All practice questions

Question 1761Question →

A security system requires a 4-digit pass code formed using the digits $1, 2, 3, 4, 5, 6,$ and $7$ , with no digit repeated within a code. If the first digit of the pass code must be an even number and the last digit must be an odd number, how many such distinct pass codes can be created?

120

240

300

504

840

Show answer & explanation

Answer: 240

Answer

240 pass codes

To form a 4-digit code with distinct digits from the set {1, 2, 3, 4, 5, 6, 7}: there are 3 options for the first digit (even: 2, 4, 6) and 4 options for the fourth digit (odd: 1, 3, 5, 7). Because the sets of even and odd numbers are disjoint, choosing the first digit does not affect the number of odd choices available for the fourth position. After placing these 2 digits, 5 digits remain from the original set of 7. The second position can be filled in 5 ways, and the third position in 4 ways. By the Fundamental Counting Principle, the total number of codes is 3 × 4 × 5 × 4 = 240.

Step-by-Step Solution

1

Identify the choices for the first digit (even restriction).

3 possible choices (2, 4, or 6).

The first digit must be even, and the available set contains three even digits: {2, 4, 6}.

2

Identify the choices for the last (fourth) digit (odd restriction).

4 possible choices (1, 3, 5, or 7).

The fourth digit must be odd, and the available set contains four odd digits: {1, 3, 5, 7}.

3

Determine the available choices for the remaining middle digits.

5 choices for the second digit and 4 choices for the third digit.

Two distinct digits have been selected for the first and fourth positions out of the 7 available digits, leaving 5 remaining digits. Since no digits may repeat, the second position has 5 choices and the third position has 4 choices.

4

Apply the Fundamental Counting Principle to compute the total number of pass codes.

3 × 4 × 5 × 4 = 240 distinct pass codes.

Multiplying the independent choices for each position gives the total valid arrangements.

Key Concept

Fundamental Counting Principle and Permutations with Positional Restrictions

Question 1762Question →

A tech company's quality assurance division needs to form a 5-member project panel selected from a pool of 6 software engineers and 4 hardware engineers. The panel must include at least 2 software engineers and at least 1 hardware engineer. However, 2 specific software engineers, Engineer $X$ and Engineer $Y$ , refuse to serve on the same panel together. How many different valid 5-member panels can be formed?

136

188

196

240

252

Show answer & explanation

Answer: 188

Answer

188 valid panels can be formed.

The total number of panels satisfying the software and hardware role distribution rules is 240. Among these, exactly 52 panels contain both Engineer X and Engineer Y. Subtracting 52 from 240 results in 188 valid panels.

Step-by-Step Solution

1

Determine valid software and hardware engineer compositions for a 5-member panel.

Three compositions satisfy the requirement of at least 2 software engineers (S) and at least 1 hardware engineer (H): (4S, 1H), (3S, 2H), and (2S, 3H).

Panels must have 5 total members adhering to the specified minimum headcount limits.

2

Calculate the total number of panels satisfying composition requirements without considering the exclusion restriction.

For (4S, 1H):

\binom{6}{4} \times \binom{4}{1} = 15 \times 4 = 60

. For (3S, 2H):

\binom{6}{3} \times \binom{4}{2} = 20 \times 6 = 120

. For (2S, 3H):

\binom{6}{2} \times \binom{4}{3} = 15 \times 4 = 60

. Total composition-valid panels =

60 + 120 + 60 = 240

.

Apply the combination formula

\binom{n}{r} = \frac{n!}{r!(n-r)!}

and the Fundamental Counting Principle.

3

Calculate the number of invalid panels that contain both Engineer X and Engineer Y.

If Engineer X and Engineer Y are both included (2 S), 3 remaining panel members must be selected from the remaining 4 software engineers and 4 hardware engineers. For (4S, 1H):

\binom{4}{2} \times \binom{4}{1} = 6 \times 4 = 24

. For (3S, 2H):

\binom{4}{1} \times \binom{4}{2} = 4 \times 6 = 24

. For (2S, 3H):

\binom{4}{0} \times \binom{4}{3} = 1 \times 4 = 4

. Total invalid panels =

24 + 24 + 4 = 52

.

Isolating combinations that contain both restricted engineers allows simple subtraction from the total.

4

Subtract invalid panels from the total composition-valid panels.

240 - 52 = 188

.

This leaves only the panels that satisfy both composition and exclusion rules.

Key Concept

Combinations with multi-group minimum constraints and pair exclusion

Question 1763Question →

A theater sells student tickets for $\$20$ each and adult tickets for $\$35$ each. For a specific performance, the theater sold a total of $150$ tickets and collected total revenue of $R$ dollars. If at least $40$ student tickets were sold and at most $90$ adult tickets were sold, which of the following values could be the total revenue $R$ ? Select all such values.

Select all that apply

\$2{}850

\$3{}450

\$4{}200

\$4{}400

\$4{}650

Show answer & explanation

Answer: $\$3{}450$ ; $\$4{}200$

Answer

The possible values for the total revenue

R

are

\$3{}450

and

\$4{}200

.

The linear revenue model is

R = 5250 - 15s

. Considering both conditions (

s \ge 40

and

a = 150 - s \le 90 \implies s \ge 60

), the valid range for student tickets is

60 \le s \le 150

. This restricts the possible revenue

R

to multiples of

\$15

between

\$3{}000

and

\$4{}350

. Both

\$3{}450

and

\$4{}200

fall within this valid range and correspond to integer ticket quantities (

s = 120, a = 30

and

s = 70, a = 80

, respectively).

Step-by-Step Solution

1

Set up equations for the total number of tickets and revenue.

Let

s

be the number of student tickets and

a

be the number of adult tickets. Then

s + a = 150

, so

a = 150 - s

. Total revenue

R = 20s + 35a = 20s + 35(150 - s) = 5250 - 15s

.

Expressing revenue in terms of a single variable

s

simplifies finding the domain and range.

2

Determine the constraints on the variable

s

.

We are given

s \ge 40

and

a \le 90

. Substituting

a = 150 - s \le 90

yields

s \ge 60

. Combining constraints gives

60 \le s \le 150

.

The number of adult tickets being at most

90

forces the number of student tickets to be at least

60

.

3

Calculate the upper and lower bounds for the revenue

R

.

Maximum revenue occurs when

s = 60

:

R_{\text{max}} = 5250 - 15(60) = \$4{}350

. Minimum revenue occurs when

s = 150

:

R_{\text{min}} = 5250 - 15(150) = \$3{}000

.

Since

R = 5250 - 15s

is a decreasing linear function of

s

, the maximum revenue occurs at the minimum valid value of

s

and vice versa.

4

Evaluate the given choices against the range and divisibility requirements.

R

must be an integer multiple of

15

subtracted from

5250

, meaning

R

must be between

\$3{}000

and

\$4{}350

inclusive, and

(5250 - R)

must be divisible by

15

.

\$3{}450

(where

s = 120

) and

\$4{}200

(where

s = 70

) are both valid.

\$2{}850

is below the minimum bound,

\$4{}400

does not yield an integer value for

s

, and

\$4{}650

violates the adult ticket upper bound.

Only options meeting both inequality constraints and integer ticket requirements are valid.

Key Concept

Linear modeling of word problems under linear system constraints and inequalities

Estimated Time:1m 30s

Question 1764Question →

A city planning board needs to form a $6$ -member advisory task force selected from a pool of $5$ architects and $5$ civil engineers. The task force must include at least $2$ architects and at least $2$ civil engineers. However, two specific architects, Architect X and Architect Y, cannot both serve on the task force together. How many different $6$ -member task forces can be formed satisfying these conditions?

Show answer & explanation

Answer: 135

Answer

The total number of different valid 6-member task forces that can be formed is 135.

To solve this problem, we apply the addition principle over mutually exclusive cases of committee composition, followed by complementary counting to enforce the exclusion restriction. First, we identify the valid breakdown of architects and engineers for a 6-member team requiring at least 2 of each profession: (4 architects, 2 engineers), (3 architects, 3 engineers), and (2 architects, 4 engineers). Calculating the combinations for each breakdown yields 50, 100, and 50 ways respectively, totaling 200 unconstrained team options. Next, we determine how many of these teams include both Architect X and Architect Y. Pre-assigning both architects reduces the remaining available architects to 3. The invalid cases for each breakdown are 30, 30, and 5 respectively, totaling 65 invalid configurations. Subtracting the 65 invalid teams from the 200 total unconstrained teams yields 135 valid task forces.

Step-by-Step Solution

1

Identify the allowed group breakdowns under the restriction of at least 2 architects and at least 2 engineers.

The valid (architect, engineer) count pairs for a 6-member task force are (4, 2), (3, 3), and (2, 4).

Choosing 5 architects would leave only 1 engineer, violating the minimum requirement of 2 engineers, and vice versa.

2

Compute the total combinations without the exclusion restriction.

Total unconstrained combinations = 200.

\binom{5}{4}\binom{5}{2} + \binom{5}{3}\binom{5}{3} + \binom{5}{2}\binom{5}{4} = (5 \times 10) + (10 \times 10) + (10 \times 5) = 50 + 100 + 50 = 200

.

3

Calculate the number of task forces that violate the restriction by including both Architect X and Architect Y.

Total invalid combinations = 65.

If Architect X and Architect Y are both included, selecting remaining architects from the other 3 yields:

\binom{3}{2}\binom{5}{2} + \binom{3}{1}\binom{5}{3} + \binom{3}{0}\binom{5}{4} = (3 \times 10) + (3 \times 10) + (1 \times 5) = 30 + 30 + 5 = 65

.

4

Subtract the invalid combinations from the total unconstrained combinations.

200 - 65 = 135.

Using the complementary counting principle provides the exact number of valid combinations where Architect X and Architect Y do not serve together.

Key Concept

Combinations with multiple category constraints and complementary counting for exclusion rules.

Question 1765Question →

A museum curator is arranging $6$ distinct paintings in a single row along a gallery wall. If $2$ specific paintings must not be placed adjacent to each other, how many different arrangements of the $6$ paintings are possible?

Show answer & explanation

Answer: 480

Answer

480

To find the number of valid arrangements where two specific paintings are not adjacent, use complementary counting. First, compute the total number of ways to arrange 6 distinct paintings without restrictions, which is

6! = 720

. Next, calculate the number of arrangements where the two specific paintings are placed adjacent to one another by treating them as a single block. There are 5 units in total to arrange (the pair block plus the remaining 4 individual paintings), which gives

5! = 120

ways. Since the two specific paintings can be arranged in

2! = 2

ways inside their block, the total number of adjacent arrangements is

120 \times 2 = 240

. Finally, subtract the adjacent arrangements from the total arrangements:

720 - 240 = 480

.

Step-by-Step Solution

1

Calculate the total number of ways to arrange all 6 paintings in a row without any restrictions.

6! = 720

There are 6 distinct items to arrange in 6 sequential positions.

2

Calculate the number of arrangements where the 2 specific paintings are adjacent (placed next to each other).

5! × 2! = 120 × 2 = 240

Treat the 2 specific paintings as a single block unit. This leaves 5 items to arrange (the block + 4 individual paintings), which can be ordered in 5! ways. Within the block, the 2 paintings can be ordered in 2! ways.

3

Subtract the number of adjacent arrangements from the total unrestricted arrangements.

720 - 240 = 480

Complementary counting dictates that valid non-adjacent arrangements equal total possible arrangements minus adjacent arrangements.

Key Concept

Permutations with Adjacency Restrictions (Complementary Counting)

Question 1766Question →

Which word best completes the sentence based on the contrast clue provided?

Fill in the blanks below

Far from being during the mid-Cretaceous epoch, the polar regions—long assumed by early geologists to have been frozen wastes—actually supported temperate, biodiverse forest ecosystems.

Show answer & explanation

Answer

The blank must be filled with 'desolate' (or a synonym such as 'barren' or 'inhospitable').

The reversal signal 'Far from being' dictates that the word in the blank must contrast directly with the subsequent clause describing 'temperate, biodiverse forest ecosystems.' Therefore, a word indicating a lack of life or extreme harshness—such as 'desolate', 'barren', or 'inhospitable'—is required.

Step-by-Step Solution

1

Identify the structural contrast signal

The opening phrase 'Far from being' serves as a reversal pivot.

This structural signal indicates that the state described in the blank is opposite to the actual reality presented later in the sentence.

2

Analyze the contextual clue following the pivot

The sentence states that the regions 'actually supported temperate, biodiverse forest ecosystems.'

This establishes that the actual condition was life-supporting, green, and lush.

3

Deduce the semantic requirement for the blank

The word in the blank must express the opposite of lush and biodiverse, such as 'desolate', 'barren', or 'inhospitable'.

Pairing 'Far from being' with a word meaning devoid of life correctly expresses the contrast with a thriving forest ecosystem.

Key Concept

Contrast and Reversal Clues

Question 1767Question →

In a quality control assessment, the weights of manufactured steel components are normally distributed with a mean of $450$ grams and a standard deviation of $12$ grams. Components weighing less than $426$ grams or more than $474$ grams are classified as defective and discarded. Of the remaining non-defective components, those weighing at least $462$ grams are classified as Premium Grade. Assuming the $68\text{--}95\text{--}99.7$ empirical rule for normal distributions, approximately how many components in a batch of $10,000$ are Premium Grade?

1,283

1,350

1,520

1,600

3,400

Show answer & explanation

Answer: $1,350$

Answer

The correct number of Premium Grade components is

1,350

.

By standardizing the given weight thresholds into z-scores (

z = -2.0

for

426\text{ g}

,

z = +1.0

for

462\text{ g}

, and

z = +2.0

for

474\text{ g}

), Premium Grade components are defined by the interval

+1.0 \le z \le +2.0

. According to the empirical rule,

95\%

of data falls within

[-2\sigma, +2\sigma]

and

68\%

falls within

[-1\sigma, +1\sigma]

. The portion in the positive tail between

+1\sigma

and

+2\sigma

is

\frac{95\% - 68\%}{2} = 13.5\%

. Multiplying

13.5\%

by the batch total of

10,000

yields

1,350

components.

Step-by-Step Solution

1

Calculate the z-scores for the defect thresholds and the Premium Grade threshold.

Lower defect limit:

z = \frac{426 - 450}{12} = -2.0

; Upper defect limit:

z = \frac{474 - 450}{12} = +2.0

; Premium Grade lower threshold:

z = \frac{462 - 450}{12} = +1.0

.

Standardizing the raw weight values into z-scores allows the application of the empirical rule.

2

Identify the z-score interval representing non-defective Premium Grade components.

The target weight interval is

462 \le \text{weight} \le 474

grams, corresponding to

+1.0 \le z \le +2.0

.

Components must weigh at least

462

grams (

z \ge +1.0

) to be Premium Grade, but must not exceed

474

grams (

z > +2.0

) because those exceeding

474

grams are defective.

3

Determine the percentage of the total distribution within

+1.0 \le z \le +2.0

using the empirical rule.

The area between

z = -1.0

and

z = +1.0

is

68\%

, and between

z = -2.0

and

z = +2.0

is

95\%

. The region between

z = +1.0

and

z = +2.0

is

\frac{95\% - 68\%}{2} = 13.5\%

.

By symmetry of the normal distribution curve, half of the difference between the

2\sigma

and

1\sigma

intervals lies in the upper tail.

4

Calculate the expected count of Premium Grade components in a batch of

10,000

.

10,000 \times 0.135 = 1,350

.

Multiplying the population proportion by the batch size gives the expected count.

Key Concept

Calculating areas under a normal curve bounded by standard deviation thresholds (z-scores) using the empirical rule.

Estimated Time:2m 30s

Question 1768Question →

For a real constant $k$ , the quadratic equation $x^2 - 2kx + (k^2 - k - 6) = 0$ has two distinct real roots $r$ and $s$ such that $r < 0 < s$ and $|r| < |s|$ . Which of the following inequalities expresses all possible values of $k$ ?

0 < k < 3

-2 < k < 3

-6 < k < 3

k < -2

or

k > 3

-2 < k < 0

Show answer & explanation

Answer: $0 < k < 3$

Answer

The inequality expressing all possible values of

k

is

0 < k < 3

.

The condition that one root is negative and one root is positive (

r < 0 < s

) requires the product of the roots

rs = k^2 - k - 6

to be negative, which resolves to

-2 < k < 3

. Furthermore, since the positive root

s

has a larger absolute magnitude than the negative root

r

(

|r| < |s|

), the sum of the roots

r + s = 2k

must be positive, requiring

k > 0

. Taking the intersection of

-2 < k < 3

and

k > 0

yields

0 < k < 3

.

Step-by-Step Solution

1

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

k

.

For

x^2 - 2kx + (k^2 - k - 6) = 0

, the sum of roots is

r + s = 2k

and the product of roots is

rs = k^2 - k - 6

.

Vieta's formulas directly relate the coefficients of a quadratic equation to the sum and product of its roots.

2

Analyze the condition

r < 0 < s

.

Since one root is negative and the other is positive, their product must be negative:

rs = k^2 - k - 6 < 0

. Factoring gives

(k - 3)(k + 2) < 0

, which yields

-2 < k < 3

.

A positive number multiplied by a negative number produces a negative product.

3

Analyze the condition

|r| < |s|

.

Since

r < 0

,

|r| = -r

. Since

s > 0

,

|s| = s

. The inequality

|r| < |s|

becomes

-r < s

, which simplifies to

r + s > 0

. Substituting

r + s = 2k

gives

2k > 0

, or

k > 0

.

The positive root having greater magnitude than the absolute value of the negative root means the sum of the roots must be positive.

4

Combine the conditions to find the valid range for

k

.

Combining

-2 < k < 3

and

k > 0

yields the intersection

0 < k < 3

. (The discriminant condition

\Delta = 4(k + 6) > 0 \implies k > -6

is satisfied for all

k \in (0, 3)

).

The parameter

k

must satisfy both root sign constraints simultaneously.

Key Concept

Using Vieta's formulas and root magnitude conditions to solve quadratic parameter inequality problems.

Question 1769Question →

A clean energy technology company manufactures two models of solar panels: Model X and Model Y. Model X produces $150$ kilowatt-hours (kWh) of electricity per day and costs $\$400$ to manufacture, while Model Y produces $200$ kWh of electricity per day and costs $\$550$ to manufacture. A solar energy project purchased a total of $50$ panels for a total manufacturing cost of $\$24,500$ . What is the total daily electricity production, in kilowatt-hours, of all $50$ panels combined?

8,500

8,750

9,000

9,500

10,000

Show answer & explanation

Answer: $9,000$

Answer

The total daily electricity production of all

50

panels combined is

9,000

kWh.

The correct option is

9,000

kWh. Defining

x

as the number of Model X panels and

y

as the number of Model Y panels gives the equations

x + y = 50

and

400x + 550y = 24,500

. Substituting

y = 50 - x

yields

400x + 550(50 - x) = 24,500

, which simplifies to

-150x = -3,000

, so

x = 20

and

y = 30

. Computing the total output yields

20(150) + 30(200) = 3,000 + 6,000 = 9,000

kWh.

Step-by-Step Solution

1

Define variables for the quantities of each panel model.

Let

x

represent the number of Model X panels and

y

represent the number of Model Y panels.

Establishing explicit variables allows formulating a system of linear equations from the word problem.

2

Set up equations for total panel quantity and total manufacturing cost.

System equations:

x + y = 50

and

400x + 550y = 24,500

.

The total number of panels is

50

, and the combined manufacturing cost equals

\$24,500

.

3

Solve the system using substitution.

Substitute

y = 50 - x

into the cost equation:

400x + 550(50 - x) = 24,500 \implies 400x + 27,500 - 550x = 24,500 \implies -150x = -3,000 \implies x = 20

. Consequently,

y = 50 - 20 = 30

.

Determining

x = 20

and

y = 30

gives the exact number of Model X and Model Y panels purchased.

4

Calculate the total daily electricity production.

Total production

= 20(150) + 30(200) = 3,000 + 6,000 = 9,000

kWh.

Multiply the quantity of each panel model by its respective daily output rate and sum the products.

Key Concept

Formulating and solving a system of two linear equations from a real-world scenario to find unknown quantities and evaluate a secondary combination function.

Estimated Time:1m 30s

Question 1770Question →

In right triangle $ABC$ , the right angle is at vertex $B$ , $AB = 6$ , and the measure of $\angle ACB$ is $30^\circ$ . Point $P$ lies on segment $BC$ such that the measure of $\angle APB$ is $60^\circ$ . Which of the following statements must be true? Select all that apply.

Select all that apply

The length of segment

AP

is

4\sqrt{3}

.

The length of segment

PC

is

4\sqrt{3}

.

The area of triangle

APC

is

18\sqrt{3}

.

The perimeter of triangle

ABC

is

18 + 6\sqrt{3}

.

The length of segment

BP

is

3\sqrt{3}

.

Show answer & explanation

Answer: The length of segment $AP$ is $4\sqrt{3}$ .; The length of segment $PC$ is $4\sqrt{3}$ .; The perimeter of triangle $ABC$ is $18 + 6\sqrt{3}$ .

Answer

The correct statements are that the length of segment

AP

is

4\sqrt{3}

, the length of segment

PC

is

4\sqrt{3}

, and the perimeter of triangle

ABC

is

18 + 6\sqrt{3}

.

The 30-60-90 triangle ratio (

1 : \sqrt{3} : 2

) establishes

AC = 12

,

BC = 6\sqrt{3}

,

AP = 4\sqrt{3}

, and

BP = 2\sqrt{3}

. Consequently,

PC = 4\sqrt{3}

and the perimeter of triangle

ABC

equals

18 + 6\sqrt{3}

. Thus, the statements asserting that

AP = 4\sqrt{3}

,

PC = 4\sqrt{3}

, and the perimeter of triangle

ABC

is

18 + 6\sqrt{3}

are all correct.

Step-by-Step Solution

1

Analyze main triangle

ABC

Hypotenuse

AC = 12

and base

BC = 6\sqrt{3}

Triangle

ABC

is a 30-60-90 triangle with side opposite

30^\circ

equal to

AB = 6

. Therefore, hypotenuse

AC = 2(6) = 12

and leg

BC = 6\sqrt{3}

.

2

Analyze sub-triangle

ABP

Leg

BP = 2\sqrt{3}

and hypotenuse

AP = 4\sqrt{3}

Triangle

ABP

is a 30-60-90 triangle with side opposite

60^\circ

equal to

AB = 6

. The shorter leg is

BP = \frac{6}{\sqrt{3}} = 2\sqrt{3}

and the hypotenuse is

AP = 2(2\sqrt{3}) = 4\sqrt{3}

.

3

Calculate segment

PC

and area of triangle

APC

PC = 4\sqrt{3}

and

\text{Area}(\triangle APC) = 12\sqrt{3}

PC = BC - BP = 6\sqrt{3} - 2\sqrt{3} = 4\sqrt{3}

. The area of

\triangle APC

is

\frac{1}{2} \times PC \times AB = \frac{1}{2}(4\sqrt{3})(6) = 12\sqrt{3}

.

4

Calculate perimeter of triangle

ABC

Perimeter

= 18 + 6\sqrt{3}

Sum of sides

AB + BC + AC = 6 + 6\sqrt{3} + 12 = 18 + 6\sqrt{3}

.

Key Concept

Properties of 30-60-90 Special Right Triangles and Side Length Ratios (

1 : \sqrt{3} : 2

)

Question 1771Question →

In convex quadrilateral $ABCD$ , the diagonals $AC$ and $BD$ intersect at point $P$ at right angles ( $AC \perp BD$ ). The length of diagonal $AC$ is $16$ and the length of diagonal $BD$ is $12$ . Points $E$ , $F$ , $G$ , and $H$ are the midpoints of sides $AB$ , $BC$ , $CD$ , and $DA$ , respectively. Which of the following statements MUST be true? Select all such statements.

Select all that apply

The perimeter of quadrilateral

EFGH

is

28

.

Quadrilateral

EFGH

is a rectangle.

The area of quadrilateral

ABCD

is

96

.

The area of quadrilateral

EFGH

is

96

.

Quadrilateral

EFGH

is a rhombus.

Show answer & explanation

Answer: The perimeter of quadrilateral $EFGH$ is $28$ .; Quadrilateral $EFGH$ is a rectangle.; The area of quadrilateral $ABCD$ is $96$ .

Answer

The true statements are that the perimeter of quadrilateral

EFGH

is

28

, quadrilateral

EFGH

is a rectangle, and the area of quadrilateral

ABCD

is

96

.

Applying the Midpoint Theorem shows that midsegments

EF

and

GH

are parallel to

AC

with length

8

, while

FG

and

HE

are parallel to

BD

with length

6

. This gives a perimeter of

8 + 6 + 8 + 6 = 28

. Because

AC \perp BD

, the adjacent midsegments meet at

90^\circ

, confirming that quadrilateral

EFGH

is a rectangle. Additionally, for any orthodiagonal quadrilateral, the area is

\frac{1}{2} d_1 d_2 = \frac{1}{2} \times 16 \times 12 = 96

.

Step-by-Step Solution

1

Determine the side lengths of midpoint quadrilateral

EFGH

using the Triangle Midpoint Theorem.

EF = GH = \frac{1}{2}AC = 8

and

FG = HE = \frac{1}{2}BD = 6

.

In any triangle, the segment connecting the midpoints of two sides is parallel to the third side and half its length.

2

Calculate the perimeter of quadrilateral

EFGH

.

Perimeter =

EF + FG + GH + HE = 8 + 6 + 8 + 6 = 28

.

The perimeter is the sum of all four side lengths of the quadrilateral.

3

Determine the shape classification of quadrilateral

EFGH

.

EFGH

is a rectangle.

Since

EF \parallel AC

and

FG \parallel BD

, the angle between

EF

and

FG

equals the angle between diagonals

AC

and

BD

. Given

AC \perp BD

, the angle

\angle EFG = 90^\circ

. Opposite sides are equal and parallel with right angles, making

EFGH

a rectangle (and not a rhombus, as adjacent sides

8 \neq 6

).

4

Calculate the area of quadrilateral

ABCD

and quadrilateral

EFGH

.

Area(

ABCD

) =

\frac{1}{2} \times AC \times BD = \frac{1}{2} \times 16 \times 12 = 96

; Area(

EFGH

) =

8 \times 6 = 48

.

The area of a quadrilateral with perpendicular diagonals is half the product of its diagonal lengths. The midpoint quadrilateral has half the area of the outer quadrilateral.

Key Concept

Midpoint Theorem (Varignon's Theorem) and Area of Orthodiagonal Quadrilaterals

Estimated Time:2m 0s

Question 1772Question →

A project manager is scheduling $7$ distinct project milestones: $4$ technical milestones and $3$ managerial milestones. The milestones must be scheduled sequentially across $7$ consecutive weeks. To avoid scheduling burnout, no two managerial milestones can be scheduled in consecutive weeks. In how many different valid sequences can all $7$ milestones be scheduled?

240

720

1,440

2,160

5,040

Show answer & explanation

Answer: $1,440$

Answer

1,440

To ensure no two managerial milestones are adjacent, first order the

4

distinct technical milestones, which can be done in

4! = 24

ways. These

4

milestones create

5

available slots (gaps before, between, and after them). To place the

3

distinct managerial milestones into these

5

slots such that no slot contains more than one managerial milestone, we calculate the permutations

P(5, 3) = 5 \times 4 \times 3 = 60

. Multiplying the independent choices yields

24 \times 60 = 1,440

valid sequences.

Step-by-Step Solution

1

Arrange the non-restricted items (the 4 distinct technical milestones)

Number of ways

= 4! = 24

Since all 4 technical milestones are distinct, they can be ordered in

4!

ways.

2

Determine the number of available slots (gaps) created for the restricted items

Number of slots

= 4 + 1 = 5

Placing 4 technical milestones in a line creates 5 potential slots (before the first, between adjacent ones, and after the last) where managerial milestones can be placed without being adjacent to each other: _ T1 _ T2 _ T3 _ T4 _

3

Place and order the 3 distinct managerial milestones into the 5 available slots

Number of ways

= P(5, 3) = 5 \times 4 \times 3 = 60

Because the managerial milestones are distinct and order matters, we choose 3 slots out of 5 and arrange them.

4

Apply the Fundamental Counting Principle to find the total valid arrangements

Total valid arrangements

= 24 \times 60 = 1,440

The placement of technical milestones and managerial milestones are independent choices in sequence.

Key Concept

Permutations with Non-Adjacency Restrictions (Gap Insertion Method)

Question 1773Question →

For all real numbers $x$ and $y$ , the custom operation $\odot$ is defined by $x \odot y = x|y| - y|x|$ . Which of the following statements must be true for all real numbers $x$ and $y$ ? Select all such statements.

Select all that apply

x \odot y = 0

whenever

x

and

y

have the same sign

x \odot y = -(y \odot x)

If

x > 0

and

y < 0

, then

x \odot y > 0

x \odot (-x) = 0

for all non-zero real numbers

x

(x \odot y) \odot z = x \odot (y \odot z)

Show answer & explanation

Answer: $x \odot y = 0$ whenever $x$ and $y$ have the same sign; $x \odot y = -(y \odot x)$ ; If $x > 0$ and $y < 0$ , then $x \odot y > 0$

Answer

The correct statements are the statement asserting

x \odot y = 0

when

x

and

y

have the same sign, the statement asserting anti-commutativity

x \odot y = -(y \odot x)

, and the statement asserting

x \odot y > 0

when

x > 0

and

y < 0

.

The operation

x \odot y = x|y| - y|x|

produces 0 whenever

x

and

y

share the same sign because terms evaluate to identical quantities. Swapping variables negates the expression, establishing anti-commutativity. When

x

is positive and

y

is negative,

x \odot y

simplifies to

-2xy

, which is strictly greater than 0 since

xy < 0

.

Step-by-Step Solution

1

Analyze the first statement regarding same-sign inputs

If

x > 0

and

y > 0

,

|x|=x

and

|y|=y

, so

x \odot y = xy - yx = 0

. If

x < 0

and

y < 0

,

|x|=-x

and

|y|=-y

, so

x \odot y = x(-y) - y(-x) = -xy + xy = 0

. Thus,

x \odot y = 0

when

x

and

y

have the same sign.

Verifying the definition under both positive and negative cases of identical sign.

2

Analyze the second statement regarding operand order reversal

y \odot x = y|x| - x|y| = -(x|y| - y|x|) = -(x \odot y)

, which holds universally for all real numbers.

Testing anti-commutativity by algebraic substitution into the custom operation.

3

Analyze the third statement for opposite signs (

x > 0

and

y < 0

)

Since

x > 0

,

|x|=x

. Since

y < 0

,

|y|=-y

. Substituting yields

x(-y) - y(x) = -xy - xy = -2xy

. Because

x > 0

and

y < 0

, the product

xy

is negative, making

-2xy

strictly positive.

Determining the overall algebraic sign of the expression when variables have opposite signs.

4

Counter-test the remaining statements to verify incorrectness

For

x \odot (-x)

with

x=1

:

1 \odot (-1) = 1|-1| - (-1)|1| = 1 - (-1) = 2 \neq 0

. For associativity with

x=2, y=-1, z=-1

:

(2 \odot -1) \odot -1 = 4 \odot -1 = 8

, but

2 \odot (-1 \odot -1) = 2 \odot 0 = 0

.

Demonstrating specific counterexamples for false generalizations.

Key Concept

Custom Binary Operations and Absolute Value Properties

Estimated Time:1m 30s

Question 1774Question →

A right rectangular prism has a square base. The diagonal of the base has a length of $6\sqrt{2}$ units. A space diagonal of the prism makes a $30^\circ$ angle with the diagonal of the base. What is the height of the prism?

2\sqrt{6}

6\sqrt{6}

6\sqrt{2}

12\sqrt{2}

4\sqrt{6}

Show answer & explanation

Answer: $2\sqrt{6}$

Answer

2\sqrt{6}

The height of a right rectangular prism is perpendicular to the base, forming a right triangle with the base diagonal as one leg and the space diagonal as the hypotenuse. Given that the angle between the space diagonal and the base diagonal is

30^\circ

, this right triangle is a

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

triangle. The base diagonal of length

6\sqrt{2}

is adjacent to the

30^\circ

angle, making it the longer leg (

x\sqrt{3}

). Dividing

6\sqrt{2}

by

\sqrt{3}

gives

h = 2\sqrt{6}

, which correctly matches the length of the leg opposite the

30^\circ

angle.

Step-by-Step Solution

1

Identify the right triangle inside the prism.

The vertical right triangle has legs

d

(base diagonal) and

h

(height), with hypotenuse

D

(space diagonal).

The height of a right prism is perpendicular to its base, forming a right angle with any line segment lying in the base, including the base diagonal.

2

Apply special right triangle ratios for a

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

triangle.

The base diagonal

d = 6\sqrt{2}

is adjacent to the

30^\circ

angle, so

d = x\sqrt{3}

, where

x = h

is the height opposite the

30^\circ

angle.

In a

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

triangle, side lengths follow the ratio

1 : \sqrt{3} : 2

.

3

Solve for height

h

.

h = \frac{6\sqrt{2}}{\sqrt{3}} = \frac{6\sqrt{6}}{3} = 2\sqrt{6}

.

Rationalize the denominator by multiplying top and bottom by

\sqrt{3}

.

Key Concept

Special Right Triangles (

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

) in 3D Space Diagonals

Estimated Time:1m 30s

Question 1775Question →

Two events $A$ and $B$ are defined on a sample space such that $P(A) = 0.60$ and $P(B) = 0.75$ . Which of the following statements must be true? Select all such statements.

Select all that apply

Events

A

and

B

cannot be mutually exclusive.

The probability that both events

A

and

B

occur,

P(A \cap B)

, is at least

0.35

.

Events

A

and

B

must be independent.

The conditional probability

P(A \mid B)

is at least

\frac{7}{15}

.

The maximum possible value of

P(A \text{ or } B)

is

0.90

.

Show answer & explanation

Answer: Events $A$ and $B$ cannot be mutually exclusive.; The probability that both events $A$ and $B$ occur, $P(A \cap B)$ , is at least $0.35$ .; The conditional probability $P(A \mid B)$ is at least $\frac{7}{15}$ .

Answer

The statements asserting that events

A

and

B

cannot be mutually exclusive, that the joint probability

P(A \cap B)

is at least

0.35

, and that the conditional probability

P(A \mid B)

is at least

\frac{7}{15}

are all correct.

The sum of the probabilities of events

A

and

B

(

1.35

) exceeds

1

, making mutual exclusivity impossible. The inclusion-exclusion principle dictates

P(A \cap B) \geq 0.60 + 0.75 - 1.00 = 0.35

. Consequently, the minimum conditional probability

P(A \mid B)

is

\frac{0.35}{0.75} = \frac{7}{15}

.

Step-by-Step Solution

1

Evaluate mutual exclusivity

If

A

and

B

were mutually exclusive,

P(A \cap B) = 0

, so

P(A \cup B) = P(A) + P(B) = 0.60 + 0.75 = 1.35

. Since probability cannot exceed

1

, the events cannot be mutually exclusive.

Verify if the sum of individual probabilities exceeds 1.

2

Determine the minimum joint probability

P(A \cap B)

Using

P(A \cup B) = P(A) + P(B) - P(A \cap B) \leq 1

, we have

0.60 + 0.75 - P(A \cap B) \leq 1 \implies P(A \cap B) \geq 0.35

.

Apply the inclusion-exclusion principle bounded by maximum total probability.

3

Test for required independence

Independence requires

P(A \cap B) = 0.60 \times 0.75 = 0.45

. Since

P(A \cap B)

can legitimately range anywhere between

0.35

and

0.60

, independence is possible but not guaranteed.

Check whether joint probability is strictly fixed at the product of individual probabilities.

4

Calculate the lower bound for conditional probability

P(A \mid B)

P(A \mid B) = \frac{P(A \cap B)}{P(B)} \geq \frac{0.35}{0.75} = \frac{35}{75} = \frac{7}{15}

.

Substitute the minimum joint probability into the conditional probability formula.

Key Concept

Probability rules governing overlap, mutual exclusivity, joint probability bounds, and conditional probability.

Question 1776Question →

A meteorologist recorded the daily minimum temperatures, in degrees Celsius, at a high-altitude research station over a 7-day period: $3$ , $-5$ , $7$ , $-2$ , $10$ , $-8$ , and $2$ .

If $M$ represents the median of these daily minimum temperatures and $A$ represents the arithmetic mean, what is the value of $M - A$ ?

-3

-1

0

1

3

Show answer & explanation

Answer: $1$

Answer

The value of

M - A

is

1

.

To evaluate

M - A

, first arrange the data set in ascending order:

-8, -5, -2, 2, 3, 7, 10

. Since there are 7 numbers, the median

M

is the middle (4th) value, which is

2

. Next, find the arithmetic mean

A

by taking the sum of the elements,

(-8) + (-5) + (-2) + 2 + 3 + 7 + 10 = 7

, and dividing by

7

, yielding

A = 1

. Finally, subtract the mean from the median:

M - A = 2 - 1 = 1

.

Step-by-Step Solution

1

Sort the dataset in ascending order to find the median

M

.

The sorted list of 7 temperatures is:

-8, -5, -2, 2, 3, 7, 10

.

The median of a set with an odd number of elements is the middle value of the ordered dataset.

2

Identify the 4th element in the sorted dataset.

M = 2

.

In a dataset of 7 ordered values, the middle (4th) position represents the median.

3

Calculate the arithmetic mean

A

by summing all temperatures and dividing by 7.

Sum

= (-8) + (-5) + (-2) + 2 + 3 + 7 + 10 = 7

. Thus,

A = \frac{7}{7} = 1

.

The arithmetic mean is defined as the total sum of observations divided by the number of observations.

4

Compute

M - A

.

M - A = 2 - 1 = 1

.

Subtracting the mean from the median yields the required target value.

Key Concept

Calculating the median of a dataset requires arranging values in numerical order before identifying the central value.

Estimated Time:1m 30s

Question 1777Question →

A financial firm has a pool of $10$ analysts, consisting of $6$ senior analysts and $4$ junior analysts. Which of the following selection procedures will yield EXACTLY $120$ unique possible groups? Select all such procedures.

Select all that apply

Forming a

5

-member committee that contains exactly

3

senior analysts and

2

junior analysts

Forming a

3

-member subcommittee from the entire pool of

10

analysts without any restrictions

Forming a

7

-member project panel from the entire pool of

10

analysts without any restrictions

Forming a

5

-member committee that contains exactly

4

senior analysts and

1

junior analyst

Forming a

4

-member task force from the entire pool of

10

analysts without any restrictions

Show answer & explanation

Answer: Forming a $5$ -member committee that contains exactly $3$ senior analysts and $2$ junior analysts; Forming a $3$ -member subcommittee from the entire pool of $10$ analysts without any restrictions; Forming a $7$ -member project panel from the entire pool of $10$ analysts without any restrictions

Answer

The procedures that yield exactly 120 unique possible groups are: forming a 5-member committee with 3 senior and 2 junior analysts, forming a 3-member subcommittee from all 10 analysts, and forming a 7-member project panel from all 10 analysts.

The correct procedures are those that evaluate to exactly 120 combinations: (1) Selecting 3 senior analysts from 6 and 2 junior analysts from 4 gives $\binom{6}{3} \times \binom{4}{2} = 20 \times 6 = 120$. (2) Choosing 3 analysts from 10 gives $\binom{10}{3} = 120$. (3) Choosing 7 analysts from 10 is symmetric to choosing 3 analysts, yielding $\binom{10}{7} = \binom{10}{3} = 120$.

Step-by-Step Solution

1

Calculate combinations for forming a 5-member committee with 3 senior and 2 junior analysts

$\binom{6}{3} \times \binom{4}{2} = 20 \times 6 = 120$

The selection of senior and junior analysts are independent decisions, so their combination values are multiplied together.

2

Calculate combinations for choosing 3 analysts out of 10 without restrictions

$\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$

Order of selection does not matter, so the standard combination formula $\binom{n}{k}$ is applied.

3

Calculate combinations for choosing 7 analysts out of 10 without restrictions

$\binom{10}{7} = \binom{10}{10-7} = \binom{10}{3} = 120$

Choosing 7 people to include is mathematically equivalent to choosing 3 people to exclude.

4

Evaluate the remaining options to verify they do not yield 120

$\binom{6}{4} \times \binom{4}{1} = 60$ and $\binom{10}{4} = 210$

Neither of these evaluations equals the target value of 120.

Key Concept

Combinations and the Fundamental Counting Principle

Question 1778Question →

In rhombus $ABCD$ , the side length is $10$ and the length of diagonal $BD$ is $12$ . Line segment $AP$ is drawn perpendicular to side $BC$ , with point $P$ lying on segment $BC$ . What is the length of segment $AP$ ?

Show answer & explanation

Answer: 9.6

Answer

9.6

The diagonals of rhombus

ABCD

intersect perpendicularly at

O

and bisect each other. Given

BD = 12

, half of the diagonal is

BO = 6

. Right triangle

AOB

has hypotenuse

AB = 10

and leg

BO = 6

, so by the Pythagorean theorem, leg

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

. Thus, diagonal

AC = 16

. The area of rhombus

ABCD

is

\frac{1}{2} \times AC \times BD = \frac{1}{2} \times 16 \times 12 = 96

. The area is also equal to

\text{base} \times \text{height} = BC \times AP = 10 \times AP

. Setting

10 \times AP = 96

gives

AP = 9.6

.

Step-by-Step Solution

1

Find half the length of diagonal BD.

Segment BO = 6.

The diagonals of a rhombus bisect each other at right angles.

2

Apply the Pythagorean theorem to right triangle AOB to determine half of diagonal AC.

AO = 8, so diagonal AC = 16.

Triangle AOB has hypotenuse 10 and leg 6, forming a 6-8-10 Pythagorean triple.

3

Calculate the total area of rhombus ABCD from its diagonal lengths.

Area = 96.

The area of a rhombus equals half the product of its two diagonals.

4

Use the alternative area formula (base × height) to solve for altitude AP.

AP = 9.6.

Base BC = 10 and height AP give Area = 10 × AP = 96.

Key Concept

Properties of rhombus diagonals, Pythagorean theorem, and dual area formulas for quadrilaterals

Question 1779Question →

A list consists of six numbers: $3$ , $7$ , $10$ , $14$ , $18$ , and $x$ . If the arithmetic mean of these six numbers is equal to their median, which of the following could be the value of $x$ ? Indicate all such values.

Select all that apply

-1

5

11

14

20

Show answer & explanation

Answer: $-1$ ; $11$ ; $20$

Answer

The valid values for

x

are

-1

,

11

, and

20

.

The values

-1

,

11

, and

20

each yield a dataset where the arithmetic mean equals the median:

-1

gives a mean and median of

8.5

,

11

gives a mean and median of

10.5

, and

20

gives a mean and median of

12

.

Step-by-Step Solution

1

Express the arithmetic mean in terms of

x

.

Mean =

\frac{3 + 7 + 10 + 14 + 18 + x}{6} = \frac{52 + x}{6}

.

The mean of a dataset of

n

numbers is the sum of all elements divided by

n

.

2

Analyze the median based on the position of

x

relative to the sorted known values

3, 7, 10, 14, 18

.

Case 1:

x \le 7 \implies

median =

\frac{7+10}{2} = 8.5

.
Case 2:

7 < x < 14 \implies

median =

\frac{x+10}{2}

(for

7 < x \le 10

) or

\frac{10+x}{2}

(for

10 < x < 14

).
Case 3:

x \ge 14 \implies

median =

\frac{10+14}{2} = 12

.

For an even number of elements (

n=6

), the median is the average of the 3rd and 4th terms in ascending order.

3

Set the mean equal to the median for each case and solve for

x

.

Case 1:

\frac{52+x}{6} = 8.5 \implies 52+x = 51 \implies x = -1

(valid since

-1 \le 7

).
Case 2:

\frac{52+x}{6} = \frac{10+x}{2} \implies 52+x = 30+3x \implies 2x = 22 \implies x = 11

(valid since

7 < 11 < 14

).
Case 3:

\frac{52+x}{6} = 12 \implies 52+x = 72 \implies x = 20

(valid since

20 \ge 14

).

This identifies all values of

x

satisfying the problem constraint.

Key Concept

Evaluating mean and median of a dataset containing an unknown variable across different intervals of the variable's possible values.

Question 1780Question →

A cybersecurity system generates 5-character identification codes consisting of 3 distinct letters followed by 2 distinct digits. The letters must be selected from the set $\{A, B, C, D, E, F\}$ and the digits from the set $\{1, 2, 3, 4\}$ . If the first character of the code must be a vowel ( $A$ or $E$ ), how many such distinct 5-character identification codes can be formed?

120

240

480

720

1,440

Show answer & explanation

Answer: 480

Answer

480 distinct 5-character identification codes can be formed.

To construct the 5-character code, break the process into sequential choices for each position: Position 1 must be a vowel (

A

or

E

), giving 2 options. Position 2 can be any of the remaining 5 distinct letters. Position 3 can be any of the remaining 4 distinct letters. Position 4 (the first digit) can be any of the 4 available digits. Position 5 (the second digit) can be any of the remaining 3 distinct digits. Multiplying these independent choices together yields

2 \times 5 \times 4 \times 4 \times 3 = 480

.

Step-by-Step Solution

1

Calculate the number of ways to choose the first letter (must be a vowel)

There are 2 choices (

A

or

E

).

The problem restricts the first character of the code to a vowel.

2

Calculate the number of choices for the second and third letters

5 choices for the second letter and 4 choices for the third letter.

The letters must be distinct, leaving 5 remaining letters from the original pool of 6 for the second position, and 4 for the third position.

3

Calculate the number of choices for the two distinct digits

4 choices for the first digit and 3 choices for the second digit.

There are 4 digits available in

\{1, 2, 3, 4\}

and they must be distinct.

4

Apply the Fundamental Counting Principle to find total codes

2 \times 5 \times 4 \times 4 \times 3 = 480

.

Multiply the number of independent choices for each sequential position.

Key Concept

Fundamental Counting Principle with Permutations and Restrictions

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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