All practice questions

Question 121Question →

In right triangle $ABC$ , the measure of angle $C$ is $90^\circ$ . Point $D$ lies on segment $AC$ such that $AD = 11$ and $BD = 13$ . If $DC = 5$ , what is the perimeter of triangle $ABD$ ?

Show answer & explanation

Answer: 44

Answer

The perimeter of triangle

ABD

is

44

.

To find the perimeter of triangle

ABD

, we need the lengths of its three sides:

AD

,

BD

, and

AB

. We are given

AD = 11

and

BD = 13

. To find

AB

, we first analyze right triangle

BCD

with right angle at

C

, hypotenuse

BD = 13

, and leg

DC = 5

. Using the Pythagorean theorem, leg

BC = \sqrt{13^2 - 5^2} = 12

. Next, the full leg

AC = AD + DC = 11 + 5 = 16

. In right triangle

ABC

, legs are

AC = 16

and

BC = 12

, making hypotenuse

AB = \sqrt{16^2 + 12^2} = 20

. Finally, the perimeter of triangle

ABD

is

11 + 13 + 20 = 44

.

Step-by-Step Solution

1

Find the length of side

BC

using right triangle

BCD

BC = 12

In right triangle

BCD

, angle

C = 90^\circ

, hypotenuse

BD = 13

, and leg

DC = 5

. By the Pythagorean theorem,

BC = \sqrt{13^2 - 5^2} = \sqrt{144} = 12

.

2

Determine the total length of side

AC

AC = 16

Since point

D

lies on segment

AC

,

AC = AD + DC = 11 + 5 = 16

.

3

Calculate the hypotenuse

AB

of the main right triangle

ABC

AB = 20

In right triangle

ABC

, legs are

AC = 16

and

BC = 12

. By the Pythagorean theorem,

AB = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20

.

4

Calculate the perimeter of triangle

ABD

Perimeter

= 44

The sides of triangle

ABD

are

AD = 11

,

BD = 13

, and

AB = 20

. Adding these gives

11 + 13 + 20 = 44

.

Key Concept

Pythagorean Theorem and Multi-Step Triangle Properties

Question 122Question →

A survey of $250$ registered voters was conducted to analyze their primary news sources: Television ( $T$ ), the Internet ( $I$ ), and Print newspapers ( $P$ ). The survey revealed the following results:

- $140$ voters get news from Television.
- $150$ voters get news from the Internet.
- $80$ voters get news from Print newspapers.
- $45$ voters get news from both Television and Print newspapers.
- $60$ voters get news from both the Internet and Print newspapers.
- $30$ voters get news from all three sources.
- $20$ voters do not get news from any of these three sources.

How many of the surveyed voters get news from Television and the Internet, but NOT from Print newspapers?

Show answer & explanation

Answer: 35

Answer

35 voters get news from Television and the Internet, but not from Print newspapers.

Using the principle of inclusion-exclusion for three sets, the total union size is 230 voters (250 total minus 20 who use none). Setting up the formula 230 = 140 + 150 + 80 - 45 - 60 - |T ∩ I| + 30 allows us to solve for |T ∩ I| = 65. To find those who use Television and Internet but NOT Print newspapers, we subtract the 30 voters who use all three sources from 65, resulting in 35 voters.

Step-by-Step Solution

1

Find the size of the union of all three sets

|T ∪ I ∪ P| = 250 - 20 = 230

Subtracting the 20 voters who use none of the three news sources from the total sample of 250 gives the total number of voters in at least one category.

2

Set up the Principle of Inclusion-Exclusion for three sets

230 = 140 + 150 + 80 - 45 - 60 - |T ∩ I| + 30

The formula sums individual set sizes, subtracts pairwise intersections, and adds back the triple intersection.

3

Solve for the total intersection of Television and Internet

|T ∩ I| = 65

Simplifying the equation gives 230 = 295 - |T ∩ I|, which yields |T ∩ I| = 65.

4

Exclude those who also read Print newspapers

|(T ∩ I) \ P| = 65 - 30 = 35

Subtracting the 30 voters who use all three sources leaves only those who use Television and Internet without Print newspapers.

Key Concept

Three-Set Principle of Inclusion-Exclusion

Question 123Question →

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 124Question →

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 125Question →

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 126Question →

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 127Question →

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 128Question →

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 129Question →

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 130Question →

In trapezoid $ABCD$ , side $AB$ is perpendicular to parallel bases $AD$ and $BC$ . Diagonal $AC$ is perpendicular to side $CD$ . If the measure of angle $CAD$ is $30^\circ$ and the length of $BC$ is $9$ , what is the length of side $CD$ ?

Show answer & explanation

Answer: 6

Answer

6

Because base

BC

and base

AD

are parallel, transversal

AC

creates equal alternate interior angles, making

\angle ACB = \angle CAD = 30^\circ

. Triangle

ABC

is a right triangle with

\angle B = 90^\circ

, so

BC

is adjacent to

30^\circ

. The ratio of the side adjacent to

30^\circ

to the hypotenuse

AC

is

\frac{\sqrt{3}}{2}

, which yields

AC = \frac{9}{\sqrt{3}/2} = 6\sqrt{3}

. Triangle

ACD

is also a

30^\circ-60^\circ-90^\circ

right triangle with right angle at

C

. Side

CD

is opposite the

30^\circ

angle and side

AC

is opposite the

60^\circ

angle. The ratio of the short leg to the long leg is

\frac{1}{\sqrt{3}}

, giving

CD = \frac{6\sqrt{3}}{\sqrt{3}} = 6

.

Step-by-Step Solution

1

Determine angle measures in right triangle ABC using parallel line properties

Angle ACB = 30 degrees and angle BAC = 60 degrees

Since base BC is parallel to base AD, alternate interior angles formed by transversal AC are equal: angle ACB = angle CAD = 30 degrees.

2

Calculate the length of diagonal AC using special right triangle ratios

AC = 6*sqrt(3)

In 30-60-90 right triangle ABC, the ratio of the side adjacent to 30 degrees (BC) to the hypotenuse (AC) is sqrt(3)/2. Therefore, 9 / AC = sqrt(3)/2, which gives AC = 18 / sqrt(3) = 6*sqrt(3).

3

Calculate the length of side CD using right triangle ACD

CD = 6

In 30-60-90 right triangle ACD, angle ACD = 90 degrees and angle CAD = 30 degrees. The ratio of the short leg opposite 30 degrees (CD) to the long leg opposite 60 degrees (AC) is 1/sqrt(3). Therefore, CD = AC / sqrt(3) = (6*sqrt(3)) / sqrt(3) = 6.

Key Concept

Side ratios of 30-60-90 special right triangles and alternate interior angles

Question 131Question →

In $\triangle ABC$ , angle $B$ is a right angle, and line segment $BD$ is an altitude drawn to side $AC$ with point $D$ lying on $AC$ . If $AD = 4$ units and $DC = 16$ units, what is the area, in square units, of $\triangle ABC$ ?

Show answer & explanation

Answer: 80

Answer

The area of triangle ABC is 80 square units.

In right triangle

ABC

with right angle at

B

, altitude

BD

drawn to hypotenuse

AC

divides the hypotenuse into segments

AD

and

DC

. By the geometric mean theorem,

BD^2 = AD \times DC = 4 \times 16 = 64

, which gives

BD = 8

units. The length of hypotenuse

AC

is

AD + DC = 4 + 16 = 20

units. The area of triangle

ABC

is

\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \times 8 = 80

square units.

Step-by-Step Solution

1

Calculate the height (altitude)

BD

of the triangle.

BD = 8

units.

In a right triangle, the altitude to the hypotenuse divides the hypotenuse into two segments such that

BD^2 = AD \times DC

. Substituting the given values yields

BD^2 = 4 \times 16 = 64

, so

BD = \sqrt{64} = 8

.

2

Calculate the total length of hypotenuse

AC

.

AC = 20

units.

Since point

D

lies on segment

AC

, the total length is the sum of its parts:

AC = AD + DC = 4 + 16 = 20

.

3

Calculate the area of

\triangle ABC

.

Area = 80 square units.

Using the triangle area formula

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

, with base

AC = 20

and altitude

BD = 8

, we get

\text{Area} = \frac{1}{2} \times 20 \times 8 = 80

.

Key Concept

Geometric mean theorem for right triangle altitude and area of a triangle

Question 132Question →

An environmental auditing agency surveyed $250$ manufacturing plants regarding their compliance with three environmental standards: Air Quality ( $A$ ), Water Discharge ( $W$ ), and Waste Management ( $M$ ). The survey yielded the following data:

- $130$ plants meet Air Quality standards ( $A$ ).
- $140$ plants meet Water Discharge standards ( $W$ ).
- $120$ plants meet Waste Management standards ( $M$ ).
- $40$ plants meet all three standards.
- $20$ plants meet none of the three standards.
- The number of plants meeting both Air Quality and Water Discharge standards is equal to the number of plants meeting both Water Discharge and Waste Management standards.
- The number of plants meeting both Air Quality and Waste Management standards is $10$ fewer than the number meeting both Air Quality and Water Discharge standards.

How many of the surveyed plants meet exactly one of the three environmental standards?

Show answer & explanation

Answer: 110

Answer

110

By setting up the 3-set inclusion-exclusion equation, the unknown pairwise intersections are found to be 70, 70, and 60. Subtracting the 40 plants that meet all three standards gives the exclusive double-overlap regions (30, 30, and 20). Subtracting these along with the central intersection from each single set yields 40 plants meeting only Air Quality, 40 meeting only Water Discharge, and 30 meeting only Waste Management, totaling 110 plants.

Step-by-Step Solution

1

Find total number of plants meeting at least one standard

|A ∪ W ∪ M| = 250 - 20 = 230

Subtracting plants that meet no standards from the total surveyed gives the union of all three sets.

2

Set up algebraic expressions for pairwise intersections

|A ∩ M| = k, |A ∩ W| = k + 10, |W ∩ M| = k + 10

Define the smallest pairwise intersection as k and express the other two based on the given relationships.

3

Apply the Principle of Inclusion-Exclusion (PIE) for three sets to solve for k

230 = 130 + 140 + 120 - (k + 10 + k + 10 + k) + 40 => k = 60

Substitute set sizes and the triple intersection into the 3-set inclusion-exclusion formula.

4

Calculate the number of plants in each exclusive region

Only (A ∩ W) = 30, Only (W ∩ M) = 30, Only (A ∩ M) = 20

Subtract the triple intersection (40) from each pairwise intersection.

5

Calculate plants meeting exactly one standard and sum them

Only A = 40, Only W = 40, Only M = 30; Total = 40 + 40 + 30 = 110

Subtract all double-overlap and triple-overlap regions from each individual set total.

Key Concept

Three-Set Principle of Inclusion-Exclusion and Venn Diagram Region Decomposition

Question 133Question →

A university surveyed a cohort of $150$ freshmen regarding their membership in three student organizations: the Art Club ( $A$ ), the Music Society ( $M$ ), and the Theater Guild ( $T$ ). The survey revealed the following data:

- $68$ students belong to the Art Club.
- $62$ students belong to the Music Society.
- $54$ students belong to the Theater Guild.
- $22$ students belong to both the Art Club and the Music Society.
- $18$ students belong to both the Music Society and the Theater Guild.
- $15$ students belong to both the Art Club and the Theater Guild.
- $8$ students belong to all three organizations.

How many of the surveyed students belong to exactly one of these three organizations?

Show answer & explanation

Answer: 98

Answer

98 students belong to exactly one of the three organizations.

To find the number of students belonging to exactly one organization, analyze the regions of a 3-set Venn diagram starting from the innermost region (all three clubs =

8

). Subtracting

8

from each pairwise intersection gives the students in exactly two clubs: Art & Music only (

14

), Music & Theater only (

10

), and Art & Theater only (

7

). Next, subtract the overlapping regions from each club total: Art only is

68 - (14 + 7 + 8) = 39

; Music only is

62 - (14 + 10 + 8) = 30

; Theater only is

54 - (7 + 10 + 8) = 29

. Summing these single-club regions gives

39 + 30 + 29 = 98

.

Step-by-Step Solution

1

Find the number of students belonging strictly to each pair of organizations (two-set intersections only).

Art and Music only =

14

; Music and Theater only =

10

; Art and Theater only =

7

.

The given pairwise totals include students who belong to all three organizations (

8

), so subtracting

8

isolates those in exactly two groups.

2

Determine the number of students belonging to each individual organization exclusively.

Art only =

39

; Music only =

30

; Theater only =

29

.

Subtract all shared membership regions (both two-group only and three-group) from each total organization membership.

3

Add the counts of students belonging to exactly one group.

39 + 30 + 29 = 98

.

The question requests the sum of all students in the non-overlapping single-set regions.

Key Concept

Three-Set Venn Diagram Region Partitioning

Question 134Question →

For any real number $x$ , the custom unary operation $\triangle$ is defined by $\triangle(x) = 2x - x^2$ . For all real numbers $u$ and $v$ , the custom binary operation $\odot$ is defined by $u \odot v = \triangle(u) + \triangle(v) + uv$ . The function $f$ is defined for all real numbers $x$ by $f(x) = x \odot (3 - x)$ . What is the maximum value of $f(x)$ ?

Show answer & explanation

Answer: 3.75

Answer

The maximum value of

f(x)

is

3.75

(or

\frac{15}{4}

).

Expanding

f(x) = x \odot (3 - x)

yields

\triangle(x) + \triangle(3 - x) + x(3 - x)

. Substituting

\triangle(t) = 2t - t^2

gives

(2x - x^2) + (6 - 2x - (9 - 6x + x^2)) + (3x - x^2) = -3x^2 + 9x - 3

. The vertex of this quadratic function occurs at

x = \frac{9}{6} = 1.5

, where

f(1.5) = -3(2.25) + 13.5 - 3 = 3.75

.

Step-by-Step Solution

1

Evaluate the custom unary operation

\triangle

for each argument

\triangle(x) = 2x - x^2

and

\triangle(3-x) = 2(3-x) - (3-x)^2 = -x^2 + 4x - 3

Substitute

x

and

3-x

into the definition

\triangle(t) = 2t - t^2

and expand carefully.

2

Compute the product term

uv

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

The binary definition

u \odot v

includes an additive product term

uv

.

3

Sum all components to construct the explicit quadratic expression for

f(x)

f(x) = (2x - x^2) + (-x^2 + 4x - 3) + (3x - x^2) = -3x^2 + 9x - 3

Combine like terms for

x^2

,

x

, and the constant.

4

Determine the vertex of the downward-opening parabola

f(x) = -3x^2 + 9x - 3

x = \frac{3}{2} = 1.5

, yielding

f(1.5) = 3.75$

Since the coefficient of

x^2

is negative (

-3 < 0

), the maximum occurs at

x = -\frac{b}{2a} = \frac{9}{6} = 1.5

.

Key Concept

Custom operations combined with quadratic function optimization

Estimated Time:2m 0s

Question 135Question →

A parabola defined by the quadratic function $f(x) = ax^2 + bx + c$ , where $a, b,$ and $c$ are real constants with $a > 0$ , has its vertex at a minimum value of $-16$ . The distance between the two $x$ -intercepts of the parabola is $8$ . If $f(1) = -7$ and the $x$ -coordinate of the vertex is positive, what is the value of $f(-2)$ ?

Show answer & explanation

Answer: 20

Answer

The value of

f(-2)

is 20.

By converting the parabola into vertex form

f(x) = a(x - h)^2 - 16

, the

x

-intercepts are found at

h \pm \frac{4}{\sqrt{a}}

. Equating their difference to

8

yields

a = 1

. Substituting

f(1) = -7

gives

(1 - h)^2 = 9

, which yields

h = 4

under the condition

h > 0

. Evaluating

f(-2) = (-2 - 4)^2 - 16

produces

20

.

Step-by-Step Solution

1

Express the quadratic function in vertex form using the minimum value

f(x) = a(x - h)^2 - 16

, where

(h, -16)

is the vertex.

Since

a > 0

, the parabola opens upwards and its minimum value occurs at the vertex

y

-coordinate,

k = -16

.

2

Determine the leading coefficient

a

using the distance between

x

-intercepts

a = 1

Setting

f(x) = 0

yields

a(x - h)^2 - 16 = 0 \implies (x - h)^2 = \frac{16}{a} \implies x = h \pm \frac{4}{\sqrt{a}}

. The distance between roots is

\frac{8}{\sqrt{a}} = 8

, which gives

\sqrt{a} = 1 \implies a = 1

.

3

Determine the vertex

x

-coordinate

h

using the point

f(1) = -7

h = 4

Substituting

a = 1

and

x = 1

into the vertex form gives

(1 - h)^2 - 16 = -7 \implies (1 - h)^2 = 9

. Taking square roots gives

1 - h = 3 \implies h = -2

or

1 - h = -3 \implies h = 4

. Since

h > 0

, we select

h = 4

.

4

Evaluate

f(-2)

using the fully specified function

f(-2) = 20

With

f(x) = (x - 4)^2 - 16

, substituting

x = -2

yields

f(-2) = (-2 - 4)^2 - 16 = (-6)^2 - 16 = 36 - 16 = 20

.

Key Concept

Quadratic Vertex Form, Root Separation, and Evaluation

Question 136Question →

A rectangular plot of land has a perimeter of 56 meters and an area of 180 square meters. A border of uniform width $x$ meters is constructed around the inside edge of the plot, reducing the remaining inner area to 84 square meters. What is the value of $x$ ?

Show answer & explanation

Answer: 2

Answer

The border width x is 2 meters.

Solving the system

L + W = 28

and

LW = 180

gives outer dimensions of 18 and 10 meters. The inner area equation

(18 - 2x)(10 - 2x) = 84

simplifies to

x^2 - 14x + 24 = 0

, which factors into

(x - 2)(x - 12) = 0

. Since

2x

must be less than the total width of 10 meters (

x < 5

), the extraneous root

x = 12

is discarded, leaving

x = 2

.

Step-by-Step Solution

1

Determine the outer dimensions of the plot.

Length = 18 meters, Width = 10 meters

The semi-perimeter is L + W = 56 / 2 = 28, and the area is L * W = 180. Forming the quadratic equation t^2 - 28t + 180 = 0 yields roots t = 18 and t = 10.

2

Write the quadratic expression for the inner area in terms of x.

(18 - 2x)(10 - 2x) = 84

A uniform border of width x along all four sides reduces both the length and width by 2x.

3

Expand and simplify the quadratic equation into standard form.

x^2 - 14x + 24 = 0

Expanding yields 180 - 56x + 4x^2 = 84. Subtracting 84 gives 4x^2 - 56x + 96 = 0, which divides by 4 to give x^2 - 14x + 24 = 0.

4

Factor the quadratic equation to solve for possible values of x.

x = 2 or x = 12

Factoring x^2 - 14x + 24 = 0 gives (x - 2)(x - 12) = 0.

5

Eliminate the extraneous root based on physical geometry constraints.

x = 2

The total width is 10 meters, so 2x must be less than 10 (x < 5). Thus, x = 12 is physically impossible and discarded.

Key Concept

Quadratic Equations, Factoring, and Extraneous Solutions

Question 137Question →

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 138Question →

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 139Question →

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 140Question →

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

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept