All practice questions

Question 1621Question →

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

15

20

22.5

25

30

Show answer & explanation

Answer: 25

Answer

Clara's original speed

c

is 25 miles per hour.

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

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

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

\frac{120}{2.4} = 50

mph. Setting up the equation

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

yields

c = 25

.

Step-by-Step Solution

1

Determine the initial combined rate of Clara and Dan.

Since they cover

120

miles in

3

hours, their combined speed is

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

miles per hour.

Distance equals rate multiplied by time (

D = r \cdot t

). When moving toward each other, combined rate is

c + d

.

2

Calculate the new travel time and the new combined rate.

The new travel time is

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

hours (or

\frac{12}{5}

hours). The new combined rate is

\frac{120}{2.4} = 50

miles per hour.

36 minutes is equivalent to

\frac{36}{60} = 0.6

hours.

3

Express the modified individual rates algebraically.

Clara's new speed is

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

. Dan's new speed is

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

.

Increasing by

20\%

multiplies rate by

1.2

, and increasing by

33\frac{1}{3}\%

multiplies rate by

\frac{4}{3}

.

4

Formulate and solve the system of linear equations.

We have the system:
1)

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

2)

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

Substituting

d = 40 - c

into equation 2:

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

Multiplying the entire equation by

15

to eliminate denominators:

18c + 20(40 - c) = 750

18c + 800 - 20c = 750

-2c = -50 \implies c = 25

Solving for

c

yields Clara's original speed in miles per hour.

Key Concept

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

Estimated Time:2m 30s

Question 1622Question →

For all real numbers

x

and

y

, the custom operation

\diamondsuit

is defined by

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

The function

f

is defined for all real numbers

t

by

f(t) = t + 1

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

a

and

b

? Select all such statements.

Select all that apply

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

If

a \neq b

, then

a \diamondsuit b = 0

if and only if

a + b = 2

.

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

a \diamondsuit (-a) = 0

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

Show answer & explanation

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

Answer

The statements 'If

a \neq b

, then

a \diamondsuit b = 0

if and only if

a + b = 2

' and '

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

' must be true.

The statement claiming that if

a \neq b

,

a \diamondsuit b = 0

if and only if

a + b = 2

is correct because factoring

a \diamondsuit b

gives

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

, which requires

a + b = 2

when

a - b \neq 0

. The statement claiming

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

is also correct because substituting

f(a) = a + 1

and

f(b) = b + 1

into the operation yields

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

.

Step-by-Step Solution

1

Factor the custom operation definition

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

Factoring groups terms into products that simplify statement testing.

2

Analyze the condition for

a \diamondsuit b = 0

when

a \neq b

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

Since

a \neq b

,

a - b \neq 0

, allowing division by

(a - b)

to isolate

a + b = 2

.

3

Evaluate

f(a) \diamondsuit f(b)

using

f(t) = t + 1

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

Direct substitution into the factored form simplifies the evaluation.

4

Evaluate remaining statements to identify counterexamples

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

;

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

;

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

Confirming false statements ensures accurate multi-selection.

Key Concept

Algebraic manipulation of custom binary operations and function evaluation

Question 1623Question →

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

Show answer & explanation

Answer: 200

Answer

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

Total revenue is given by

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

, and total cost is given by

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

. The profit equation

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

yields

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

. Rearranging gives

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

, which simplifies to

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

. Factoring gives

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

, leading to prices of

\$200

and

\$300

. The smaller price is

\$200

.

Step-by-Step Solution

1

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

P

.

Revenue

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

and Total Cost

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

.

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

2

Formulate the net profit function

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

and equate it to the target profit of

\$45,000

.

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

, which simplifies to

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

Net profit is defined as total revenue minus total cost.

3

Divide the resulting quadratic equation by

-3

to normalize coefficients and solve for

P

.

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

, giving roots

P = 200

and

P = 300$.

Simplifying the quadratic equation reveals two factored linear roots.

4

Identify the smaller of the two valid price solutions.

P = 200 dollars.

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

Key Concept

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

Question 1624Question →

A beverage manufacturing company tracks the monthly production counts (in thousands of units) for its facilities located in two separate regions. Region A consists of $6$ facilities with monthly production counts of $12$ , $15$ , $15$ , $18$ , $20$ , and $28$ . Region B consists of $4$ facilities with monthly production counts of $10$ , $14$ , $22$ , and $34$ .

Which of the following statements must be true regarding the individual regional datasets and the combined dataset of all $10$ facilities? Select all such statements.

Select all that apply

The arithmetic mean of the combined dataset of all

10

facilities is greater than the arithmetic mean of Region A.

The median of the combined dataset of all

10

facilities is equal to the median of Region A.

The median of the combined dataset of all

10

facilities is equal to the arithmetic mean of the median of Region A and the median of Region B.

The mode of the combined dataset of all

10

facilities is greater than the mode of Region A.

The arithmetic mean of the combined dataset of all

10

facilities is equal to the simple average of the mean of Region A and the mean of Region B.

Show answer & explanation

Answer: The arithmetic mean of the combined dataset of all $10$ facilities is greater than the arithmetic mean of Region A.; The median of the combined dataset of all $10$ facilities is equal to the median of Region A.

Answer

The correct statements are the one asserting that the combined arithmetic mean is greater than Region A's mean, and the one asserting that the combined median is equal to Region A's median.

The mean of the combined dataset (

18.8

) is greater than the mean of Region A (

18.0

), making the statement comparing the combined mean to Region A's mean correct. Additionally, both Region A's median and the combined dataset's median evaluate to

16.5

, making the statement asserting equality between these two medians correct.

Step-by-Step Solution

1

Calculate the sum, mean, and median for Region A.

Region A sum

= 12 + 15 + 15 + 18 + 20 + 28 = 108

. Mean

= \frac{108}{6} = 18

. Sorted values are

12, 15, 15, 18, 20, 28

, so Median

= \frac{15 + 18}{2} = 16.5

. Mode

= 15

.

Establishing base central metrics for the first subgroup.

2

Calculate the sum, mean, and median for Region B.

Region B sum

= 10 + 14 + 22 + 34 = 80

. Mean

= \frac{80}{4} = 20

. Sorted values are

10, 14, 22, 34

, so Median

= \frac{14 + 22}{2} = 18

.

Establishing base central metrics for the second subgroup.

3

Combine and order all 10 values to calculate combined metrics.

Combined ordered set:

10, 12, 14, 15, 15, 18, 20, 22, 28, 34

. Total sum

= 108 + 80 = 188

. Combined Mean

= \frac{188}{10} = 18.8

. Combined Median

= \frac{15 + 18}{2} = 16.5

. Combined Mode

= 15

.

Necessary to evaluate combined properties accurately against subgroup properties.

4

Evaluate each candidate statement against calculated values.

Combined mean (

18.8

) > Region A mean (

18

) is TRUE. Combined median (

16.5

) = Region A median (

16.5

) is TRUE. Average of medians (

17.25

) = Combined median (

16.5

) is FALSE. Combined mode (

15

) > Region A mode (

15

) is FALSE. Simple average of means (

19

) = Combined mean (

18.8

) is FALSE.

Determining which statements satisfy the required conditions.

Key Concept

Combining Datasets and Weighted Measures of Central Tendency

Question 1625Question →

The table below shows the frequency distribution of order fulfillment times, in minutes, for a sample of $50$ online orders processed at a warehouse facility.

Fulfillment Time $t$ (minutes)	Number of Orders
$10 \le t < 20$	$10$
$20 \le t < 30$	$15$
$30 \le t < 40$	$15$
$40 \le t < 50$	$7$
$50 \le t < 60$	$3$

An order is categorized as "delayed" if its fulfillment time is at least $30$ minutes. What percentage of all orders that took at least $20$ minutes to fulfill were categorized as "delayed"?

30.0%

37.5%

50.0%

62.5%

75.0%

Show answer & explanation

Answer: 62.5%

Answer

62.5%

To find the required percentage, first determine the subset of orders that took at least

20

minutes to fulfill. Summing the frequencies of the last four intervals (

20 \le t < 30

,

30 \le t < 40

,

40 \le t < 50

, and

50 \le t < 60

) yields

15 + 15 + 7 + 3 = 40

orders. Next, find how many of these orders took at least

30

minutes to fulfill (the delayed orders), which corresponds to the last three intervals:

15 + 7 + 3 = 25

orders. Finally, calculate the percentage:

\frac{25}{40} \times 100\% = 62.5\%

.

Step-by-Step Solution

1

Identify the sub-population that defines the base (denominator) of the percentage.

The target base is all orders with fulfillment times of at least

20

minutes (

t \ge 20

). Summing the frequencies for intervals

20 \le t < 30

,

30 \le t < 40

,

40 \le t < 50

, and

50 \le t < 60

gives

15 + 15 + 7 + 3 = 40

orders.

The question specifically asks 'of all orders that took at least 20 minutes to fulfill', establishing a conditional base.

2

Determine the number of orders meeting the numerator condition.

The delayed orders are those taking at least

30

minutes (

t \ge 30

). Summing frequencies for

30 \le t < 40

,

40 \le t < 50

, and

50 \le t < 60

gives

15 + 7 + 3 = 25

orders.

Orders in all intervals starting from

30

minutes onwards qualify as delayed.

3

Compute the required percentage.

\left(\frac{25}{40}\right) \times 100\% = 0.625 \times 100\% = 62.5\%

.

Dividing the target count by the restricted base count gives the conditional proportion.

Key Concept

Grouped Frequency Distributions and Conditional Percentages

Question 1626Question →

A market research firm surveyed $150$ venture capital investors regarding their investments in three emerging sectors: Artificial Intelligence ( $A$ ), Biotechnology ( $B$ ), and Clean Energy ( $C$ ). The survey revealed the following data:

- $15$ investors did not invest in any of the three sectors.
- $78$ investors invested in Artificial Intelligence.
- $65$ investors invested in Biotechnology.
- $62$ investors invested in Clean Energy.
- $10$ investors invested in all three sectors.

If $50$ investors invested in exactly two of the three sectors, how many investors invested in exactly one of the three sectors?

65

70

75

85

90

Show answer & explanation

Answer: 75

Answer

75 investors invested in exactly one of the three sectors.

The total number of investors who invested in at least one of the three sectors is

150 - 15 = 135

. In a 3-set Venn diagram, the total union is the sum of those who belong to exactly one sector (

E_1

), exactly two sectors (

E_2

), and all three sectors (

E_3

). We are given

E_2 = 50

and

E_3 = 10

. Setting up the region equation

E_1 + E_2 + E_3 = 135

gives

E_1 + 50 + 10 = 135

, which yields

E_1 = 75

.

Step-by-Step Solution

1

Calculate the total number of investors who invested in at least one sector (the union of sets A, B, and C).

|A \cup B \cup C| = 150 - 15 = 135

Subtracting the investors who invested in none of the sectors from the total population yields the size of the union.

2

Express the union as the sum of non-overlapping region counts.

|A \cup B \cup C| = E_1 + E_2 + E_3

, where

E_1

is the number of investors in exactly 1 sector,

E_2

is the number in exactly 2 sectors, and

E_3

is the number in all 3 sectors.

Any element in the union belongs to exactly one, exactly two, or all three sets.

3

Substitute the known values into the partition equation and solve for

E_1

.

135 = E_1 + 50 + 10 \implies E_1 = 135 - 60 = 75

We are given

E_2 = 50

and

E_3 = 10

, leaving

E_1

as the remaining unknown value.

Key Concept

Three-Set Inclusion-Exclusion and Region Partitioning

Estimated Time:2m 0s

Question 1627Question →

A right circular cylinder has a base radius of $r$ and a height of $h$ . If the base radius is doubled and the height is halved to create a new right circular cylinder, which of the following statements must be true? Select all that apply.

Select all that apply

The volume of the new cylinder is twice the volume of the original cylinder.

The lateral surface area of the new cylinder is equal to the lateral surface area of the original cylinder.

The total surface area of the new cylinder is twice the total surface area of the original cylinder.

The volume of the new cylinder is four times the volume of the original cylinder.

The area of the circular base of the new cylinder is four times the area of the circular base of the original cylinder.

Show answer & explanation

Answer: The volume of the new cylinder is twice the volume of the original cylinder.; The lateral surface area of the new cylinder is equal to the lateral surface area of the original cylinder.; The area of the circular base of the new cylinder is four times the area of the circular base of the original cylinder.

Answer

The correct statements are: the volume of the new cylinder is twice the volume of the original cylinder, the lateral surface area of the new cylinder is equal to the lateral surface area of the original cylinder, and the area of the circular base of the new cylinder is four times the area of the circular base of the original cylinder.

The volume formula

V = \pi r^2 h

scales by

2^2 \times \frac{1}{2} = 2

, so volume doubles. The lateral surface area formula

L = 2\pi r h

scales by

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

, remaining unchanged. The base area formula

A = \pi r^2

scales by

2^2 = 4

, quadrupling the area.

Step-by-Step Solution

1

Write down the original formulas for a cylinder with radius

r

and height

h

.

Base Area

A_1 = \pi r^2

, Lateral Surface Area

L_1 = 2\pi r h

, Total Surface Area

T_1 = 2\pi r h + 2\pi r^2

, Volume

V_1 = \pi r^2 h

.

Establish baseline geometric properties.

2

Substitute the new dimensions

r' = 2r

and

h' = \frac{h}{2}

into each formula.

New Base Area

A_2 = \pi (2r)^2 = 4\pi r^2 = 4A_1

.
New Lateral Surface Area

L_2 = 2\pi (2r) \left(\frac{h}{2}\right) = 2\pi r h = L_1

.
New Volume

V_2 = \pi (2r)^2 \left(\frac{h}{2}\right) = \pi (4r^2) \left(\frac{h}{2}\right) = 2\pi r^2 h = 2V_1

.

Calculate the updated values for each geometric property.

3

Evaluate each statement against the calculated values.

Volume doubles (

V_2 = 2V_1

), lateral surface area remains equal (

L_2 = L_1

), and base area quadruples (

A_2 = 4A_1

).

Determine which options state correct mathematical relationships.

Key Concept

Scaling effects on 3D geometry formulas for right circular cylinders

Question 1628Question →

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

1

-\frac{13}{9}

-1

\frac{5}{3}

3

Show answer & explanation

Answer: $1$

Answer

1

Expanding the absolute value inequality

|4 - 3x| \le 11

gives

-11 \le 4 - 3x \le 11

. Subtracting

4

from all sides yields

-15 \le -3x \le 7

. Dividing by

-3

and reversing the inequality signs gives

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

. Simplifying the target expression gives

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

. Because

\frac{x - 2}{3}

increases with

x

, substituting the maximum allowed value

x = 5

gives

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

.

Step-by-Step Solution

1

Unfold the absolute value inequality into a compound inequality

-11 \le 4 - 3x \le 11

For any real number

k \ge 0

,

|A| \le k

is equivalent to

-k \le A \le k

.

2

Isolate the variable term by subtracting 4 from all parts

-15 \le -3x \le 7

Subtracting a constant maintains the direction of the inequality.

3

Divide by -3 and flip the inequality signs

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

, or equivalently

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

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

4

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

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

. Maximum occurs at

x = 5

:

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

Since

\frac{x - 2}{3}

increases as

x

increases, its maximum occurs at the largest possible value of

x

.

Key Concept

Linear Inequalities and Absolute Value Bounds

Question 1629Question →

A specialty coffee roastery surveyed $200$ customers to determine their preferences among three origin beans: Ethiopian ( $E$ ), Colombian ( $C$ ), and Guatemalan ( $G$ ). Every customer surveyed preferred at least one of the three origins. The survey revealed that $105$ customers preferred Ethiopian, $90$ preferred Colombian, and $85$ preferred Guatemalan. Additionally, $35$ customers preferred both Ethiopian and Colombian, $30$ preferred both Colombian and Guatemalan, and $40$ preferred both Ethiopian and Guatemalan. How many customers preferred all three origin beans?

15

20

25

35

50

Show answer & explanation

Answer: $25$

Answer

25 customers preferred all three origin beans.

The correct answer is obtained by setting up the three-set inclusion-exclusion formula

|A \cup B \cup C| = |A| + |B| + |C| - (|A \cap B| + |B \cap C| + |A \cap C|) + |A \cap B \cap C|

. Substituting the given values gives

200 = 105 + 90 + 85 - (35 + 30 + 40) + x

. Solving

200 = 175 + x

yields

x = 25

.

Step-by-Step Solution

1

State the 3-set Inclusion-Exclusion Principle formula.

|E \cup C \cup G| = |E| + |C| + |G| - (|E \cap C| + |C \cap G| + |E \cap G|) + |E \cap C \cap G|

This fundamental formula relates the total union of three overlapping sets to their individual sizes, pairwise intersections, and three-set intersection.

2

Substitute the given values into the formula.

200 = 105 + 90 + 85 - (35 + 30 + 40) + x

, where

x = |E \cap C \cap G|

Since every customer preferred at least one origin, the union of all three sets equals the total number of customers surveyed (

200

).

3

Simplify the sum of individual sets and pairwise intersections.

Sum of individual sets =

105 + 90 + 85 = 280

. Sum of pairwise intersections =

35 + 30 + 40 = 105

.

Grouping terms makes evaluating the equation easier and reduces calculation errors.

4

Solve for the unknown three-set intersection

x

.

200 = 280 - 105 + x \implies 200 = 175 + x \implies x = 25

Subtracting

175

from

200

yields the number of customers who prefer all three origin beans.

Key Concept

Three-Set Principle of Inclusion-Exclusion

Question 1630Question →

Consider the following system of linear equations in variables

x

,

y

, and

z

, where

a

,

b

, and

c

are real constants:

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

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

Select all that apply

If

c = 2a + 3b

, the system has infinitely many solutions.

There exist no real values of

a

,

b

, and

c

for which the system has a unique solution.

If

a = 0

,

b = 0

, and

c = 0

, the system has

(0, 0, 0)

as its only solution.

If

c \neq 2a + 3b

, the system has infinitely many solutions.

If

a = 1

,

b = 2

, and

c = 7

, the system has at least one solution.

Show answer & explanation

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

Answer

The correct statements are: 'If

c = 2a + 3b

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

a

,

b

, and

c

for which the system has a unique solution.'

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

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

. Comparing this with the third equation,

7x + 4y + 3z = c

, shows that when

c = 2a + 3b

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

a

,

b

, and

c

.

Step-by-Step Solution

1

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

Observe that

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

.

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

2

Determine the condition for consistency.

The system is consistent if and only if

2a + 3b = c

.

If

c = 2a + 3b

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

3

Evaluate the possibility of a unique solution.

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

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

a, b, c

can produce a unique solution.

4

Verify specific numerical options.

For

a=0, b=0, c=0

,

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

, giving infinitely many solutions. For

a=1, b=2, c=7

,

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

, giving zero solutions.

Testing specific constant values confirms consistency or inconsistency based on whether

c = 2a + 3b

is satisfied.

Key Concept

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

Question 1631Question →

In the $xy$ -plane, line $L_1$ has a positive slope $m$ and a $y$ -intercept of $(0, 4)$ . The perpendicular distance from the origin $(0, 0)$ to line $L_1$ is $2\sqrt{2}$ . Line $L_2$ is perpendicular to line $L_1$ and passes through the points $(2, 1)$ and $(6, k)$ . What is the value of $k$ ?

-7

-5

-3

3

5

Show answer & explanation

Answer: $-3$

Answer

-3

The equation of line

L_1

is

mx - y + 4 = 0

. Using the distance formula from the origin to a line,

\frac{4}{\sqrt{m^2 + 1}} = 2\sqrt{2}

, which simplifies to

m^2 + 1 = 2

, yielding

m = 1

. The slope of line

L_2

is the negative reciprocal of

1

, which is

-1

. Using the slope equation

\frac{k - 1}{6 - 2} = -1

, we find

k - 1 = -4

, giving

k = -3

.

Step-by-Step Solution

1

Express line

L_1

in standard form and set up the distance formula from the origin.

Line

L_1

has slope

m

and

y

-intercept

(0,4)

, so its equation is

y = mx + 4

, or

mx - y + 4 = 0

. The perpendicular distance

d

from

(0,0)

to

Ax + By + C = 0

is

d = \frac{|C|}{\sqrt{A^2 + B^2}} = \frac{4}{\sqrt{m^2 + 1}}

.

The shortest distance from a point to a line is given by the perpendicular distance formula.

2

Solve for the positive slope

m

of line

L_1

.

Setting

\frac{4}{\sqrt{m^2 + 1}} = 2\sqrt{2}

yields

\sqrt{m^2 + 1} = \sqrt{2}

. Squaring both sides gives

m^2 + 1 = 2

, so

m^2 = 1

. Since

m > 0

,

m = 1

.

The problem specifies that line

L_1

has a positive slope.

3

Determine the slope of line

L_2

.

Since line

L_2

is perpendicular to line

L_1

, its slope

m_2

is the negative reciprocal of

m

:

m_2 = -\frac{1}{1} = -1

.

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

4

Use the slope formula with points

(2, 1)

and

(6, k)

to solve for

k

.

m_2 = \frac{k - 1}{6 - 2} = \frac{k - 1}{4} = -1 \implies k - 1 = -4 \implies k = -3

.

Equating the slope calculated from two points to the known perpendicular slope solves for the missing coordinate.

Key Concept

Perpendicular Distance and Perpendicular Lines in Coordinate Geometry

Question 1632Question →

An aviation safety committee audited $240$ international flights for operational delays caused by Weather ( $W$ ), Mechanical issues ( $M$ ), and Air Traffic Control ( $T$ ). The audit revealed the following data:

- $112$ flights experienced Weather delays.
- $98$ flights experienced Mechanical delays.
- $104$ flights experienced Air Traffic Control delays.
- $28$ flights experienced both Weather and Mechanical delays.
- $34$ flights experienced both Mechanical and Air Traffic Control delays.
- $42$ flights experienced both Weather and Air Traffic Control delays.
- The number of flights that experienced none of these three types of delays was equal to twice the number of flights that experienced all three types of delays.

How many flights experienced exactly one of these three types of delays?

Show answer & explanation

Answer: 136

Answer

The number of flights that experienced exactly one of the three types of delays is 136.

The total number of audited flights is partitioned into those experiencing at least one delay and those experiencing none. By applying the Inclusion-Exclusion formula

|W \cup M \cup T| = 112 + 98 + 104 - (28 + 34 + 42) + x = 210 + x

, and setting the total audited flights to

240 = (210 + x) + 2x

, we solve for the triple intersection

x = 10

. Isolating the exclusive single-delay regions yields

52

flights for Weather only,

46

for Mechanical only, and

38

for ATC only, giving a total of

136

flights.

Step-by-Step Solution

1

Define the unknown intersection and external set quantities in terms of a single variable.

Let

x = |W \cap M \cap T|

represent the number of flights experiencing all three types of delays. Consequently, the number of flights experiencing none of the delays is

2x

.

Relating the unknown triple intersection to the 'neither' category allows all regions to be expressed in terms of

x

.

2

Apply the Principle of Inclusion-Exclusion for three overlapping sets to find the total number of flights with at least one delay.

|W \cup M \cup T| = |W| + |M| + |T| - (|W \cap M| + |M \cap T| + |W \cap T|) + |W \cap M \cap T| = 112 + 98 + 104 - (28 + 34 + 42) + x = 210 + x

.

Subtracting pairwise intersections corrects for double-counting, and adding back the triple intersection corrects for over-subtraction.

3

Formulate and solve the linear equation for the total sample population of 240 flights.

\text{Total Flights} = |W \cup M \cup T| + |\text{Neither}| \implies 240 = (210 + x) + 2x \implies 240 = 210 + 3x \implies 3x = 30 \implies x = 10

.

The sum of flights in the union of the sets and flights outside all three sets must equal the total number of flights audited.

4

Determine the number of flights in the two-set intersection regions excluding the triple intersection.

Weather and Mechanical only:

28 - 10 = 18

; Mechanical and ATC only:

34 - 10 = 24

; Weather and ATC only:

42 - 10 = 32

.

Each pairwise intersection given in the stem includes the triple intersection of 10 flights.

5

Compute the number of flights belonging exclusively to single categories and sum them.

Weather only:

112 - (18 + 32 + 10) = 52

; Mechanical only:

98 - (18 + 24 + 10) = 46

; ATC only:

104 - (32 + 24 + 10) = 38

. Sum of single categories:

52 + 46 + 38 = 136

.

Subtracting all overlapping components from each set total isolates the exclusive single-category elements.

Key Concept

Three-Set Inclusion-Exclusion Principle and Venn Diagram Region Partitioning

Estimated Time:2m 30s

Question 1633Question →

In the $xy$ -plane, the parabola defined by $y = 2x^2 - 8x + 5$ is reflected across the line $y = x$ . The resulting curve is then translated $3$ units to the right and $4$ units downward to produce curve $C$ . If the point $(a, -1)$ lies on curve $C$ , what is the value of $a$ ?

Show answer & explanation

Answer: 2

Answer

The value of

a

is

2

.

Completing the square yields

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

. Reflecting across

y = x

interchanges

x

and

y

, producing

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

. Shifting this curve

3

units to the right replaces

x

with

x - 3

, and shifting

4

units down replaces

y

with

y + 4

. The equation simplifies to

x = 2(y + 2)^2

. Setting

y = -1

gives

a = 2(-1 + 2)^2 = 2

.

Step-by-Step Solution

1

Rewrite the original parabola equation in vertex form.

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

Expressing the parabola in vertex form makes tracking coordinate transformations straightforward.

2

Apply the reflection across the line

y = x

.

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

Reflecting across

y = x

interchanges the

x

and

y

coordinates of every point on the graph.

3

Apply the translations to find the equation of curve

C

.

x = 2(y + 2)^2

Translating right by

3

replaces

x

with

x - 3

, and translating down by

4

replaces

y

with

y + 4

.

4

Substitute

y = -1

into the equation for curve

C

to calculate

a

.

a = 2

Since the point

(a, -1)

lies on curve

C

, substituting

x = a

and

y = -1

yields

a = 2(1)^2 = 2

.

Key Concept

Reflection across

y = x

and rigid horizontal/vertical translations of conic sections

Question 1634Question →

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

Show answer & explanation

Answer: 21

Answer

The sum of all possible integer values of

x

is 21.

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

|7 - 2x| \le 9

, express it as

-9 \le 7 - 2x \le 9

. Subtracting

7

yields

-16 \le -2x \le 2

. Dividing all parts by

-2

requires reversing the inequality symbols, giving

8 \ge x \ge -1

, or

-1 \le x \le 8

. For

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

, multiplying by

-2

reverses the inequality symbol to

1 - x < -4

. Subtracting

1

gives

-x < -5

, and dividing by

-1

reverses the symbol again to

x > 5

. The integer values satisfying both

-1 \le x \le 8

and

x > 5

are

6, 7,

and

8

. Summing these integer values yields

6 + 7 + 8 = 21

.

Step-by-Step Solution

1

Solve the absolute value inequality

|7 - 2x| \le 9

-1 \le x \le 8

Unfolding the absolute value gives

-9 \le 7 - 2x \le 9

. Subtracting 7 yields

-16 \le -2x \le 2

, and dividing by

-2

reverses the inequalities to give

-1 \le x \le 8

.

2

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

x > 5

Multiplying by

-2

flips the inequality sign to yield

1 - x < -4

. Subtracting 1 gives

-x < -5

, and dividing by

-1

flips the inequality sign again to yield

x > 5

.

3

Find the overlapping integer values for x

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

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

4

Sum the integer solutions

21

Calculating 6 + 7 + 8 = 21.

Key Concept

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

Question 1635Question →

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

\frac{9}{4}

\frac{11}{4}

\frac{7}{2}

\frac{31}{8}

\frac{31}{4}

Show answer & explanation

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

Answer

The value of the constant

k

is

\frac{11}{4}

.

Using Vieta's formulas for the quadratic equation

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

, we have

\alpha + \beta = k - 2

and

\alpha\beta = k - 5

. The given condition

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

simplifies to

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

. Substituting the expressions in terms of

k

yields

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

, which expands to

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

. Subtracting

k^2

from both sides gives

4k = 11

, so

k = \frac{11}{4}

.

Step-by-Step Solution

1

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

k

\alpha + \beta = k - 2

and

\alpha\beta = k - 5

For any standard quadratic

ax^2 + bx + c = 0

, the sum of roots is

-\frac{b}{a}

and the product of roots is

\frac{c}{a}

.

2

Rewrite the given sum of reciprocal squares using algebraic identities

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

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

\alpha+\beta

and product

\alpha\beta

.

3

Substitute the expressions for sum and product into the equation

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

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

4

Expand both numerator and denominator and solve for

k

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

Equating numerator and denominator eliminates the quadratic

k^2

terms, yielding a linear equation in

k

.

5

Verify that

k = \frac{11}{4}

yields real, distinct, non-zero roots

Discriminant

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

, and

\alpha\beta = -2.25 \neq 0

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

Key Concept

Vieta's Formulas and Symmetric Polynomial Transformations of Quadratic Roots

Question 1636Question →

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

Show answer & explanation

Answer: 8.2

Answer

The value of

a

is 8.2 (or

\frac{41}{5}

).

Converting

2x - 5y = 10

to slope-intercept form yields

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

, so the slope of line

\ell_2

is

\frac{2}{5}

. Line

\ell_1

is perpendicular to

\ell_2

, giving it a slope of

m_1 = -\frac{5}{2}

. Using the point

(3, 7)

, the line equation for

\ell_1

is

y - 7 = -2.5(x - 3)

, which simplifies to

y = -2.5x + 14.5

. Line

\ell_3

is parallel to

\ell_1

, so

m_3 = -2.5

, and its

y

-intercept is

14.5 + 6 = 20.5

. Writing the equation for line

\ell_3

as

y = -2.5x + 20.5

and setting

y = 0

gives

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

.

Step-by-Step Solution

1

Determine the slope of line

\ell_2

The slope of line

\ell_2

is

\frac{2}{5}

.

Convert

2x - 5y = 10

into slope-intercept form

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

.

2

Determine the slope of line

\ell_1

The slope of line

\ell_1

is

-\frac{5}{2}

.

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

3

Find the

y

-intercept of line

\ell_1

The

y

-intercept of

\ell_1

is

\frac{29}{2} = 14.5

.

Apply point-slope form with point

(3, 7)

:

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

.

4

Construct the equation for line

\ell_3

The equation of

\ell_3

is

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

.

Line

\ell_3

has the same slope as

\ell_1

(

-\frac{5}{2}

) and its

y

-intercept is

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

.

5

Calculate the

x

-intercept coordinate

a

of line

\ell_3

a = 8.2

.

Substitute

y = 0

into the equation for

\ell_3

:

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

.

Key Concept

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

Estimated Time:2m 30s

Question 1637Question →

Data set $A$ consists of 7 numbers that form an arithmetic progression with a non-zero common difference. Data set $B$ is formed by removing the smallest number and the largest number from Data set $A$ . If $\sigma_A^2$ represents the population variance of Data set $A$ and $\sigma_B^2$ represents the population variance of Data set $B$ , what is the value of $\frac{\sigma_A^2}{\sigma_B^2}$ ?

Show answer & explanation

Answer: 2

Answer

2

By writing the 7 terms of Data set A symmetrically about their mean

\mu

with common difference

d

, the sum of squared deviations is

28d^2

, giving a population variance of

\sigma_A^2 = \frac{28d^2}{7} = 4d^2

. Removing the minimum and maximum values leaves 5 terms centered at

\mu

with a sum of squared deviations of

10d^2

, resulting in a population variance of

\sigma_B^2 = \frac{10d^2}{5} = 2d^2

. The ratio of the variance of Data set A to the variance of Data set B is therefore

\frac{4d^2}{2d^2} = 2

.

Step-by-Step Solution

1

Represent the elements of Data set A using its mean

\mu

and common difference

d

.

Let the 7 elements of Data set A be

\mu - 3d, \mu - 2d, \mu - d, \mu, \mu + d, \mu + 2d, \mu + 3d

, where

d > 0

.

Symmetrical representation around the mean simplifies variance computation for arithmetic sequences.

2

Calculate the population variance

\sigma_A^2

of Data set A.

\sigma_A^2 = \frac{1}{7} \sum_{k=-3}^{3} (kd)^2 = \frac{d^2}{7}(9 + 4 + 1 + 0 + 1 + 4 + 9) = \frac{28d^2}{7} = 4d^2.

Variance is defined as the mean of the squared deviations from the dataset mean.

3

Determine the elements and population variance

\sigma_B^2

of Data set B.

Data set B contains the 5 remaining central elements, having mean

\mu

. Thus, $\sigma_B^2 = \frac{1}{5} \sum_{k=-2}^{2} (kd)^2 = \frac{d^2}{5}(4 + 1 + 0 + 1 + 4) = \frac{10d^2}{5} = 2d^2.

Removing the minimum and maximum terms leaves a symmetric 5-element dataset with unchanged mean

\mu

.

4

Compute the ratio

\frac{\sigma_A^2}{\sigma_B^2}

.

\frac{\sigma_A^2}{\sigma_B^2} = \frac{4d^2}{2d^2} = 2.

The parameter

d^2

cancels out completely, yielding a constant numerical ratio.

Key Concept

Effect of trimming extreme symmetric values on the population variance of an arithmetic sequence

Question 1638Question →

In the $xy$ -plane, point $M(3, 3)$ is the midpoint of line segment $AB$ , where point $A$ has coordinates $(1, 2)$ . Line $L$ is perpendicular to line segment $AB$ and passes through point $B$ . What is the $x$ -intercept of line $L$ ?

Show answer & explanation

Answer: 7

Answer

7

Using the midpoint formula with

A(1, 2)

and

M(3, 3)

gives endpoint

B(5, 4)

. The slope of segment

AB

is

\frac{4-2}{5-1} = \frac{1}{2}

, so line

L

, being perpendicular to

AB

, has slope

-2

. The equation of line

L

passing through

(5, 4)

is

y - 4 = -2(x - 5)

, which simplifies to

y = -2x + 14

. Setting

y = 0

gives

0 = -2x + 14

, so the

x

-intercept is

7

.

Step-by-Step Solution

1

Calculate the coordinates of endpoint B using the midpoint formula.

Point B has coordinates

(5, 4)

.

Midpoint

M(x_m, y_m) = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right)

. Solving

\frac{1 + x_B}{2} = 3

yields

x_B = 5

, and solving

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

yields

y_B = 4

.

2

Determine the slope of segment AB.

The slope

m_{AB} = \frac{1}{2}

.

Using the slope formula

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

between

(1, 2)

and

(5, 4)

gives

\frac{4 - 2}{5 - 1} = \frac{2}{4} = \frac{1}{2}

.

3

Determine the slope of line L.

The slope of line L is

-2

.

Perpendicular lines have negative reciprocal slopes. The negative reciprocal of

\frac{1}{2}

is

-2

.

4

Find the equation of line L and solve for its x-intercept.

The x-intercept is 7.

Using point-slope form with

B(5, 4)

and

m = -2

:

y - 4 = -2(x - 5) \implies y = -2x + 14

. Setting

y = 0

gives

0 = -2x + 14 \implies x = 7

.

Key Concept

Midpoint Formula, Perpendicular Slopes, and Line Intercepts

Question 1639Question →

A dataset $W$ consists of 25 distinct positive numbers with mean $M$ , standard deviation $s > 0$ , and interquartile range $IQR > 0$ . Dataset $V$ is constructed by adding $2M$ to every number in $W$ that is strictly greater than the median of $W$ , and subtracting $2M$ from every number in $W$ that is strictly less than the median of $W$ . The value equal to the median of $W$ itself remains unchanged. Which of the following statements MUST be true regarding dataset $V$ compared to dataset $W$ ?

Both the standard deviation and the interquartile range of dataset

V

are greater than those of dataset

W

.

The standard deviation of dataset

V

is greater than

s

, but the interquartile range of dataset

V

is equal to

IQR

.

The standard deviation of dataset

V

is equal to

s

, but the interquartile range of dataset

V

is greater than

IQR

.

Both the standard deviation and the interquartile range of dataset

V

are equal to those of dataset

W

.

The standard deviation of dataset

V

is less than

s

, while the interquartile range of dataset

V

is greater than

IQR

.

Show answer & explanation

Answer: Both the standard deviation and the interquartile range of dataset $V$ are greater than those of dataset $W$ .

Answer

Both the standard deviation and the interquartile range of dataset

V

are greater than those of dataset

W

.

The correct response identifies that both standard deviation and interquartile range increase. In dataset

W

, the 12 elements strictly below the median are reduced by

2M

, and the 12 elements strictly above the median are increased by

2M

. This leaves the mean

M

unchanged but increases the distance of every non-median element from

M

, thereby strictly increasing the standard deviation. Furthermore, the first quartile

Q_1

shifts down by

2M

and the third quartile

Q_3

shifts up by

2M

, expanding the interquartile range from

IQR

to

IQR + 4M

.

Step-by-Step Solution

1

Analyze the impact of the transformation on the first and third quartiles (

Q_1

and

Q_3

).

In an ordered dataset of 25 distinct values, the median is the 13th element. The first quartile

Q_1

is in the lower half (below the median) and the third quartile

Q_3

is in the upper half (above the median).

Understanding where

Q_1

and

Q_3

fall relative to the median determines how their values change.

2

Calculate the new Interquartile Range (

IQR_V

).

Since

Q_1

is in the lower half, its new value is

Q_1 - 2M

. Since

Q_3

is in the upper half, its new value is

Q_3 + 2M

. Thus,

IQR_V = (Q_3 + 2M) - (Q_1 - 2M) = (Q_3 - Q_1) + 4M = IQR + 4M > IQR

because

M > 0

.

Evaluating the difference between the transformed 75th and 25th percentiles.

3

Analyze the impact on the mean and standard deviation.

Equal numbers of values (12 values) are shifted down by

2M

and shifted up by

2M

, so the mean of dataset

V

remains equal to

M

. For every shifted point

x_i

, its squared distance from the mean

(x_i - M)^2

strictly increases. Therefore, the variance and standard deviation strictly increase (

s_V > s

).

Standard deviation measures dispersion relative to the mean; pushing points further from the mean increases standard deviation.

Key Concept

Effect of non-uniform linear shifts on measures of dispersion (Standard Deviation and IQR)

Estimated Time:2m 0s

Question 1640Question →

An urban forestry department measured the trunk diameters, in inches, of a sample of $60$ oak trees in a municipal park. The table below summarizes the frequency distribution of the measurements.

Trunk Diameter (inches)	Frequency
$10$ to $14$	$8$
$15$ to $19$	$15$
$20$ to $24$	$22$
$25$ to $29$	$11$
$30$ to $34$	$4$

Based on the table, what fraction of the trees with a trunk diameter of at least $15$ inches have a trunk diameter of less than $25$ inches?

\frac{37}{60}

\frac{37}{52}

\frac{45}{52}

\frac{11}{26}

\frac{37}{15}

Show answer & explanation

Answer: $\frac{37}{52}$

Answer

The fraction of trees with a trunk diameter of at least 15 inches that have a trunk diameter of less than 25 inches is 37/52.

To calculate the required fraction, first identify the subset of trees that have a trunk diameter of at least 15 inches. Summing the frequencies of the intervals 15–19, 20–24, 25–29, and 30–34 gives 15 + 22 + 11 + 4 = 52 trees. Next, identify how many of these 52 trees have a trunk diameter of less than 25 inches. These are the trees in the intervals 15–19 and 20–24, giving 15 + 22 = 37 trees. Therefore, the fraction of trees meeting the condition is 37/52.

Step-by-Step Solution

1

Determine the total number of trees satisfying the condition 'at least 15 inches'.

Sum the frequencies for intervals 15–19, 20–24, 25–29, and 30–34:

15 + 22 + 11 + 4 = 52

trees.

This establishes the restricted denominator (base group) specified by the conditional statement.

2

Determine the number of trees within this group that satisfy the condition 'less than 25 inches'.

Sum the frequencies for intervals 15–19 and 20–24:

15 + 22 = 37

trees.

These are the target intervals that are both at least 15 inches and strictly less than 25 inches.

3

Calculate the required fraction.

\frac{37}{52}

.

Divide the target count by the conditional total count to form the part-to-whole ratio.

Key Concept

Conditional relative frequencies and part-to-whole calculations from grouped data tables.

Estimated Time:1m 30s

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