Algebra Practice Questions — GRE General Test — Page 14

Question 261Question →

A theater sells student tickets for $\$20$ each and adult tickets for $\$35$ each. For a specific performance, the theater sold a total of $150$ tickets and collected total revenue of $R$ dollars. If at least $40$ student tickets were sold and at most $90$ adult tickets were sold, which of the following values could be the total revenue $R$ ? Select all such values.

Select all that apply

\$2{}850

\$3{}450

\$4{}200

\$4{}400

\$4{}650

Show answer & explanation

Answer: $\$3{}450$ ; $\$4{}200$

Answer

The possible values for the total revenue

R

are

\$3{}450

and

\$4{}200

.

The linear revenue model is

R = 5250 - 15s

. Considering both conditions (

s \ge 40

and

a = 150 - s \le 90 \implies s \ge 60

), the valid range for student tickets is

60 \le s \le 150

. This restricts the possible revenue

R

to multiples of

\$15

between

\$3{}000

and

\$4{}350

. Both

\$3{}450

and

\$4{}200

fall within this valid range and correspond to integer ticket quantities (

s = 120, a = 30

and

s = 70, a = 80

, respectively).

Step-by-Step Solution

1

Set up equations for the total number of tickets and revenue.

Let

s

be the number of student tickets and

a

be the number of adult tickets. Then

s + a = 150

, so

a = 150 - s

. Total revenue

R = 20s + 35a = 20s + 35(150 - s) = 5250 - 15s

.

Expressing revenue in terms of a single variable

s

simplifies finding the domain and range.

2

Determine the constraints on the variable

s

.

We are given

s \ge 40

and

a \le 90

. Substituting

a = 150 - s \le 90

yields

s \ge 60

. Combining constraints gives

60 \le s \le 150

.

The number of adult tickets being at most

90

forces the number of student tickets to be at least

60

.

3

Calculate the upper and lower bounds for the revenue

R

.

Maximum revenue occurs when

s = 60

:

R_{\text{max}} = 5250 - 15(60) = \$4{}350

. Minimum revenue occurs when

s = 150

:

R_{\text{min}} = 5250 - 15(150) = \$3{}000

.

Since

R = 5250 - 15s

is a decreasing linear function of

s

, the maximum revenue occurs at the minimum valid value of

s

and vice versa.

4

Evaluate the given choices against the range and divisibility requirements.

R

must be an integer multiple of

15

subtracted from

5250

, meaning

R

must be between

\$3{}000

and

\$4{}350

inclusive, and

(5250 - R)

must be divisible by

15

.

\$3{}450

(where

s = 120

) and

\$4{}200

(where

s = 70

) are both valid.

\$2{}850

is below the minimum bound,

\$4{}400

does not yield an integer value for

s

, and

\$4{}650

violates the adult ticket upper bound.

Only options meeting both inequality constraints and integer ticket requirements are valid.

Key Concept

Linear modeling of word problems under linear system constraints and inequalities

Estimated Time:1m 30s

Question 262Question →

For a real constant $k$ , the quadratic equation $x^2 - 2kx + (k^2 - k - 6) = 0$ has two distinct real roots $r$ and $s$ such that $r < 0 < s$ and $|r| < |s|$ . Which of the following inequalities expresses all possible values of $k$ ?

0 < k < 3

-2 < k < 3

-6 < k < 3

k < -2

or

k > 3

-2 < k < 0

Show answer & explanation

Answer: $0 < k < 3$

Answer

The inequality expressing all possible values of

k

is

0 < k < 3

.

The condition that one root is negative and one root is positive (

r < 0 < s

) requires the product of the roots

rs = k^2 - k - 6

to be negative, which resolves to

-2 < k < 3

. Furthermore, since the positive root

s

has a larger absolute magnitude than the negative root

r

(

|r| < |s|

), the sum of the roots

r + s = 2k

must be positive, requiring

k > 0

. Taking the intersection of

-2 < k < 3

and

k > 0

yields

0 < k < 3

.

Step-by-Step Solution

1

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

k

.

For

x^2 - 2kx + (k^2 - k - 6) = 0

, the sum of roots is

r + s = 2k

and the product of roots is

rs = k^2 - k - 6

.

Vieta's formulas directly relate the coefficients of a quadratic equation to the sum and product of its roots.

2

Analyze the condition

r < 0 < s

.

Since one root is negative and the other is positive, their product must be negative:

rs = k^2 - k - 6 < 0

. Factoring gives

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

, which yields

-2 < k < 3

.

A positive number multiplied by a negative number produces a negative product.

3

Analyze the condition

|r| < |s|

.

Since

r < 0

,

|r| = -r

. Since

s > 0

,

|s| = s

. The inequality

|r| < |s|

becomes

-r < s

, which simplifies to

r + s > 0

. Substituting

r + s = 2k

gives

2k > 0

, or

k > 0

.

The positive root having greater magnitude than the absolute value of the negative root means the sum of the roots must be positive.

4

Combine the conditions to find the valid range for

k

.

Combining

-2 < k < 3

and

k > 0

yields the intersection

0 < k < 3

. (The discriminant condition

\Delta = 4(k + 6) > 0 \implies k > -6

is satisfied for all

k \in (0, 3)

).

The parameter

k

must satisfy both root sign constraints simultaneously.

Key Concept

Using Vieta's formulas and root magnitude conditions to solve quadratic parameter inequality problems.

Question 263Question →

A clean energy technology company manufactures two models of solar panels: Model X and Model Y. Model X produces $150$ kilowatt-hours (kWh) of electricity per day and costs $\$400$ to manufacture, while Model Y produces $200$ kWh of electricity per day and costs $\$550$ to manufacture. A solar energy project purchased a total of $50$ panels for a total manufacturing cost of $\$24,500$ . What is the total daily electricity production, in kilowatt-hours, of all $50$ panels combined?

8,500

8,750

9,000

9,500

10,000

Show answer & explanation

Answer: $9,000$

Answer

The total daily electricity production of all

50

panels combined is

9,000

kWh.

The correct option is

9,000

kWh. Defining

x

as the number of Model X panels and

y

as the number of Model Y panels gives the equations

x + y = 50

and

400x + 550y = 24,500

. Substituting

y = 50 - x

yields

400x + 550(50 - x) = 24,500

, which simplifies to

-150x = -3,000

, so

x = 20

and

y = 30

. Computing the total output yields

20(150) + 30(200) = 3,000 + 6,000 = 9,000

kWh.

Step-by-Step Solution

1

Define variables for the quantities of each panel model.

Let

x

represent the number of Model X panels and

y

represent the number of Model Y panels.

Establishing explicit variables allows formulating a system of linear equations from the word problem.

2

Set up equations for total panel quantity and total manufacturing cost.

System equations:

x + y = 50

and

400x + 550y = 24,500

.

The total number of panels is

50

, and the combined manufacturing cost equals

\$24,500

.

3

Solve the system using substitution.

Substitute

y = 50 - x

into the cost equation:

400x + 550(50 - x) = 24,500 \implies 400x + 27,500 - 550x = 24,500 \implies -150x = -3,000 \implies x = 20

. Consequently,

y = 50 - 20 = 30

.

Determining

x = 20

and

y = 30

gives the exact number of Model X and Model Y panels purchased.

4

Calculate the total daily electricity production.

Total production

= 20(150) + 30(200) = 3,000 + 6,000 = 9,000

kWh.

Multiply the quantity of each panel model by its respective daily output rate and sum the products.

Key Concept

Formulating and solving a system of two linear equations from a real-world scenario to find unknown quantities and evaluate a secondary combination function.

Estimated Time:1m 30s

Question 264Question →

For all real numbers $x$ and $y$ , the custom operation $\odot$ is defined by $x \odot y = x|y| - y|x|$ . Which of the following statements must be true for all real numbers $x$ and $y$ ? Select all such statements.

Select all that apply

x \odot y = 0

whenever

x

and

y

have the same sign

x \odot y = -(y \odot x)

If

x > 0

and

y < 0

, then

x \odot y > 0

x \odot (-x) = 0

for all non-zero real numbers

x

(x \odot y) \odot z = x \odot (y \odot z)

Show answer & explanation

Answer: $x \odot y = 0$ whenever $x$ and $y$ have the same sign; $x \odot y = -(y \odot x)$ ; If $x > 0$ and $y < 0$ , then $x \odot y > 0$

Answer

The correct statements are the statement asserting

x \odot y = 0

when

x

and

y

have the same sign, the statement asserting anti-commutativity

x \odot y = -(y \odot x)

, and the statement asserting

x \odot y > 0

when

x > 0

and

y < 0

.

The operation

x \odot y = x|y| - y|x|

produces 0 whenever

x

and

y

share the same sign because terms evaluate to identical quantities. Swapping variables negates the expression, establishing anti-commutativity. When

x

is positive and

y

is negative,

x \odot y

simplifies to

-2xy

, which is strictly greater than 0 since

xy < 0

.

Step-by-Step Solution

1

Analyze the first statement regarding same-sign inputs

If

x > 0

and

y > 0

,

|x|=x

and

|y|=y

, so

x \odot y = xy - yx = 0

. If

x < 0

and

y < 0

,

|x|=-x

and

|y|=-y

, so

x \odot y = x(-y) - y(-x) = -xy + xy = 0

. Thus,

x \odot y = 0

when

x

and

y

have the same sign.

Verifying the definition under both positive and negative cases of identical sign.

2

Analyze the second statement regarding operand order reversal

y \odot x = y|x| - x|y| = -(x|y| - y|x|) = -(x \odot y)

, which holds universally for all real numbers.

Testing anti-commutativity by algebraic substitution into the custom operation.

3

Analyze the third statement for opposite signs (

x > 0

and

y < 0

)

Since

x > 0

,

|x|=x

. Since

y < 0

,

|y|=-y

. Substituting yields

x(-y) - y(x) = -xy - xy = -2xy

. Because

x > 0

and

y < 0

, the product

xy

is negative, making

-2xy

strictly positive.

Determining the overall algebraic sign of the expression when variables have opposite signs.

4

Counter-test the remaining statements to verify incorrectness

For

x \odot (-x)

with

x=1

:

1 \odot (-1) = 1|-1| - (-1)|1| = 1 - (-1) = 2 \neq 0

. For associativity with

x=2, y=-1, z=-1

:

(2 \odot -1) \odot -1 = 4 \odot -1 = 8

, but

2 \odot (-1 \odot -1) = 2 \odot 0 = 0

.

Demonstrating specific counterexamples for false generalizations.

Key Concept

Custom Binary Operations and Absolute Value Properties

Estimated Time:1m 30s

Question 265Question →

For all positive real numbers $a$ and $b$ , the custom operation $\diamondsuit$ is defined by $a \diamondsuit b = \frac{a^2 + b^2}{ab}$ . The function $f$ is defined for all $x > 0$ by $f(x) = x \diamondsuit 4$ . If $f(x) = 2.5$ , what is the value of $x$ that is greater than $4$ ?

Show answer & explanation

Answer: 8

Answer

The value of

x

greater than

4

is

8

.

Applying the custom operator gives

f(x) = \frac{x^2 + 16}{4x}

. Setting this equal to

2.5

yields

\frac{x^2 + 16}{4x} = \frac{5}{2}

, which simplifies to

x^2 - 10x + 16 = 0

. The roots are

x = 2

and

x = 8

. Since

x

must be greater than

4

, the only valid answer is

8

.

Step-by-Step Solution

1

Substitute

a = x

and

b = 4

into the custom operation definition

a \diamondsuit b = \frac{a^2 + b^2}{ab}

.

f(x) = \frac{x^2 + 16}{4x}

This establishes the explicit algebraic rule for the function

f(x)

.

2

Set

f(x)

equal to

2.5

and clear the fraction.

\frac{x^2 + 16}{4x} = 2.5 \implies x^2 + 16 = 10x

Multiplying both sides by

4x

converts the rational equation into a standard polynomial equation.

3

Rearrange into standard quadratic form and solve by factoring.

x^2 - 10x + 16 = 0 \implies (x - 2)(x - 8) = 0 \implies x = 2

or

x = 8

Factoring determines all potential positive real solutions for

x

.

4

Select the solution satisfying the constraint

x > 4

.

x = 8

The question explicitly specifies that

x

must be greater than

4

, eliminating

x = 2

.

Key Concept

Evaluating custom binary operations and solving algebraic function equations involving quadratic constraints.

Question 266Question →

A courier service calculates the shipping cost for a package based on its weight $w$ , in pounds. For packages weighing up to $20$ pounds, the cost is a flat fee of $\$12.00$ plus $\$2.50$ per pound. For packages weighing more than $20$ pounds, the cost is a flat fee of $\$20.00$ plus $\$2.00$ per pound. If the total shipping cost for a package was strictly greater than $\$45.00$ and at most $\$70.00$ , which of the following could be the weight of the package, in pounds? Select all such weights.

Select all that apply

12

pounds

15

pounds

20

pounds

24

pounds

26

pounds

Show answer & explanation

Answer: $15$ pounds; $20$ pounds; $24$ pounds

Answer

The valid weights of the package are 15 pounds, 20 pounds, and 24 pounds.

The weight of the package must result in a total cost

C(w)

such that

\$45.00 < C(w) \le \$70.00

. Evaluating the options:
- For 15 pounds:

12 + 2.50(15) = \$49.50

, which falls within the range.
- For 20 pounds:

12 + 2.50(20) = \$62.00

, which falls within the range.
- For 24 pounds:

20 + 2.00(24) = \$68.00

, which falls within the range.

Step-by-Step Solution

1

Formulate the cost function

C(w)

as a piecewise model.

C(w) = 12 + 2.50w

for

0 < w \le 20

, and

C(w) = 20 + 2.00w

for

w > 20

.

The cost structure changes depending on whether the weight exceeds

20

pounds.

2

Set up and solve the inequality

45 < C(w) \le 70

for packages up to

20

pounds.

45 < 12 + 2.50w \le 70 \implies 33 < 2.50w \le 58 \implies 13.2 < w \le 23.2

. Restricting to

w \le 20

gives

13.2 < w \le 20

.

This determines the valid weight interval for packages billed under the first tier.

3

Set up and solve the inequality

45 < C(w) \le 70

for packages over

20

pounds.

45 < 20 + 2.00w \le 70 \implies 25 < 2.00w \le 50 \implies 12.5 < w \le 25

. Restricting to

w > 20

gives

20 < w \le 25

.

This determines the valid weight interval for packages billed under the second tier.

4

Combine the valid intervals and evaluate each option.

The overall valid weight interval is

13.2 < w \le 25

. Testing the values: 12 pounds is outside the range; 15 pounds, 20 pounds, and 24 pounds are within the range; 26 pounds is outside the range.

Any weight strictly greater than

13.2

pounds and up to

25

pounds produces a cost between

\$45.00

and

\$70.00

.

Key Concept

Algebraic Modeling of Piecewise Functions and Compound Inequalities

Question 267Question →

A logistics company operates two delivery vans, Van A and Van B. Van A consumes fuel at a rate of $1$ gallon for every $15$ miles driven, and Van B consumes fuel at a rate of $1$ gallon for every $25$ miles driven. On a specific trip, the two vans were driven a total combined distance of $450$ miles and together consumed exactly $22$ gallons of fuel. What was the total distance, in miles, driven by Van A?

100 miles

150 miles

225 miles

250 miles

300 miles

Show answer & explanation

Answer: 150 miles

Answer

The total distance driven by Van A was 150 miles.

The correct answer is 150 miles. By setting up the total fuel consumption equation as

\frac{d_A}{15} + \frac{450 - d_A}{25} = 22

, multiplying by the least common multiple

75

gives

5d_A + 1350 - 3d_A = 1650

, which simplifies to

2d_A = 300

and

d_A = 150

miles.

Step-by-Step Solution

1

Define variables for the distance traveled by each van.

Let

d_A

be the distance driven by Van A. Then the distance driven by Van B is

450 - d_A

.

The total combined distance for both vans is

450

miles.

2

Express the total fuel consumed in terms of

d_A

.

\frac{d_A}{15} + \frac{450 - d_A}{25} = 22

Fuel used equals distance divided by miles per gallon for each vehicle.

3

Clear denominators by multiplying the entire equation by the common denominator 75.

5d_A + 3(450 - d_A) = 1650

Simplifies fractions to solve the linear algebraic equation easily.

4

Solve for

d_A

.

5d_A + 1350 - 3d_A = 1650 \implies 2d_A = 300 \implies d_A = 150

Isolates the target variable

d_A

to find the distance driven by Van A.

Key Concept

Linear Equations and Rate-Distance Modeling

Question 268Question →

In the $xy$ -plane, line $k$ has an $x$ -intercept of $(8, 0)$ and a $y$ -intercept of $(0, 6)$ . Line $p$ is perpendicular to line $k$ and intersects line $k$ at its $y$ -intercept. What is the $x$ -intercept of line $p$ ?

-8

-\frac{9}{2}

\frac{9}{2}

8

-\frac{32}{3}

Show answer & explanation

Answer: $-\frac{9}{2}$

Answer

-\frac{9}{2}

The line

k

passes through

(8, 0)

and

(0, 6)

, giving a slope of

m_k = \frac{6 - 0}{0 - 8} = -\frac{3}{4}

. A line perpendicular to

k

must have a slope equal to the negative reciprocal of

-\frac{3}{4}

, which is

\frac{4}{3}

. Because line

p

intersects line

k

at

(0, 6)

, its

y

-intercept is also

6

, making its equation

y = \frac{4}{3}x + 6

. Setting

y = 0

gives

0 = \frac{4}{3}x + 6

, which solves to

x = -\frac{9}{2}

.

Step-by-Step Solution

1

Calculate the slope of line

k

Slope of line

k

is

m_k = \frac{6 - 0}{0 - 8} = -\frac{6}{8} = -\frac{3}{4}

The slope formula is

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

using given intercepts

(8, 0)

and

(0, 6)

.

2

Determine the slope of line

p

Slope of line

p

is

m_p = -\frac{1}{m_k} = \frac{4}{3}

Perpendicular lines in the coordinate plane have slopes that are negative reciprocals of each other.

3

Write the equation of line

p

Equation of line

p

is

y = \frac{4}{3}x + 6

Line

p

intersects line

k

at its

y

-intercept

(0, 6)

, so line

p

has a

y

-intercept of

6

.

4

Find the

x

-intercept of line

p

x = -\frac{9}{2}

(or

-4.5

)

Set

y = 0

in the line equation:

0 = \frac{4}{3}x + 6 \implies \frac{4}{3}x = -6 \implies x = -6 \cdot \frac{3}{4} = -\frac{9}{2}

.

Key Concept

Perpendicular Slopes and Line Intercepts

Estimated Time:1m 30s

Question 269Question →

A laboratory technician has $12$ liters of a solution containing $15\%$ salt by weight. The technician wants to increase the salt concentration to $25\%$ by adding a second solution that contains $40\%$ salt by weight. How many liters of the $40\%$ solution must be added?

Show answer & explanation

Answer: 8

Answer

8 liters of the 40% solution must be added.

The total amount of salt contributed by the two solutions is

0.15(12) + 0.40x = 1.8 + 0.40x

. The final mixture volume is

(12 + x)

liters, and its concentration must be

25\%

. Equating the total salt to

25\%

of total volume yields

1.8 + 0.40x = 0.25(12 + x)

. Expanding gives

1.8 + 0.40x = 3.0 + 0.25x

, which simplifies to

0.15x = 1.2

, giving

x = 8

liters.

Step-by-Step Solution

1

Define the unknown variable and calculate the initial solute quantity.

Let

x

be the volume of the

40\%

solution added (in liters). Salt in initial solution =

0.15 \times 12 = 1.8

liters.

Establishing the mass balance of the salt solute is necessary to construct the algebraic equation.

2

Set up the algebraic concentration equation.

1.8 + 0.40x = 0.25(12 + x)

The combined salt from both solutions must equal

25\%

of the total combined liquid volume

(12 + x)

liters.

3

Solve the linear equation for

x

.

1.8 + 0.40x = 3.0 + 0.25x \implies 0.15x = 1.2 \implies x = 8

Isolating

x

yields the exact number of liters required.

Key Concept

Algebraic mixture problems using mass balance equations.

Estimated Time:1m 30s

Question 270Question →

In the $xy$ -plane, line $L$ has a slope of $-\frac{3}{4}$ and passes through the point $(4, 1)$ . Which of the following statements about line $L$ must be true? Select all such statements.

Select all that apply

Line

L

passes through the point

(-4, 7)

.

Line

L

passes through Quadrant III.

Line

L

is perpendicular to the line defined by

4x - 3y = 12

.

The

x

-intercept of line

L

is

\left(\frac{4}{3}, 0\right)

.

Line

L

is parallel to the line defined by

3x + 4y = 25

.

Show answer & explanation

Answer: Line $L$ passes through the point $(-4, 7)$ .; Line $L$ is perpendicular to the line defined by $4x - 3y = 12$ .; Line $L$ is parallel to the line defined by $3x + 4y = 25$ .

Answer

The true statements are that line L passes through (-4, 7), is perpendicular to 4x - 3y = 12, and is parallel to 3x + 4y = 25.

The correct statements are those identifying that

(-4, 7)

satisfies

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

, that

4x - 3y = 12

has a perpendicular slope of

\frac{4}{3}

, and that

3x + 4y = 25

has an identical parallel slope of

-\frac{3}{4}

.

Step-by-Step Solution

1

Find the equation of line L using point-slope form.

y - 1 = -\frac{3}{4}(x - 4) \implies y = -\frac{3}{4}x + 4

, or in standard form

3x + 4y = 16

.

Establishing the linear equation allows systematic verification of points, intercepts, and slope properties.

2

Verify point

(-4, 7)

on line L.

3(-4) + 4(7) = -12 + 28 = 16

, which satisfies the equation.

Checking if the point coordinates satisfy

3x + 4y = 16

confirms it lies on the line.

3

Determine quadrant coverage.

The

y

-intercept is

(0, 4)

and the

x

-intercept is

\left(\frac{16}{3}, 0\right)

. For

x < 0

,

y > 4

(Quadrant II); for

0 < x < \frac{16}{3}

,

y > 0

(Quadrant I); for

x > \frac{16}{3}

,

y < 0

(Quadrant IV).

Line L passes through Quadrants I, II, and IV only, so it does not enter Quadrant III.

4

Analyze slope relationships for perpendicularity and parallelism.

Line

4x - 3y = 12

has slope

\frac{4}{3}

(negative reciprocal of

-\frac{3}{4}

). Line

3x + 4y = 25

has slope

-\frac{3}{4}

(identical slope).

Perpendicular lines have slopes that multiply to

-1

, while parallel lines have equal slopes and different intercepts.

Key Concept

Linear line equations, parallel/perpendicular slopes, and point/intercept evaluations in coordinate geometry.

Estimated Time:1m 30s

Question 271Question →

A municipal water treatment facility uses two intake pipes, Pipe $X$ and Pipe $Y$ , to fill a main reservoir. Operating alone at its constant rate, Pipe $X$ can fill the empty reservoir in $10$ hours. Operating alone at its constant rate, Pipe $Y$ can fill the empty reservoir in $15$ hours. Pipe $X$ is turned on first and operates alone for $3$ hours. Then, Pipe $Y$ is also turned on, and both pipes operate together until the reservoir is completely full. What is the total number of hours Pipe $X$ operates from the moment it is turned on until the reservoir is completely filled?

4.2

4.8

6.0

7.2

7.5

Show answer & explanation

Answer: $7.2$

Answer

The total number of hours Pipe X operates is

7.2

hours.

The correct answer is

7.2

hours. Pipe X completes

3/10

of the reservoir in 3 hours, leaving

7/10

of the reservoir to be filled. The combined rate of Pipe X and Pipe Y is

1/10 + 1/15 = 1/6

per hour. The joint time required to fill the remaining

7/10

is

(7/10) / (1/6) = 4.2

hours. Adding the initial 3 hours Pipe X worked alone yields a total of

3 + 4.2 = 7.2

hours.

Step-by-Step Solution

1

Determine the individual hourly work rates of Pipe X and Pipe Y.

Rate of Pipe X =

\frac{1}{10}

reservoir per hour; Rate of Pipe Y =

\frac{1}{15}

reservoir per hour.

The rate is the reciprocal of the total time required to complete the job individually.

2

Calculate the fraction of the reservoir filled by Pipe X during its initial 3-hour solo operation.

Work done in first 3 hours =

3 \times \frac{1}{10} = \frac{3}{10}

of the reservoir.

Multiplying Pipe X's rate by its solo operating time gives the completed portion of the work.

3

Find the remaining fraction of the reservoir that needs to be filled.

Remaining work =

1 - \frac{3}{10} = \frac{7}{10}

of the reservoir.

Subtracting the completed portion from the whole (1) leaves the uncompleted portion.

4

Determine the combined hourly rate when both pipes operate simultaneously.

Combined rate =

\frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6}

of the reservoir per hour.

Simultaneous operation rates are additive.

5

Calculate the time

t

during which both pipes operate together to finish the remaining work.

t = \frac{\frac{7}{10}}{\frac{1}{6}} = \frac{7}{10} \times 6 = \frac{42}{10} = 4.2

hours.

Dividing the remaining work by the combined rate yields the joint operation time.

6

Add Pipe X's solo time to the joint operation time to get Pipe X's total operating time.

Total time =

3 + 4.2 = 7.2

hours.

Pipe X was active during both the initial solo period and the joint operating period.

Key Concept

Combined Rate and Staggered Work Modeling

Estimated Time:2m 0s

Question 272Question →

In the $xy$ -plane, line $k$ is defined by the equation $2x - y = 8$ . Line $m$ is perpendicular to line $k$ and passes through the points $(-1, 9)$ and $(t, 5)$ . What is the value of $t$ ?

-9

-3

1

7

9

Show answer & explanation

Answer: $7$

Answer

The value of

t

is

7

.

The line

k

has equation

y = 2x - 8

, giving a slope of

2

. A line perpendicular to it must have a slope of

-\frac{1}{2}

. Using the slope formula for line

m

passing through

(-1, 9)

and

(t, 5)

yields

\frac{5 - 9}{t - (-1)} = -\frac{1}{2}

, which simplifies to

\frac{-4}{t + 1} = -\frac{1}{2}

, leading directly to

t = 7

.

Step-by-Step Solution

1

Find the slope of line

k

.

Converting

2x - y = 8

to slope-intercept form gives

y = 2x - 8

, so the slope of line

k

is

m_k = 2

.

The slope-intercept form

y = mx + b

directly identifies the slope

m

of a line.

2

Determine the slope of line

m

.

Since line

m

is perpendicular to line

k

, its slope is

m_m = -\frac{1}{m_k} = -\frac{1}{2}

.

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

3

Set up the slope formula for line

m

using the given points and solve for

t

.

\frac{5 - 9}{t - (-1)} = -\frac{1}{2} \implies \frac{-4}{t + 1} = -\frac{1}{2} \implies \frac{4}{t + 1} = \frac{1}{2} \implies t + 1 = 8 \implies t = 7

.

The slope between two points

(x_1, y_1)

and

(x_2, y_2)

is given by

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

.

Key Concept

Perpendicular lines in the coordinate plane have slopes that are negative reciprocals of each other.

Estimated Time:1m 30s

Question 273Question →

An investor allocated a total principal of $\$10,000$ between two accounts, Account X and Account Y. Account X pays simple annual interest at a rate of $6\%$ , while Account Y pays simple annual interest at a rate of $8\%$ . Let $x$ represent the amount of money, in dollars, invested in Account X, and let $d$ represent the total annual interest earned, in dollars, from both accounts after one year. Which of the following statements must be true? Select all that apply.

Select all that apply

The total annual interest

d

satisfies

600 \le d \le 800

.

The amount invested in Account X can be represented as

x = \frac{800 - d}{0.02}

.

If

d = 750

, the amount invested in Account X is greater than the amount invested in Account Y.

If equal amounts were invested in both accounts, the total annual interest earned is

\$700

.

If

d = 680

, the ratio of the amount invested in Account X to the amount invested in Account Y is

2

to

3

.

Show answer & explanation

Answer: The total annual interest $d$ satisfies $600 \le d \le 800$ .; The amount invested in Account X can be represented as $x = \frac{800 - d}{0.02}$ .; If equal amounts were invested in both accounts, the total annual interest earned is $\$700$ .

Answer

The true statements are that the total annual interest

d

satisfies

600 \le d \le 800

, the amount in Account X is given by

x = \frac{800 - d}{0.02}

, and equal investment in both accounts yields

\$700

in total interest.

The total interest equation

d = 800 - 0.02x

dictates all valid relationships. Because

0 \le x \le 10,000

, the bounds for

d

are strictly between

600

and

800

. Rearranging the equation yields

x = \frac{800 - d}{0.02}

, which correctly models the Account X investment. Substituting

x = 5,000

yields

d = 700

, confirming the equal-allocation scenario.

Step-by-Step Solution

1

Formulate the total interest model as a linear equation in terms of

x

.

d = 0.06x + 0.08(10,000 - x) = 800 - 0.02x

The interest earned from Account X is

0.06x

and the interest from Account Y is

0.08(10,000 - x)

.

2

Determine the range of possible interest values for

0 \le x \le 10,000

.

When

x = 10,000

,

d = 600

. When

x = 0

,

d = 800

. Thus

600 \le d \le 800

.

The minimum and maximum interest values occur at the extreme allocation bounds.

3

Rearrange the interest equation to express

x

in terms of

d

.

0.02x = 800 - d \implies x = \frac{800 - d}{0.02}

Isolating

x

provides a formula for calculating the Account X principal directly from the total interest.

4

Evaluate the specific numerical cases given in the options.

Equal investment (

x = 5,000

) gives

d = 800 - 100 = 700

. For

d = 750

,

x = 2,500

(Account Y =

7,500

). For

d = 680

,

x = 6,000

(Account Y =

4,000

, ratio

3:2

).

Substituting specific values tests the validity of each conditional statement.

Key Concept

Linear Modeling and Algebraic Rate Allocation

Question 274Question →

If $3x + y = 14$ and $x - 2y = -7$ , what is the value of $x + y$ ?

3

5

8

11

2

Show answer & explanation

Answer: 8

Answer

8

Solving the linear system yields

x = 3

and

y = 5

. Adding these values together gives

3 + 5 = 8

, which is the correct answer.

Step-by-Step Solution

1

Express

y

in terms of

x

using the first equation

y = 14 - 3x

Isolating one variable allows for direct substitution into the second equation.

2

Substitute

y = 14 - 3x

into the second equation

x - 2y = -7

x - 2(14 - 3x) = -7 \implies x - 28 + 6x = -7 \implies 7x = 21 \implies x = 3

Solving the single-variable equation gives the exact value for

x

.

3

Calculate

y

using

x = 3

y = 14 - 3(3) = 5

Substituting

x = 3

back into

y = 14 - 3x

gives the value for

y

.

4

Compute the sum

x + y

x + y = 3 + 5 = 8

The question specifically asks for the sum of both variables.

Key Concept

Solving a 2x2 System of Linear Equations by Substitution or Elimination

Estimated Time:45s

Question 275Question →

In the $xy$ -plane, line $L_1$ is defined by the equation $y = 3x - 4$ . Line $L_2$ is perpendicular to line $L_1$ and passes through the point $(6, 2)$ . What is the $x$ -intercept of line $L_2$ ?

Show answer & explanation

Answer: 12

Answer

12

Line

L_1

has a slope of

3

. A line perpendicular to

L_1

must have a slope equal to the negative reciprocal of

3

, which is

-\frac{1}{3}

. Substituting the point

(6, 2)

into the point-slope form gives

y - 2 = -\frac{1}{3}(x - 6)

, which simplifies to

y = -\frac{1}{3}x + 4

. Setting

y = 0

yields

0 = -\frac{1}{3}x + 4

, giving

x = 12

as the

x

-intercept.

Step-by-Step Solution

1

Find the slope of line

L_1

Slope

m_1 = 3

The equation

y = 3x - 4

is in slope-intercept form

y = mx + b

, where

m = 3

.

2

Determine the slope of perpendicular line

L_2

Slope

m_2 = -\frac{1}{3}

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

m_1 \cdot m_2 = -1

).

3

Find the equation of line

L_2

y = -\frac{1}{3}x + 4

Substitute point

(6, 2)

into point-slope formula

y - 2 = -\frac{1}{3}(x - 6)

.

4

Calculate the

x

-intercept of line

L_2

x = 12

Set

y = 0

in the line equation and solve for

x

:

0 = -\frac{1}{3}x + 4 \implies x = 12

.

Key Concept

Perpendicular lines have negative reciprocal slopes (

m_1 \cdot m_2 = -1

). The

x

-intercept is the point where

y = 0

.

Question 276Question →

In the $xy$ -plane, line $k$ passes through the points $(-2, 5)$ and $(4, -7)$ . Which of the following statements about line $k$ must be true? Select all that apply.

Select all that apply

Line

k

has a slope of

-2

.

Line

k

passes through the point

(3, -5)

.

Line

k

does not pass through Quadrant III.

Line

k

has an

x

-intercept at

(-2, 0)

.

A line perpendicular to line

k

has a slope of

2

.

Show answer & explanation

Answer: Line $k$ has a slope of $-2$ .; Line $k$ passes through the point $(3, -5)$ .; Line $k$ does not pass through Quadrant III.

Answer

The correct statements are: line

k

has a slope of

-2

, line

k

passes through the point

(3, -5)

, and line

k

does not pass through Quadrant III.

The slope of line

k

is

m = \frac{-7 - 5}{4 - (-2)} = -2

. The line equation is

y = -2x + 1

. Substituting

x = 3

yields

y = -5

, confirming

(3, -5)

is on the line. For all negative

x

-values (

x < 0

),

y = -2x + 1

remains strictly positive (

y > 1

), so no point on the line has both negative coordinates; hence line

k

never passes through Quadrant III.

Step-by-Step Solution

1

Calculate the slope of line

k

slope

m = \frac{-7 - 5}{4 - (-2)} = \frac{-12}{6} = -2

The slope formula is

m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}

.

2

Determine the equation of line

k

y = -2x + 1

Using point-slope form

y - y_1 = m(x - x_1)

with point

(-2, 5)

gives

y - 5 = -2(x + 2) \implies y = -2x + 1

.

3

Test point

(3, -5)

and calculate the

x

-intercept

Point

(3, -5)

lies on the line since

-2(3) + 1 = -5

. Setting

y = 0

gives

0 = -2x + 1 \implies x = \frac{1}{2}

.

A point lies on a line if its coordinates satisfy the line equation. The

x

-intercept occurs where

y = 0

.

4

Analyze quadrant passage and perpendicular slope

For

x < 0

,

y = -2x + 1 > 0

, so points in Quadrant III (

x < 0, y < 0

) are never reached. Perpendicular slope is

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

.

Quadrant III requires both coordinates to be negative. Perpendicular lines have negative reciprocal slopes.

Key Concept

Coordinate Geometry of Lines: Slope, Line Equations, Intercepts, Quadrant Passage, and Perpendicular Slopes

Question 277Question →

A technician charges a one-time fixed diagnostic fee of $\$35$ plus $\$25$ for each hour of repair work. If the total bill for a repair job was $\$160$ , how many hours of repair work were performed?

3

4

5

6

7

Show answer & explanation

Answer: $5$

Answer

5 hours

Let

h

represent the number of hours of repair work. The total charge is given by the sum of the fixed fee (

\$35

) and the hourly rate times hours worked (

25h

). Setting up the linear equation

35 + 25h = 160

and isolating

h

gives

25h = 125

, which simplifies to

h = 5

. Thus, 5 hours of work were performed.

Step-by-Step Solution

1

Define the variable and set up the linear equation

35 + 25h = 160

, where

h

is the number of repair hours.

The total cost is the sum of the fixed diagnostic fee and the hourly charge multiplied by hours worked.

2

Subtract the fixed fee from both sides of the equation

25h = 160 - 35 \implies 25h = 125

Isolate the variable term containing

h

.

3

Divide both sides by the hourly rate coefficient

h = \frac{125}{25} = 5

Solve for

h

.

Key Concept

Formulating and solving linear equations in one variable from real-world scenarios

Question 278Question →

If

x

satisfies the linear equation

\frac{3(x - 4)}{4} - \frac{2x + 1}{6} = \frac{x + 2}{3} - 5

what is the value of

2x + 5

?

-31

-23

-14

41

87

Show answer & explanation

Answer: $-23$

Answer

The value of

2x + 5

is

-23

.

Clearing denominators by multiplying the entire equation by

12

gives

9(x - 4) - 2(2x + 1) = 4(x + 2) - 60

. Expanding both sides yields

9x - 36 - 4x - 2 = 4x + 8 - 60

, which simplifies to

5x - 38 = 4x - 52

. Isolating

x

gives

x = -14

. Substituting

x = -14

into the target expression

2x + 5

yields

2(-14) + 5 = -23

.

Step-by-Step Solution

1

Clear the denominators by multiplying every term on both sides of the equation by the least common multiple of

4, 6,

and

3

, which is

12

.

12 \cdot \frac{3(x - 4)}{4} - 12 \cdot \frac{2x + 1}{6} = 12 \cdot \frac{x + 2}{3} - 12 \cdot 5

, leading to

9(x - 4) - 2(2x + 1) = 4(x + 2) - 60

.

Clearing denominators simplifies the multi-step fractional equation into an integer linear equation.

2

Expand all grouping symbols and combine like terms on both sides.

9x - 36 - 4x - 2 = 4x + 8 - 60

, which simplifies to

5x - 38 = 4x - 52

.

Distributing terms carefully ensures proper sign distribution, especially for negative signs across parentheses.

3

Isolate the variable

x

on one side of the equation.

5x - 4x = -52 + 38

, giving

x = -14

.

Subtracting

4x

and adding

38

isolates

x

.

4

Evaluate the target expression

2x + 5

using the solved value

x = -14

.

2(-14) + 5 = -28 + 5 = -23

.

The question asks for the value of

2x + 5

, not

x

.

Key Concept

Solving linear equations in one variable with fractional coefficients and evaluating algebraic expressions

Estimated Time:1m 30s

Question 279Question →

A water reservoir initially contains $450$ liters of water. Water drains out of the reservoir at a constant rate of $r$ liters per hour, while an inlet pipe supplies water at a constant rate of $18$ liters per hour. If the reservoir contains $390$ liters of water after $6$ hours, what is the value of $r$ ?

Show answer & explanation

Answer: 28

Answer

The value of

r

is

28

.

The reservoir starts with

450

liters. Over

6

hours, water enters at

18

liters/hour and leaves at

r

liters/hour, giving a net volume equation of

450 + 6(18 - r) = 390

. Simplifying yields

558 - 6r = 390

, which subtracts to

-6r = -168

, giving

r = 28

.

Step-by-Step Solution

1

Set up the linear equation representing the net change in water volume over time.

450 + 6(18 - r) = 390

The final volume equals the initial volume plus the net water added (inflow rate minus outflow rate multiplied by hours).

2

Expand and simplify the linear expression.

558 - 6r = 390

Distribute

6

across

(18 - r)

to obtain

108 - 6r

, then add to

450

.

3

Isolate the variable term

-6r

.

-6r = -168

Subtract

558

from both sides of the equation.

4

Solve for the rate

r

.

r = 28

Divide both sides by

-6

.

Key Concept

Formulating and solving a linear equation in one variable from a rate problem context.

Estimated Time:1m 30s

Question 280Question →

If

x

satisfies the linear equation

4(x - 2) - 15 = 2x + 3

which of the following statements must be true? Select all that apply.

Select all that apply

x

is a prime number

2x - 5

is equal to

21

x + 7

is a multiple of

5

x

is an even integer

\frac{x + 1}{2} = 6

Show answer & explanation

Answer: $x$ is a prime number; $2x - 5$ is equal to $21$ ; $x + 7$ is a multiple of $5$

Answer

The correct statements are that

x

is a prime number,

2x - 5

is equal to

21

, and

x + 7

is a multiple of

5

.

Solving the equation

4(x - 2) - 15 = 2x + 3

yields

x = 13

. Evaluating the choices with

x = 13

confirms that

13

is a prime number,

2(13) - 5 = 21

, and

13 + 7 = 20

(a multiple of

5

).

Step-by-Step Solution

1

Expand the left side of the linear equation by distributing the constant term.

4x - 8 - 15 = 2x + 3

Distribution removes parentheses so like terms can be combined.

2

Combine constant terms on the left side of the equation.

4x - 23 = 2x + 3

Simplifying

-8 - 15

yields

-23

.

3

Isolate variable terms on the left side and constant terms on the right side by subtracting

2x

and adding

23

to both sides.

2x = 26

Subtracting

2x

gives

2x

on the left, and adding

23

gives

26

on the right.

4

Divide both sides of the equation by

2

to solve for

x

.

x = 13

Dividing

26

by

2

determines the unique value of

x

.

5

Test each statement using

x = 13

.

13

is prime (True);

2(13) - 5 = 21

(True);

13 + 7 = 20

which is a multiple of

5

(True);

13

is even (False);

\frac{13 + 1}{2} = 7 \neq 6

(False).

Direct evaluation identifies which statements hold true.

Key Concept

Solving linear equations in one variable and evaluating numerical properties of the solution.

Estimated Time:1m 0s

Algebra

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