All practice questions

Question 1681Question →

A theater sold a total of $500$ tickets for an evening performance, consisting of VIP tickets priced at $\$80$ each and General Admission tickets priced at $\$50$ each. On the day of the show, a promotional discount of $20\%$ was applied to all General Admission tickets, while VIP ticket prices remained unchanged. If the total revenue collected from ticket sales was $\$31,000$ , which of the following statements must be true? Select all such statements.

Select all that apply

More than 70% of the total revenue was generated from VIP ticket sales.

The number of VIP tickets sold was 50 greater than the number of General Admission tickets sold.

The ratio of VIP tickets sold to General Admission tickets sold was 11 to 7.

Applying the 20% discount to General Admission tickets reduced the total potential revenue by 20%.

The number of General Admission tickets sold was 250.

Show answer & explanation

Answer: More than 70% of the total revenue was generated from VIP ticket sales.; The number of VIP tickets sold was 50 greater than the number of General Admission tickets sold.

Answer

The statements confirming that more than 70% of total revenue came from VIP ticket sales and that 50 more VIP tickets were sold than General Admission tickets are correct.

The system of equations

V + G = 500

and

80V + 40G = 31,000

yields

V = 275

VIP tickets and

G = 225

General Admission tickets. The VIP revenue is

22,000, which is approximately 70.97% of the total

31,000 revenue (greater than 70%). Additionally, the difference

275 - 225 = 50

confirms that 50 more VIP tickets were sold than General Admission tickets.

Step-by-Step Solution

1

Define variables and determine the discounted price of General Admission tickets.

Let

V

be the number of VIP tickets and

G

be the number of General Admission tickets. The discounted price for General Admission tickets is

\$50 \times (1 - 0.20) = \$40

.

Establishing correct variable representations and effective unit prices is required to build the revenue model.

2

Set up and solve the system of linear equations.

V + G = 500

and

80V + 40G = 31,000

. Substituting

G = 500 - V

gives

80V + 40(500 - V) = 31,000 \implies 40V = 11,000 \implies V = 275

. Thus,

G = 225

.

Solving the linear system determines the exact quantity of each ticket type sold.

3

Calculate revenue shares and verify each statement.

VIP revenue =

275 \times \$80 = \$22,000

. Revenue share of VIP =

\frac{22,000}{31,000} \approx 70.97\% > 70\%

. Ticket difference =

275 - 225 = 50

. Ratio

V:G = 275:225 = 11:9

.

Evaluating each calculated metric against the given statements determines which statements must be true.

Key Concept

Linear word problems involving systems of equations and percentage modifications.

Estimated Time:2m 0s

Question 1682Question →

A 6-digit security code is to be formed using distinct digits chosen from the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$ . The code must satisfy the following conditions:
1. The code must be an even number (its final digit must be $2$ , $4$ , $6$ , or $8$ ).
2. Both digits $1$ and $2$ must be included in the 6-digit code.
3. Digits $1$ and $2$ cannot occupy adjacent positions in the code.

Which of the following statements regarding the number of possible 6-digit security codes must be true? Select all such statements.

Select all that apply

The total number of valid 6-digit security codes that satisfy all conditions is

3,600

.

The number of valid security codes in which the final digit is

2

is

1,440

.

The number of valid security codes in which the final digit is an even digit other than

2

is

2,160

.

If the non-adjacency constraint between

1

and

2

were removed, the total number of valid security codes would be

4,320

.

There are

20

valid ways to place digits

1

and

2

non-adjacently in the first 5 positions of the code.

Show answer & explanation

Answer: The total number of valid 6-digit security codes that satisfy all conditions is $3,600$ .; The number of valid security codes in which the final digit is $2$ is $1,440$ .; The number of valid security codes in which the final digit is an even digit other than $2$ is $2,160$ .

Answer

The statements confirming that the total number of codes is 3,600, that 1,440 codes end in 2, and that 2,160 codes end in an even digit other than 2 are all correct.

The solution requires partitioning into two mutually exclusive scenarios depending on whether digit 2 occupies the final position. When digit 2 is at the end, digit 1 can occupy any of the first 4 positions (excluding position 5 to avoid adjacency), yielding

4 \times P(6,4) = 1,440

codes. When the final position is occupied by 4, 6, or 8 (3 choices), digits 1 and 2 have

P(5,2) - 8 = 12

valid non-adjacent placements across the first 5 positions, and the remaining 3 positions can be filled in

P(5,3) = 60

ways, yielding

3 \times 12 \times 60 = 2,160

codes. The sum of these two cases gives

3,600

total valid codes. Thus, the three statements asserting total codes of 3,600, 1,440 ending in 2, and 2,160 ending in 4, 6, or 8 are all correct.

Step-by-Step Solution

1

Analyze the conditions and split into two disjoint cases based on the last digit.

Case 1: The last digit (6th position) is

2

. Case 2: The last digit (6th position) is

4

,

6

, or

8

.

Digit

2

plays a dual role: it satisfies the even-ending condition and is one of the restricted digits.

2

Calculate Case 1 (last digit is 2).

Position 6 is fixed as

2

(

1

choice). Digit

1

must be in the first 5 positions but cannot be adjacent to position 6 (so position 5 is excluded). Thus digit

1

has

4

choices (positions 1, 2, 3, 4). The remaining

4

positions are filled from the remaining

6

available digits

\{3, 4, 5, 6, 7, 8\}

in

P(6, 4) = 6 \times 5 \times 4 \times 3 = 360

ways. Total for Case 1:

1 \times 4 \times 360 = 1,440

.

This determines the valid arrangements when 2 is forced to the end of the code.

3

Calculate Case 2 (last digit is 4, 6, or 8).

There are

3

choices for the 6th position. Digits

1

and

2

must be placed among the first 5 positions non-adjacently. Total ordered placements of

1

and

2

in 5 positions is

P(5, 2) = 20

. The number of adjacent position pairs is

4

(positions (1,2), (2,3), (3,4), (4,5)), with

2! = 2

orderings per pair, giving

4 \times 2 = 8

adjacent placements. Non-adjacent placements =

20 - 8 = 12

. The remaining 3 open positions are filled from the remaining 5 available digits in

P(5, 3) = 60

ways. Total for Case 2:

3 \times 12 \times 60 = 2,160

.

This accounts for codes ending in 4, 6, or 8 while respecting the non-adjacency of 1 and 2.

4

Combine the cases and verify statements.

Total valid codes =

1,440 + 2,160 = 3,600

. The statements asserting totals of 3,600, 1,440 ending in 2, and 2,160 ending in 4, 6, or 8 are true.

Summing mutually exclusive cases gives the overall number of valid outcomes.

Key Concept

Fundamental Counting Principle with Permutations under Restricted Adjacency

Question 1683Question →

In the $xy$ -plane, line $k$ passes through the points $(-1, 2)$ and $(3, 10)$ . Line $m$ is perpendicular to line $k$ and passes through the point $(2, 4)$ . Line $m$ intersects the $x$ -axis at point $P$ and the $y$ -axis at point $Q$ . What is the distance between point $P$ and point $Q$ ?

3\sqrt{5}

4\sqrt{5}

5\sqrt{3}

5\sqrt{5}

15

Show answer & explanation

Answer: $5\sqrt{5}$

Answer

5\sqrt{5}

The answer

5\sqrt{5}

is obtained by taking the negative reciprocal of the slope of line

k

(

m = 2

) to find line

m

's slope (

m = -1/2

), establishing line

m

's equation as

y = -1/2 x + 5

, finding the intercepts

P(10, 0)

and

Q(0, 5)

, and applying the distance formula.

Step-by-Step Solution

1

Calculate the slope of line

k

.

The slope of line

k

is

m_k = \frac{10 - 2}{3 - (-1)} = \frac{8}{4} = 2

.

The slope formula is

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

.

2

Find the slope of line

m

, which is perpendicular to line

k

.

The slope of line

m

is

m_m = -\frac{1}{2}

.

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

m_1 \cdot m_2 = -1

).

3

Determine the equation of line

m

using point-slope form with point

(2, 4)

.

y - 4 = -\frac{1}{2}(x - 2) \implies y = -\frac{1}{2}x + 5

.

Substituting slope

m_m = -\frac{1}{2}

and point

(x_1, y_1) = (2, 4)

yields line

m

's slope-intercept equation.

4

Find the coordinates of the

x

-intercept

P

and

y

-intercept

Q

.

P = (10, 0)

and

Q = (0, 5)

.

Setting

y = 0

gives

0 = -\frac{1}{2}x + 5 \implies x = 10

. Setting

x = 0

gives

y = 5

.

5

Calculate the distance between

P(10, 0)

and

Q(0, 5)

.

d = \sqrt{(10 - 0)^2 + (0 - 5)^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5}

.

Apply the distance formula

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

.

Key Concept

Perpendicular Line Slopes and Coordinate Distance

Estimated Time:2m 0s

Question 1684Question →

A manufacturing plant operates two automated production lines, Assembly Line X and Assembly Line Y. Assembly Line Y operates at a standard rate that is $25\%$ faster than the standard rate of Assembly Line X. During a specific shift, Assembly Line X operated for $4$ hours at its standard rate, after which its processing rate decreased by $20\%$ for an additional $2$ hours due to maintenance. Assembly Line Y began operating $1$ hour after Line X started; it operated at its standard rate for $4$ hours, and then operated for another $2$ hours at $80\%$ of its standard rate. If the two lines produced a combined total of $2,772$ units during this shift, what was the standard operating rate of Assembly Line X, in units per hour?

Show answer & explanation

Answer: 220

Answer

The standard operating rate of Assembly Line X is 220 units per hour.

Assembly Line X produces

4r

units in its first 4 hours and

1.6r

units in its final 2 hours, totaling

5.6r

units. Assembly Line Y produces

4 \times 1.25r = 5r

units in its first 4 hours and

2 \times (0.80 \times 1.25r) = 2r

units in its final 2 hours, totaling

7r

units. The sum of their outputs is

5.6r + 7r = 12.6r = 2,772

. Dividing

2,772

by

12.6

gives

r = 220

.

Step-by-Step Solution

1

Set up rate expressions for both assembly lines using a single variable

Standard rate of Line X =

r

; Standard rate of Line Y =

1.25r

Line Y is 25% faster than Line X, so its rate is

r + 0.25r = 1.25r

.

2

Calculate the total work done by Assembly Line X

Line X output =

4(r) + 2(0.80r) = 5.6r

units

Line X worked 4 hours at 100% rate and 2 hours at 80% rate.

3

Calculate the total work done by Assembly Line Y

Line Y output =

4(1.25r) + 2(0.80 \times 1.25r) = 5r + 2r = 7r

units

Line Y worked 4 hours at full rate

1.25r

and 2 hours at 80% of

1.25r

, which equals rate

r

.

4

Equate combined production to 2,772 units and solve for

r

5.6r + 7r = 12.6r = 2,772 \implies r = 220

Dividing the total combined output by the total rate multiplier 12.6 yields the baseline standard rate.

Key Concept

Algebraic Modeling of Staggered Work and Variable Production Rates

Estimated Time:2m 30s

Question 1685Question →

For all positive real numbers $x$ and $y$ , the custom operation $\star$ is defined by $x \star y = \frac{x^2 + y^2}{xy}$ . The function $g$ is defined by $g(t) = 4(t \star 3) - 7$ . If $g(k) = 3$ , where $k > 3$ , what is the value of $k$ ?

Show answer & explanation

Answer: 6

Answer

The value of

k

is 6.

Solving

g(k) = 3

yields

4(k \star 3) - 7 = 3

, so

k \star 3 = \frac{5}{2}

. Replacing

k \star 3

with

\frac{k^2+9}{3k}

results in the quadratic equation

2k^2 - 15k + 18 = 0

. The roots are

1.5

and

6

. Because

k

must be strictly greater than

3

, the only valid solution is

6

.

Step-by-Step Solution

1

Isolate the custom operation term in the function equation

k \star 3 = \frac{5}{2}

Setting

g(k) = 4(k \star 3) - 7 = 3

allows us to find the required value of

k \star 3

by adding 7 to both sides and dividing by 4.

2

Express the custom operation algebraically using its definition

\frac{k^2 + 9}{3k} = \frac{5}{2}

By definition,

x \star y = \frac{x^2 + y^2}{xy}

, so

k \star 3 = \frac{k^2 + 3^2}{3(k)} = \frac{k^2 + 9}{3k}

.

3

Form and solve the quadratic equation

k = 1.5 \text{ or } k = 6

Cross-multiplying gives

2(k^2 + 9) = 5(3k)

, which simplifies to

2k^2 - 15k + 18 = 0

. Factoring as

(2k - 3)(k - 6) = 0

yields two roots:

k = 1.5

and

k = 6

.

4

Apply the domain constraint

k = 6

The problem specifies that

k > 3

. Since

1.5 \le 3

,

k = 1.5

is rejected, leaving

k = 6

as the unique correct answer.

Key Concept

Custom Binary Symbol Operations and Composite Functions

Question 1686Question →

The frequency distribution table below summarizes the monthly water consumption, $c$ (in cubic meters, $\text{m}^3$ ), recorded for a sample of $150$ municipal water accounts.

Monthly Water Consumption ( $\text{m}^3$ )	Number of Accounts
$0 \le c < 10$	$25$
$10 \le c < 20$	$45$
$20 \le c < 30$	$50$
$30 \le c < 40$	$20$
$40 \le c < 50$	$10$

If one water account is selected at random from among all accounts with a monthly water consumption of at least $10\text{ m}^3$ , what is the probability that the selected account has a monthly water consumption of less than $30\text{ m}^3$ ? (Give your answer as a decimal rounded to two decimal places.)

Show answer & explanation

Answer: 0.76

Answer

0.76

To calculate the required probability, first restrict the sample space to accounts with a monthly consumption of at least

10\text{ m}^3

. Summing the frequencies for the intervals

10 \le c < 20

,

20 \le c < 30

,

30 \le c < 40

, and

40 \le c < 50

gives

45 + 50 + 20 + 10 = 125

accounts. Among these

125

accounts, those with a consumption of less than

30\text{ m}^3

fall into the intervals

10 \le c < 20

and

20 \le c < 30

, giving a count of

45 + 50 = 95

accounts. Dividing the favorable outcomes by the total outcomes in the restricted sample space yields

\frac{95}{125} = 0.76

.

Step-by-Step Solution

1

Determine the total number of accounts meeting the condition of having consumption of at least

10\text{ m}^3

.

Total eligible accounts =

45 + 50 + 20 + 10 = 125

.

Accounts with consumption of at least

10\text{ m}^3

fall into the intervals

10 \le c < 20

,

20 \le c < 30

,

30 \le c < 40

, and

40 \le c < 50

.

2

Determine the number of accounts among the eligible set with consumption less than

30\text{ m}^3

.

Number of favorable accounts =

45 + 50 = 95

.

Within the eligible set, accounts with consumption less than

30\text{ m}^3

fall into the intervals

10 \le c < 20

and

20 \le c < 30

.

3

Calculate the conditional probability as a decimal.

\frac{95}{125} = 0.76

Dividing the favorable outcomes (

95

) by the total possible outcomes in the restricted sample space (

125

) yields

0.76

.

Key Concept

Conditional probability and sample space restriction in grouped frequency tables

Question 1687Question →

A company has 8 departments. The dataset of the number of employees in these 8 departments has a median of 42, a range of 25, and a unique mode of 38, which appears exactly 3 times. If no department has more than 55 employees, what is the maximum possible arithmetic mean of the number of employees across all 8 departments?

Show answer & explanation

Answer: 44.25

Answer

The maximum possible arithmetic mean of the number of employees across all 8 departments is 44.25.

To maximize the mean, the sum of the 8 department sizes must be maximized under all given constraints. Ordering the dataset as

x_1 \le x_2 \le x_3 \le x_4 \le x_5 \le x_6 \le x_7 \le x_8

, the median requirement gives

x_4 + x_5 = 84

. Since no element exceeds 55 and the range is 25,

x_1

cannot be 38 because

38 + 25 = 63 > 55

. Hence, the three 38s must be

x_2 = x_3 = x_4 = 38

, which forces

x_5 = 46

. To maximize the sum,

x_8

is set to its maximum limit of 55, forcing

x_1 = 55 - 25 = 30

. Next,

x_7

is set to 55, and

x_6

is set to 54 so that 55 appears only twice and 38 remains the unique mode. The maximum sum is

30 + 38 + 38 + 38 + 46 + 54 + 55 + 55 = 354

, giving a maximum mean of

354 / 8 = 44.25

.

Step-by-Step Solution

1

Order the dataset variables and define constraints.

Let the department sizes be ordered as

x_1 \le x_2 \le x_3 \le x_4 \le x_5 \le x_6 \le x_7 \le x_8

.

Arranging values in ascending order allows direct analysis of median, range, and mode bounds.

2

Use the median to form an equation for the middle two elements.

\frac{x_4 + x_5}{2} = 42 \implies x_4 + x_5 = 84

.

For

n=8

elements, the median is the arithmetic mean of the 4th and 5th terms.

3

Determine the exact position of the three occurrences of 38.

x_2 = x_3 = x_4 = 38

, forcing

x_5 = 84 - 38 = 46

.

If

x_1 = 38

, then

x_8 = 38 + 25 = 63

, exceeding the upper bound of 55. Thus 38 cannot start at

x_1

, so it must occupy

x_2, x_3, x_4

.

4

Maximize the remaining elements

x_1, x_6, x_7, x_8

.

x_8 = 55

,

x_1 = 30

,

x_7 = 55

, and

x_6 = 54

.

To maximize the sum, set

x_8 = 55

, which fixes

x_1 = 55 - 25 = 30

. Set

x_7 = 55

.

x_6

can be at most 54 because setting

x_6 = 55

would give 55 a frequency of 3, violating the unique mode requirement.

5

Calculate the maximum sum and arithmetic mean.

Sum

= 30 + 38 + 38 + 38 + 46 + 54 + 55 + 55 = 354

; Mean

= 354 / 8 = 44.25

.

Dividing the maximum total sum of 354 by 8 gives the maximum possible arithmetic mean.

Key Concept

Optimization of Means Subject to Central Tendency and Range Constraints

Estimated Time:2m 30s

Question 1688Question →

A bookshelf holds $4$ distinct fiction novels and $3$ distinct non-fiction books. If a reader chooses exactly $1$ fiction novel and $1$ non-fiction book to take on a trip, how many different pairs of books can the reader select?

Show answer & explanation

Answer: 12

Answer

The total number of different pairs of books that can be selected is

12

.

According to the Fundamental Counting Principle, if one task can be performed in

m

ways and a second task can be performed in

n

ways, the two tasks together can be performed in

m \times n

ways. Choosing a fiction novel (

4

options) and a non-fiction book (

3

options) results in

4 \times 3 = 12

unique pairs.

Step-by-Step Solution

1

Determine the number of ways to choose one fiction novel

There are

4

possible choices.

The shelf contains

4

distinct fiction novels.

2

Determine the number of ways to choose one non-fiction book

There are

3

possible choices.

The shelf contains

3

distinct non-fiction books.

3

Calculate total pairs using the Fundamental Counting Principle

4 \times 3 = 12

The selection of a fiction novel and a non-fiction book are independent decisions, so the number of outcomes is the product of the number of choices for each decision.

Key Concept

Fundamental Counting Principle

Question 1689Question →

The frequency distribution table below summarizes the calibration offset errors, $x$ (in microvolts, $\mu\text{V}$ ), measured for a sample of $200$ precision voltage sensors in a robotics laboratory.

Offset Error Interval ( $\mu\text{V}$ )	Frequency
$0 \le x < 10$	$32$
$10 \le x < 20$	$f_1$
$20 \le x < 30$	$68$
$30 \le x < 40$	$f_2$
$40 \le x < 50$	$24$

The estimated mean offset error calculated using the midpoints of the five class intervals is equal to $24.8\ \mu\text{V}$ . If a sensor is selected at random from among those with an offset error of at least $20\ \mu\text{V}$ , what is the probability that its offset error is less than $40\ \mu\text{V}$ ?

\frac{14}{17}

\frac{14}{25}

\frac{25}{31}

\frac{11}{34}

\frac{17}{25}

Show answer & explanation

Answer: $\frac{14}{17}$

Answer

The correct answer is

\frac{14}{17}

, which represents the conditional probability that a sensor's offset error is less than

40\ \mu\text{V}

given that it is at least

20\ \mu\text{V}

.

The correct answer is

\frac{14}{17}

. Solving the system of equations formed by the total sample size (

f_1 + f_2 = 76

) and the estimated midpoint mean (

3f_1 + 7f_2 = 404

) yields

f_1 = 32

and

f_2 = 44

. The number of sensors with offset

\ge 20\ \mu\text{V}

is

68 + 44 + 24 = 136

. Among these, the number of sensors with offset

< 40\ \mu\text{V}

is

68 + 44 = 112

. Thus, the conditional probability is

\frac{112}{136} = \frac{14}{17}

.

Step-by-Step Solution

1

Set up an equation for the total frequency of the sample.

32 + f_1 + 68 + f_2 + 24 = 200 \implies f_1 + f_2 = 76

The sum of all class frequencies must equal the given total sample size of 200.

2

Set up an equation for the estimated mean using interval midpoints.

Midpoints are

5, 15, 25, 35, 45

. Total weighted sum

= 32(5) + f_1(15) + 68(25) + f_2(35) + 24(45) = 160 + 15f_1 + 1700 + 35f_2 + 1080 = 2940 + 15f_1 + 35f_2

. Mean

= \frac{2940 + 15f_1 + 35f_2}{200} = 24.8 \implies 15f_1 + 35f_2 = 2020 \implies 3f_1 + 7f_2 = 404

.

The estimated mean of grouped data is the sum of products of interval midpoints and frequencies divided by total sample size.

3

Solve the linear system of equations for

f_1

and

f_2

.

Multiply

f_1 + f_2 = 76

by

3

to get

3f_1 + 3f_2 = 228

. Subtracting from

3f_1 + 7f_2 = 404

gives

4f_2 = 176 \implies f_2 = 44

. Then

f_1 = 76 - 44 = 32

.

Elimination yields exact unknown frequencies for the remaining intervals.

4

Determine the conditional sample space and target frequency.

Condition (offset

\ge 20\ \mu\text{V}

): Intervals

[20, 30), [30, 40), [40, 50)

with total frequency

68 + 44 + 24 = 136

. Target condition (offset

< 40\ \mu\text{V}

within condition): Intervals

[20, 30)

and

[30, 40)

with frequency

68 + 44 = 112

.

Conditional probability restricts the denominator to sensors meeting the given condition.

5

Calculate the final probability fraction.

Probability

= \frac{112}{136} = \frac{14}{17}

.

Dividing target count by conditional total count yields the simplified fraction.

Key Concept

Grouped Data Mean Estimation & Conditional Probability from Frequency Distributions

Estimated Time:2m 30s

Question 1690Question →

An agricultural facility uses two automated irrigation systems, System P and System Q, to water a field. System P pumps water at a constant rate of $120$ gallons per hour after requiring an initial setup overhead of $30$ minutes ( $0.5$ hours). System Q pumps water at a constant rate of $180$ gallons per hour after requiring an initial setup overhead of $45$ minutes ( $0.75$ hours). Both systems begin their setup process at the exact same time and operate continuously until a combined total of $1,605$ gallons of water has been pumped. Which of the following statements must be true? Select all such statements.

Select all that apply

The total elapsed time from when setup began until the target volume was reached is

6

hours.

System Q pumped

285

more gallons of water than System P.

System P was actively pumping water for exactly

5

hours and

15

minutes.

System Q accounted for more than

60\%

of the total water pumped.

The ratio of the volume of water pumped by System P to that pumped by System Q is

44:63

.

Show answer & explanation

Answer: The total elapsed time from when setup began until the target volume was reached is $6$ hours.; System Q pumped $285$ more gallons of water than System P.; The ratio of the volume of water pumped by System P to that pumped by System Q is $44:63$ .

Answer

The correct statements are: the total elapsed time from when setup began is 6 hours, System Q pumped 285 more gallons than System P, and the ratio of water pumped by System P to System Q is 44:63.

Solving the combined linear rate model yields a total elapsed time of 6 hours. With this duration, System P operates for 5.5 hours producing 660 gallons, and System Q operates for 5.25 hours producing 945 gallons. This confirms that the total time is 6 hours, System Q produces 285 more gallons than System P (945 - 660 = 285), and the ratio of System P's volume to System Q's volume is 660:945, which simplifies to 44:63.

Step-by-Step Solution

1

Define the variable for total time and establish time expressions for active pumping.

Let

t

be the total elapsed time in hours since both systems began setup (

t \geq 0.75

). System P actively pumps for

(t - 0.5)

hours, and System Q actively pumps for

(t - 0.75)

hours.

Setup overhead delays the start of pumping, so active pumping duration equals total elapsed time minus setup time.

2

Set up and solve the linear rate equation for combined total volume.

120(t - 0.5) + 180(t - 0.75) = 1605 \implies 120t - 60 + 180t - 135 = 1605 \implies 300t - 195 = 1605 \implies 300t = 1800 \implies t = 6

hours.

The sum of the volumes produced by both systems must equal the target total of 1,605 gallons.

3

Calculate individual pumping times and volumes produced by each system.

System P: active time

= 5.5

hours, volume

= 120 \times 5.5 = 660

gallons. System Q: active time

= 5.25

hours, volume

= 180 \times 5.25 = 945

gallons.

Individual volumes are required to evaluate statements regarding volume differences, percentages, and ratios.

4

Evaluate each given statement against the calculated values.

Elapsed time is

6

hours (True). Volume difference is

945 - 660 = 285

gallons (True). Active pumping time for P is

5.5

hours, not

5.25

hours (False). Percentage for Q is

945 / 1605 \approx 58.88\% < 60\%

(False). Ratio P to Q is

660 : 945 = 44 : 63

(True).

Determines which of the statements must be selected.

Key Concept

Linear Modeling with Combined Work Rates and Staggered Start Times

Question 1691Question →

In the $xy$ -plane, line $L_1$ is given by the equation $ax + 3y = 12$ , where $a$ is a constant. Line $L_2$ is perpendicular to line $L_1$ and passes through the point $(4, -1)$ . If the $x$ -intercept of line $L_2$ is double its $y$ -intercept, and both intercepts of line $L_2$ are non-zero, what is the value of $a$ ?

-6

-\frac{3}{2}

\frac{3}{2}

3

6

Show answer & explanation

Answer: $-6$

Answer

-6

Rearranging

L_1: ax + 3y = 12

into slope-intercept form gives

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

, yielding a slope of

m_1 = -\frac{a}{3}

. The perpendicular line

L_2

has a slope of

m_2 = \frac{3}{a}

. Using the point

(4, -1)

, the equation for

L_2

is

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

, which simplifies to

y = \frac{3}{a}x - \frac{12+a}{a}

. Setting

x = 0

gives the

y

-intercept

y_0 = -\frac{12+a}{a}

, and setting

y = 0

gives the

x

-intercept

x_0 = \frac{12+a}{3}

. Equating

x_0 = 2 y_0

yields

\frac{12+a}{3} = -2\frac{12+a}{a}

. Canceling the non-zero term

(12+a)

gives

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

, which leads to

a = -6

.

Step-by-Step Solution

1

Determine the slope of line

L_1

and line

L_2

Slope of

L_1

is

m_1 = -\frac{a}{3}

, so the slope of perpendicular line

L_2

is

m_2 = \frac{3}{a}

.

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

2

Write the equation of line

L_2

using point-slope form

y - (-1) = \frac{3}{a}(x - 4) \implies y = \frac{3}{a}x - \frac{12}{a} - 1 = \frac{3}{a}x - \left(\frac{12 + a}{a}\right)

.

Line

L_2

passes through

(4, -1)

with slope

m_2 = \frac{3}{a}

.

3

Find the

x

-intercept and

y

-intercept of line

L_2

The

y

-intercept is

y_0 = -\frac{12 + a}{a}

. Setting

y = 0

gives

0 = \frac{3}{a}x - \frac{12 + a}{a} \implies x_0 = \frac{12 + a}{3}

.

Intercepts occur where

x = 0

(for

y

-intercept) and

y = 0

(for

x

-intercept).

4

Set up and solve the given relation

x_0 = 2 y_0

\frac{12 + a}{3} = 2 \left( -\frac{12 + a}{a} \right)

. Since the intercepts are non-zero,

12 + a \neq 0

. Dividing both sides by

(12 + a)

gives

\frac{1}{3} = -\frac{2}{a} \implies a = -6

.

The problem states the

x

-intercept is double the

y

-intercept.

Key Concept

Perpendicular Slopes and Line Intercepts

Estimated Time:2m 0s

Question 1692Question →

Let $f(x) = x^2 + bx + c$ , where $b$ and $c$ are integers. The equation $f(x) = 0$ has two distinct real roots, $\alpha$ and $\beta$ . If $\alpha + \beta = \alpha\beta$ and $c > 0$ , which of the following statements must be true? Indicate all such statements.

Select all that apply

The constant

c

is strictly greater than 4.

Both roots

\alpha

and

\beta

are positive.

The vertex of the parabola

y = f(x)

lies in Quadrant IV of the xy-plane.

The coefficient

b

is positive.

The difference between the two roots,

|\alpha - \beta|

, can equal 2.

Show answer & explanation

Answer: The constant $c$ is strictly greater than 4.; Both roots $\alpha$ and $\beta$ are positive.; The vertex of the parabola $y = f(x)$ lies in Quadrant IV of the xy-plane.

Answer

The statements asserting that the constant

c

is strictly greater than 4, both roots are positive, and the vertex lies in Quadrant IV must all be true.

By Vieta's formulas,

\alpha + \beta = -b

and

\alpha\beta = c

. Equating sum and product gives

b = -c

. The discriminant

\Delta = c^2 - 4c > 0

requires

c > 4

because

c > 0

. Since sum and product of the roots equal

c > 0

, both roots are positive. The vertex coordinates

(c/2, -c(c-4)/4)

have a positive x-value and a negative y-value, placing the vertex in Quadrant IV.

Step-by-Step Solution

1

Apply Vieta's formulas and establish the relationship between coefficients

\alpha + \beta = -b

and

\alpha\beta = c

. Setting them equal gives

-b = c \implies b = -c

.

The stem specifies that the sum of the roots equals their product.

2

Analyze the discriminant for distinct real roots

\Delta = b^2 - 4c = (-c)^2 - 4c = c(c - 4) > 0

. Since

c > 0

,

c - 4 > 0 \implies c > 4

.

Two distinct real roots require a strictly positive discriminant.

3

Determine the signs of the roots

\alpha + \beta = c > 0

and

\alpha\beta = c > 0

, implying

\alpha > 0

and

\beta > 0

.

If the sum and product of two real numbers are both positive, both numbers must be positive.

4

Find the location of the parabola's vertex

x_v = \frac{c}{2} > 0

and

y_v = -\frac{c(c-4)}{4} < 0

, placing the vertex in Quadrant IV.

A point with a positive x-coordinate and negative y-coordinate resides in the fourth quadrant.

5

Evaluate the remaining options regarding

b

and

|\alpha - \beta|

b = -c < -4

(negative), and

|\alpha - \beta| = 2 \implies c = 2 + 2\sqrt{2}

, which is not an integer.

These evaluations disprove the statements that

b

is positive and that

|\alpha - \beta|

can equal 2.

Key Concept

Quadratic Root Properties and Vieta's Formulas

Question 1693Question →

Based on the explicit continuation signal in the sentence below, fill in the blank with the appropriate word that completes the sentence logically.

Fill in the blanks below

In cognitive psychology, researchers noted that severe sleep deprivation does not merely impair immediate attentional control; indeed, by disrupting key stages of sleep architecture, it consistently the consolidation of complex spatial memories.

Show answer & explanation

Answer

A word meaning 'impedes', 'hinders', 'hampers', or 'thwarts'.

The phrase 'does not merely...; indeed,' acts as an explicit continuation and intensification signal. Because the initial clause establishes that sleep deprivation has a harmful effect ('impair'), the second clause must amplify this negative effect. Thus, a word meaning to hinder or obstruct, such as 'impedes', 'hinders', or 'hampers', is required to maintain contextual and structural coherence.

Step-by-Step Solution

1

Analyze the sentence structure and identify transition signals.

The construct 'does not merely...; indeed,' serves as an explicit continuation and intensification signal. This shows that the second clause must extend and reinforce the idea introduced in the first clause.

Continuation signals maintain the semantic direction of the sentence and amplify the established tone.

2

Determine the required meaning and tone for the blank.

The first clause notes a negative effect ('impair immediate attentional control'). Therefore, the blank following 'indeed, by disrupting...' must express a negative outcome of similar or greater intensity regarding memory consolidation.

The word filling the blank must signify obstruction, hindrance, or damage to align logically with the causal clause.

Key Concept

Interpreting Continuation and Causal Structural Signals

Estimated Time:1m 0s

Question 1694Question →

In the $xy$ -plane, line $k$ has a slope of $\frac{3}{4}$ and passes through the point $(2, 1)$ . Line $k$ intersects the line $x = 10$ at point $P$ . What is the distance between point $P$ and the point $(2, 1)$ ?

Show answer & explanation

Answer: 10

Answer

10

Point P lies on the line x = 10, so its x-coordinate is 10. The horizontal change from x = 2 to x = 10 is 8 units. Given that line k has slope 3/4, the corresponding vertical change is (3/4)(8) = 6 units. Therefore, point P has coordinates (10, 7). The distance between (2, 1) and (10, 7) is sqrt((10 - 2)^2 + (7 - 1)^2) = sqrt(64 + 36) = sqrt(100) = 10.

Step-by-Step Solution

1

Determine the coordinates of point P

P is located at (10, 7)

Since P lies on the line x = 10, its x-coordinate is 10. The horizontal distance from (2, 1) to P is 10 - 2 = 8. Using the slope m = 3/4, the vertical change is (3/4) * 8 = 6, so the y-coordinate of P is 1 + 6 = 7.

2

Calculate the distance between (2, 1) and (10, 7)

The distance is 10

Using the distance formula sqrt((10 - 2)^2 + (7 - 1)^2) = sqrt(8^2 + 6^2) = sqrt(64 + 36) = sqrt(100) = 10.

Key Concept

Slope definition and distance formula in coordinate geometry

Estimated Time:1m 30s

Question 1695Question →

A laboratory processes two types of chemical samples, Type A and Type B. Processing each Type A sample requires $40$ minutes and costs $\$30$ , while processing each Type B sample requires $30$ minutes and costs $\$50$ . On a given day, the laboratory spent a total of $20$ hours processing these two types of samples at a total cost of $\$1,450$ . How many more Type B samples were processed than Type A samples?

5

10

15

20

35

Show answer & explanation

Answer: 5

Answer

5 more Type B samples were processed than Type A samples.

The correct answer is 5. By defining

x

as the number of Type A samples and

y

as the number of Type B samples, we construct the time equation

40x + 30y = 1,200

and cost equation

30x + 50y = 1,450

. Solving this system yields

x = 15

and

y = 20

. The difference

y - x = 20 - 15 = 5

.

Step-by-Step Solution

1

Define variables and convert units to maintain consistency.

Let

x

be the number of Type A samples and

y

be the number of Type B samples. Total time available is

20 \text{ hours} \times 60 \text{ minutes/hour} = 1,200 \text{ minutes}

.

Time specifications for individual samples are given in minutes, so total time must also be in minutes.

2

Set up a system of two linear equations representing total time and total cost.

Time equation:

40x + 30y = 1,200 \implies 4x + 3y = 120

. Cost equation:

30x + 50y = 1,450 \implies 3x + 5y = 145

.

The total processing time is the sum of time spent on each sample type, and total cost is the sum of costs for each sample type.

3

Solve the system of equations using elimination.

Multiply the time equation by 3:

12x + 9y = 360

. Multiply the cost equation by 4:

12x + 20y = 580

. Subtracting the first from the second gives

11y = 220 \implies y = 20

. Substitute

y = 20

into

4x + 3(20) = 120 \implies 4x = 60 \implies x = 15

.

Eliminating

x

allows direct calculation of

y

, which then yields

x

.

4

Calculate the difference requested by the problem.

Difference =

y - x = 20 - 15 = 5

.

The question specifically asks for how many more Type B samples were processed than Type A samples.

Key Concept

Formulating and solving systems of linear equations from real-world rate and budget constraints.

Estimated Time:1m 45s

Question 1696Question →

For all real numbers $x$ , the function $g$ is defined by $g(x) = cx + 5$ , where $c$ is a constant. The custom operation $\diamond$ is defined for all real numbers $a$ and $b$ by $a \diamond b = g(a+b) - g(a-b)$ . If $3 \diamond 1 = 12$ , what is the value of $g(4)$ ?

Show answer & explanation

Answer: 29

Answer

29

Applying the function definition

g(x) = cx + 5

to the custom operation yields

a \diamond b = [c(a+b)+5] - [c(a-b)+5] = 2bc

. Substituting

a=3

and

b=1

into

3 \diamond 1 = 12

gives

2(1)c = 12

, so

c = 6

. Consequently,

g(x) = 6x + 5

, and evaluating at

x = 4

yields

g(4) = 6(4) + 5 = 29

.

Step-by-Step Solution

1

Substitute

(a+b)

and

(a-b)

into the function definition

g(x) = cx + 5

to simplify

a \diamond b

.

a \diamond b = [c(a+b) + 5] - [c(a-b) + 5] = 2bc

.

Applying the definition of the custom binary operation in terms of function

g

eliminates the constant term

5

.

2

Use the given equality

3 \diamond 1 = 12

to determine the constant

c

.

2(1)c = 12 \implies 2c = 12 \implies c = 6

.

Plugging

a=3

and

b=1

into

2bc = 12

yields an equation in terms of

c

.

3

Evaluate

g(4)

using

c = 6

.

g(4) = 6(4) + 5 = 29

.

Substituting

x = 4

into

g(x) = 6x + 5

calculates the required numerical value.

Key Concept

Functions and Custom Symbol Operations

Estimated Time:1m 30s

Question 1697Question →

A environmental monitoring group collected $200$ soil samples from a nature reserve and recorded their pH levels in the frequency table below:

pH Level Range	Number of Samples
$5.0 – 5.9$	$40$
$6.0 – 6.9$	$70$
$7.0 – 7.9$	$50$
$8.0 – 8.9$	$30$
$9.0 – 9.9$	$10$

What percent of the soil samples with a pH level of at least $6.0$ have a pH level in the range $6.0$ to $7.9$ ?

43.75\%

60.00\%

75.00\%

80.00\%

68.75\%

Show answer & explanation

Answer: $75.00\%$

Answer

75.00\%

To find the desired percentage, first restrict the sample space to all soil samples having a pH level of at least

6.0

. Adding the frequencies for

6.0–6.9

(

70

),

7.0–7.9

(

50

),

8.0–8.9

(

30

), and

9.0–9.9

(

10

) gives a base total of

160

samples. Next, find the number of samples within that group that fall in the range

6.0

to

7.9

, which is

70 + 50 = 120

. The percentage is calculated as

(120 / 160) \times 100\% = 75.00\%

.

Step-by-Step Solution

1

Determine the conditional base population (samples with pH

\ge 6.0

).

Sum of frequencies for ranges

6.0-6.9

,

7.0-7.9

,

8.0-8.9

, and

9.0-9.9

:

70 + 50 + 30 + 10 = 160

.

The question asks 'of the soil samples with a pH level of at least 6.0', which restricts the total base to these four intervals.

2

Determine the target frequency (samples with pH in range

6.0

to

7.9

).

Sum of frequencies for ranges

6.0-6.9

and

7.0-7.9

:

70 + 50 = 120

.

This captures all qualifying samples within the specified target interval.

3

Calculate the conditional percentage.

\frac{120}{160} \times 100\% = 0.75 \times 100\% = 75.00\%

.

Divide the target sub-group count by the conditional base count and convert to a percentage.

Key Concept

Conditional Relative Frequency in Grouped Data Tables

Estimated Time:1m 30s

Question 1698Question →

An academic conference has 7 consecutive presentation time slots. A committee must assign 7 presentations—3 in Biology ( $B$ ), 2 in Chemistry ( $C$ ), and 2 in Physics ( $P$ )—to these slots subject to the following conditions:

1. Presentations of the same discipline are indistinguishable (only the subject sequence matters).
2. No two Biology presentations may be scheduled in consecutive time slots.
3. The two Physics presentations must be scheduled in consecutive time slots.

How many different subject-sequence schedules for the 7 presentation time slots satisfy all of these conditions?

24

12

30

60

72

Show answer & explanation

Answer: $12$

Answer

12

To satisfy the condition that the two Physics presentations are consecutive, we combine them into a single block

(PP)

. We then arrange the non-Biology items—two Chemistry presentations

C

and one block

(PP)

—which can be ordered in

\frac{3!}{2!} = 3

distinct ways. These 3 items create 4 potential gaps (before the first item, between items, and after the last item). To ensure no two Biology presentations are adjacent, we place one Biology presentation into each of 3 chosen gaps out of the 4 available spaces, which can be done in

\binom{4}{3} = 4

ways. By the Fundamental Counting Principle, the total number of valid subject-sequence schedules is

3 \times 4 = 12

.

Step-by-Step Solution

1

Group the adjacent Physics presentations into a single block

The two Physics presentations

(P, P)

form a single block

(PP)

. The set of non-Biology items consists of two indistinguishable

C

's and one

(PP)

block, making 3 items in total.

Since the two Physics presentations must be scheduled in consecutive slots, treating them as a single unit guarantees they stay adjacent.

2

Calculate the number of distinct arrangements of the non-Biology items

The number of distinct arrangements of

\{C, C, (PP)\}

is

\frac{3!}{2!1!} = 3

ways.

The 2 Chemistry presentations are identical, so we divide the total permutations (

3!

) by

2!

to account for indistinguishability.

3

Apply the gap method to place the non-adjacent Biology presentations

Any sequence of the 3 non-Biology items creates 4 available spaces (gaps), including the two ends:

\_ \text{item}_1 \_ \text{item}_2 \_ \text{item}_3 \_

. Choosing 3 gaps out of 4 yields

\binom{4}{3} = 4

ways to insert the 3 identical

B

presentations.

Placing at most one Biology presentation in each gap guarantees no two Biology presentations are consecutive.

4

Apply the Fundamental Counting Principle

Total valid schedules =

3 \times 4 = 12

.

The choices of arranging non-Biology items and selecting gaps for Biology items are independent.

Key Concept

Combinatorics with Adjacency Restrictions and Indistinguishable Objects

Question 1699Question →

An executive education program surveyed $400$ professional candidates regarding their enrollment in three specialized tracks: Artificial Intelligence ( $A$ ), Financial Technology ( $F$ ), and Sustainable Energy ( $S$ ).

The survey revealed the following data:
- $40$ candidates were enrolled in none of the three tracks.
- $210$ candidates were enrolled in Artificial Intelligence.
- $180$ candidates were enrolled in Financial Technology.
- $150$ candidates were enrolled in Sustainable Energy.
- $90$ candidates were enrolled in both Artificial Intelligence and Financial Technology.
- $70$ candidates were enrolled in both Financial Technology and Sustainable Energy.
- $60$ candidates were enrolled in both Artificial Intelligence and Sustainable Energy.

How many candidates were enrolled in exactly one of the three tracks?

100

140

180

220

260

Show answer & explanation

Answer: $220$

Answer

The number of candidates enrolled in exactly one of the three tracks is

220

.

To find the number of candidates enrolled in exactly one track, we first apply the Principle of Inclusion-Exclusion to find the triple intersection

|A \cap F \cap S|

:

|A \cup F \cup S| = |A| + |F| + |S| - (|A \cap F| + |F \cap S| + |A \cap S|) + |A \cap F \cap S|

. Substituting known values gives

360 = 210 + 180 + 150 - 220 + |A \cap F \cap S|

, so

|A \cap F \cap S| = 40

. Isolating mutually exclusive regions yields

100

candidates in Artificial Intelligence only,

60

in Financial Technology only, and

60

in Sustainable Energy only. Summing these gives

100 + 60 + 60 = 220

.

Step-by-Step Solution

1

Determine the number of candidates enrolled in at least one track.

|A \cup F \cup S| = 400 - 40 = 360

Candidates in none of the tracks must be subtracted from the total survey population.

2

Calculate the number of candidates enrolled in all three tracks (

x = |A \cap F \cap S|

) using the Principle of Inclusion-Exclusion.

360 = 210 + 180 + 150 - (90 + 70 + 60) + x \implies 360 = 320 + x \implies x = 40

The sum of individual sets overcounts pairwise intersections and undercounts the triple intersection.

3

Calculate the counts for regions representing enrollment in exactly two tracks.

A \cap F \text{ only} = 90 - 40 = 50

,

F \cap S \text{ only} = 70 - 40 = 30

,

A \cap S \text{ only} = 60 - 40 = 20

Subtracting the triple intersection from each pairwise intersection isolates candidates enrolled in precisely two tracks.

4

Calculate the counts for candidates enrolled in exactly one track and sum them.

A \text{ only} = 210 - (50 + 20 + 40) = 100

;

F \text{ only} = 180 - (50 + 30 + 40) = 60

;

S \text{ only} = 150 - (20 + 30 + 40) = 60

. Total

= 100 + 60 + 60 = 220

.

Subtracting all overlap regions from each set total yields the single-track enrollment.

Key Concept

Principle of Inclusion-Exclusion for Three Sets

Question 1700Question →

A fair six-sided die with faces numbered $1$ through $6$ is rolled twice. What is the probability of rolling a $4$ on the first roll and an odd number on the second roll?

\frac{1}{12}

\frac{2}{3}

\frac{1}{3}

\frac{1}{4}

\frac{1}{36}

Show answer & explanation

Answer: $\frac{1}{12}$

Answer

\frac{1}{12}

Because the outcome of the first die roll does not affect the outcome of the second die roll, the two events are independent. The probability of rolling a

4

on the first roll is

\frac{1}{6}

, and the probability of rolling an odd number (

1, 3,

or

5

) on the second roll is

\frac{3}{6} = \frac{1}{2}

. Multiplying these individual probabilities together yields

\frac{1}{6} \times \frac{1}{2} = \frac{1}{12}

.

Step-by-Step Solution

1

Determine the probability of the first event (rolling a 4).

There is

1

favorable outcome out of

6

possible outcomes, so

P(\text{First roll is } 4) = \frac{1}{6}

.

Each face of a fair six-sided die is equally likely to land face up.

2

Determine the probability of the second event (rolling an odd number).

The odd numbers are

1, 3,

and

5

, giving

3

favorable outcomes out of

6

total outcomes, so

P(\text{Second roll is odd}) = \frac{3}{6} = \frac{1}{2}

.

Half of the outcomes on a standard six-sided die are odd.

3

Calculate the joint probability of both independent events occurring.

P(\text{4 on first AND odd on second}) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}

.

For independent events

A

and

B

, the multiplication rule states that

P(A \text{ and } B) = P(A) \times P(B)

.

Key Concept

Probability of Independent Events

Estimated Time:1m 0s

Answer

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