 5| = 1

. The maximum possible value is 8.

Step-by-Step Solution

1

Solve the absolute value inequality

|3 - 2x| \le 9

.

-9 \le 3 - 2x \le 9

By definition of absolute value,

|u| \le k

means

-k \le u \le k

for

k \ge 0

.

2

Isolate

-2x

by subtracting 3 from each part.

-12 \le -2x \le 6

Subtracting a constant from all parts preserves the inequality direction.

3

Isolate

x

by dividing each part by

-2

.

-3 \le x \le 6

Dividing an inequality by a negative number flips the inequality signs.

4

Determine the maximum value of

|x - 5|

over the interval

[-3, 6]

.

Maximum occurs at

x = -3

, where

|-3 - 5| = 8

.

The expression

|x - 5|

represents the distance from

x

to 5 on the number line. The endpoint

x = -3

is furthest from 5.

Key Concept

Linear inequalities and absolute value as distance on a number line

Question 191Question →

If $x$ is an integer that satisfies both $|2x + 1| \le 7$ and $\frac{5 - x}{-2} < -1$ , how many possible values of $x$ are there?

0

3

7

8

11

Show answer & explanation

Answer: 7

Answer

There are 7 possible integer values of

x

.

Solving

|2x + 1| \le 7

yields

-4 \le x \le 3

. Solving

\frac{5 - x}{-2} < -1

requires flipping the inequality symbol when multiplying by

-2

, resulting in

5 - x > 2

, which simplifies to

x < 3

. Taking the intersection of

-4 \le x \le 3

and

x < 3

gives the interval

-4 \le x < 3

. The integers in this range are

-4, -3, -2, -1, 0, 1, 2

, totaling 7 integers.

Step-by-Step Solution

1

Solve the absolute value inequality

|2x + 1| \le 7

.

-7 \le 2x + 1 \le 7 \implies -8 \le 2x \le 6 \implies -4 \le x \le 3

.

An absolute value inequality of the form

|u| \le k

(for

k \ge 0

) expands to

-k \le u \le k

.

2

Solve the rational inequality

\frac{5 - x}{-2} < -1

.

5 - x > (-1)(-2) \implies 5 - x > 2 \implies -x > -3 \implies x < 3

.

Multiplying or dividing an inequality by a negative number requires reversing the direction of the inequality sign.

3

Find the intersection of the two solution sets and count integer solutions.

-4 \le x < 3

. The integer values satisfying this inequality are

-4, -3, -2, -1, 0, 1, 2

, which gives a total of 7 integer values.

The solution must satisfy both conditions simultaneously.

Key Concept

Solving combined linear and absolute value inequalities while correctly reversing inequality signs when multiplying or dividing by negative values.

Estimated Time:1m 30s

Question 192Question →

Two water pumps, Pump A and Pump B, working simultaneously at their respective constant rates, can fill an empty storage tank in $4$ hours. If Pump A operates alone at its constant rate for $2$ hours and then Pump B operates alone at its constant rate for $7$ hours, the tank is also filled completely. How many hours would it take Pump A, working alone at its constant rate, to fill the entire storage tank?

\frac{10}{3}

hours

\frac{20}{3}

hours

10

hours

12

hours

15

hours

Show answer & explanation

Answer: $\frac{20}{3}$ hours

Answer

\frac{20}{3}

hours (or

6\frac{2}{3}

hours)

The correct answer is derived by setting up two linear equations representing the total work accomplished:

4r_A + 4r_B = 1

and

2r_A + 7r_B = 1

. Solving this system gives Pump A's rate

r_A = \frac{3}{20}

tanks per hour. The time required for Pump A working alone is the reciprocal of its rate, which equals

\frac{20}{3}

hours.

Step-by-Step Solution

1

Define variables and formulate the system of linear equations

Let

r_A

be the rate of Pump A (tanks/hour) and

r_B

be the rate of Pump B (tanks/hour).
Combined work equation:

4(r_A + r_B) = 1 \implies 4r_A + 4r_B = 1

Sequential work equation:

2r_A + 7r_B = 1

Work done equals rate multiplied by time, and completing one full tank corresponds to total work

= 1

.

2

Express

r_B

in terms of

r_A

using the first equation

r_A + r_B = \frac{1}{4} \implies r_B = \frac{1}{4} - r_A

Simplifying the combined rate equation allows substitution into the second linear equation.

3

Substitute

r_B

into the second equation and solve for

r_A

2r_A + 7\left(\frac{1}{4} - r_A\right) = 1 \implies 2r_A + \frac{7}{4} - 7r_A = 1 \implies -5r_A = 1 - \frac{7}{4} = -\frac{3}{4} \implies r_A = \frac{3}{20}

Eliminating

r_B

isolates

r_A

as a single-variable linear equation.

4

Calculate the time required for Pump A to fill the tank alone

\text{Time}_A = \frac{1}{r_A} = \frac{1}{\frac{3}{20}} = \frac{20}{3}\text{ hours}

The total time to complete 1 unit of work is the reciprocal of the unit work rate.

Key Concept

Formulating and solving systems of two linear equations in two variables derived from work-rate relationships.

Question 193Question →

In the $xy$ -plane, line $\ell_1$ is defined by the equation $3x - 4y = 12$ . Line $\ell_2$ is perpendicular to line $\ell_1$ and intersects the positive $y$ -axis at the point $(0, d)$ . If the area of the triangular region bounded by line $\ell_1$ , line $\ell_2$ , and the $y$ -axis is $24$ , what is the value of $d$ ?

5

6

7

9

13

Show answer & explanation

Answer: $7$

Answer

The value of

d

is

7

.

The line

\ell_1

has equation

y = \frac{3}{4}x - 3

, placing its

y

-intercept at

(0, -3)

. Line

\ell_2

is perpendicular, so its slope is

-\frac{4}{3}

, giving the equation

y = -\frac{4}{3}x + d

. The vertical base of the triangle along the

y

-axis spans from

(0, -3)

to

(0, d)

, with a length of

d + 3

. The intersection of the two lines occurs at an

x

-coordinate of

\frac{12(d+3)}{25}

, which serves as the height of the triangle. Setting the area

\frac{1}{2} \times (d+3) \times \frac{12(d+3)}{25} = 24

simplifies to

(d+3)^2 = 100

. Because

d > 0

,

d + 3 = 10

, giving

d = 7

.

Step-by-Step Solution

1

Find the slope and

y

-intercept of line

\ell_1

.

Converting

3x - 4y = 12

into slope-intercept form yields

y = \frac{3}{4}x - 3

. The slope of

\ell_1

is

m_1 = \frac{3}{4}

and its

y

-intercept is

(0, -3)

.

Knowing the slope and

y

-intercept of

\ell_1

is essential to determine the equation of line

\ell_2

and the vertices of the triangular region along the

y

-axis.

2

Determine the equation of line

\ell_2

.

Since

\ell_2

is perpendicular to

\ell_1

, its slope is the negative reciprocal of

\frac{3}{4}

, which is

m_2 = -\frac{4}{3}

. Given that

\ell_2

intersects the

y

-axis at

(0, d)

, its equation is

y = -\frac{4}{3}x + d

.

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

3

Find the

x

-coordinate of the intersection point of

\ell_1

and

\ell_2

.

Set the two equations equal:

\frac{3}{4}x - 3 = -\frac{4}{3}x + d \implies \left(\frac{3}{4} + \frac{4}{3}\right)x = d + 3 \implies \frac{25}{12}x = d + 3 \implies x = \frac{12(d+3)}{25}

.

The

x

-coordinate of the intersection point represents the horizontal altitude (height) of the triangle with respect to the vertical base along the

y

-axis.

4

Express the area of the triangular region in terms of

d

and solve for

d

.

The vertical base along the

y

-axis stretches from

(0, -3)

to

(0, d)

, having length

d - (-3) = d + 3

. The height is

h = \frac{12(d+3)}{25}

. Using the area formula

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}

, we set

\frac{1}{2}(d+3)\left(\frac{12(d+3)}{25}\right) = 24 \implies \frac{6(d+3)^2}{25} = 24 \implies (d+3)^2 = 100

. Since

d > 0

,

d + 3 = 10

, so

d = 7

.

The problem states that the area of the triangle is

24

.

Key Concept

Perpendicular line slopes and geometric area calculations using coordinate geometry

Estimated Time:2m 30s

Question 194Question →

A coffee roaster creates a custom blend by combining Grade X beans costing $\$p$ per pound with Grade Y beans costing $\$q$ per pound, where $0 < p < q$ . The batch contains a total of $M$ pounds, consisting of $x$ pounds of Grade X and $y$ pounds of Grade Y. The total cost of the batch is $C$ dollars, and the average cost per pound of the blend is $A$ dollars. Which of the following algebraic relationships MUST be true? Indicate all such statements.

Select all that apply

The weight of Grade X beans in pounds is given by

x = \frac{qM - C}{q - p}

.

The weight of Grade X beans in pounds is given by

x = \frac{C - pM}{q - p}

.

The average cost per pound of the blend satisfies

A = q - \frac{x(q - p)}{M}

.

If the weight of Grade X in the blend is increased by

20\%

while keeping the weight of Grade Y constant, the total cost of the batch increases by

0.20C

dollars.

The ratio of the weight of Grade X to the weight of Grade Y in the blend is equal to

\frac{q - A}{A - p}

.

Show answer & explanation

Answer: The weight of Grade X beans in pounds is given by $x = \frac{qM - C}{q - p}$ .; The average cost per pound of the blend satisfies $A = q - \frac{x(q - p)}{M}$ .; The ratio of the weight of Grade X to the weight of Grade Y in the blend is equal to $\frac{q - A}{A - p}$ .

Answer

The true statements are that the weight of Grade X beans is

x = \frac{qM - C}{q - p}

, the average cost per pound satisfies

A = q - \frac{x(q - p)}{M}

, and the ratio of Grade X weight to Grade Y weight is

\frac{q - A}{A - p}

.

The correct statements correctly model the system of equations

x + y = M

and

C = px + qy

. Solving for

x

in terms of total cost gives

x = \frac{qM - C}{q - p}

. Dividing the expression for total cost by total mass

M

yields

A = q - \frac{x(q - p)}{M}

. Finally, setting total cost

C = A(x+y) = px + qy

and rearranging gives the ratio

\frac{x}{y} = \frac{q - A}{A - p}

.

Step-by-Step Solution

1

Set up equations for total mass

M

and total cost

C

using individual weights

x

and

y

.

x + y = M \implies y = M - x

and

C = px + qy = px + q(M - x)

.

Relating the two variables through total weight eliminates

y

to express cost purely in terms of

x

.

2

Solve the total cost equation for

x

.

C = qM - (q - p)x \implies (q - p)x = qM - C \implies x = \frac{qM - C}{q - p}

.

Isolating

x

confirms the valid algebraic formula for the weight of Grade X.

3

Calculate average cost

A = \frac{C}{M}

by substituting the simplified expression for

C

.

A = \frac{qM - (q - p)x}{M} = q - \frac{x(q - p)}{M}

.

Dividing total cost by total weight

M

yields the weighted average cost per pound.

4

Determine the ratio of component weights

\frac{x}{y}

in terms of unit costs

p, q

and average cost

A

.

A(x + y) = px + qy \implies Ax + Ay = px + qy \implies y(q - A) = x(A - p) \implies \frac{x}{y} = \frac{q - A}{A - p}

.

Rearranging the weighted average equation isolates the ratio of the quantities of the two components.

Key Concept

Linear weighted averages and multi-variable system modeling in mixture word problems.

Estimated Time:2m 30s

Question 195Question →

A municipal utility company charges residential customers according to two seasonal rate structures. During the peak summer season, electricity costs $\$0.24$ per kilowatt-hour ( $\text{kWh}$ ). During the off-peak season, electricity costs $\$0.15$ per $\text{kWh}$ . A homeowner consumed a total of $14,000\text{ kWh}$ of electricity over the entire year, resulting in a total annual bill of $\$2,640$ . How many kilowatt-hours of electricity did the homeowner consume during the peak summer season?

6,000

8,000

2,250

11,000

7,000

Show answer & explanation

Answer: $6,000$

Answer

6,000\text{ kWh}

The correct answer of

6,000\text{ kWh}

is found by setting up a linear model for total energy consumption (

x + y = 14,000

) and total cost (

0.24x + 0.15y = 2,640

). Substituting

y = 14,000 - x

gives

0.09x = 540

, which solves to

x = 6,000\text{ kWh}

.

Step-by-Step Solution

1

Define variables for the two seasons.

Let

x

be the number of kilowatt-hours consumed during the peak summer season, and let

y

be the number of kilowatt-hours consumed during the off-peak season.

Establishing clear variables allows modeling the problem as a system of linear equations.

2

Set up the total consumption equation and express

y

in terms of

x

.

x + y = 14,000 \implies y = 14,000 - x

The total consumption across both seasons equals

14,000\text{ kWh}

.

3

Set up the total cost equation using the respective seasonal rates.

0.24x + 0.15y = 2,640

Peak electricity costs

\$0.24/\text{kWh}

and off-peak costs

\$0.15/\text{kWh}

, summing to the total annual cost of

\$2,640

.

4

Substitute

y = 14,000 - x

into the cost equation and solve for

x

.

0.24x + 0.15(14,000 - x) = 2,640 \implies 0.24x + 2,100 - 0.15x = 2,640 \implies 0.09x = 540 \implies x = 6,000

Simplifying the single-variable linear equation isolates the peak season consumption.

Key Concept

Linear Modeling and Systems of Linear Equations

Estimated Time:1m 30s

Question 196Question →

The quadratic equation $x^2 - px + q = 0$ , where $p$ and $q$ are real constants with $p > 0$ , has two real roots $r_1$ and $r_2$ . If $r_1^2 + r_2^2 = 29$ and $r_1^3 + r_2^3 = 117$ , what is the value of $q$ ?

Show answer & explanation

Answer: -10

Answer

The value of

q

is

-10

.

By Vieta's formulas,

r_1 + r_2 = p

and

r_1 r_2 = q

. Using the identity

r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2 = 29

, we get

q = \frac{p^2 - 29}{2}

. Using the sum of cubes identity

r_1^3 + r_2^3 = (r_1 + r_2)(r_1^2 - r_1 r_2 + r_2^2) = 117

, we obtain

p(29 - q) = 117

. Substituting

q

into this equation yields

p^3 - 87p + 234 = 0

. Factoring this cubic equation yields

p = 3

as a valid root, leading to

q = -10

. Evaluating the discriminant

p^2 - 4q = 9 - 4(-10) = 49 > 0

confirms that real roots exist. The second positive root for

p

gives a negative discriminant, making

-10

the unique correct value of

q

.

Step-by-Step Solution

1

Express symmetric sums of roots in terms of equation coefficients

p

and

q

using Vieta's formulas.

r_1 + r_2 = p

and

r_1 r_2 = q

. Therefore,

r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2 = p^2 - 2q = 29

, which gives

q = \frac{p^2 - 29}{2}

.

Relating root sums and products directly to coefficients simplifies the algebraic system.

2

Express the sum of cubes in terms of

p

and

q

and substitute

q

.

r_1^3 + r_2^3 = (r_1 + r_2)(r_1^2 - r_1 r_2 + r_2^2) = p(29 - q) = 117

. Substituting

q = \frac{p^2 - 29}{2}

yields

p\left(29 - \frac{p^2 - 29}{2}\right) = 117 \implies p(87 - p^2) = 234 \implies p^3 - 87p + 234 = 0

.

Combining symmetric expressions forms a single polynomial equation in variable

p

.

3

Factor the cubic equation

p^3 - 87p + 234 = 0

to find candidate values for

p > 0

.

Testing factor

p = 3

gives

3^3 - 87(3) + 234 = 27 - 261 + 234 = 0

. Factoring out

(p - 3)

gives

(p - 3)(p^2 + 3p - 78) = 0

.

Finding the roots of the polynomial identifies candidate values for coefficient

p

.

4

Verify the real root constraint

p^2 - 4q > 0

for candidates of

p

.

For

p = 3

,

q = \frac{3^2 - 29}{2} = -10

, giving discriminant

p^2 - 4q = 9 - 4(-10) = 49 > 0

(real roots

5

and

-2

). For the positive root

p = \frac{\sqrt{321}-3}{2} \approx 7.458

of

p^2 + 3p - 78 = 0

, the discriminant is

p^2 - 4q = 3p - 29 \approx -6.63 < 0

, which yields non-real complex roots.

The question specifies real roots, so candidate solutions resulting in negative discriminants must be discarded.

Key Concept

Quadratic Equations and Factoring (Vieta's Formulas, Symmetric Polynomials, and Real Root Conditions)

Question 197Question →

For all real numbers $x$ , the function $f$ is defined by $f(x) = x^2 - 4x + 7$ and the function $g$ is defined by $g(x) = 2x + 1$ . If $f(g(k)) = 19$ and $k < 0$ , what is the value of $k$ ?

Show answer & explanation

Answer: -1.5

Answer

The value of

k

is

-1.5

.

To solve for

k

, substitute

g(k) = 2k + 1

into

f(x)

to get

f(g(k)) = (2k + 1)^2 - 4(2k + 1) + 7

. Expanding and combining like terms yields

4k^2 - 4k + 4

. Setting this equal to

19

gives

4k^2 - 4k - 15 = 0

. Factoring the quadratic expression yields

(2k - 5)(2k + 3) = 0

, which gives

k = \frac{5}{2} = 2.5

or

k = -\frac{3}{2} = -1.5

. Since the problem specifies that

k < 0

, the value of

k

must be

-1.5

.

Step-by-Step Solution

1

Express the nested composite function

f(g(k))

in terms of

k

f(g(k)) = (2k + 1)^2 - 4(2k + 1) + 7

Substituting

g(k) = 2k + 1

for every occurrence of

x

in

f(x)

sets up the composite function evaluation.

2

Expand terms and simplify the algebraic expression

f(g(k)) = 4k^2 - 4k + 4

Expanding

(2k + 1)^2 = 4k^2 + 4k + 1

and distributing

-4(2k + 1) = -8k - 4

, then combining like terms yields

4k^2 - 4k + 4

.

3

Set the composite function expression equal to 19 and form a standard quadratic equation

4k^2 - 4k - 15 = 0

Subtracting

19

from both sides produces a standard quadratic equation

ax^2 + bx + c = 0

.

4

Solve the quadratic equation and apply the condition

k < 0

k = -1.5

Factoring

(2k - 5)(2k + 3) = 0

gives

k = 2.5

or

k = -1.5

. The constraint

k < 0

selects

k = -1.5

.

Key Concept

Composite function evaluation

f(g(x))

combined with quadratic equation solving under domain constraints.

Question 198Question →

An industrial facility has three supply pipes—Pipe A, Pipe B, and Pipe C—that can fill a storage reservoir. Pipe A operating alone can fill the empty reservoir in $a$ hours. Pipe B operating alone takes $50\%$ longer than Pipe A to fill the empty reservoir. Pipe C operating alone fills the empty reservoir at a rate equal to the combined filling rate of Pipe A and Pipe B.

Initially, the reservoir is empty. Pipe A and Pipe B are opened simultaneously. After $2$ hours, Pipe B is closed and Pipe C is opened, while Pipe A remains open. The reservoir becomes completely full exactly $4$ hours after Pipe A and Pipe B were initially opened.

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

Select all that apply

Pipe A operating alone would fill the empty reservoir in

8

hours and

40

minutes.

During the first

2

hours of the process, exactly

\frac{5}{13}

of the total capacity of the reservoir is filled.

Pipe C operating alone would fill the empty reservoir in

5

hours and

12

minutes.

Pipe B operating alone would fill the empty reservoir in

11

hours and

30

minutes.

If all three pipes were opened simultaneously from the beginning, the empty reservoir would be filled in

3

hours and

15

minutes.

Show answer & explanation

Answer: Pipe A operating alone would fill the empty reservoir in $8$ hours and $40$ minutes.; During the first $2$ hours of the process, exactly $\frac{5}{13}$ of the total capacity of the reservoir is filled.; Pipe C operating alone would fill the empty reservoir in $5$ hours and $12$ minutes.

Answer

The statements asserting that Pipe A operating alone takes 8 hours and 40 minutes, that 5/13 of the capacity is filled in the first 2 hours, and that Pipe C operating alone takes 5 hours and 12 minutes are all true.

Using the rate relationships

R_A = \frac{1}{a}

,

R_B = \frac{2}{3a}

, and

R_C = \frac{5}{3a}

, the equation

2(R_A + R_B) + 2(R_A + R_C) = 1

simplifies to

\frac{26}{3a} = 1

, giving

a = \frac{26}{3}

hours. This verifies that Pipe A takes 8 hours 40 minutes alone, Phase 1 fills 5/13 of the reservoir capacity, and Pipe C takes 5.2 hours (5 hours 12 minutes) alone.

Step-by-Step Solution

1

Express the individual work rates of Pipe A, Pipe B, and Pipe C in terms of parameter

a

.

R_A = \frac{1}{a}

,

R_B = \frac{1}{1.5a} = \frac{2}{3a}

, and

R_C = R_A + R_B = \frac{1}{a} + \frac{2}{3a} = \frac{5}{3a}

.

Work rate is defined as the reciprocal of the time required to complete one unit of work.

2

Formulate an equation for total work completed over the two 2-hour phases.

2(R_A + R_B) + 2(R_A + R_C) = 1 \implies 2\left(\frac{5}{3a}\right) + 2\left(\frac{8}{3a}\right) = 1 \implies \frac{10}{3a} + \frac{16}{3a} = 1

.

Pipes A and B operate for the first 2 hours, followed by Pipes A and C operating for the next 2 hours to complete 1 full reservoir.

3

Solve the work equation for

a

and determine Pipe A's solo time.

\frac{26}{3a} = 1 \implies a = \frac{26}{3} = 8\frac{2}{3} \text{ hours} = 8 \text{ hours } 40 \text{ minutes}.

Fractional hours are converted to minutes by multiplying

\frac{2}{3}

by

60

.

4

Evaluate the volume filled in Phase 1 and the solo filling time for Pipe C.

\text{Phase 1 volume} = \frac{10}{3(26/3)} = \frac{5}{13}. \text{ Pipe C solo time} = \frac{1}{R_C} = \frac{3a}{5} = \frac{26}{5} = 5.2 \text{ hours} = 5 \text{ hours } 12 \text{ minutes}.

Substituting

a = \frac{26}{3}

into the respective rate expressions yields the exact time and volume parameters.

Key Concept

Formulating combined rate models and solving sequential multi-stage work problems.

Question 199Question →

For what values of the real constant $k$ does the quadratic equation $(k-2)x^2 - 2kx + (2k - 3) = 0$ have two distinct real roots $r_1$ and $r_2$ such that $r_1 < 1 < r_2$ ?

1 < k < 6

2 < k < 5

k < 2

or

k > 5

1 < k < 5

k > 2

Show answer & explanation

Answer: $2 < k < 5$

Answer

The correct inequality expressing all possible values of

k

is

2 < k < 5

.

The correct inequality

2 < k < 5

is obtained by requiring

(k-2)f(1) < 0

, which ensures

x = 1

falls between the two distinct real roots. Evaluating

f(1) = k - 5

yields

(k-2)(k-5) < 0

, giving

2 < k < 5

. The discriminant condition

\Delta = -4(k-1)(k-6) > 0

gives

1 < k < 6

, which fully encompasses

(2, 5)

.

Step-by-Step Solution

1

Define the quadratic function and state the conditions for

r_1 < 1 < r_2

.

Let

f(x) = (k-2)x^2 - 2kx + (2k - 3)

. For a quadratic function to have two real roots with

x = 1

located between them, the product of the leading coefficient

(k-2)

and

f(1)

must be strictly negative, i.e.,

(k-2)f(1) < 0

.

If a parabola opens upwards (

k-2 > 0

), its value at a point between its roots must be negative (

f(1) < 0

). If it opens downwards (

k-2 < 0

), its value at a point between its roots must be positive (

f(1) > 0

).

2

Evaluate

f(1)

in terms of

k

.

f(1) = (k-2)(1)^2 - 2k(1) + (2k - 3) = k - 2 - 2k + 2k - 3 = k - 5

.

Substitute

x = 1

directly into the expression for

f(x)

.

3

Solve the inequality

(k-2)f(1) < 0

.

(k-2)(k-5) < 0 \implies 2 < k < 5

.

The product of two linear factors

(k-2)

and

(k-5)

is negative between their roots,

k = 2

and

k = 5

.

4

Verify discriminant condition

\Delta > 0

for real roots.

\Delta = (-2k)^2 - 4(k-2)(2k-3) = 4k^2 - 4(2k^2 - 7k + 6) = -4k^2 + 28k - 24 = -4(k-1)(k-6) > 0 \implies 1 < k < 6

.

Since the interval

(2, 5)

is entirely contained within

(1, 6)

, any

k \in (2, 5)

automatically guarantees two distinct real roots.

Key Concept

Location of roots of quadratic equations and sign analysis of quadratic functions.

Estimated Time:2m 0s

Question 200Question →

For all real numbers $x$ and $y$ , the custom operation $\odot$ is defined by $x \odot y = x^2 y - y^2 x$ . The function $f$ is defined by $f(t) = t \odot 3$ . If $t$ is a positive real number such that $f(f(t)) = 0$ and $f(t) \neq 0$ , what is the value of $t$ ?

3

\frac{3 + \sqrt{5}}{2}

\frac{3 + \sqrt{13}}{2}

\frac{3 + \sqrt{13}}{4}

\frac{9 + \sqrt{13}}{2}

Show answer & explanation

Answer: $\frac{3 + \sqrt{13}}{2}$

Answer

The correct value of

t

is

\frac{3 + \sqrt{13}}{2}

.

Applying the custom binary operator gives

f(t) = 3t^2 - 9t

. Substituting

u = f(t)

into

f(u) = 0

yields

3u(u - 3) = 0

, so

u = 0

or

u = 3

. Because

f(t) \neq 0

, it must be that

f(t) = 3

. Setting

3t^2 - 9t = 3

leads to

t^2 - 3t - 1 = 0

. Applying the quadratic formula yields the positive value

\frac{3 + \sqrt{13}}{2}

.

Step-by-Step Solution

1

Express

f(t)

using the custom symbol definition.

f(t) = t \odot 3 = t^2(3) - (3)^2 t = 3t^2 - 9t

.

Apply the rule

x \odot y = x^2 y - y^2 x

with

x = t

and

y = 3

.

2

Analyze the nested function condition

f(f(t)) = 0

.

Let

u = f(t)

. Then

f(u) = 3u^2 - 9u = 3u(u - 3) = 0

, which yields

u = 0

or

u = 3

.

Evaluate the outer function

f

at the argument

u = f(t)

.

3

Apply the problem constraints to determine the exact value of

f(t)

.

Since

f(t) \neq 0

,

u = f(t) = 3

. Thus,

3t^2 - 9t = 3

.

Eliminate

f(t) = 0

based on the explicit condition given in the problem.

4

Solve the quadratic equation for

t > 0

.

Dividing

3t^2 - 9t - 3 = 0

by 3 gives

t^2 - 3t - 1 = 0

. Using the quadratic formula,

t = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)} = \frac{3 \pm \sqrt{13}}{2}

. Since

t > 0

,

t = \frac{3 + \sqrt{13}}{2}

.

Find the positive real root of the simplified quadratic equation.

Key Concept

Nested Function Evaluation and Custom Symbol Operations

Estimated Time:2m 30s

Algebra

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept