All practice questions

Question 341Question →

In the $xy$ -plane, line $L_1$ passes through the origin $(0,0)$ and the point $(4, 3)$ . Line $L_2$ is formed by reflecting line $L_1$ across the vertical line $x = 2$ and then translating the resulting line downward by $5$ units. If line $L_2$ intersects the $y$ -axis at the point $(0, k)$ , what is the value of $k$ ?

-5

-2

\frac{1}{3}

3

8

Show answer & explanation

Answer: $-2$

Answer

The correct value of

k

is

-2

.

Line

L_1

has slope

\frac{3}{4}

and equation

y = \frac{3}{4}x

. Reflecting across

x = 2

replaces

x

with

4 - x

, transforming the equation into

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

. Translating downward by

5

units gives

y = -\frac{3}{4}x - 2

. Setting

x = 0

gives the

y

-intercept

(0, -2)

, making

-2

the correct value.

Step-by-Step Solution

1

Find the equation of line

L_1

.

The slope of line

L_1

passing through

(0,0)

and

(4,3)

is

m = \frac{3 - 0}{4 - 0} = \frac{3}{4}

. Thus, the equation is

y = \frac{3}{4}x

.

Establishing the initial linear equation is necessary before applying coordinate transformations.

2

Apply the reflection across the line

x = 2

.

Reflecting any point

(x, y)

across

x = 2

transforms its x-coordinate to

2(2) - x = 4 - x

. Substituting

4 - x

into the equation gives

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

.

Reflection across a vertical line

x = a

preserves the y-values while mapping

x \mapsto 2a - x

.

3

Apply the downward vertical translation by

5

units.

Subtracting

5

from the equation yields

y = \left(3 - \frac{3}{4}x\right) - 5 = -\frac{3}{4}x - 2

.

Translating a graph downward by

c

units subtracts

c

from the output

y

.

4

Determine the y-intercept of line

L_2

.

Setting

x = 0

in

y = -\frac{3}{4}x - 2

yields

y = -2

, so

k = -2

.

The y-intercept of a line occurs where

x = 0

.

Key Concept

Coordinate Transformations of Lines: Reflection across

x = a

and Vertical Translations

Estimated Time:2m 0s

Question 342Question →

In right triangle $ABC$ , the measure of $\angle B$ is $90^\circ$ , the measure of $\angle A$ is $60^\circ$ , and the length of side $AB$ is $6$ . Square $BDEF$ is inscribed in $\triangle ABC$ such that vertex $D$ lies on side $AB$ , vertex $E$ lies on hypotenuse $AC$ , and vertex $F$ lies on side $BC$ . What is the length of segment $BE$ ?

9\sqrt{2} - 3\sqrt{6}

9 - 3\sqrt{3}

3\sqrt{6} - 3\sqrt{2}

9\sqrt{3} - 9

9\sqrt{2} - 3\sqrt{3}

Show answer & explanation

Answer: $9\sqrt{2} - 3\sqrt{6}$

Answer

The length of segment

BE

is

9\sqrt{2} - 3\sqrt{6}

.

The correct answer is

9\sqrt{2} - 3\sqrt{6}

. Since

\triangle ADE

is a

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

triangle with short leg

AD = 6 - x

and long leg

DE = x

, the side length of the square is

x = 9 - 3\sqrt{3}

. Segment

BE

is the diagonal of the square, which equals

x\sqrt{2} = (9 - 3\sqrt{3})\sqrt{2} = 9\sqrt{2} - 3\sqrt{6}

.

Step-by-Step Solution

1

Analyze the properties of right triangle

ABC

and the inscribed square

BDEF

.

In

\triangle ABC

,

\angle B = 90^\circ

and

\angle A = 60^\circ

, so

\angle C = 30^\circ

. The side

AB = 6

. Let

x

be the side length of square

BDEF

.

Establishing the variable

x

allows us to express the dimensions of the smaller triangles formed by the square.

2

Identify the side lengths and angles of the smaller triangle

\triangle ADE

.

Since

BD = x

lies on side

AB

,

AD = AB - BD = 6 - x

. Because

DE \parallel BC

,

\angle ADE = 90^\circ

and

\angle A = 60^\circ

, making

\triangle ADE

a

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

right triangle with

DE = x

.

In a

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

triangle, the side opposite the

60^\circ

angle is

\sqrt{3}

times the side adjacent to the

60^\circ

angle.

3

Set up and solve the equation for the side length

x

of the square.

Since

DE = AD \cdot \sqrt{3}

, we have

x = (6 - x)\sqrt{3} \implies x(1 + \sqrt{3}) = 6\sqrt{3} \implies x = \frac{6\sqrt{3}}{\sqrt{3} + 1} = 3(3 - \sqrt{3}) = 9 - 3\sqrt{3}

.

Rationalizing the denominator

\frac{6\sqrt{3}(\sqrt{3}-1)}{2}

yields the exact side length of the square.

4

Calculate the diagonal length

BE

of square

BDEF

.

Segment

BE

is the diagonal of square

BDEF

. In a

45^\circ-45^\circ-90^\circ

right triangle

\triangle BDE

, the hypotenuse is

x\sqrt{2}

. Thus,

BE = (9 - 3\sqrt{3})\sqrt{2} = 9\sqrt{2} - 3\sqrt{6}

.

Applying the special right triangle ratio

1:1:\sqrt{2}

for the square's diagonal gives the required segment length.

Key Concept

Combining

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

side ratios (

1:\sqrt{3}:2

) and

45^\circ-45^\circ-90^\circ

hypotenuse ratios (

1:1:\sqrt{2}

) to solve composite geometric figures.

Estimated Time:2m 30s

Question 343Question →

In $\triangle ABC$ , point $D$ lies on side $BC$ such that the ratio of the length of segment $BD$ to the length of segment $DC$ is $3 : 2$ . Point $E$ is the midpoint of line segment $AD$ . If the area of $\triangle ABC$ is $60$ square units, which of the following statements must be true? Select all that apply.

Select all that apply

The area of

\triangle ABD

is

36

square units.

The area of

\triangle ABE

is

18

square units.

The area of

\triangle BEC

is

30

square units.

The area of

\triangle ADC

is

30

square units.

The area of

\triangle ECD

is

15

square units.

Show answer & explanation

Answer: The area of $\triangle ABD$ is $36$ square units.; The area of $\triangle ABE$ is $18$ square units.; The area of $\triangle BEC$ is $30$ square units.

Answer

The statements asserting that the area of triangle ABD is 36 square units, the area of triangle ABE is 18 square units, and the area of triangle BEC is 30 square units are all true.

Triangles sharing a vertex and having bases along the same straight line share the same height. Thus, their areas are in the exact ratio of their bases. Since segment BD is 3/5 of BC, triangle ABD has an area of (3/5) * 60 = 36 square units. Median BE divides triangle ABD into two equal areas of 18 square units each. Median CE divides triangle ADC (area 24) into two equal areas of 12 square units each. Summing triangles EBD (18) and ECD (12) gives an area of 30 square units for triangle BEC.

Step-by-Step Solution

1

Determine the areas of triangles ABD and ADC using the base ratio.

Area of triangle ABD = 36 square units, and Area of triangle ADC = 24 square units.

Triangles ABD and ADC share the same altitude from vertex A to line BC. Therefore, their areas are directly proportional to their base lengths BD and DC. Since BD : DC = 3 : 2, BD is 3/5 of BC and DC is 2/5 of BC.

2

Calculate the area of triangle ABE.

Area of triangle ABE = 18 square units.

Point E is the midpoint of segment AD. In triangle ABD, line segment BE is a median from vertex B to side AD. A median bisects a triangle into two region of equal area, so Area(ABE) = 1/2 * Area(ABD) = 1/2 * 36 = 18 square units.

3

Calculate the area of triangle ECD.

Area of triangle ECD = 12 square units.

Similarly, segment CE is a median in triangle ADC from vertex C to side AD. Thus, Area(ECD) = 1/2 * Area(ADC) = 1/2 * 24 = 12 square units.

4

Calculate the area of triangle BEC.

Area of triangle BEC = 30 square units.

Triangle BEC is formed by combining triangles EBD and ECD. Since Area(EBD) = 18 and Area(ECD) = 12, Area(BEC) = 18 + 12 = 30 square units.

Key Concept

Triangles sharing a vertex and altitude have areas proportional to their bases; a median divides a triangle into two equal areas.

Estimated Time:2m 0s

Question 344Question →

What is the sum of all real solutions to the equation $x - 3 = \sqrt{x + 3}$ ?

1

4

5

6

7

Show answer & explanation

Answer: 6

Answer

The sum of all real solutions to the equation is 6.

Squaring both sides of

x - 3 = \sqrt{x + 3}

gives

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

, which simplifies to

x^2 - 7x + 6 = 0

. Factoring yields

(x - 6)(x - 1) = 0

, giving candidate solutions

x = 6

and

x = 1

. Substituting

x = 6

into the original equation gives

3 = \sqrt{9} = 3

, which is valid. Substituting

x = 1

gives

-2 = \sqrt{4} = 2

, which is false, making

x = 1

an extraneous solution. Therefore,

x = 6

is the single valid solution, and its sum is 6.

Step-by-Step Solution

1

Eliminate the radical by squaring both sides of the equation.

(x - 3)^2 = (\sqrt{x + 3})^2 \implies x^2 - 6x + 9 = x + 3

Squaring both sides converts the radical equation into a standard quadratic equation.

2

Rearrange the terms into standard quadratic form

ax^2 + bx + c = 0

and factor.

x^2 - 7x + 6 = 0 \implies (x - 6)(x - 1) = 0

Subtracting

x

and

3

from both sides isolates zero on one side so the quadratic expression can be factored.

3

Solve for potential values of

x

.

x = 6

or

x = 1

Setting each factor to zero yields the algebraic solutions to the quadratic equation.

4

Check each potential solution in the original radical equation

x - 3 = \sqrt{x + 3}

.

For

x = 6

:

6 - 3 = 3

and

\sqrt{6 + 3} = 3

(Valid). For

x = 1

:

1 - 3 = -2

and

\sqrt{1 + 3} = 2

(Invalid, since

-2 \neq 2

).

Squaring an equation can introduce extraneous solutions that do not satisfy the original principal square root relationship.

5

Calculate the sum of all valid real solutions.

The only valid real solution is

6

, so the sum is

6

.

Extraneous solutions must be discarded before computing the final sum.

Key Concept

Solving radical equations using quadratic factoring and identifying extraneous solutions

Estimated Time:2m 0s

Question 345Question →

A train travels from Station A to Station B, a distance of 180 miles, at a constant speed of $v$ miles per hour. On the return trip from Station B to Station A, the train travels the first half of the distance at a constant speed that is $20\%$ less than $v$ , and the remaining half of the distance at a constant speed that is $25\%$ greater than $v$ . If the total time for the return trip is 6 minutes longer than the total time for the trip from Station A to Station B, what is the value of $v$ ?

Show answer & explanation

Answer: 45

Answer

The value of

v

is

45

.

The outbound travel time for 180 miles at speed

v

is

\frac{180}{v}

hours. On the return trip, the first 90 miles at speed

0.80v

require

\frac{90}{0.80v} = \frac{112.5}{v}

hours, while the second 90 miles at speed

1.25v

require

\frac{90}{1.25v} = \frac{72}{v}

hours. The total return duration is

\frac{112.5 + 72}{v} = \frac{184.5}{v}

hours. Setting the difference between the return time and outbound time equal to 6 minutes (

0.1

hours) gives

\frac{184.5}{v} - \frac{180}{v} = 0.1

, which simplifies to

\frac{4.5}{v} = 0.1

, yielding

v = 45

.

Step-by-Step Solution

1

Write the expression for the outbound trip duration in terms of

v

.

T_{\text{outbound}} = \frac{180}{v}

hours.

Time is equal to total distance divided by constant speed.

2

Calculate the duration for each half of the return trip in terms of

v

.

The first 90 miles take

\frac{90}{0.80v} = \frac{112.5}{v}

hours, and the second 90 miles take

\frac{90}{1.25v} = \frac{72}{v}

hours, giving a total return duration of

\frac{184.5}{v}

hours.

The return trip consists of two 90-mile segments driven at

0.80v

and

1.25v

respectively.

3

Equate the difference between return and outbound times to 0.1 hours and solve for

v

.

\frac{184.5}{v} - \frac{180}{v} = 0.1 \implies \frac{4.5}{v} = 0.1 \implies v = 45

.

The time difference of 6 minutes is equal to

\frac{6}{60} = 0.1

hours.

Key Concept

Distance, Rate, and Time Modeling with Piecewise Speed Changes

Question 346Question →

In 2023, the city of Alton introduced a high-capacity organic waste composting program that provided all households with specialized food scrap collection bins. Municipal authorities projected a significant decrease in the total tonnage of municipal solid waste sent to local landfills. However, official waste management reports from 2025 revealed that despite a 90 percent household participation rate in the composting program, the total tonnage of municipal waste deposited in local landfills actually increased by 12 percent over the two-year period.

Which of the following, if true, help to resolve the apparent discrepancy described above? Select all that apply.

Select all that apply

Concurrently with the residential initiative, a major expansion in regional industrial manufacturing generated a massive volume of non-organic solid waste that far surpassed the tonnage of residential food scraps diverted from landfills.

The municipal government spent significantly more money operating the new composting facility than it had originally budgeted for the program.

Prior to the composting program, the vast majority of households routinely ground their food scraps in sink garbage disposals rather than placing them in standard trash bins.

Surrounding municipalities that lacked organic composting programs also experienced an increase in landfill waste during the same two-year period.

The majority of residents reported that using the new food scrap bins required less effort than separating plastic and glass recyclables.

Show answer & explanation

Answer: Concurrently with the residential initiative, a major expansion in regional industrial manufacturing generated a massive volume of non-organic solid waste that far surpassed the tonnage of residential food scraps diverted from landfills.; Prior to the composting program, the vast majority of households routinely ground their food scraps in sink garbage disposals rather than placing them in standard trash bins.

Answer

The statements explaining that an expansion in industrial manufacturing generated non-organic waste exceeding residential scrap diversion, and that households previously disposed of food scraps via sink garbage disposals rather than landfill trash bins, together resolve the discrepancy.

To resolve the paradox, an option must explain how landfill waste tonnage increased even though 90% of households diverted food scraps. The statement about expanding industrial manufacturing identifies an external waste source whose massive volume offset residential scrap diversion. The statement about household garbage disposals shows that food scraps were previously washed down sinks rather than put in landfill trash, meaning their diversion did not reduce landfill volume. Both statements allow both facts to be true.

Step-by-Step Solution

1

Identify the two contradictory facts in the passage.

Fact 1: 90% of households participated in diverting food scraps into composting bins.
Fact 2: Total tonnage of municipal solid waste sent to landfills increased by 12% rather than decreasing.

Resolving a paradox requires finding statements that allow both facts to remain true simultaneously.

2

Evaluate statements to see which provide a viable mechanism for landfill waste increasing despite high residential food scrap diversion.

The statement regarding industrial manufacturing introduces a separate, non-residential waste stream that grew enough to outweigh food scrap diversion. The statement regarding sink garbage disposals reveals that diverted food scraps were never in the landfill waste stream to begin with.

Both statements account for why landfill waste tonnage rose without denying high residential participation in composting.

3

Eliminate choices that fail to resolve the paradox.

Statements about operating budgets, parallel regional trends, and resident convenience ratings do not explain why landfill waste volume increased.

Distractors bring in out-of-scope details, correlate regional trends without causal explanation, or make unwarranted assumptions.

Key Concept

Resolving Paradoxes and Discrepancies

Question 347Question →

An agricultural research station evaluated $300$ soil plots for the presence of three specific mineral deficiencies: Nitrogen ( $N$ ), Phosphorus ( $P$ ), and Potassium ( $K$ ). The survey revealed the following data:
- Exactly $60$ plots exhibited none of the three deficiencies.
- $140$ plots exhibited Nitrogen deficiency ( $N$ ).
- $130$ plots exhibited Phosphorus deficiency ( $P$ ).
- $120$ plots exhibited Potassium deficiency ( $K$ ).
- Exactly $70$ plots exhibited exactly two of the three deficiencies.

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

Select all that apply

Exactly

40

plots exhibited all three mineral deficiencies.

Exactly

130

plots exhibited exactly one mineral deficiency.

The number of plots exhibiting at least two mineral deficiencies is

110

.

Exactly

170

plots exhibited at most two mineral deficiencies.

The number of plots with Nitrogen deficiency only is

70

.

Show answer & explanation

Answer: Exactly $40$ plots exhibited all three mineral deficiencies.; Exactly $130$ plots exhibited exactly one mineral deficiency.; The number of plots exhibiting at least two mineral deficiencies is $110$ .

Answer

The statements confirming that exactly 40 plots exhibited all three deficiencies, exactly 130 plots exhibited exactly one deficiency, and 110 plots exhibited at least two deficiencies are all correct.

The system of set equations shows that

n_3 = 40

plots have all three deficiencies,

n_1 = 130

plots have exactly one deficiency, and

n_2 + n_3 = 70 + 40 = 110

plots have at least two deficiencies. Therefore, the statements asserting

40

plots for all three deficiencies,

130

plots for exactly one deficiency, and

110

plots for at least two deficiencies are all guaranteed to be true.

Step-by-Step Solution

1

Determine the total number of plots exhibiting at least one deficiency.

The total number of plots with at least one deficiency is

300 - 60 = 240

.

Plots with no deficiencies are excluded from the set union

|N \cup P \cup K|

.

2

Set up the inclusion-exclusion equations for set membership.

Let

n_1

be the number of plots with exactly one deficiency,

n_2 = 70

be the number of plots with exactly two deficiencies, and

n_3

be the number of plots with all three deficiencies.
Equation 1 (Total elements in union):

n_1 + n_2 + n_3 = 240 \implies n_1 + 70 + n_3 = 240 \implies n_1 + n_3 = 170

.
Equation 2 (Sum of individual set cardinalities):

|N| + |P| + |K| = n_1 + 2n_2 + 3n_3 \implies 140 + 130 + 120 = n_1 + 2(70) + 3n_3 \implies 390 = n_1 + 140 + 3n_3 \implies n_1 + 3n_3 = 250

.

Each element in an individual set sum is counted once for single-set membership, twice for double-set membership, and three times for triple-set membership.

3

Solve the system of linear equations for

n_1

and

n_3

.

Subtract Equation 1 from Equation 2:

(n_1 + 3n_3) - (n_1 + n_3) = 250 - 170 \implies 2n_3 = 80 \implies n_3 = 40

.
Substitute

n_3 = 40

into Equation 1:

n_1 + 40 = 170 \implies n_1 = 130

.

This yields

40

plots with all three deficiencies and

130

plots with exactly one deficiency.

4

Evaluate each option statement against the computed set cardinalities.

1) All three deficiencies (

n_3

) =

40

(True).
2) Exactly one deficiency (

n_1

) =

130

(True).
3) At least two deficiencies (

n_2 + n_3

) =

70 + 40 = 110

(True).
4) At most two deficiencies = Total -

n_3 = 300 - 40 = 260 \neq 170

(False).
5) Nitrogen deficiency only depends on how the

70

dual-deficiency plots are split between

N \cap P

,

N \cap K

, and

P \cap K

, which is not uniquely determined by the given data (False).

Only statements meeting the exact numerical bounds and logical guarantees must be true.

Key Concept

Three-Set Principle of Inclusion-Exclusion and Partitioning into Disjoint Regions

Question 348Question →

An agricultural research station evaluated a sample of $120$ organic farm plots for contamination by three types of crop fungi: *Fusarium* ( $F$ ), *Rhizoctonia* ( $R$ ), and *Sclerotinia* ( $S$ ). The audit revealed the following data:

- $55$ plots were contaminated with *Fusarium*.
- $50$ plots were contaminated with *Rhizoctonia*.
- $45$ plots were contaminated with *Sclerotinia*.
- $20$ plots were contaminated with both *Fusarium* and *Rhizoctonia*.
- $15$ plots were contaminated with both *Rhizoctonia* and *Sclerotinia*.
- $18$ plots were contaminated with both *Fusarium* and *Sclerotinia*.
- $8$ plots were contaminated with all three fungi.

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

Select all that apply

Exactly 15 farm plots were free from all three types of crop fungi.

Exactly 29 farm plots were contaminated by exactly two of the three crop fungi.

Exactly 37 farm plots were contaminated by *Fusarium* only.

Exactly 53 farm plots were contaminated by at least two of the three crop fungi.

Exactly 50 percent of all plots contaminated by *Rhizoctonia* were also contaminated by *Fusarium*.

Show answer & explanation

Answer: Exactly 15 farm plots were free from all three types of crop fungi.; Exactly 29 farm plots were contaminated by exactly two of the three crop fungi.

Answer

The correct statements are that exactly 15 farm plots were free from all three types of crop fungi, and exactly 29 farm plots were contaminated by exactly two of the three crop fungi.

The statement regarding 15 farm plots being free from all three fungi is correct because the total number of plots contaminated by at least one fungus is calculated using inclusion-exclusion: 55 + 50 + 45 - 20 - 15 - 18 + 8 = 105, leaving 120 - 105 = 15 plots clean. The statement regarding 29 farm plots contaminated by exactly two fungi is correct because summing the mutually exclusive regions containing exactly two fungi yields (20 - 8) + (15 - 8) + (18 - 8) = 12 + 7 + 10 = 29.

Step-by-Step Solution

1

Calculate the non-overlapping regions for each intersection.

Determining exact two-set regions requires removing elements that belong to all three sets.

2

Calculate single-fungus-only regions.

|F only| = 55 - (12 + 10 + 8) = 25.
|R only| = 50 - (12 + 7 + 8) = 23.
|S only| = 45 - (10 + 7 + 8) = 20.

Subtract all double and triple overlaps from total individual set counts.

3

Determine union of all three sets and the neither region.

Total contaminated = 25 + 23 + 20 + 12 + 7 + 10 + 8 = 105.
Plots free from all fungi = 120 - 105 = 15.

Subtracting the total contaminated plots from the overall sample size gives the plots outside all sets.

4

Verify statement validity.

Plots free of all fungi = 15 (Valid).
Plots with exactly two fungi = 12 + 7 + 10 = 29 (Valid).

Match calculated regional counts against the presented options.

Key Concept

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

Question 349Question →

In right triangle $ABC$ , the measure of $\angle B$ is $90^\circ$ , the measure of $\angle A$ is $30^\circ$ , and hypotenuse $AC = 12$ . Point $D$ lies on hypotenuse $AC$ such that $BD$ is perpendicular to $AC$ , and point $E$ lies on leg $BC$ such that $DE$ is perpendicular to $BC$ . What is the area of quadrilateral $ABDE$ ?

\frac{135\sqrt{3}}{8}

\frac{117\sqrt{3}}{8}

\frac{27\sqrt{3}}{2}

\frac{63\sqrt{3}}{4}

\frac{144\sqrt{3} - 9}{8}

Show answer & explanation

Answer: $\frac{135\sqrt{3}}{8}$

Answer

The area of quadrilateral

ABDE

is

\frac{135\sqrt{3}}{8}

.

The area of quadrilateral

ABDE

is obtained by subtracting the area of the smaller right triangle

\triangle DEC

from the area of the outer right triangle

\triangle ABC

. Since both triangles share the

60^\circ

angle at vertex

C

, repeated application of the

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

ratio (

1:\sqrt{3}:2

) gives

\text{Area}(\triangle ABC) = 18\sqrt{3}

and

\text{Area}(\triangle DEC) = \frac{9\sqrt{3}}{8}

. Subtraction yields

\frac{135\sqrt{3}}{8}

.

Step-by-Step Solution

1

Calculate side lengths and area of

\triangle ABC

.

BC = 6

,

AB = 6\sqrt{3}

, and

\text{Area}(\triangle ABC) = 18\sqrt{3}

.

In

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

triangle

ABC

with hypotenuse

AC = 12

, the side opposite

30^\circ

(

BC

) is half the hypotenuse (

BC = 6

), and the side opposite

60^\circ

(

AB

) is

6\sqrt{3}

. The area is

\frac{1}{2} \times 6 \times 6\sqrt{3} = 18\sqrt{3}

.

2

Find the hypotenuse

DC

of

\triangle BDC

.

DC = 3

.

Since

BD \perp AC

,

\triangle BDC

is a right triangle with

\angle BDC = 90^\circ

and

\angle C = 60^\circ

, making

\angle DBC = 30^\circ

. Its hypotenuse is

BC = 6

. The side opposite

30^\circ

is

DC = \frac{1}{2} BC = 3

.

3

Determine the sides and area of

\triangle DEC

.

EC = \frac{3}{2}

,

DE = \frac{3\sqrt{3}}{2}

, and

\text{Area}(\triangle DEC) = \frac{9\sqrt{3}}{8}

.

Since

DE \perp BC

,

\triangle DEC

is another

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

right triangle with hypotenuse

DC = 3

. Thus

EC = \frac{3}{2}

and

DE = \frac{3\sqrt{3}}{2}

. Its area is

\frac{1}{2} \times \frac{3}{2} \times \frac{3\sqrt{3}}{2} = \frac{9\sqrt{3}}{8}

.

4

Subtract

\text{Area}(\triangle DEC)

from

\text{Area}(\triangle ABC)

to get the area of quadrilateral

ABDE

.

\text{Area}(ABDE) = 18\sqrt{3} - \frac{9\sqrt{3}}{8} = \frac{135\sqrt{3}}{8}

.

Quadrilateral

ABDE

is formed by removing

\triangle DEC

from

\triangle ABC

.

Key Concept

Iterative application of

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

special right triangle side ratios (

1:\sqrt{3}:2

) along dropped altitudes.

Estimated Time:2m 30s

Question 350Question →

Pipeline A operating alone can fill a storage tank in $x$ hours, whereas Pipeline B operating alone takes $50\%$ longer than Pipeline A to fill the same tank. Drainage Pipe C operating alone can empty a full tank in $2x$ hours. If all three pipes are opened simultaneously when the tank is empty, the tank becomes completely full in $12$ hours. What is the value of $x$ ?

6

12

14

24

26

Show answer & explanation

Answer: $14$

Answer

The value of

x

is

14

.

To find

x

, calculate the hourly rate of each pipe: Pipeline A fills

\frac{1}{x}

of the tank per hour, Pipeline B takes

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

hours so it fills

\frac{2}{3x}

per hour, and Pipe C empties

\frac{1}{2x}

per hour. The combined rate is

\frac{1}{x} + \frac{2}{3x} - \frac{1}{2x} = \frac{7}{6x}

. Since the tank fills in

12

hours, the net rate is

\frac{1}{12}

. Setting

\frac{7}{6x} = \frac{1}{12}

gives

6x = 84

, so

x = 14

.

Step-by-Step Solution

1

Express the individual work rates per hour in terms of

x

.

Pipeline A rate =

\frac{1}{x}

, Pipeline B time =

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

Pipeline B rate =

\frac{2}{3x}

, Pipe C rate =

-\frac{1}{2x}

.

Work rate is the reciprocal of the total time required to complete the job alone, with drainage represented as a negative rate.

2

Formulate the equation for the combined rate of all three pipes operating together.

Combined Rate =

\frac{1}{x} + \frac{2}{3x} - \frac{1}{2x} = \frac{1}{x} \left(1 + \frac{2}{3} - \frac{1}{2}\right) = \frac{1}{x} \left(\frac{6 + 4 - 3}{6}\right) = \frac{7}{6x}

.

Simultaneous operation means summing the individual rates.

3

Equate the combined rate to the overall rate required to fill the tank in 12 hours and solve for

x

.

\frac{7}{6x} = \frac{1}{12} \implies 6x = 84 \implies x = 14.

The tank is filled in 12 hours, so the net rate per hour is

\frac{1}{12}

.

Key Concept

Combined Work Rates with Inflow and Outflow

Estimated Time:2m 0s

Question 351Question →

An isosceles triangle has two sides of length $10$ units each and a base of length $12$ units. A line segment parallel to the base cuts through the triangle, creating a smaller top triangle with an area of $12$ square units. What is the perimeter of the resulting trapezoid?

Show answer & explanation

Answer: 28

Answer

The perimeter of the trapezoid is 28.

First, find the height of the original isosceles triangle with sides 10, 10, and base 12 by applying the Pythagorean theorem to half of the base:

h = \sqrt{10^2 - 6^2} = 8

. The area of the original triangle is

\frac{1}{2} \times 12 \times 8 = 48

square units. Because the segment is parallel to the base, the smaller top triangle is similar to the original triangle. The ratio of their areas is

\frac{12}{48} = \frac{1}{4}

, which means the linear scale factor is

\sqrt{\frac{1}{4}} = \frac{1}{2}

. Thus, the top triangle has legs of length

5

and a base of length

6

. The remaining non-parallel sides of the trapezoid each measure

10 - 5 = 5

units, and its bottom base is

12

units. Summing these four side lengths gives

5 + 5 + 6 + 12 = 28

.

Step-by-Step Solution

1

Calculate the height and area of the original isosceles triangle

The altitude to the base bisects the base into two segments of length

6

. The altitude length is

h = \sqrt{10^2 - 6^2} = 8

. The area of the original triangle is

\frac{1}{2} \times 12 \times 8 = 48

square units.

Splitting the isosceles triangle along its altitude creates two right triangles with hypotenuse 10 and base leg 6.

2

Determine the linear scale factor of the smaller top triangle

The ratio of the area of the smaller triangle to the original triangle is

\frac{12}{48} = \frac{1}{4}

. Taking the square root yields a linear scale factor of

k = \sqrt{\frac{1}{4}} = \frac{1}{2}

.

A line parallel to the base forms a smaller triangle similar to the original triangle, and the ratio of areas of similar triangles is equal to the square of their linear scale factor.

3

Find the side lengths of the smaller triangle and the remaining side segments

The sides of the smaller triangle are

\frac{1}{2} \times 10 = 5

,

\frac{1}{2} \times 10 = 5

, and base

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

. The non-parallel side segments of the trapezoid are each

10 - 5 = 5

.

Multiplying the dimensions of the original triangle by the linear scale factor gives the side lengths of the top triangle.

4

Calculate the perimeter of the trapezoid

Perimeter =

5 + 5 + 6 + 12 = 28

.

Sum the lengths of the four boundary segments of the trapezoid.

Key Concept

Properties of isosceles triangles, Pythagorean theorem, area calculations, and similar triangle area ratios

Question 352Question →

An urban planning department surveyed a total of $360$ commercial buildings regarding three sustainability certifications: LEED ( $L$ ), BREEAM ( $B$ ), and WELL ( $W$ ). Exactly $50$ of the surveyed buildings held none of the three certifications. The survey revealed the following data:

• $180$ buildings held the LEED certification.
• $140$ buildings held the BREEAM certification.
• $130$ buildings held the WELL certification.
• $45$ buildings held both the LEED and BREEAM certifications.
• $40$ buildings held both the BREEAM and WELL certifications.
• $25$ buildings held all three certifications.

How many of the surveyed buildings held EXACTLY TWO of the three certifications?

Show answer & explanation

Answer: 90

Answer

90

To find the number of buildings with exactly two certifications, we first determine the total number of buildings holding at least one certification:

360 - 50 = 310

. Applying the three-set inclusion-exclusion formula,

310 = 180 + 140 + 130 - (45 + 40 + |L \cap W|) + 25

, which simplifies to

310 = 390 - |L \cap W|

, giving

|L \cap W| = 80

. Next, we isolate the regions holding exactly two certifications by subtracting the 25 triple-certified buildings from each pairwise intersection: LEED & BREEAM only =

45 - 25 = 20

, BREEAM & WELL only =

40 - 25 = 15

, and LEED & WELL only =

80 - 25 = 55

. Summing these three exclusive regions gives

20 + 15 + 55 = 90

.

Step-by-Step Solution

1

Determine the total size of the union of all three sets

|L \cup B \cup W| = 360 - 50 = 310

Buildings holding at least one certification represent the entire surveyed population minus those holding no certifications.

2

Solve for the unknown overlap of LEED and WELL certifications using the inclusion-exclusion formula

|L \cap W| = 80

Using

|L \cup B \cup W| = |L| + |B| + |W| - |L \cap B| - |B \cap W| - |L \cap W| + |L \cap B \cap W|

, we have

310 = 180 + 140 + 130 - 45 - 40 - |L \cap W| + 25 = 390 - |L \cap W|

.

3

Calculate the number of buildings in each region corresponding to exactly two certifications

LEED and BREEAM only:

45 - 25 = 20

; BREEAM and WELL only:

40 - 25 = 15

; LEED and WELL only:

80 - 25 = 55

Each total pairwise intersection includes the 25 buildings that hold all three certifications, so subtracting 25 isolates those holding exclusively two certifications.

4

Sum the three isolated regions

20 + 15 + 55 = 90

The total number of buildings holding exactly two certifications is the sum of the three non-overlapping regions representing two certifications.

Key Concept

Three-Set Inclusion-Exclusion Principle and Partitioning Venn Diagrams

Estimated Time:2m 0s

Question 353Question →

Events $E$ and $F$ are two events in a sample space such that $P(E) = 0.30$ and $P(F) = 0.40$ . Which of the following statements must be true? Select all that apply.

Select all that apply

If

E

and

F

are mutually exclusive, then

P(E \text{ and } F) = 0

.

If

E

and

F

are independent, then

P(E \text{ and } F) = 0.12

.

If

E

and

F

are independent, then

P(E \text{ or } F) = 0.70

.

If

E

and

F

are mutually exclusive, then

E

and

F

are independent.

If

E

and

F

are independent, then

P(E \text{ given } F) = 0.12

.

Show answer & explanation

Answer: If $E$ and $F$ are mutually exclusive, then $P(E \text{ and } F) = 0$ .; If $E$ and $F$ are independent, then $P(E \text{ and } F) = 0.12$ .

Answer

The correct statements are that if

E

and

F

are mutually exclusive, then

P(E \text{ and } F) = 0

, and if

E

and

F

are independent, then

P(E \text{ and } F) = 0.12

.

The statement asserting that

P(E \text{ and } F) = 0

for mutually exclusive events is correct because mutually exclusive events by definition cannot occur together. The statement asserting that

P(E \text{ and } F) = 0.12

for independent events is correct because independent events satisfy the multiplication rule

P(E \text{ and } F) = P(E) \times P(F) = 0.30 \times 0.40 = 0.12

.

Step-by-Step Solution

1

Evaluate the statement regarding mutually exclusive events.

P(E \text{ and } F) = 0

.

By definition, mutually exclusive events cannot both happen at the same time, so

P(E \text{ and } F) = 0

is true.

2

Evaluate the joint probability for independent events.

P(E \text{ and } F) = P(E) \times P(F) = 0.30 \times 0.40 = 0.12

.

The multiplication rule for independent events states that

P(E \text{ and } F) = P(E) \cdot P(F)

.

3

Evaluate the union probability

P(E \text{ or } F)

for independent events.

P(E \text{ or } F) = P(E) + P(F) - P(E \text{ and } F) = 0.30 + 0.40 - 0.12 = 0.58

.

Simply adding

P(E) + P(F) = 0.70

fails to subtract the intersection that is counted twice.

4

Evaluate conditional probability for independent events.

P(E \text{ given } F) = P(E) = 0.30

.

Independence implies that knowing event

F

occurred does not change the probability of event

E

.

Key Concept

Probability rules for mutually exclusive and independent events

Question 354Question →

For all non-zero real numbers

u

and

v

, the custom operation

\Delta

is defined by

u \Delta v = \frac{u}{v} - \frac{v}{u}

The function

f

is defined for all non-zero real numbers

x

by

f(x) = x \Delta 2

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

a

and

b

? Select all such statements.

Select all that apply

a \Delta b = -(b \Delta a)

f(-a) = -f(a)

f(a^2) = (f(a))^2

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

Show answer & explanation

Answer: $a \Delta b = -(b \Delta a)$ ; $f(-a) = -f(a)$

Answer

The statements

a \Delta b = -(b \Delta a)

and

f(-a) = -f(a)

must be true.

The custom operation is anti-commutative because swapping variables negates the difference of the fractions, making the statement

a \Delta b = -(b \Delta a)

true. Additionally, substituting

-a

into

f(x) = \frac{x}{2} - \frac{2}{x}

yields

-f(a)

, showing

f(-a) = -f(a)

is also true.

Step-by-Step Solution

1

Analyze the first identity

a \Delta b = -(b \Delta a)

.

a \Delta b = \frac{a}{b} - \frac{b}{a}

and

b \Delta a = \frac{b}{a} - \frac{a}{b} = -\left(\frac{a}{b} - \frac{b}{a}\right) = -(a \Delta b)

.

Factoring

-1

out of

b \Delta a

proves anti-commutativity.

2

Evaluate

f(x)

explicitly and test

f(-a)

.

f(x) = \frac{x}{2} - \frac{2}{x}

. Substituting

-a

gives

f(-a) = \frac{-a}{2} - \frac{2}{-a} = -\left(\frac{a}{2} - \frac{2}{a}\right) = -f(a)

.

This confirms that

f

is an odd function.

3

Compare

f(a^2)

with

(f(a))^2

.

f(a^2) = \frac{a^2}{2} - \frac{2}{a^2}

, but

(f(a))^2 = \frac{a^2}{4} - 2 + \frac{4}{a^2}

. Testing

a = 1

yields

f(1) = -1.5 \implies (f(1))^2 = 2.25

, whereas

f(1^2) = -1.5 \neq 2.25

.

Counterexample shows the equality does not hold in general.

4

Compare

f(a+b)

with

f(a) + f(b)

.

For

a=1, b=1

,

f(1+1) = f(2) = \frac{2}{2} - \frac{2}{2} = 0

, but

f(1) + f(1) = -1.5 + (-1.5) = -3 \neq 0

.

Functions do not in general preserve additivity

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

unless they are linear through the origin.

Key Concept

Properties of algebraic custom operations and odd functions

Question 355Question →

If $x$ is a real number that satisfies the inequality $\left|\frac{5 - 2x}{3}\right| \le 3$ , what is the minimum possible value of $4 - 3x$ ?

-23

-17

-2

10

25

Show answer & explanation

Answer: $-17$

Answer

-17

To find the minimum possible value of

4 - 3x

, we first solve the inequality

\left|\frac{5 - 2x}{3}\right| \le 3

. Multiplying by

3

gives

|5 - 2x| \le 9

. This unfolds into the compound inequality

-9 \le 5 - 2x \le 9

. Subtracting

5

gives

-14 \le -2x \le 4

. Dividing all parts by

-2

requires flipping the inequality signs, resulting in

7 \ge x \ge -2

, or

-2 \le x \le 7

. Because

4 - 3x

decreases as

x

increases, the expression reaches its minimum when

x

is at its maximum value of

7

. Substituting

x = 7

yields

4 - 3(7) = -17

.

Step-by-Step Solution

1

Clear the denominator from the absolute value inequality

|5 - 2x| \le 9

Multiplying both sides of the inequality by the positive number 3 preserves the inequality direction.

2

Express the absolute value inequality as a compound inequality

-9 \le 5 - 2x \le 9

For any non-negative constant

c

,

|u| \le c

is equivalent to

-c \le u \le c

.

3

Isolate the variable term by subtracting 5 from all parts

-14 \le -2x \le 4

Subtracting a constant from all parts of a compound inequality maintains the inequality relationships.

4

Divide by -2 and reverse the inequality signs

7 \ge x \ge -2 \implies -2 \le x \le 7

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

5

Determine which bound of

x

minimizes

4 - 3x

and evaluate

Minimum value

= 4 - 3(7) = -17

The expression

4 - 3x

has a negative coefficient for

x

, making it a decreasing function. Therefore, the minimum value of

4 - 3x

occurs when

x

takes its maximum possible value (

x = 7

).

Key Concept

Linear Inequalities and Absolute Value

Question 356Question →

A logistics company packages cargo using three types of containers: small, medium, and large.

- A shipment of $3$ small, $2$ medium, and $1$ large container has a total weight of $130$ kilograms.
- A shipment of $1$ small, $4$ medium, and $2$ large containers has a total weight of $185$ kilograms.
- A shipment of $2$ small, $1$ medium, and $3$ large containers has a total weight of $160$ kilograms.

What is the weight, in kilograms, of one large container?

Show answer & explanation

Answer: 35

Answer

The weight of one large container is 35 kilograms.

Representing the weights of small, medium, and large containers as variables

S

,

M

, and

L

yields the 3x3 system of linear equations:
1)

3S + 2M + L = 130

2)

S + 4M + 2L = 185

3)

2S + M + 3L = 160

Solving for

S

in equation (2) gives

S = 185 - 4M - 2L

. Substituting this into equations (1) and (3) reduces the system to:
-

2M + L = 85

-

7M + L = 210

Subtracting the first equation from the second yields

5M = 125

, so

M = 25

. Substituting

M = 25

into

2M + L = 85

gives

50 + L = 85

, which simplifies to

L = 35

.

Step-by-Step Solution

1

Set up a system of three linear equations based on the shipment descriptions

3S + 2M + L = 130

,

S + 4M + 2L = 185

, and

2S + M + 3L = 160

Translate the physical constraints of the three shipments into mathematical relationships

2

Isolate variable

S

in the second equation and substitute into the first and third equations

Two equations in two variables:

2M + L = 85

and

7M + L = 210

Reduce the 3x3 system to a 2x2 system to eliminate variable

S

3

Subtract the two simplified equations to solve for

M

5M = 125 \implies M = 25

Eliminate variable

L

to obtain the value of

M

4

Substitute

M = 25

back into

2M + L = 85

to solve for

L

L = 35

Find the requested value for the weight of one large container

Key Concept

Solving 3x3 Systems of Linear Equations using Substitution and Elimination

Question 357Question →

A pharmaceutical laboratory produces a custom solution in a mixing tank using two automated pumps, Pump A and Pump B. Initially, the tank contains $600$ liters of a solution that is $20\%$ active reagent by volume. Pump A adds pure active reagent to the tank at a constant rate of $15$ liters per minute, while Pump B simultaneously adds distilled water ( $0\%$ active reagent) to the tank at a constant rate of $25$ liters per minute. At the exact instant when the mixture in the tank reaches $35\%$ active reagent by volume, Pump B is turned off while Pump A continues to add pure active reagent at $15$ liters per minute. How many total minutes, from the moment both pumps were initially started, does it take for the solution in the tank to reach $50\%$ active reagent by volume?

Show answer & explanation

Answer: 174

Answer

The total elapsed time is 174 minutes.

The problem requires setting up a two-stage rational rate model. In the first stage, initial reagent volume is

120

liters (

20\%

of

600

). With Pump A adding

15

L/min of reagent and Pump B adding

25

L/min of water, the mixture concentration after

t_1

minutes is

\frac{120 + 15t_1}{600 + 40t_1} = 0.35

, yielding

t_1 = 90

minutes. At

t_1 = 90

, the total volume is

4,200

liters and the reagent volume is

1,470

liters. In the second stage, only Pump A is active, so concentration after

t_2

additional minutes is

\frac{1470 + 15t_2}{4200 + 15t_2} = 0.50

, yielding

t_2 = 84

minutes. Summing both intervals gives

90 + 84 = 174

total minutes.

Step-by-Step Solution

1

Determine initial quantities of active reagent and total solution.

Initial total volume

V_0 = 600

liters. Initial volume of active reagent

A_0 = 0.20 \times 600 = 120

liters.

Establishing the baseline volume and solute concentration before any pumps are turned on.

2

Model Phase 1 where both pumps operate concurrently for

t_1

minutes to reach

35\%

concentration.

Total volume equation:

V(t_1) = 600 + (15 + 25)t_1 = 600 + 40t_1

. Reagent volume equation:

A(t_1) = 120 + 15t_1

. Set concentration to

0.35

:

\frac{120 + 15t_1}{600 + 40t_1} = 0.35 \implies t_1 = 90

minutes.

Both pumps add volume, but only Pump A adds active reagent. Solving

\frac{120 + 15t_1}{600 + 40t_1} = \frac{7}{20}

gives

20(120 + 15t_1) = 7(600 + 40t_1) \implies 2400 + 300t_1 = 4200 + 280t_1 \implies 20t_1 = 1800 \implies t_1 = 90

minutes.

3

Calculate the accumulated total volume and reagent volume at the end of Phase 1.

Total volume

V_1 = 600 + 40(90) = 4,200

liters. Active reagent volume

A_1 = 120 + 15(90) = 1,470

liters.

These amounts serve as the starting parameters for Phase 2.

4

Model Phase 2 where only Pump A operates for

t_2

additional minutes to reach

50\%

concentration.

In Phase 2, volume increases at

15

L/min and reagent increases at

15

L/min. Set concentration to

0.50

:

\frac{1470 + 15t_2}{4200 + 15t_2} = 0.50 \implies t_2 = 84

minutes.

Solving

2(1470 + 15t_2) = 4200 + 15t_2 \implies 2940 + 30t_2 = 4200 + 15t_2 \implies 15t_2 = 1260 \implies t_2 = 84

minutes.

5

Sum the durations of Phase 1 and Phase 2 to find the total time.

Total time

T = t_1 + t_2 = 90 + 84 = 174

minutes.

The question asks for the total minutes from the initial start of both pumps.

Key Concept

Multi-Phase Algebraic Rate and Mixture Modeling

Question 358Question →

In the $xy$ -plane, $A(1, 2)$ , $B(7, 4)$ , and $D(3, 8)$ are three vertices of rhombus $ABCD$ . A line $k$ passes through the origin $(0, 0)$ and bisects the area of rhombus $ABCD$ . What is the slope of line $k$ ?

\frac{6}{5}

\frac{5}{6}

1

-1

\frac{3}{4}

Show answer & explanation

Answer: $\frac{6}{5}$

Answer

The slope of line

k

is

\frac{6}{5}

.

Any line that divides a parallelogram or rhombus into two equal areas must pass through its center of symmetry, which is the midpoint of its diagonals. The midpoint of diagonal

BD

with endpoints

(7, 4)

and

(3, 8)

is

\left(\frac{7+3}{2}, \frac{4+8}{2}\right) = (5, 6)

. Since line

k

passes through the origin

(0, 0)

and

(5, 6)

, its slope is

\frac{6 - 0}{5 - 0} = \frac{6}{5}

.

Step-by-Step Solution

1

Identify the key geometric property of area-bisecting lines for parallelograms and rhombuses.

Any line that bisects the area of a rhombus must pass through its center of symmetry (the intersection point of its diagonals).

A rhombus is centrally symmetric about the intersection point of its diagonals, so any line through this point divides the rhombus into two congruent regions.

2

Find the coordinates of the center of symmetry by calculating the midpoint of diagonal

BD

.

Midpoint

M = \left(\frac{7+3}{2}, \frac{4+8}{2}\right) = (5, 6)

.

Opposite vertices

B(7,4)

and

D(3,8)

define one of the diagonals of rhombus

ABCD

.

3

Calculate the slope of line

k

passing through the origin

(0, 0)

and center point

M(5, 6)

.

\text{Slope } m = \frac{6 - 0}{5 - 0} = \frac{6}{5}.

The slope formula between

(x_1, y_1)

and

(x_2, y_2)

is

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

.

Key Concept

Center of Symmetry and Area Bisectors of Quadrilaterals

Estimated Time:2m 0s

Question 359Question →

In right trapezoid $ABCD$ , segment $AB$ is parallel to segment $CD$ , $\angle DAB = 90^\circ$ , $AD = 12$ , $CD = 15$ , and $BC = 13$ . Point $E$ lies on segment $CD$ such that quadrilateral $ABCE$ is a parallelogram. What is the perimeter of triangle $ADE$ ?

Show answer & explanation

Answer: 30

Answer

30

Decomposing right trapezoid

ABCD

by dropping a perpendicular from

B

to

CD

forms a right triangle with height

12

and hypotenuse

13

. The Pythagorean theorem gives the base of this right triangle as

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

. Subtracting this from

CD = 15

yields

AB = 10

. Because

ABCE

is a parallelogram,

CE = AB = 10

, which leaves

DE = CD - CE = 15 - 10 = 5

. Triangle

ADE

is a right triangle with legs

AD = 12

and

DE = 5

, giving hypotenuse

AE = \sqrt{12^2 + 5^2} = 13

. The perimeter of triangle

ADE

is

12 + 5 + 13 = 30

.

Step-by-Step Solution

1

Calculate the horizontal projection of segment

BC

onto base

CD

The length of the horizontal projection is

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

Segment

AD = 12

defines the perpendicular distance between parallel lines

AB

and

CD

2

Determine the length of parallel base

AB

AB = 15 - 5 = 10

The total length of base

CD = 15

is the sum of

AB

and the horizontal projection of slant side

BC

3

Calculate the length of segment

DE

DE = 15 - 10 = 5

Quadrilateral

ABCE

is a parallelogram, which implies

CE = AB = 10

4

Compute the hypotenuse

AE

and the total perimeter of triangle

ADE

AE = \sqrt{12^2 + 5^2} = 13

, so

\text{Perimeter} = 12 + 5 + 13 = 30

Triangle

ADE

is a right-angled triangle with right angle at vertex

D

Key Concept

Trapezoid height decomposition, parallelogram side properties, and Pythagorean theorem application

Question 360Question →

The frequency distribution table below summarizes the scores of 20 students on a statistics test.

Score	Number of Students
5	3
6	5
7	$x$
8	$y$
9	4

If the arithmetic mean score of the 20 students is $7.15$ , what is the median score of the 20 students?

6.5

7.0

7.15

7.5

8.0

Show answer & explanation

Answer: 7.5

Answer

The median score of the 20 students is 7.5.

To find the median score of 20 students, we first determine the missing student counts

x

and

y

. From the total count of 20 students,

x + y = 20 - (3 + 5 + 4) = 8

. Using the given mean of 7.15, the sum of all scores is

20 \times 7.15 = 143

. The sum expressed in terms of

x

and

y

is

3(5) + 5(6) + 7x + 8y + 4(9) = 81 + 7x + 8y

. Setting

81 + 7x + 8y = 143

gives

7x + 8y = 62

. Solving

x + y = 8

and

7x + 8y = 62

yields

x = 2

and

y = 6

. Listing scores in non-decreasing order, the 10th student scored 7 and the 11th student scored 8. Because the total number of students is even, the median is the average of these two middle values:

(7 + 8) / 2 = 7.5

.

Step-by-Step Solution

1

Set up an equation for the total number of students to find the sum of

x

and

y

.

3 + 5 + x + y + 4 = 20 \implies x + y + 12 = 20 \implies x + y = 8

.

The total frequency of all score categories must equal the total number of students (20).

2

Set up an equation for the sum of all scores using the given arithmetic mean.

Total sum

= 20 \times 7.15 = 143

. Also, total sum

= 3(5) + 5(6) + 7x + 8y + 4(9) = 81 + 7x + 8y

. Therefore,

81 + 7x + 8y = 143 \implies 7x + 8y = 62

.

The sum of all data values equals the mean multiplied by the total number of observations.

3

Solve the system of linear equations for

x

and

y

.

Multiply

x + y = 8

by 7 to get

7x + 7y = 56

. Subtract this from

7x + 8y = 62

to get

y = 6

. Then

x = 8 - 6 = 2

.

Determining the exact frequencies

x=2

and

y=6

is required to locate the middle values.

4

Determine the cumulative frequencies and find the 10th and 11th values.

Score 5: positions 1 to 3 (3 students); Score 6: positions 4 to 8 (5 students); Score 7: positions 9 to 10 (2 students); Score 8: positions 11 to 16 (6 students); Score 9: positions 17 to 20 (4 students). The 10th value is 7 and the 11th value is 8.

For an even number of data points (

N=20

), the median is the arithmetic mean of the

(N/2)

-th and

(N/2 + 1)

-th values.

5

Calculate the median score.

\text{Median} = \frac{7 + 8}{2} = 7.5

.

Average the two middle values.

Key Concept

Calculating the median from a frequency distribution table with missing counts by combining system of linear equations with cumulative frequency analysis.

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