All practice questions

Question 1701Question →

In the $xy$ -plane, line $m$ has an $x$ -intercept of $-6$ and a $y$ -intercept of $3$ . Line $k$ is perpendicular to line $m$ and passes through the point $(4, 9)$ . What is the $x$ -intercept of line $k$ ?

Show answer & explanation

Answer: 8.5

Answer

8.5

First, determine the slope of line

m

using the given intercepts

(-6,0)

and

(0,3)

, yielding

\frac{3-0}{0-(-6)} = \frac{1}{2}

. Since line

k

is perpendicular to line

m

, its slope is the negative reciprocal of

\frac{1}{2}

, which is

-2

. Using the point-slope equation with point

(4,9)

, the equation of line

k

is

y - 9 = -2(x - 4)

, simplifying to

y = -2x + 17

. Finding the

x

-intercept by setting

y = 0

yields

0 = -2x + 17

, giving

x = 8.5

.

Step-by-Step Solution

1

Calculate the slope of line

m

Slope of line

m

is

\frac{1}{2}

Line

m

passes through the points

(-6,0)

and

(0,3)

.

2

Determine the slope of perpendicular line

k

Slope of line

k

is

-2

Perpendicular lines have negative reciprocal slopes.

3

Derive the equation of line

k

y = -2x + 17

Use point-slope form with given point

(4,9)

and slope

-2

.

4

Solve for the

x

-intercept of line

k

x = 8.5

Set

y = 0

in the linear equation

y = -2x + 17

.

Key Concept

Perpendicular lines, slope calculation from intercepts, and line equations

Question 1702Question →

A commercial bakery uses two automated ovens, Oven X and Oven Y, to bake identical orders of bread. Working alone at its constant rate, Oven X bakes a full order of bread in $8$ hours, while Oven Y, working alone at its constant rate, bakes the same order in $12$ hours. Both ovens begin baking a full order together at 8:00 a.m. At 10:00 a.m., Oven X shuts down due to a maintenance alert, and Oven Y continues working alone at its constant rate until the order is completed. How many total hours, from 8:00 a.m. until completion, does it take to finish the order?

Show answer & explanation

Answer: 9

Answer

The total time required from 8:00 a.m. to complete the order is

9

hours.

Working together for

2

hours at a combined rate of

\frac{1}{8} + \frac{1}{12} = \frac{5}{24}

per hour completes

\frac{5}{12}

of the order. The remaining

\frac{7}{12}

of the order takes Oven Y

7

hours to complete at its rate of

\frac{1}{12}

per hour. Adding the initial

2

hours gives a total time of

9

hours.

Step-by-Step Solution

1

Determine the individual hourly work rates.

Oven X completes

\frac{1}{8}

of the job per hour, and Oven Y completes

\frac{1}{12}

of the job per hour.

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

2

Calculate the fraction of the job completed in the first

2

hours.

Combined rate is

\frac{1}{8} + \frac{1}{12} = \frac{5}{24}

job per hour. In

2

hours, they complete

2 \times \frac{5}{24} = \frac{5}{12}

of the job.

Both ovens work simultaneously for

2

hours before Oven X stops.

3

Determine the remaining fraction of the job.

1 - \frac{5}{12} = \frac{7}{12}

of the job remains.

The full order represents

1

whole unit of work.

4

Find the additional time required for Oven Y to finish the remaining job alone.

\text{Time} = \frac{7/12}{1/12} = 7

hours.

Time equals remaining work divided by Oven Y's individual work rate.

5

Calculate the total time elapsed from start to completion.

2 + 7 = 9

hours.

The total time includes the

2

hours of combined work plus the

7

hours Oven Y worked alone.

Key Concept

Combined Work Rates and Modeling Staggered Work

Question 1703Question →

A cybersecurity firm audited a sample of $300$ corporate networks for compliance across three security standards: Network Encryption ( $E$ ), Multi-Factor Authentication ( $M$ ), and Access Logging ( $L$ ). The audit revealed that $160$ networks met standard $E$ , $140$ met standard $M$ , and $120$ met standard $L$ . Exactly $25$ networks met all three standards, while $45$ networks met none of the three standards. How many of the audited networks met exactly two of the three security standards?

90

115

140

165

190

Show answer & explanation

Answer: 115

Answer

115 networks met exactly two of the three security standards.

The total number of networks meeting at least one standard is

300 - 45 = 255

. By the three-set inclusion-exclusion principle,

255 = 160 + 140 + 120 - S_2 + 25

, where

S_2

is the sum of the pairwise intersections

|E \cap M| + |M \cap L| + |E \cap L|

. Solving for

S_2

yields

S_2 = 190

. Since each pairwise intersection includes the 25 networks that met all three standards, the number of networks meeting exactly two standards is

190 - 3(25) = 115

.

Step-by-Step Solution

1

Calculate the total number of networks that met at least one security standard.

|E \cup M \cup L| = 300 - 45 = 255

Subtracting networks that met none of the standards from the total sample size gives the union of the three sets.

2

Apply the Principle of Inclusion-Exclusion for three sets to find the sum of pairwise overlapping regions.

255 = 160 + 140 + 120 - (|E \cap M| + |M \cap L| + |E \cap L|) + 25

, so

|E \cap M| + |M \cap L| + |E \cap L| = 445 - 255 = 190

The formula

|E \cup M \cup L| = |E| + |M| + |L| - (|E \cap M| + |M \cap L| + |E \cap L|) + |E \cap M \cap L|

relates all known quantities.

3

Calculate the number of networks that met exactly two standards.

\text{Exactly 2} = (|E \cap M| + |M \cap L| + |E \cap L|) - 3|E \cap M \cap L| = 190 - 3(25) = 190 - 75 = 115

Each of the three pairwise intersections includes the triple intersection. Subtracting three times the triple intersection isolates the regions representing membership in exactly two sets.

Key Concept

Three-Set Principle of Inclusion-Exclusion

Question 1704Question →

A technology company manages data storage across three servers: Server X, Server Y, and Server Z. Initially, the amount of data stored on Server Y is $25\%$ greater than the amount stored on Server X, and Server Z stores $40$ gigabytes less data than Server Y. During a system reorganization, the data on Server X increases by $20\%$ , the data on Server Y decreases by $20\%$ , and the data on Server Z increases by $60$ gigabytes. If the total amount of data stored across all three servers after the reorganization is $10\%$ greater than the total initial amount of data, what was the initial amount of data, in gigabytes, stored on Server X?

120

150

160

192

200

Show answer & explanation

Answer: 160

Answer

160 gigabytes

Let

x

be the initial storage on Server X in gigabytes. Then Server Y initially stores

1.25x

gigabytes and Server Z stores

1.25x - 40

gigabytes. The initial total storage is

T_1 = x + 1.25x + (1.25x - 40) = 3.5x - 40

. After reorganization, Server X stores

1.20x

, Server Y stores

0.80(1.25x) = 1.00x

, and Server Z stores

(1.25x - 40) + 60 = 1.25x + 20

. The new total storage is

T_2 = 1.20x + 1.00x + 1.25x + 20 = 3.45x + 20

. Since

T_2

is

10\%

greater than

T_1

, we have

3.45x + 20 = 1.10(3.5x - 40)

. Expanding the right side gives

3.45x + 20 = 3.85x - 44

. Rearranging terms yields

0.40x = 64

, which simplifies to

x = 160

gigabytes.

Step-by-Step Solution

1

Define variables for the initial storage on each server in terms of the initial storage on Server X,

x

.

Server X =

x

, Server Y =

1.25x

, Server Z =

1.25x - 40

.

Server Y is 25% greater than Server X (

1 + 0.25 = 1.25

), and Server Z is 40 gigabytes less than Server Y.

2

Sum the initial storage values to find the total initial storage

T_1

.

T_1 = x + 1.25x + (1.25x - 40) = 3.5x - 40

.

Combining like terms gives the overall starting storage equation.

3

Express the storage on each server after the reorganization in terms of

x

.

Server X' =

1.20x

, Server Y' =

0.80(1.25x) = 1.00x

, Server Z' =

(1.25x - 40) + 60 = 1.25x + 20

.

Server X increases by 20%, Server Y decreases by 20% of its initial value, and Server Z gains 60 gigabytes.

4

Sum the updated storage values to find the new total storage

T_2

.

T_2 = 1.20x + 1.00x + (1.25x + 20) = 3.45x + 20

.

Adding the three updated amounts gives the new total expression.

5

Set up and solve the equation representing the

10\%

overall increase (

T_2 = 1.10 T_1

).

3.45x + 20 = 1.10(3.5x - 40) \implies 3.45x + 20 = 3.85x - 44 \implies 64 = 0.40x \implies x = 160

.

Solving the linear algebraic model yields the value for initial storage on Server X.

Key Concept

Linear algebraic modeling of dynamic multi-variable systems involving percentage changes.

Estimated Time:2m 30s

Question 1705Question →

For all real numbers

x

, the function

f

is defined by

f(x) = x^2 - k

, where

k

is a constant. The custom operation

\star

is defined for all real numbers

a

and

b

by

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

If

5 \star (-3) = -12

, what is the value of

k

?

18

12

-12

0

-42

Show answer & explanation

Answer: 18

Answer

The value of

k

is 18.

Evaluating

f(2) = 4 - k

,

f(5) = 25 - k

, and

f(-3) = 9 - k

yields

5 \star (-3) = (4 - k) - (34 - 2k) = -30 + k

. Setting

-30 + k = -12

correctly gives

k = 18

.

Step-by-Step Solution

1

Evaluate

f(a+b)

for

a=5

and

b=-3

a + b = 5 + (-3) = 2

, so

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

.

Substitute the input

a+b = 2

into the function definition

f(x) = x^2 - k

.

2

Evaluate

f(a)

and

f(b)

individually

f(5) = 5^2 - k = 25 - k

and

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

.

Apply the function rule to inputs 5 and -3.

3

Substitute the evaluated expressions into the custom operation definition

5 \star (-3) = (4 - k) - [(25 - k) + (9 - k)] = (4 - k) - (34 - 2k) = -30 + k

.

Simplify the algebraic expression by combining like terms and distributing the negative sign.

4

Solve for

k

using the given equation

5 \star (-3) = -12

-30 + k = -12 \implies k = 18

.

Add 30 to both sides of the equation to isolate

k

.

Key Concept

Evaluating custom binary operations by substituting function definitions and simplifying algebraic expressions.

Estimated Time:1m 30s

Question 1706Question →

The quadratic equation $x^2 - 2mx + (m^2 - 4m + 12) = 0$ , where $m$ is a real constant, has two distinct real roots $x_1$ and $x_2$ . If the distance between the two roots on the real number line is $4\sqrt{3}$ , what is the value of $x_1^2 + x_2^2$ ?

48

96

120

144

192

Show answer & explanation

Answer: 96

Answer

96

By Vieta's formulas, the sum of the roots is

x_1 + x_2 = 2m

and the product of the roots is

x_1 x_2 = m^2 - 4m + 12

. The distance between the roots is given by

|x_1 - x_2| = \sqrt{(x_1 + x_2)^2 - 4x_1 x_2}

. Squaring both sides gives

(4\sqrt{3})^2 = 48 = (2m)^2 - 4(m^2 - 4m + 12) = 16m - 48

. Solving

16m - 48 = 48

yields

m = 6

. Substituting

m = 6

into the expressions for the sum and product of the roots gives

x_1 + x_2 = 12

and

x_1 x_2 = 24

. Finally, applying the identity

x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2

yields

12^2 - 2(24) = 144 - 48 = 96

.

Step-by-Step Solution

1

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

m

.

For

x^2 - 2mx + (m^2 - 4m + 12) = 0

, the sum of roots is

x_1 + x_2 = 2m

and the product of roots is

x_1 x_2 = m^2 - 4m + 12

.

Vieta's relations relate quadratic coefficients directly to root sums and products.

2

Express the distance between the roots

|x_1 - x_2|

in terms of

m

and solve for

m

.

|x_1 - x_2|^2 = (x_1 + x_2)^2 - 4x_1 x_2 = (2m)^2 - 4(m^2 - 4m + 12) = 16m - 48

. Given

|x_1 - x_2| = 4\sqrt{3}

, we have

16m - 48 = (4\sqrt{3})^2 = 48 \implies 16m = 96 \implies m = 6

.

The difference between two roots of a quadratic is linked to the discriminant via

|x_1 - x_2| = \frac{\sqrt{b^2 - 4ac}}{|a|}

.

3

Calculate the numerical values of the root sum

x_1 + x_2

and root product

x_1 x_2

using

m = 6

.

x_1 + x_2 = 2(6) = 12

and

x_1 x_2 = 6^2 - 4(6) + 12 = 24

.

Substituting

m = 6

gives the exact values needed for algebraic evaluation.

4

Compute

x_1^2 + x_2^2

using the identity

x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2

.

x_1^2 + x_2^2 = 12^2 - 2(24) = 144 - 48 = 96

.

The sum of squares identity isolates

x_1^2 + x_2^2

without requiring explicit calculation of individual root values.

Key Concept

Quadratic Root Relationships and Vieta's Formulas

Estimated Time:2m 30s

Question 1707Question →

A retail analyst recorded the number of online orders fulfilled per day over a period of $9$ days. The data points, listed in non-decreasing order, are:

3, 5, 8, x, y, 17, 21, 24, z

The median of the $9$ daily order counts is $14$ , the arithmetic mean is $16$ , and the dataset has a unique mode of $8$ . If $z$ represents the highest number of orders fulfilled in a single day, what is the value of $z$ ?

28

38

44

52

54

Show answer & explanation

Answer: 44

Answer

44

For a sorted set of 9 values, the median is the 5th value, giving

y = 14

. To make 8 the unique mode, 8 must appear at least twice, which forces

x = 8

. With an arithmetic mean of 16 across 9 numbers, the total sum is

9 \times 16 = 144

. Subtracting the sum of the eight known numbers (

3+5+8+8+14+17+21+24 = 100

) gives

z = 44

.

Step-by-Step Solution

1

Determine the value of

y

using the median definition.

y = 14

For a dataset of

n = 9

numbers arranged in non-decreasing order, the median is the

\frac{9+1}{2} = 5\text{th}

element. Thus,

y = 14

.

2

Determine the value of

x

using the unique mode condition.

x = 8

The dataset is ordered as

3 \le 5 \le 8 \le x \le 14 \le 17 \le 21 \le 24 \le z

. For

8

to be a unique mode, it must appear more than once. Since all other given numbers are distinct,

x

must equal

8

so that

8

occurs twice.

3

Calculate the required total sum of all 9 data points from the given arithmetic mean.

\text{Total Sum} = 144

\text{Mean} = \frac{\text{Total Sum}}{9} \implies \text{Total Sum} = 9 \times 16 = 144.

4

Sum the known values and solve for

z

.

z = 44

3 + 5 + 8 + 8 + 14 + 17 + 21 + 24 + z = 144 \implies 100 + z = 144 \implies z = 44

.

Key Concept

Combining mean, median, and mode definitions to solve for unknown elements in a ordered dataset.

Question 1708Question →

In the $xy$ -coordinate plane, point $A$ has coordinates $(-9, 0)$ and point $C$ has coordinates $(0, 12)$ . Point $B$ lies on the positive $x$ -axis such that line segment $BD$ is perpendicular to segment $AC$ , with point $D$ lying on segment $AC$ . If the area of right triangle $ABD$ is $54$ , what is the length of segment $OB$ , where $O$ is the origin $(0,0)$ ?

Show answer & explanation

Answer: 6

Answer

The length of segment

OB

is

6

.

The length of segment

OB

is

6

. Using the Pythagorean theorem on

\triangle AOC

, hypotenuse

AC = 15

, establishing a

3:4:5

side ratio for

\triangle AOC

. Because

\triangle ABD

shares acute angle

\angle A

with

\triangle AOC

and has a right angle at

D

,

\triangle ABD

is also a

3:4:5

right triangle with hypotenuse

AB

. Expressing the area

\frac{1}{2} \times \left(\frac{3}{5}AB\right) \times \left(\frac{4}{5}AB\right) = 54

yields

AB = 15

. Since

A

is at

(-9,0)

, point

B

is at

(6,0)

, making

OB = 6

.

Step-by-Step Solution

1

Find the side lengths and hypotenuse of right triangle

AOC

.

Leg

AO = 9

, leg

OC = 12

, and by the Pythagorean theorem, hypotenuse

AC = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15

.

The coordinates of

A(-9,0)

and

C(0,12)

form a right triangle at the origin

O(0,0)

.

2

Use angle similarity to determine the side ratio of right triangle

ABD

.

Triangle

ABD

is similar to triangle

AOC

because both contain a right angle and share

\angle DAO

. Thus, the sides of

\triangle ABD

maintain the ratio

AD : BD : AB = 3 : 4 : 5

.

Right triangles with a shared acute angle are similar.

3

Express legs

AD

and

BD

in terms of hypotenuse

AB

and set up the area equation.

AD = \frac{3}{5}AB

and

BD = \frac{4}{5}AB

. The area of

\triangle ABD = \frac{1}{2} \times AD \times BD = \frac{1}{2} \times \frac{3}{5}AB \times \frac{4}{5}AB = \frac{6}{25}AB^2

. Setting

\frac{6}{25}AB^2 = 54

yields

AB^2 = 225

, so

AB = 15

.

The area of a right triangle is half the product of its perpendicular legs.

4

Calculate the length of segment

OB

.

Since point

A

is at

(-9,0)

and

B

lies on the positive

x

-axis,

AB = x_B - (-9) = 15 \implies x_B = 6

. Therefore, the length of

OB

is

6

.

The distance from the origin

(0,0)

to

(6,0)

on the

x

-axis is equal to the

x

-coordinate

6

.

Key Concept

Applying Pythagorean triples (3-4-5 right triangle family) and similar right triangles in coordinate geometry.

Question 1709Question →

A reliability study recorded the operating lifespan, $t$ (in thousands of hours), for a sample of $250$ semiconductor laser diodes. The results are summarized in the grouped frequency table below.

Lifespan $t$ (thousands of hours)	Frequency
$0 \le t < 4$	$35$
$4 \le t < 8$	$55$
$8 \le t < 12$	$80$
$12 \le t < 16$	$50$
$16 \le t < 20$	$30$

Laser diodes with an operating lifespan of at least $6{}000$ hours ( $t \ge 6$ ) but less than $14{}000$ hours ( $t < 14$ ) are designated as high-efficiency units. Assuming that the values within each class interval are uniformly distributed, what percentage of the $250$ laser diodes in the sample are designated as high-efficiency units?

Show answer & explanation

Answer: 53

Answer

53%

To find the percentage of diodes with lifespans between

6{}000

and

14{}000

hours (

6 \le t < 14

), evaluate the relevant intervals. For

4 \le t < 8

, the sub-interval

[6, 8)

represents

\frac{8-6}{8-4} = 0.5

of the interval width, containing

0.5 \times 55 = 27.5

diodes. The interval

[8, 12)

is fully contained, contributing

80

diodes. For

12 \le t < 16

, the sub-interval

[12, 14)

represents

\frac{14-12}{16-12} = 0.5

of the interval width, containing

0.5 \times 50 = 25

diodes. Totaling these gives

27.5 + 80 + 25 = 132.5

diodes. Expressed as a percentage of the total

250

diodes,

\frac{132.5}{250} \times 100\% = 53\%

.

Step-by-Step Solution

1

Determine the estimated number of diodes in the partial interval

6 \le t < 8

.

The target range

[6, 8)

covers half of the interval

[4, 8)

width of

4

units. With uniform distribution, the count is

0.5 \times 55 = 27.5

diodes.

Linear interpolation estimates frequencies for sub-intervals within grouped data.

2

Include the count for the complete interval

8 \le t < 12

.

All

80

diodes in this interval fall within

6 \le t < 14

.

The entire interval is fully contained within the target upper and lower bounds.

3

Determine the estimated number of diodes in the partial interval

12 \le t < 14

.

The target range

[12, 14)

covers half of the interval

[12, 16)

width of

4

units. With uniform distribution, the count is

0.5 \times 50 = 25

diodes.

Linear interpolation estimates frequencies for the upper partial boundary.

4

Sum the target diode counts and convert to a percentage of the total sample.

Total target count =

27.5 + 80 + 25 = 132.5

diodes. Percentage =

\frac{132.5}{250} \times 100\% = 53\%

.

Divide the calculated frequency sum by the total sample size of 250 and multiply by 100.

Key Concept

Grouped Frequency Distribution and Linear Interpolation

Estimated Time:2m 0s

Question 1710Question →

In the $xy$ -plane, line $L$ passes through the points $(-2, 5)$ and $(4, 1)$ . Line $M$ is the perpendicular bisector of the line segment connecting these two points. What is the $y$ -intercept of line $M$ ?

-\frac{3}{2}

\frac{1}{2}

\frac{3}{2}

\frac{7}{3}

\frac{11}{3}

Show answer & explanation

Answer: $\frac{3}{2}$

Answer

\frac{3}{2}

The midpoint of the segment with endpoints

(-2, 5)

and

(4, 1)

is

\left(\frac{-2+4}{2}, \frac{5+1}{2}\right) = (1, 3)

. The slope of the segment is

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

. Therefore, the perpendicular bisector (line

M

) has a slope equal to the negative reciprocal,

\frac{3}{2}

. Using point-slope form with point

(1, 3)

, line

M

has the equation

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

, which simplifies to

y = \frac{3}{2}x + \frac{3}{2}

. The

y

-intercept is

\frac{3}{2}

.

Step-by-Step Solution

1

Calculate the midpoint of the line segment with endpoints

(-2, 5)

and

(4, 1)

Midpoint

(x_m, y_m) = \left(\frac{-2 + 4}{2}, \frac{5 + 1}{2}\right) = (1, 3)

The perpendicular bisector passes through the midpoint of the segment.

2

Determine the slope of line

L

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

Slope is calculated as change in

y

divided by change in

x

.

3

Determine the slope of line

M

(perpendicular bisector)

m_M = -\frac{1}{m_L} = \frac{3}{2}

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

4

Find the equation of line

M

using point-slope form through

(1, 3)

and evaluate its

y

-intercept

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

. Setting

x = 0

gives

y = \frac{3}{2}

.

The

y

-intercept is the value of

y

when

x = 0

.

Key Concept

Perpendicular Bisectors and Slope-Intercept Form

Question 1711Question →

For all real numbers $x$ , the function $f$ is defined by $f(x) = 3 - x^2$ . For all real numbers $x \neq -2$ , the function $g$ is defined by $g(x) = \frac{2x - 1}{x + 2}$ . What is the value of $g(f(-3))$ ?

\frac{13}{4}

\frac{23}{14}

-46

-42

-\frac{1}{2}

Show answer & explanation

Answer: $\frac{13}{4}$

Answer

The correct answer is

\frac{13}{4}

.

First, evaluate the inner expression

f(-3) = 3 - (-3)^2 = 3 - 9 = -6

. Next, evaluate the outer expression at this value:

g(-6) = \frac{2(-6) - 1}{-6 + 2} = \frac{-13}{-4} = \frac{13}{4}

.

Step-by-Step Solution

1

Evaluate the inner function

f(-3)

f(-3) = 3 - (-3)^2 = 3 - 9 = -6

Applying the definition of

f(x)

where the negative base is squared to yield positive

9

.

2

Substitute the result into the outer function

g(x)

g(-6) = \frac{2(-6) - 1}{-6 + 2} = \frac{-12 - 1}{-4} = \frac{-13}{-4} = \frac{13}{4}

Evaluating

g(x)

at

x = -6

and simplifying the fraction.

Key Concept

Nested Function Evaluation

g(f(x))

Estimated Time:1m 30s

Question 1712Question →

A community health center conducted a study of $250$ adults regarding their participation in three wellness programs: Nutrition Counseling ( $N$ ), Exercise Coaching ( $E$ ), and Stress Management ( $S$ ). The survey revealed the following data:

• $125$ adults participate in Nutrition Counseling.
• $105$ adults participate in Exercise Coaching.
• $85$ adults participate in Stress Management.
• $15$ adults participate in all three programs.
• $30$ adults participate in none of the three programs.

How many adults participate in exactly two of the three programs?

Show answer & explanation

Answer: 65

Answer

65

The total number of surveyed adults is

250

, and

30

participate in none of the programs, meaning

220

adults participate in at least one program. Let

x_1

be the number of adults in exactly one program,

x_2

in exactly two programs, and

x_3 = 15

in all three programs. We have

x_1 + x_2 + 15 = 220

, which simplifies to

x_1 + x_2 = 205

. Additionally, summing the individual program participants yields

|N| + |E| + |S| = 125 + 105 + 85 = 315

. By region expansion,

|N| + |E| + |S| = x_1 + 2x_2 + 3x_3

. Substituting

x_3 = 15

gives

x_1 + 2x_2 + 45 = 315

, or

x_1 + 2x_2 = 270

. Subtracting

x_1 + x_2 = 205

from

x_1 + 2x_2 = 270

gives

x_2 = 65

.

Step-by-Step Solution

1

Determine the number of adults in the union of all three set categories

|N \cup E \cup S| = 250 - 30 = 220

Subtracting the individuals participating in none of the programs from the total surveyed yields the total count of individuals participating in at least one program.

2

Set up an equation for the total unique participants using disjoint region variables

x_1 + x_2 = 205

The union equals

x_1 + x_2 + x_3 = 220

, where

x_1

represents adults in exactly 1 program,

x_2

in exactly 2, and

x_3 = 15

in all 3 programs.

3

Set up an equation using the sum of the individual program totals

x_1 + 2x_2 = 270

The sum

|N| + |E| + |S| = 125 + 105 + 85 = 315

counts single-program participants once, double-program participants twice, and triple-program participants three times (

x_1 + 2x_2 + 3(15) = 315

).

4

Solve the system of linear equations for

x_2

x_2 = 270 - 205 = 65

Subtracting

(x_1 + x_2 = 205)

from

(x_1 + 2x_2 = 270)

isolates

x_2

, which is the exact number of adults participating in exactly two programs.

Key Concept

3-Set Inclusion-Exclusion Principle & Venn Diagram Region Partitioning

Estimated Time:1m 45s

Question 1713Question →

A committee of $8$ people consists of $4$ men and $4$ women. A subcommittee of $4$ people is to be selected from this group such that the subcommittee contains at least one man and at least one woman. If two specific members, one man and one woman, refuse to serve together on the same subcommittee, how many different valid subcommittees of $4$ people can be formed?

Show answer & explanation

Answer: 53

Answer

53

The total number of ways to choose 4 people out of 8 is

\binom{8}{4} = 70

. Removing the 2 single-gender subcommittees (4 men or 4 women) leaves 68 gender-valid subcommittees. Among these 68 subcommittees, exactly

\binom{6}{2} = 15

contain both of the two conflicting individuals. Subtracting these 15 forbidden subcommittees gives

68 - 15 = 53

valid subcommittees.

Step-by-Step Solution

1

Calculate total ways to pick 4 people out of 8 without restrictions

\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70

This establishes the total baseline sample space of possible 4-person groups.

2

Exclude single-gender groups to satisfy the gender balance constraint

70 - \binom{4}{4} - \binom{4}{4} = 70 - 1 - 1 = 68

Groups with 0 men or 0 women are invalid.

3

Count the forbidden groups that contain both of the conflicting individuals

\binom{6}{2} = 15

Fixing the 2 specific individuals in the subcommittee requires selecting 2 additional members from the remaining 6 people.

4

Subtract forbidden groups from gender-valid groups

68 - 15 = 53

Every group containing both conflicting individuals already satisfies the gender constraint, so exactly 15 invalid groups must be removed from the 68 gender-valid groups.

Key Concept

Combinations with multiple overlapping constraints (complementary counting)

Question 1714Question →

In the $xy$ -plane, line $L_1$ passes through the points $(2, 5)$ and $(6, -3)$ . Line $L_2$ is perpendicular to line $L_1$ and passes through the point $(1, 2)$ . Line $L_3$ is parallel to line $L_2$ and has a $y$ -intercept that is $5$ units greater than the $y$ -intercept of line $L_2$ . What is the $x$ -intercept of line $L_3$ ?

-13

-\frac{5}{2}

\frac{9}{2}

13

15

Show answer & explanation

Answer: $-13$

Answer

The

x

-intercept of line

L_3

is

-13

.

The slope of line

L_1

is

\frac{-3 - 5}{6 - 2} = -2

. The negative reciprocal slope for line

L_2

is

\frac{1}{2}

. Using point-slope form with point

(1,2)

, line

L_2

is

y = \frac{1}{2}x + \frac{3}{2}

, giving a

y

-intercept of

\frac{3}{2}

. Line

L_3

, being parallel to

L_2

, shares slope

\frac{1}{2}

and has

y

-intercept

\frac{3}{2} + 5 = \frac{13}{2}

. Finding the

x

-intercept of

y = \frac{1}{2}x + \frac{13}{2}

by setting

y=0

yields

x = -13

.

Step-by-Step Solution

1

Calculate the slope of line

L_1

The slope

m_1 = \frac{-3 - 5}{6 - 2} = \frac{-8}{4} = -2

.

The slope formula is

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

.

2

Determine the slope and equation of perpendicular line

L_2

The slope

m_2 = -\frac{1}{-2} = \frac{1}{2}

. Since

L_2

passes through

(1, 2)

, its equation is

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

, which simplifies to

y = \frac{1}{2}x + \frac{3}{2}

.

Perpendicular lines have negative reciprocal slopes.

3

Determine the equation of line

L_3

The

y

-intercept of

L_2

is

\frac{3}{2}

. The

y

-intercept of

L_3

is

\frac{3}{2} + 5 = \frac{13}{2}

. Since

L_3

is parallel to

L_2

, its slope is

m_3 = \frac{1}{2}

. Thus, the equation of

L_3

is

y = \frac{1}{2}x + \frac{13}{2}

.

Parallel lines have equal slopes.

4

Find the

x

-intercept of line

L_3

Setting

y = 0

gives

0 = \frac{1}{2}x + \frac{13}{2} \implies \frac{1}{2}x = -\frac{13}{2} \implies x = -13

.

The

x

-intercept is the value of

x

when

y = 0

.

Key Concept

Slope of parallel and perpendicular lines, line equations, and intercept calculations

Question 1715Question →

A chord $AB$ of length $24$ is drawn in a circle with center $O$ and radius $13$ . A point $P$ lies on the circle such that the area of triangle $ABP$ is maximized. What is the perimeter of triangle $ABP$ ?

24 + 8\sqrt{13}

12 + 12\sqrt{13}

24 + 12\sqrt{13}

24 + 6\sqrt{13}

24 + 12\sqrt{17}

Show answer & explanation

Answer: $24 + 12\sqrt{13}$

Answer

The perimeter of triangle

ABP

is

24 + 12\sqrt{13}

.

The distance from center O to chord AB forms a 5-12-13 right triangle with half the chord length (12) and the radius (13), giving OM = 5. To maximize triangle area, P must be on the major arc, making the altitude PM = 13 + 5 = 18. Using the Pythagorean theorem in right triangle AMP with legs 12 and 18 gives AP = sqrt(12^2 + 18^2) = 6sqrt(13). The total perimeter is base AB plus twice AP, which equals 24 + 12sqrt(13).

Step-by-Step Solution

1

Find the distance from the center

O

to the chord

AB

.

Let

M

be the midpoint of chord

AB

. Since

AB = 24

,

AM = 12

. Triangle

OMA

is a right triangle with hypotenuse

OA = 13

and leg

AM = 12

. By the Pythagorean theorem:

OM = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5

.

The perpendicular from the center of a circle to a chord bisects the chord and forms a right triangle with the radius.

2

Determine the location of

P

that maximizes the area of

\triangle ABP

and calculate the altitude.

The area of

\triangle ABP

is

\frac{1}{2} \times AB \times h

, where

h

is the perpendicular distance from

P

to segment

AB

. Area is maximized when

h

is maximized. Point

P

must lie on the major arc along the diameter perpendicular to

AB

. Thus, the maximum height is

PM = PO + OM = 13 + 5 = 18

.

Since the base

AB

is fixed, maximizing the area requires maximizing the height perpendicular to

AB

.

3

Calculate the length of side

AP

(and

BP

) using the Pythagorean theorem.

In right triangle

AMP

,

AM = 12

and

PM = 18

. Therefore:

AP = \sqrt{12^2 + 18^2} = \sqrt{144 + 324} = \sqrt{468} = 6\sqrt{13}

. By symmetry,

BP = AP = 6\sqrt{13}

.

The perpendicular bisector of a chord creates two congruent right triangles for any point

P

lying on it.

4

Calculate the total perimeter of

\triangle ABP

.

\text{Perimeter} = AB + AP + BP = 24 + 6\sqrt{13} + 6\sqrt{13} = 24 + 12\sqrt{13}.

The perimeter of a triangle is the sum of all three side lengths.

Key Concept

Pythagorean Theorem and Special Right Triangles in Circle Geometry

Question 1716Question →

A dataset $S$ consists of $5$ positive integers. The dataset has a unique mode of $15$ , a median of $15$ , and an arithmetic mean of $12$ .

Which of the following could be the range of dataset $S$ ? Indicate all such values.

Select all that apply

8

10

18

26

28

Show answer & explanation

Answer: 10; 18; 26

Answer

The values 10, 18, and 26 are all possible values for the range of dataset

S

.

The range of dataset S is defined as x_5 - x_1. Based on the mean, median, and unique mode conditions, the 5 positive integers must take the form x_1 < x_2 < 15 = 15 < x_5 with x_1 + x_2 + x_5 = 30. Testing valid integer assignments shows the range can be any integer from 10 to 26 inclusive. Therefore, 10, 18, and 26 are all valid possible range values.

Step-by-Step Solution

1

Express the sum of elements using the mean

Sum of 5 elements = 5 × 12 = 60

Since the mean of 5 numbers is 12, their total sum must equal 60.

2

Set up the ordered elements and apply the median and unique mode conditions

Elements in ascending order: x_1 ≤ x_2 ≤ x_3 ≤ x_4 ≤ x_5, with x_3 = 15 and x_4 = 15.

The median (middle element x_3) is 15. For 15 to be the unique mode, 15 must appear at least twice. Since x_3 = 15, either x_2 = 15 or x_4 = 15. If x_2 = x_3 = x_4 = 15, then x_1 + x_5 = 15, which forces x_1 < 1 and violates positive integer constraints. Thus, 15 appears exactly twice: x_3 = 15 and x_4 = 15 (with x_1 < x_2 < 15 and x_5 > 15).

3

Determine the constraints on x_1, x_2, and x_5

x_1 + x_2 + x_5 = 30, where 1 ≤ x_1 < x_2 ≤ 14 and x_5 > 15.

Substituting x_3 = 15 and x_4 = 15 into the sum gives x_1 + x_2 + 15 + 15 + x_5 = 60, simplifying to x_1 + x_2 + x_5 = 30.

4

Calculate the bounds for the Range R = x_5 - x_1

Minimum range = 10, Maximum range = 26

To maximize R = x_5 - x_1 = 30 - 2x_1 - x_2, choose minimum x_1 = 1 and minimum x_2 = 2, yielding x_5 = 27 and R = 26. To minimize R, maximize x_1 = 6 and x_2 = 8, yielding x_5 = 16 and R = 10.

Key Concept

Measures of Central Tendency (Mean, Median, Mode) and Data Range Constraints

Question 1717Question →

A container holds 5 red marbles, 4 blue marbles, and 3 green marbles. Two marbles are drawn sequentially at random without replacement. Let event $A$ be the event that at least one of the drawn marbles is red, and let event $B$ be the event that the second marble drawn is green. What is the conditional probability $P(B \mid A)$ ?

\frac{1}{6}

\frac{1}{4}

\frac{3}{22}

\frac{3}{19}

\frac{5}{22}

Show answer & explanation

Answer: $\frac{1}{6}$

Answer

The conditional probability

P(B \mid A)

is

\frac{1}{6}

.

To find

P(B \mid A)

, we evaluate

\frac{P(A \cap B)}{P(A)}

. The probability of at least one red marble

P(A)

is

1 - P(\text{no red}) = 1 - \frac{7 \times 6}{12 \times 11} = \frac{90}{132}

. The intersection event

A \cap B

requires the second marble to be green and at least one marble to be red, which means the first marble must be red and the second green. The probability of this is

\frac{5 \times 3}{12 \times 11} = \frac{15}{132}

. Taking the ratio

\frac{15/132}{90/132}

yields

\frac{15}{90} = \frac{1}{6}

.

Step-by-Step Solution

1

Calculate the total number of ordered outcomes and the probability of event A using the complementary event.

Total outcomes drawing 2 marbles from 12 without replacement is

12 \times 11 = 132

. The complement

A^c

(no red marbles selected from the 7 non-red marbles) has

7 \times 6 = 42

outcomes. Thus,

P(A^c) = \frac{42}{132} = \frac{7}{22}

, which means

P(A) = 1 - \frac{7}{22} = \frac{15}{22} = \frac{90}{132}

.

Using the complement rule is the most efficient way to compute 'at least one' probabilities.

2

Determine the intersection event

A \cap B

and calculate its probability.

Event

B

specifies that the second marble is green. For event

A

(at least one marble is red) to also occur, the first marble must be red. Thus,

A \cap B

is equivalent to 'the first marble is red AND the second marble is green'. The number of favorable outcomes is

5 \times 3 = 15

. So

P(A \cap B) = \frac{15}{132}

.

Mutual exclusivity between red and green on the second draw simplifies the intersection logic.

3

Apply the conditional probability formula

P(B \mid A) = \frac{P(A \cap B)}{P(A)}

.

P(B \mid A) = \frac{\frac{15}{132}}{\frac{90}{132}} = \frac{15}{90} = \frac{1}{6}

.

Evaluating the ratio yields the exact conditional probability requested.

Key Concept

Conditional Probability of Dependent Events

Estimated Time:3m 0s

Question 1718Question →

In the $xy$ -plane, line $K$ is defined by the equation $3x - 4y = 12$ . Line $M$ is perpendicular to line $K$ and passes through the point $(6, -1)$ . Which of the following statements regarding line $M$ must be true? Select all that apply.

Select all that apply

Line

M

has a

y

-intercept of

(0, 7)

.

Line

M

passes through the point

(-3, 11)

.

Line

M

has an

x

-intercept of

(7, 0)

.

Line

M

passes through Quadrant III.

The point of intersection between line

K

and line

M

lies in Quadrant IV.

Show answer & explanation

Answer: Line $M$ has a $y$ -intercept of $(0, 7)$ .; Line $M$ passes through the point $(-3, 11)$ .

Answer

The statements asserting that Line

M

has a

y

-intercept of

(0, 7)

and that Line

M

passes through the point

(-3, 11)

are correct.

Line

K

has a slope of

\frac{3}{4}

, making the perpendicular slope of line

M

equal to

-\frac{4}{3}

. Using the point

(6, -1)

, the equation of line

M

is

y = -\frac{4}{3}x + 7

. Evaluating the options: setting

x = 0

gives

y = 7

, confirming the

y

-intercept is

(0, 7)

; substituting

x = -3

yields

y = 11

, confirming

(-3, 11)

lies on line

M

. Both of these statements are true.

Step-by-Step Solution

1

Determine the slope of line

K

and the perpendicular slope of line

M

.

Line

K

in slope-intercept form is

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

, so its slope is

m_K = \frac{3}{4}

. The perpendicular slope for line

M

is the negative reciprocal:

m_M = -\frac{4}{3}

.

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

2

Find the equation of line

M

using point-slope form with point

(6, -1)

.

y - (-1) = -\frac{4}{3}(x - 6) \implies y + 1 = -\frac{4}{3}x + 8 \implies y = -\frac{4}{3}x + 7

.

Knowing the slope and a point on the line allows determination of the line's exact linear equation.

3

Evaluate the statements using the equation of line

M

.

1.

y

-intercept: set

x = 0 \implies y = 7

, so

(0,7)

is correct.
2. Point

(-3, 11)

:

y = -\frac{4}{3}(-3) + 7 = 4 + 7 = 11

, so

(-3, 11)

is on the line.
3.

x

-intercept: set

y = 0 \implies -\frac{4}{3}x + 7 = 0 \implies x = \frac{21}{4} = 5.25

, so

(7,0)

is incorrect.
4. Quadrants: A line with negative slope and positive

y

-intercept covers Quadrants I, II, and IV only, so passing through Quadrant III is false.
5. Intersection with line

K

: set

\frac{3}{4}x - 3 = -\frac{4}{3}x + 7 \implies \frac{25}{12}x = 10 \implies x = 4.8

,

y = 0.6

, which is in Quadrant I, not Quadrant IV.

Direct algebraic verification confirms which geometric properties hold true for line

M

.

Key Concept

Perpendicular Slopes and Linear Properties in Coordinate Geometry

Question 1719Question →

A logistics company operates a delivery van and a cargo drone along a straight route of $270$ kilometers connecting Hub X and Hub Y. The delivery van departs from Hub X toward Hub Y at a constant speed of $60$ kilometers per hour. Exactly $2$ hours later, the cargo drone departs from Hub Y toward Hub X along the same route at a constant speed of $90$ kilometers per hour. How many hours after the delivery van departs will the van and the cargo drone meet?

1.8 hours

2.6 hours

3.0 hours

4.5 hours

5.0 hours

Show answer & explanation

Answer: 3.0 hours

Answer

3.0 hours

The option specifying '3.0 hours' is correct. During the 2 hours before the drone departs, the van covers

60 \text{ km/h} \times 2 \text{ h} = 120 \text{ km}

. The remaining distance between the vehicles is

270 \text{ km} - 120 \text{ km} = 150 \text{ km}

. Once the drone departs, the two vehicles close the gap at a combined rate of

60 + 90 = 150 \text{ km/h}

. The time needed to cover the remaining

150 \text{ km}

is

\frac{150 \text{ km}}{150 \text{ km/h}} = 1 \text{ hour}

. Adding the van's initial

2

hours gives a total travel time of

3.0

hours.

Step-by-Step Solution

1

Define variables for travel time and compute distance covered by the van before the drone departs.

In

2

hours at

60

km/h, the van covers

60 \times 2 = 120

kilometers.

The van travels alone for the first two hours.

2

Determine the remaining distance separating the two vehicles when the drone starts moving.

Remaining distance =

270 - 120 = 150

kilometers.

The total distance between the hubs is

270

kilometers.

3

Calculate the relative speed of approach and the additional time required to meet.

Combined speed =

60 + 90 = 150

km/h. Additional time =

\frac{150}{150} = 1

hour.

Since the vehicles travel toward each other, their speeds add up.

4

Find the total time elapsed since the van departed.

Total time =

2 + 1 = 3.0

hours.

The question asks for total hours since the van's departure.

Key Concept

Distance-Rate-Time linear modeling with staggered start times and opposing directions

Estimated Time:1m 45s

Question 1720Question →

A jar contains $3$ red marbles and $7$ blue marbles. A marble is drawn at random from the jar, its color is noted, and it is returned to the jar. A second marble is then drawn at random. What is the probability that both drawn marbles are red?

\frac{9}{100}

\frac{3}{10}

\frac{3}{5}

\frac{9}{49}

\frac{6}{7}

Show answer & explanation

Answer: $\frac{9}{100}$

Answer

The probability that both drawn marbles are red is

\frac{9}{100}

.

The option stating

\frac{9}{100}

is correct because the two draws are independent due to replacement. The probability of getting a red marble on any single draw is

\frac{3}{10}

. By the multiplication rule for independent events, the probability of both events occurring is

\frac{3}{10} \times \frac{3}{10} = \frac{9}{100}

.

Step-by-Step Solution

1

Calculate the total number of marbles in the jar.

3 \text{ red} + 7 \text{ blue} = 10 \text{ total marbles}

.

Probability requires knowing the size of the full sample space.

2

Find the probability of drawing a red marble on a single draw.

P(\text{Red}) = \frac{3}{10}

.

There are

3

favorable outcomes (red marbles) out of

10

total possible outcomes.

3

Apply the multiplication rule for independent events.

P(\text{Red}_1 \text{ and } \text{Red}_2) = \frac{3}{10} \times \frac{3}{10} = \frac{9}{100}

.

Because the first marble is replaced, the second draw is independent of the first, so joint probability is the product of individual probabilities.

Key Concept

Multiplication Rule for Independent Events:

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

when events

A

and

B

are independent.

Estimated Time:45s

Answer

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