All practice questions

Question 1561Question →

In a circle centered at point $O$ , the radius is $18$ units. Points $P$ and $Q$ lie on the circle such that the area of sector $POQ$ is $54\pi$ square units. What is the ratio of the length of minor arc $PQ$ to the total perimeter of sector $POQ$ ?

\frac{\pi}{\pi + 6}

\frac{1}{6}

\frac{\pi}{\pi + 3}

\frac{6\pi}{\pi + 6}

\frac{9\pi}{\pi + 6}

Show answer & explanation

Answer: $\frac{\pi}{\pi + 6}$

Answer

The ratio of the length of minor arc

PQ

to the total perimeter of sector

POQ

is

\frac{\pi}{\pi + 6}

.

The full circle area is

324\pi

, making the sector

54\pi / 324\pi = 1/6

of the circle. The arc length is

1/6 \times 36\pi = 6\pi

. The sector perimeter is the arc length plus two radii (

6\pi + 36

). Taking the ratio of arc length to sector perimeter yields

6\pi / (6\pi + 36) = \pi / (\pi + 6)

.

Step-by-Step Solution

1

Calculate the total area of the circle and determine the fractional size of sector

POQ

.

Total area =

\pi(18)^2 = 324\pi

. Sector fraction =

\frac{54\pi}{324\pi} = \frac{1}{6}

.

Determining the fraction of the circle represented by the sector is required to find the arc length.

2

Calculate the length of minor arc

PQ

and the total perimeter of sector

POQ

.

Minor arc

PQ = \frac{1}{6} \times 2\pi(18) = 6\pi

. Sector perimeter =

6\pi + 2(18) = 6\pi + 36

.

The sector perimeter consists of the curved arc length plus the two straight radii

OP

and

OQ

.

3

Form and simplify the ratio of arc length to sector perimeter.

\frac{6\pi}{6\pi + 36} = \frac{6\pi}{6(\pi + 6)} = \frac{\pi}{\pi + 6}

.

Factoring out 6 from the numerator and denominator simplifies the expression to its lowest form.

Key Concept

Arc Length and Sector Perimeter Calculations

Question 1562Question →

A financial analyst models a company's weekly metrics using three variables—revenue

R

, operating cost

C

, and advertising expenditure

A

, measured in thousands of dollars. The metrics satisfy the following system of linear equations, where

k

is a real constant:

\begin{aligned} 4R - 3C + 2A &= 18 \\ R + 5C - 6A &= 14 \\ 5R + 2C - 4A &= k \end{aligned}

If this system of linear equations is consistent (has at least one solution), what is the value of

k

?

4

16

32

-32

252

Show answer & explanation

Answer: 32

Answer

32

Adding the left-hand sides of the first two equations yields

(4R - 3C + 2A) + (R + 5C - 6A) = 5R + 2C - 4A

, which is identical to the left-hand side of the third equation. For the linear system to have at least one solution (to be consistent), the right-hand side constant must satisfy the exact same linear combination:

k = 18 + 14 = 32

.

Step-by-Step Solution

1

Examine the linear combination of the left-hand sides of the first two equations

(4R - 3C + 2A) + (R + 5C - 6A) = 5R + 2C - 4A

Observing that the sum of the coefficients of the first two equations matches the left-hand side of the third equation.

2

Apply the condition for system consistency

Right-hand side of Equation 3 must equal Right-hand side of Equation 1 + Right-hand side of Equation 2

For a system with linearly dependent left-hand sides to be consistent, the same linear combination must hold for the right-hand constants.

3

Calculate the value of

k

k = 18 + 14 = 32

Adding the constants from the right-hand side of the first two equations gives the consistent value for

k

.

Key Concept

Consistency and Linear Dependence in Systems of Linear Equations

Question 1563Question →

A hollow metallic spherical shell has an inner radius of $3$ centimeters and an outer radius of $r$ centimeters, where $r > 3$ . If the spherical shell is melted down and completely recast into a solid right circular cylinder with base radius $r$ centimeters and height $7$ centimeters, what is the value of $r$ ?

4.5

5.25

6

7.5

9

Show answer & explanation

Answer: $6$

Answer

6

The volume of metal in the hollow spherical shell is

\frac{4}{3}\pi(r^3 - 3^3) = \frac{4}{3}\pi(r^3 - 27)

. The volume of the recast cylinder is

\pi r^2 h = 7\pi r^2

. Equating these volumes gives

\frac{4}{3}(r^3 - 27) = 7r^2

, which simplifies to

4r^3 - 21r^2 - 108 = 0

. Factoring this cubic equation gives

(r - 6)(4r^2 + 3r + 18) = 0

. Since the quadratic term has no real roots, the only real solution is

r = 6

.

Step-by-Step Solution

1

Set up the formula for the volume of metal in the hollow spherical shell.

V_{\text{shell}} = \frac{4}{3}\pi \left(r^3 - 3^3\right) = \frac{4}{3}\pi \left(r^3 - 27\right)

The metal occupies only the region between the inner sphere of radius 3 cm and outer sphere of radius r cm.

2

Set up the formula for the volume of the recast solid right circular cylinder.

V_{\text{cylinder}} = \pi r^2 h = 7\pi r^2

The cylinder has base radius r cm and height 7 cm.

3

Equate the two volumes since no metal is lost during melting and recasting.

\frac{4}{3}\pi \left(r^3 - 27\right) = 7\pi r^2

Conservation of volume during recasting.

4

Simplify the equation and solve for r.

4(r^3 - 27) = 21r^2 \implies 4r^3 - 21r^2 - 108 = 0

Divide both sides by

\pi

and multiply by 3 to clear the fraction.

5

Factor the cubic polynomial

4r^3 - 21r^2 - 108 = 0

.

(r - 6)(4r^2 + 3r + 18) = 0 \implies r = 6

Testing

r = 6

gives

4(216) - 21(36) - 108 = 864 - 756 - 108 = 0

. The quadratic factor

4r^2 + 3r + 18

has a negative discriminant and produces no real roots.

Key Concept

Volume formulas for hollow spheres and right circular cylinders

Question 1564Question →

For how many real values of the constant

a

does the following system of linear equations in

x

,

y

, and

z

have no solution?

\begin{aligned} x + y - z &= 3 \\ x + (a-1)y + 3z &= 5 \\ x + 4y + (a+1)z &= a + 2 \end{aligned}

None

Exactly one

Exactly two

Exactly three

Infinitely many

Show answer & explanation

Answer: Exactly one

Answer

Exactly one

To find when the system has no solution, we first eliminate

x

by subtracting the first equation from the second and third equations. This produces a two-variable system in

y

and

z

:

(a-2)y + 4z = 2

and

3y + (a+2)z = a - 1

. The determinant of this system's coefficients is

(a-2)(a+2) - 12 = a^2 - 16

. Setting the determinant to zero yields two critical values:

a = 4

and

a = -4

. Testing

a = 4

simplifies both reduced equations to

y + 2z = 1

, which means the system is consistent with infinitely many solutions. Testing

a = -4

yields

-3y + 2z = 1

and

-3y + 2z = 5

, which is impossible (

1 = 5

), making the system inconsistent. Thus, there is exactly one real value of

a

(

a = -4

) for which the system has no solution.

Step-by-Step Solution

1

Eliminate the variable

x

from the second and third equations using the first equation.

Subtracting the first equation

x + y - z = 3

from the second equation yields:

(a-2)y + 4z = 2

Subtracting the first equation from the third equation yields:

3y + (a+2)z = a - 1

Reducing the

3 \times 3

system to a

2 \times 2

system in

y

and

z

simplifies the analysis of linear dependence and consistency.

2

Determine the values of

a

for which the reduced

2 \times 2

system lacks a unique solution by setting its coefficient determinant to zero.

The determinant of the coefficient matrix is:

D = (a-2)(a+2) - (3)(4) = a^2 - 4 - 12 = a^2 - 16

Setting

D = 0

yields

a^2 = 16

, which gives

a = 4

or

a = -4

.

A system of linear equations has either a unique solution (when the determinant is non-zero) or non-unique behavior—either no solution or infinitely many solutions—when the determinant is zero.

3

Test

a = 4

in the reduced system.

Substituting

a = 4

into the reduced equations gives:

2y + 4z = 2 \implies y + 2z = 1

3y + 6z = 3 \implies y + 2z = 1

Since both equations are identical, the system is consistent and has infinitely many solutions.

When equation ratios match completely including constant terms, the equations represent identical hyperplanes, yielding infinitely many solutions.

4

Test

a = -4

in the reduced system.

Substituting

a = -4

into the reduced equations gives:

-6y + 4z = 2 \implies -3y + 2z = 1

3y - 2z = -5 \implies -3y + 2z = 5

Comparing these gives

1 = 5

, which is a contradiction. Thus, for

a = -4

, the system has no solution.

When parallel equations have equal coefficient ratios but unequal constant ratios, the system is inconsistent.

Key Concept

Parametric Systems of Linear Equations and Consistency Conditions

Question 1565Question →

Dataset $P$ consists of the five numbers $10, 20, 30, 40,$ and $50$ . Dataset $Q$ is created by replacing the minimum value in Dataset $P$ with $18$ and the maximum value with $42$ , leaving the remaining three numbers unchanged. Which of the following statements correctly compares the mean and standard deviation of Dataset $Q$ to those of Dataset $P$ ?

The mean of Dataset

Q

is equal to the mean of Dataset

P

, and the standard deviation of Dataset

Q

is less than the standard deviation of Dataset

P

.

The mean of Dataset

Q

is equal to the mean of Dataset

P

, and the standard deviation of Dataset

Q

is greater than the standard deviation of Dataset

P

.

The mean of Dataset

Q

is equal to the mean of Dataset

P

, and the standard deviation of Dataset

Q

is equal to the standard deviation of Dataset

P

.

The mean of Dataset

Q

is greater than the mean of Dataset

P

, and the standard deviation of Dataset

Q

is less than the standard deviation of Dataset

P

.

The mean of Dataset

Q

is less than the mean of Dataset

P

, and the standard deviation of Dataset

Q

is greater than the standard deviation of Dataset

P

.

Show answer & explanation

Answer: The mean of Dataset $Q$ is equal to the mean of Dataset $P$ , and the standard deviation of Dataset $Q$ is less than the standard deviation of Dataset $P$ .

Answer

The mean of Dataset

Q

is equal to the mean of Dataset

P

, and the standard deviation of Dataset

Q

is less than the standard deviation of Dataset

P

.

The mean of both datasets is 30 because the decrease of 8 from 50 balances the increase of 8 to 10. Standard deviation quantifies how far data points deviate from the mean. Because 18 and 42 are closer to the mean of 30 than 10 and 50 are, the spread of Dataset Q around the mean is strictly smaller, making its standard deviation smaller.

Step-by-Step Solution

1

Calculate the mean of Dataset

P

.

\text{Mean of } P = \frac{10 + 20 + 30 + 40 + 50}{5} = \frac{150}{5} = 30

.

To find the baseline central tendency before the dataset values are modified.

2

Calculate the mean of Dataset

Q

.

\text{Mean of } Q = \frac{18 + 20 + 30 + 40 + 42}{5} = \frac{150}{5} = 30

.

Replacing

10

with

18

(+8) and

50

with

42

(-8) results in a net change of zero to the sum, so the mean remains unchanged.

3

Compare the dispersion of Dataset

Q

relative to Dataset

P

.

In Dataset

P

, the squared deviations of the modified points from the mean are

(10 - 30)^2 = 400

and

(50 - 30)^2 = 400

. In Dataset

Q

, the squared deviations of these points are

(18 - 30)^2 = 144

and

(42 - 30)^2 = 144

.

Standard deviation measures the average distance of data points from the mean. Since the outer values in Dataset

Q

are closer to the mean than in Dataset

P

, the overall dispersion and standard deviation decrease.

Key Concept

Standard deviation measures the spread of data points around their mean; bringing extreme values closer to the mean reduces the standard deviation.

Question 1566Question →

Dataset $W$ consists of $15$ distinct real numbers with mean $\mu$ , standard deviation $\sigma > 0$ , interquartile range $I$ , and range $R$ . A 16th numerical value equal to the mean $\mu$ is added to dataset $W$ to form a new dataset $W'$ . Which of the following statements MUST be true regarding dataset $W'$ compared to dataset $W$ ? Select all that apply.

Select all that apply

The mean of dataset

W'

is equal to the mean of dataset

W

.

The standard deviation of dataset

W'

is strictly less than the standard deviation of dataset

W

.

The range of dataset

W'

is equal to the range of dataset

W

.

The interquartile range of dataset

W'

must be strictly greater than the interquartile range of dataset

W

.

The standard deviation of dataset

W'

is strictly greater than the standard deviation of dataset

W

.

Show answer & explanation

Answer: The mean of dataset $W'$ is equal to the mean of dataset $W$ .; The standard deviation of dataset $W'$ is strictly less than the standard deviation of dataset $W$ .; The range of dataset $W'$ is equal to the range of dataset $W$ .

Answer

The statements asserting that the mean of dataset

W'

equals the mean of dataset

W

, the standard deviation of dataset

W'

is strictly less than that of dataset

W

, and the range of dataset

W'

equals the range of dataset

W

must all be true.

Adding an element equal to the mean leaves the total sum of squared deviations from the mean unchanged while increasing the sample size by 1. Consequently, the mean remains unchanged, the standard deviation decreases by a factor of

\sqrt{15/16}

, and because the mean lies strictly inside the range of distinct values, the minimum and maximum remain unchanged, preserving the range.

Step-by-Step Solution

1

Analyze the impact on the mean when adding

x_{16} = \mu

.

The sum of elements in

W'

is

15\mu + \mu = 16\mu

. The new mean is

\frac{16\mu}{16} = \mu

.

Adding a value equal to the mean preserves the mean value.

2

Analyze the impact on the range.

Because all 15 elements are distinct real numbers,

\min(W) < \mu < \max(W)

. Adding

\mu

does not change the minimum or maximum values, so

\text{Range}(W') = \max(W) - \min(W) = R

.

The range depends solely on the maximum and minimum elements.

3

Analyze the impact on the standard deviation.

The sum of squared deviations for

W'

is

\sum_{i=1}^{16} (x_i - \mu)^2 = \sum_{i=1}^{15} (x_i - \mu)^2 + (\mu - \mu)^2 = 15\sigma^2

. The new variance is

\sigma'^2 = \frac{15\sigma^2}{16}

, so

\sigma' = \sigma \sqrt{\frac{15}{16}} < \sigma

.

Increasing the count

n

without increasing the total squared deviation reduces overall dispersion around the mean.

Key Concept

Effect of adding central summary values on measures of dispersion and central tendency.

Question 1567Question →

If $x$ is a real number satisfying the exponential equation $9^{x+1} - 3^{2x+1} = 162$ , what is the value of $4^x$ ?

4

6

8

16

64

Show answer & explanation

Answer: 8

Answer

8

By converting

9^{x+1}

to

3^{2x+2}

and factoring out

3^{2x}

, the equation simplifies to

6 \cdot 3^{2x} = 162

. Dividing by 6 gives

3^{2x} = 27

, so

2x = 3

and

x = \frac{3}{2}

. Raising 4 to the power of

\frac{3}{2}

yields

(\sqrt{4})^3 = 8

.

Step-by-Step Solution

1

Rewrite terms with a common base of 3.

9^{x+1} = (3^2)^{x+1} = 3^{2(x+1)} = 3^{2x+2}

Since 9 is a power of 3 (

9 = 3^2

), applying exponent rules converts the equation to base 3.

2

Factor out the common exponential expression

3^{2x}

.

3^{2x+2} - 3^{2x+1} = 3^{2x} \cdot 3^2 - 3^{2x} \cdot 3^1 = 3^{2x}(9 - 3) = 6 \cdot 3^{2x}

Using product rule of exponents (

3^{a+b} = 3^a \cdot 3^b

) allows combining like terms.

3

Solve for

x

.

6 \cdot 3^{2x} = 162 \implies 3^{2x} = 27 \implies 3^{2x} = 3^3 \implies 2x = 3 \implies x = \frac{3}{2}

Dividing both sides by 6 yields

3^{2x} = 27

, and equating exponents of matching bases gives

x = \frac{3}{2}

.

4

Evaluate the target expression

4^x

.

4^{3/2} = (4^{1/2})^3 = 2^3 = 8

Substituting

x = \frac{3}{2}

into

4^x

means taking the square root of 4 and raising it to the third power.

Key Concept

Solving exponential equations using common bases and exponent properties

Question 1568Question →

A triangle has side lengths of $8$ , $11$ , and $x$ , where $x$ is an integer. If the perimeter of the triangle is a positive integer multiple of $5$ , which of the following could be the value of $x$ ? Select all such values.

Select all that apply

1

6

11

16

21

Show answer & explanation

Answer: $6$ ; $11$ ; $16$

Answer

The possible values for

x

are

6

,

11

, and

16

.

According to the Triangle Inequality Theorem, the third side

x

must be strictly greater than

11 - 8 = 3

and strictly less than

11 + 8 = 19

. The perimeter of the triangle is

8 + 11 + x = 19 + x

. For

19 + x

to be a positive multiple of

5

,

19 + x

can be

25

,

30

, or

35

within the allowed range for

x

, giving

x = 6

,

x = 11

, and

x = 16

.

Step-by-Step Solution

1

Apply the Triangle Inequality Theorem to determine the valid range for the unknown side

x

.

The length of

x

must satisfy

11 - 8 < x < 11 + 8

, which simplifies to

3 < x < 19

.

The sum of any two side lengths of a non-degenerate triangle must be strictly greater than the third side length.

2

Set up an expression for the perimeter

P

of the triangle.

P = 8 + 11 + x = 19 + x

.

Perimeter is the total sum of all three side lengths.

3

Determine which values of

x

within the range

3 < x < 19

make

P = 19 + x

a multiple of

5

.

If

x = 6

,

P = 25

(multiple of

5

). If

x = 11

,

P = 30

(multiple of

5

). If

x = 16

,

P = 35

(multiple of

5

).

Adding

19

to

6

,

11

, and

16

gives multiples of

5

within the strict inequality bounds.

4

Test boundary values outside the inequality bounds.

x = 1

gives

P = 20

, but

1 \le 3

(invalid).

x = 21

gives

P = 40

, but

21 \ge 19

(invalid).

Values outside

3 < x < 19

cannot form a valid triangle.

Key Concept

Triangle Inequality Theorem and Perimeter Constraints

Question 1569Question →

In the $xy$ -plane, line $L$ is given by the equation $3x + 2y = 18$ . Line $N$ is perpendicular to line $L$ and passes through the point $(-1, 4)$ . If line $L$ and line $N$ intersect at the point $(p, q)$ , what is the value of $p + q$ ?

Show answer & explanation

Answer: 8

Answer

8

Converting the given equation

3x + 2y = 18

into slope-intercept form yields

y = -\frac{3}{2}x + 9

, so line

L

has slope

-\frac{3}{2}

. A perpendicular line must have a slope equal to the negative reciprocal, which is

\frac{2}{3}

. Using point

(-1, 4)

in the point-slope form gives

y - 4 = \frac{2}{3}(x + 1)

, simplifying to

2x - 3y = -14

. Solving the system of equations formed by line

L

(

3x + 2y = 18

) and line

N

(

2x - 3y = -14

) via elimination yields

x = 2

and

y = 6

. Therefore, the intersection point is

(2, 6)

, and

p + q = 2 + 6 = 8

.

Step-by-Step Solution

1

Determine the slope of line

L

Slope of line

L

is

m_L = -\frac{3}{2}

Convert

3x + 2y = 18

to

y = -\frac{3}{2}x + 9

to identify the slope coefficient of

x

.

2

Calculate the perpendicular slope for line

N

Slope of line

N

is

m_N = \frac{2}{3}

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

3

Derive the equation for line

N

Equation of line

N

is

2x - 3y = -14

Apply point-slope form

y - 4 = \frac{2}{3}(x + 1)

and rearrange into standard linear form.

4

Solve the system of equations to find the intersection point

(p, q)

p = 2

and

q = 6

, giving point

(2, 6)

Eliminate variable

y

by adding

3 \times (3x + 2y = 18)

and

2 \times (2x - 3y = -14)

to get

13x = 26

.

5

Sum the coordinates

p

and

q

p + q = 8

Evaluate

2 + 6 = 8

as required by the stem.

Key Concept

Perpendicular line slope relationships and linear system intersection

Estimated Time:2m 0s

Question 1570Question →

An executive chartered a private aircraft to complete a trip between two cities separated by a non-stop distance of $600$ miles. On the return flight along the exact same route, strong headwinds reduced the aircraft's average ground speed by $50$ miles per hour compared to its outbound speed. As a result, the return flight took $2$ hours longer than the outbound flight. What was the average speed, in miles per hour, of the aircraft on the outbound flight?

Show answer & explanation

Answer: 150

Answer

150 miles per hour

By setting the outbound speed to

v

and return speed to

v - 50

, the relationship between outbound flight time

\frac{600}{v}

and return flight time

\frac{600}{v - 50}

yields the rational equation

\frac{600}{v - 50} - \frac{600}{v} = 2

. Solving this equation gives the quadratic

v^2 - 50v - 15,000 = 0

, which factors as

(v - 150)(v + 100) = 0

. The positive root gives an outbound speed of 150 miles per hour.

Step-by-Step Solution

1

Define variables for the unknown outbound rate and express travel times for both legs.

Let

v

be the outbound speed in miles per hour. Outbound time is

\frac{600}{v}

hours and return time is

\frac{600}{v - 50}

hours.

Distance equals speed multiplied by time (

d = v \cdot t

), so time equals distance divided by speed.

2

Formulate the algebraic equation using the given difference in flight durations.

\frac{600}{v - 50} - \frac{600}{v} = 2

The return flight took

2

hours longer than the outbound flight.

3

Clear denominators and simplify into standard quadratic form.

v^2 - 50v - 15,000 = 0

Multiplying both sides by

v(v - 50)

yields

600v - 600v + 30,000 = 2(v^2 - 50v)

, which simplifies to

2v^2 - 100v - 30,000 = 0

or

v^2 - 50v - 15,000 = 0

.

4

Solve the quadratic equation for

v

.

v = 150

Factoring

(v - 150)(v + 100) = 0

gives solutions

v = 150

or

v = -100

. Physical speed must be positive.

Key Concept

Distance-Rate-Time Quadratic Algebraic Modeling

Question 1571Question →

In triangle $ABC$ , the ratio of the side lengths $AB : BC : AC$ is $3 : 4 : 5$ , and the total area of triangle $ABC$ is $24$ square units. A line segment $DE$ is drawn parallel to side $BC$ , where point $D$ lies on side $AB$ and point $E$ lies on side $AC$ . If the perimeter of triangle $ADE$ is exactly half the perimeter of triangle $ABC$ , what is the area of trapezoid $DBCE$ in square units?

6

12

16

18

20

Show answer & explanation

Answer: 18

Answer

The area of trapezoid

DBCE

is

18

square units.

Because line segment

DE

is parallel to side

BC

, triangle

ADE

is similar to triangle

ABC

. Given that the perimeter of triangle

ADE

is half the perimeter of triangle

ABC

, the ratio of their side lengths (the linear scale factor) is

1/2

. The area ratio of similar triangles is the square of the linear scale factor, which is

(1/2)^2 = 1/4

. Thus, the area of triangle

ADE

is

1/4 \times 24 = 6

square units. Subtracting this from the total area gives the area of trapezoid

DBCE

:

24 - 6 = 18

square units.

Step-by-Step Solution

1

Determine the linear scale factor between triangle

ADE

and triangle

ABC

.

Linear scale factor

k = \frac{\text{Perimeter}(ADE)}{\text{Perimeter}(ABC)} = \frac{1}{2}

.

Since

DE \parallel BC

, triangle

ADE

is similar to triangle

ABC

, so the ratio of their perimeters equals the ratio of corresponding side lengths.

2

Calculate the area of triangle

ADE

using the area scale factor

k^2

.

\text{Area}(ADE) = k^2 \times \text{Area}(ABC) = \left(\frac{1}{2}\right)^2 \times 24 = \frac{1}{4} \times 24 = 6

square units.

The ratio of the areas of two similar figures is the square of their linear scale factor.

3

Subtract the area of triangle

ADE

from the area of triangle

ABC

to find the area of trapezoid

DBCE

.

\text{Area}(DBCE) = \text{Area}(ABC) - \text{Area}(ADE) = 24 - 6 = 18

square units.

Trapezoid

DBCE

is formed by removing triangle

ADE

from triangle

ABC

.

Key Concept

Properties of Similar Triangles and Area Scaling

Estimated Time:1m 30s

Question 1572Question →

Trapezoid $PQRS$ has parallel sides $PQ$ and $RS$ with lengths of $7$ centimeters and $13$ centimeters, respectively. If the perpendicular height between these parallel sides is $4$ centimeters, what is the area of trapezoid $PQRS$ , in square centimeters?

24

28

40

52

80

Show answer & explanation

Answer: $40$

Answer

The area of trapezoid

PQRS

is

40

square centimeters.

The area of a trapezoid is found by averaging the lengths of the two parallel bases and multiplying by the perpendicular height:

\text{Area} = \frac{7 + 13}{2} \times 4 = 10 \times 4 = 40

square centimeters.

Step-by-Step Solution

1

Identify the formula for the area of a trapezoid

\text{Area} = \frac{b_1 + b_2}{2} \times h

, where

b_1

and

b_2

are the lengths of the parallel bases and

h

is the height.

The area of any trapezoid is equal to the average of its parallel bases multiplied by its perpendicular height.

2

Substitute the given dimensions into the formula

\text{Area} = \frac{7 + 13}{2} \times 4

The given bases are

b_1 = 7\text{ cm}

and

b_2 = 13\text{ cm}

, and the height is

h = 4\text{ cm}

.

3

Calculate the average base length and multiply by the height

\frac{20}{2} \times 4 = 10 \times 4 = 40\text{ cm}^2

Simplifying

\frac{7 + 13}{2}

gives

10

, and

10 \times 4 = 40

.

Key Concept

Trapezoid Area Formula

Estimated Time:45s

Question 1573Question →

In the $xy$ -plane, line $p$ is defined by the equation $y = -\frac{3}{4}x + 3$ . Which of the following statements about line $p$ must be true? Select all that apply.

Select all that apply

Line

p

passes through the point

(4, 0)

.

Line

p

is perpendicular to any line with a slope of

\frac{4}{3}

.

The

y

-intercept of line

p

is

(0, -3)

.

Line

p

passes through Quadrant III.

The slope of line

p

is

-\frac{4}{3}

.

Show answer & explanation

Answer: Line $p$ passes through the point $(4, 0)$ .; Line $p$ is perpendicular to any line with a slope of $\frac{4}{3}$ .

Answer

The statements asserting that line

p

passes through the point

(4, 0)

and that line

p

is perpendicular to any line with a slope of

\frac{4}{3}

are correct.

Substituting

x = 4

into the line equation gives

y = 0

, showing that

(4, 0)

lies on line

p

. Additionally, the slope of line

p

is

-\frac{3}{4}

, and the negative reciprocal of

-\frac{3}{4}

is

\frac{4}{3}

, making any line with slope

\frac{4}{3}

perpendicular to line

p

.

Step-by-Step Solution

1

Identify the slope and

y

-intercept directly from the equation

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

.

The slope is

m = -\frac{3}{4}

and the

y

-intercept is

(0, 3)

.

The equation is given in standard slope-intercept form

y = mx + b

.

2

Verify point inclusion and perpendicular slope relationship.

Substituting

x = 4

yields

y = -\frac{3}{4}(4) + 3 = 0

, confirming

(4, 0)

is on the line. The negative reciprocal of

-\frac{3}{4}

is

\frac{4}{3}

, confirming the perpendicular line slope.

A point lies on a line if its coordinates satisfy the equation, and perpendicular lines have slopes that multiply to

-1

.

3

Analyze quadrant coverage.

Line

p

connects

(0, 3)

on the positive

y

-axis to

(4, 0)

on the positive

x

-axis, covering Quadrant I. For

x < 0

,

y > 3

(Quadrant II). For

x > 4

,

y < 0

(Quadrant IV). It never enters Quadrant III where both coordinates are negative.

A line with a positive

y

-intercept and negative slope crosses Quadrants I, II, and IV only.

Key Concept

Line properties in coordinate geometry including slope, intercepts, perpendicularity, and quadrant passage.

Question 1574Question →

In a circle centered at point $O$ , the radius is $6$ units and central angle $\angle POQ$ measures $120^\circ$ . Which of the following statements regarding sector $POQ$ and minor arc $PQ$ must be true? Select all such statements.

Select all that apply

The length of minor arc

PQ

is

4\pi

units.

The area of sector

POQ

is

12\pi

square units.

The area of sector

POQ

is

36\pi

square units.

The length of minor arc

PQ

is

12\pi

units.

The perimeter of sector

POQ

is

4\pi + 6

units.

Show answer & explanation

Answer: The length of minor arc $PQ$ is $4\pi$ units.; The area of sector $POQ$ is $12\pi$ square units.

Answer

The correct statements are that the length of minor arc

PQ

is

4\pi

units and the area of sector

POQ

is

12\pi

square units.

The correct statements accurately apply the arc length and sector area formulas by multiplying the full circumference (

12\pi

) and full area (

36\pi

) by the central angle fraction

\frac{120^\circ}{360^\circ} = \frac{1}{3}

, yielding an arc length of

4\pi

units and a sector area of

12\pi

square units.

Step-by-Step Solution

1

Calculate the central angle fraction of the circle.

The fraction of the circle represented by sector

POQ

is

\frac{120^\circ}{360^\circ} = \frac{1}{3}

.

Arc length and sector area are proportional to the ratio of the central angle to

360^\circ

.

2

Calculate the arc length of minor arc

PQ

.

\text{Arc length} = \frac{1}{3} \times 2\pi(6) = 4\pi \text{ units}.

Arc length equals the central angle fraction times the total circumference

2\pi r

.

3

Calculate the area of sector

POQ

.

\text{Sector area} = \frac{1}{3} \times \pi(6^2) = 12\pi \text{ square units}.

Sector area equals the central angle fraction times the total circle area

\pi r^2

.

4

Evaluate the given statements against the calculated values.

Statements asserting an arc length of

4\pi

units and a sector area of

12\pi

square units are true. Other statements miscalculate by omitting the fraction or adding incorrect boundary components.

Comparing calculated values confirms the valid choices.

Key Concept

Arc Length and Sector Area Formulas

Question 1575Question →

In the $xy$ -plane, line $k$ is defined by the equation $y = 3x - 4$ . Line $L$ is perpendicular to line $k$ and passes through the point $(6, 2)$ . What is the $y$ -intercept of line $L$ ?

-16

-4

4

12

20

Show answer & explanation

Answer: $4$

Answer

The

y

-intercept of line

L

is

4

.

Line

k

has a slope of

3

. Since line

L

is perpendicular to line

k

, the slope of line

L

is

-\frac{1}{3}

. Substituting the point

(6, 2)

into the slope-intercept equation

y = mx + b

gives

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

, which simplifies to

2 = -2 + b

, so

b = 4

. Therefore, the

y

-intercept is

4

.

Step-by-Step Solution

1

Find the slope of line

L

The slope of line

L

is

-\frac{1}{3}

.

Line

k

has equation

y = 3x - 4

, so its slope is

3

. Perpendicular lines have slopes that are negative reciprocals.

2

Substitute the point

(6, 2)

and slope

-\frac{1}{3}

into the slope-intercept form

2 = -\frac{1}{3}(6) + b \implies 2 = -2 + b

The equation of a line is

y = mx + b

, where

m

is the slope and

b

is the

y

-intercept.

3

Solve for

b

b = 4

Adding

2

to both sides yields the

y

-intercept.

Key Concept

Perpendicular Line Slopes and Slope-Intercept Form

Question 1576Question →

In the $xy$ -plane, triangle $PQR$ has vertices at $P(0, 0)$ , $Q(8, 0)$ , and $R(2, 6)$ . Point $S$ lies on segment $PQ$ such that segment $RS$ divides triangle $PQR$ into two regions of equal area. Point $T$ lies on segment $QR$ such that segment $ST$ is parallel to segment $PR$ . What is the area of triangle $QST$ ?

Show answer & explanation

Answer: 6

Answer

6

The area of triangle

PQR

is calculated using base

PQ = 8

and height

h = 6

, yielding

\frac{1}{2} \times 8 \times 6 = 24

. Since segment

RS

divides triangle

PQR

into two regions of equal area that share the height from vertex

R

, point

S

must be the midpoint of

PQ

, making

QS = 4

. Because segment

ST

is parallel to segment

PR

, triangle

QST

is similar to triangle

QPR

with a side length ratio of

\frac{QS}{QP} = \frac{4}{8} = \frac{1}{2}

. The ratio of the areas of similar triangles is the square of the side ratio,

\left(\frac{1}{2}\right)^2 = \frac{1}{4}

. Multiplying the total area

24

by

\frac{1}{4}

gives the area of triangle

QST

as

6

.

Step-by-Step Solution

1

Calculate the area of the main triangle

PQR

.

\text{Area}(PQR) = \frac{1}{2} \times 8 \times 6 = 24

.

The base

PQ

lies along the x-axis with length

8 - 0 = 8

, and the perpendicular height from vertex

R(2,6)

to the base is

6

.

2

Determine the length of segment

QS

.

QS = 4

.

Line segment

RS

splits

\triangle PQR

into two smaller triangles,

\triangle PSR

and

\triangle QSR

, which share the same altitude from vertex

R

. For their areas to be equal, their base lengths

PS

and

SQ

must be equal. Therefore,

S

is the midpoint of

PQ

, giving

QS = \frac{8}{2} = 4

.

3

Establish the similarity relationship and scale factor between

\triangle QST

and

\triangle QPR

.

\triangle QST \sim \triangle QPR

with scale factor

k = \frac{1}{2}

.

Because segment

ST

is parallel to segment

PR

, corresponding angles are equal (

\angle QST = \angle QPR

and

\angle QTS = \angle QRP

). Thus,

\triangle QST

is similar to

\triangle QPR

. The ratio of corresponding side lengths is

\frac{QS}{QP} = \frac{4}{8} = \frac{1}{2}

.

4

Compute the area of triangle

QST

.

\text{Area}(QST) = 6

.

The ratio of the areas of similar triangles is equal to the square of their linear scale factor:

\text{Area}(QST) = \left(\frac{1}{2}\right)^2 \times \text{Area}(PQR) = \frac{1}{4} \times 24 = 6

.

Key Concept

Area of triangles, midpoint area partitioning, and area ratio scaling in similar triangles

Estimated Time:2m 0s

Question 1577Question →

Dataset $S$ consists of $100$ distinct positive numbers. The 25th percentile of dataset $S$ is $40$ , the median is $60$ , and the 75th percentile is $80$ . A new dataset $T$ is formed by multiplying every number in dataset $S$ that is strictly greater than the 80th percentile by $2$ , while keeping all other numbers unchanged. Which of the following statistics MUST be identical for dataset $S$ and dataset $T$ ?

The mean

The standard deviation

The interquartile range

The range

The 90th percentile

Show answer & explanation

Answer: The interquartile range

Answer

The interquartile range

The interquartile range is the difference between the 75th percentile (

Q_3

) and the 25th percentile (

Q_1

). Because only values strictly greater than the 80th percentile are scaled, the values defining

Q_1

and

Q_3

are unaffected. Therefore,

Q_1 = 40

and

Q_3 = 80

remain the same in both datasets, making the interquartile range (

80 - 40 = 40

) identical.

Step-by-Step Solution

1

Identify which values in the dataset are modified by the transformation

Only numbers strictly greater than the 80th percentile are multiplied by 2. All numbers at or below the 80th percentile remain unchanged.

The transformation criteria specifies that values below or equal to the 80th percentile threshold undergo no change.

2

Evaluate the effect on the 25th percentile (

Q_1

) and 75th percentile (

Q_3

)

Since

Q_1

(25th percentile) and

Q_3

(75th percentile) are position metrics located at or below the 80th percentile mark, neither

Q_1

nor

Q_3

changes in value.

Modifying elements only above the 80th percentile leaves the ordering and exact values of elements up to the 80th percentile completely intact.

3

Determine the impact on the Interquartile Range (IQR)

The interquartile range is defined as

IQR = Q_3 - Q_1

. Because both

Q_1

and

Q_3

remain constant,

IQR

is unchanged.

A metric defined solely by unchanged percentile boundaries must itself remain constant.

Key Concept

Effect of Upper-Tail Data Transformations on Measures of Position and Dispersion

Estimated Time:2m 0s

Question 1578Question →

A community library purchases two types of books: hardcover books for $\$24$ each and paperback books for $\$15$ each. The library spends a total of $\$360$ on these books and purchases at least one book of each type. Which of the following could be the total number of books purchased? Select all such numbers.

Select all that apply

15

18

20

21

24

Show answer & explanation

Answer: 18; 21

Answer

18 and 21

The linear modeling equation

24x + 15y = 360

simplifies to

8x + 5y = 120

. Since

x

and

y

must be positive integers,

x

must be a multiple of 5. The only valid solutions satisfying

x \ge 1

and

y \ge 1

are

(x=5, y=16)

and

(x=10, y=8)

, which yield total book counts of 21 and 18, respectively.

Step-by-Step Solution

1

Set up the linear equation from the given word problem context.

24x + 15y = 360

, where

x \ge 1

is the number of hardcover books and

y \ge 1

is the number of paperback books, with

x, y \in \mathbb{Z}^+

.

Total expenditure is the sum of cost per hardcover times number of hardcovers plus cost per paperback times number of paperbacks.

2

Simplify the equation by dividing both sides by the greatest common divisor, 3.

8x + 5y = 120

Simplifying coefficients reduces arithmetic complexity and isolates integer conditions.

3

Express

y

in terms of

x

to identify valid integer pairs

(x, y)

.

y = \frac{120 - 8x}{5} = 24 - \frac{8x}{5}

For

y

to be an integer,

8x

must be divisible by 5, meaning

x

must be a positive multiple of 5.

4

Test valid positive integer values for

x

such that

y \ge 1

.

If

x = 5

,

y = 24 - 8 = 16

, giving total books

x + y = 21

. If

x = 10

,

y = 24 - 16 = 8

, giving total books

x + y = 18

. If

x \ge 15

,

y \le 0

, which is invalid.

These are the only integer solutions satisfying

x \ge 1

and

y \ge 1

.

Key Concept

Linear Diophantine Equations in Word Problems

Estimated Time:1m 30s

Question 1579Question →

In $\triangle ABC$ , the measure of $\angle A$ is $45^\circ$ , the measure of $\angle C$ is $30^\circ$ , and the length of side $AC$ is $6 + 2\sqrt{3}$ . What is the length of side $AB$ ?

2\sqrt{6}

6\sqrt{2}

4\sqrt{3}

2\sqrt{3}

3\sqrt{2} + \sqrt{6}

Show answer & explanation

Answer: $2\sqrt{6}$

Answer

The length of side

AB

is

2\sqrt{6}

.

The correct answer is

2\sqrt{6}

. By drawing altitude

BD

from vertex

B

to side

AC

,

\triangle ABC

is decomposed into right triangle

BDA

(a

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

triangle) and right triangle

BDC

(a

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

triangle). Setting altitude

BD = x

gives

AD = x

and

CD = x\sqrt{3}

. Combining these gives

AC = x + x\sqrt{3} = x(1 + \sqrt{3})

. Equating this to

6 + 2\sqrt{3} = 2\sqrt{3}(1 + \sqrt{3})

gives

x = 2\sqrt{3}

. The hypotenuse

AB

of the

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

triangle is

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

.

Step-by-Step Solution

1

Draw altitude

BD

perpendicular to side

AC

with point

D

lying on segment

AC

.

\triangle ABC

is partitioned into two adjacent right triangles:

\triangle BDA

and

\triangle BDC

.

Constructing an interior altitude allows the application of special right triangle ratio rules.

2

Analyze

\triangle BDC

(

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

right triangle).

If

BD = x

, then

CD = x\sqrt{3}

and

BC = 2x

.

In a

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

triangle, sides opposite the angles are in the ratio

1 : \sqrt{3} : 2

.

3

Analyze

\triangle BDA

(

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

right triangle).

AD = BD = x

, and hypotenuse

AB = x\sqrt{2}

.

In a

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

isosceles right triangle, legs are equal and the hypotenuse is

\text{leg} \times \sqrt{2}

.

4

Set up an equation for total side length

AC = AD + CD

.

x + x\sqrt{3} = 6 + 2\sqrt{3} \implies x(1 + \sqrt{3}) = 2\sqrt{3}(1 + \sqrt{3}) \implies x = 2\sqrt{3}

.

Segment addition postulate combines

AD

and

CD

to match given total length

AC

.

5

Calculate requested side length

AB

.

AB = x\sqrt{2} = (2\sqrt{3})(\sqrt{2}) = 2\sqrt{6}

.

Substitute

x = 2\sqrt{3}

into the expression for hypotenuse

AB

.

Key Concept

Partitioning non-right triangles into

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

and

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

special right triangles by constructing an altitude.

Question 1580Question →

Two automated assembly lines, Line A and Line B, produce components at constant individual rates. Under normal operating conditions, Line A operating for $3$ hours and Line B operating for $4$ hours together produce a combined total of $1,400$ units. Under adjusted operating conditions, Line A operates at a rate $20\%$ higher than its normal rate, while Line B operates at a rate $10\%$ lower than its normal rate. Operating together under these adjusted conditions for $5$ hours, the two lines produce a total of $2,100$ units. What is the normal rate of Line A, in units per hour?

160

200

240

250

300

Show answer & explanation

Answer: $200$

Answer

The normal rate of Line A is

200

units per hour.

Let

r_A

and

r_B

represent the normal production rates in units per hour for Line A and Line B, respectively. From the first condition,

3r_A + 4r_B = 1400

. From the second condition, operating for

5

hours at rates

1.20r_A

and

0.90r_B

yields

5(1.20r_A + 0.90r_B) = 2100

, which simplifies to

1.20r_A + 0.90r_B = 420

, or

4r_A + 3r_B = 1400

. Subtracting

3r_A + 4r_B = 1400

from

4r_A + 3r_B = 1400

gives

r_A - r_B = 0

, meaning

r_A = r_B

. Substituting

r_B = r_A

into

3r_A + 4r_A = 1400

gives

7r_A = 1400

, so

r_A = 200

units per hour.

Step-by-Step Solution

1

Define variables and set up the equation for normal operating conditions.

3r_A + 4r_B = 1400

Line A operates for

3

hours at rate

r_A

and Line B operates for

4

hours at rate

r_B

to produce

1,400

units.

2

Set up the equation for adjusted operating conditions.

5(1.20r_A + 0.90r_B) = 2100 \implies 1.20r_A + 0.90r_B = 420

Line A's rate increases by

20\%

(

1.20r_A

) and Line B's rate decreases by

10\%

(

0.90r_B

). Divided by

5

hours, their combined hourly adjusted rate is

420

units per hour.

3

Multiply the simplified adjusted equation by

10

to clear decimals.

12r_A + 9r_B = 4200 \implies 4r_A + 3r_B = 1400

Dividing all terms by

3

simplifies the linear equation for easier elimination.

4

Solve the system of equations for

r_A

.

r_A = 200

From Step 1,

4r_B = 1400 - 3r_A \implies r_B = 350 - 0.75r_A

. Substituting into

4r_A + 3(350 - 0.75r_A) = 1400

gives

4r_A + 1050 - 2.25r_A = 1400 \implies 1.75r_A = 350 \implies r_A = 200

.

Key Concept

Linear Modeling of Combined Work and Rates

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

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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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

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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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