All practice questions

Question 1421Question →

Two concentric circles centered at point $O$ have radii $r$ and $R$ , where $r < R$ . A sector bounded by radii $OA$ and $OB$ of the outer circle has central angle $\theta^\circ$ . Region $S$ is the region lying inside sector $AOB$ but outside the inner circle. The area of region $S$ is equal to $3$ times the area of the sector of the inner circle bounded by central angle $\theta^\circ$ . If the perimeter of region $S$ is equal to $\frac{11}{6}$ times the length of arc $AB$ , what is the value of $\theta$ ?

\frac{60}{\pi}

\frac{270}{\pi}

\frac{480}{\pi}

\frac{540}{\pi}

\frac{1080}{\pi}

Show answer & explanation

Answer: $\frac{540}{\pi}$

Answer

\frac{540}{\pi}

The area of region

S

is

\frac{\theta}{360}\pi(R^2 - r^2)

, which is equal to

3

times the area of the inner sector

\frac{\theta}{360}\pi r^2

. This simplifies to

R^2 - r^2 = 3r^2 \implies R = 2r

. The perimeter of region

S

consists of outer arc length

L_{\text{outer}}

, inner arc length

L_{\text{inner}} = \frac{1}{2} L_{\text{outer}}

, and two straight line segments each of length

R - r = r

. Setting the total perimeter

\frac{3}{2} L_{\text{outer}} + 2r

equal to

\frac{11}{6} L_{\text{outer}}

yields

2r = \frac{1}{3} L_{\text{outer}}

, or

L_{\text{outer}} = 6r

. Substituting

L_{\text{outer}} = \frac{\theta}{360}(2\pi R) = \frac{4\pi r \theta}{360}

gives

\frac{4\pi r \theta}{360} = 6r

, which solves to

\theta = \frac{540}{\pi}

.

Step-by-Step Solution

1

Relate outer radius

R

to inner radius

r

using sector areas.

R = 2r

The area of region

S

is

\frac{\theta}{360}\pi(R^2 - r^2)

and the inner sector area is

\frac{\theta}{360}\pi r^2

. Setting

\frac{\theta}{360}\pi(R^2 - r^2) = 3 \cdot \frac{\theta}{360}\pi r^2

gives

R^2 - r^2 = 3r^2

, so

R^2 = 4r^2

and

R = 2r

.

2

Express the perimeter of region

S

in terms of outer arc length

L_{\text{outer}}

and radius

r

.

\text{Perimeter}(S) = \frac{3}{2} L_{\text{outer}} + 2r

Region

S

is bounded by outer arc

AB

(

L_{\text{outer}}

), inner arc

CD

(

L_{\text{inner}}

), and two straight segments

AC

and

BD

of length

R - r = 2r - r = r

. Since

R = 2r

,

L_{\text{inner}} = \frac{1}{2} L_{\text{outer}}

, making total perimeter

L_{\text{outer}} + \frac{1}{2} L_{\text{outer}} + 2(r) = \frac{3}{2} L_{\text{outer}} + 2r

.

3

Use the given perimeter relationship to express

L_{\text{outer}}

in terms of

r

.

L_{\text{outer}} = 6r

Setting

\frac{3}{2} L_{\text{outer}} + 2r = \frac{11}{6} L_{\text{outer}}

yields

2r = \left(\frac{11}{6} - \frac{9}{6}\right) L_{\text{outer}} = \frac{1}{3} L_{\text{outer}}

, so

L_{\text{outer}} = 6r

.

4

Solve for

\theta

using the definition of outer arc length.

\theta = \frac{540}{\pi}

L_{\text{outer}} = \frac{\theta}{360}(2\pi R) = \frac{\theta}{360}(4\pi r)

. Setting

\frac{4\pi r \theta}{360} = 6r

simplifies to

\frac{\pi \theta}{90} = 6

, giving

\theta = \frac{540}{\pi}

.

Key Concept

Annular sector area and perimeter relations combining arc length formulas and concentric circle geometry.

Estimated Time:2m 30s

Question 1422Question →

A circle has a radius of $9$ inches. What is the area, in square inches, of a sector of this circle formed by a central angle of $40^\circ$ ?

2\pi

9\pi

18\pi

27\pi

81\pi

Show answer & explanation

Answer: $9\pi$

Answer

The area of the sector is

9\pi

square inches.

The area of a sector with radius

r

and central angle

\theta

is calculated as

\text{Area} = \pi r^2 \left(\frac{\theta}{360^\circ}\right)

. Substituting

r = 9

and

\theta = 40^\circ

yields

\pi (9^2) \left(\frac{40^\circ}{360^\circ}\right) = 81\pi \left(\frac{1}{9}\right) = 9\pi

square inches.

Step-by-Step Solution

1

Calculate the area of the entire circle using the formula

A_{\text{circle}} = \pi r^2

.

A_{\text{circle}} = \pi (9^2) = 81\pi

square inches.

The full circle area provides the total measure from which the sector fraction is calculated.

2

Determine the fraction of the circle represented by the central angle of

40^\circ

.

\frac{40^\circ}{360^\circ} = \frac{1}{9}

.

A circle contains

360^\circ

in total, so the central angle over

360^\circ

gives the proportion of the circle covered by the sector.

3

Multiply the total area of the circle by the fraction of the sector.

\text{Sector Area} = 81\pi \times \frac{1}{9} = 9\pi

square inches.

Applying the fraction to the total area yields the specific sector area.

Key Concept

Sector Area of a Circle

Question 1423Question →

Circle $C_1$ has radius $r_1$ and Circle $C_2$ has radius $r_2$ . An arc on Circle $C_1$ subtended by a central angle of $\theta_1^\circ$ has the exact same length as an arc on Circle $C_2$ subtended by a central angle of $\theta_2^\circ$ . The sector formed by this arc in Circle $C_1$ has an area of $54\pi$ , and the sector formed by this arc in Circle $C_2$ has an area of $36\pi$ . If $\theta_1 + \theta_2 = 150$ , what is the value of $r_1$ ?

Show answer & explanation

Answer: 18

Answer

The radius

r_1

of Circle

C_1

is 18.

Using the relationship

A = \frac{1}{2} L r

, the ratio of the two sector areas gives

\frac{r_1}{r_2} = \frac{54\pi}{36\pi} = 1.5

, so

r_2 = \frac{2}{3} r_1

. Expressing arc length as

L = \frac{108\pi}{r_1}

, the central angles are

\theta_1 = \frac{19440}{r_1^2}

and

\theta_2 = \frac{29160}{r_1^2}

. Setting their sum equal to

150

leads to

\frac{48600}{r_1^2} = 150

, giving

r_1^2 = 324

and

r_1 = 18

.

Step-by-Step Solution

1

Relate sector area to arc length and radius

A_1 = \frac{1}{2} L r_1 = 54\pi

and

A_2 = \frac{1}{2} L r_2 = 36\pi

The area of a sector with arc length

L

and radius

r

is given by

A = \frac{\theta}{360}\pi r^2 = \frac{1}{2} L r

.

2

Find the ratio of

r_1

to

r_2

\frac{r_1}{r_2} = \frac{54\pi}{36\pi} = \frac{3}{2} \implies r_2 = \frac{2}{3}r_1

Dividing the first area equation by the second cancels out

\frac{1}{2}L

.

3

Express central angles in terms of

r_1

\theta_1 = \frac{19440}{r_1^2}

and

\theta_2 = \frac{29160}{r_1^2}

Since

L = \frac{108\pi}{r_1}

, substituting into

\theta_1 = \frac{L}{2\pi r_1} \times 360

yields

\theta_1 = \frac{19440}{r_1^2}

, and substituting into

\theta_2 = \frac{L}{2\pi (\frac{2}{3}r_1)} \times 360

yields

\theta_2 = \frac{29160}{r_1^2}

.

4

Solve for

r_1

using the angle sum equation

r_1 = 18

Summing the angles gives

\frac{19440 + 29160}{r_1^2} = \frac{48600}{r_1^2} = 150 \implies r_1^2 = 324 \implies r_1 = 18

.

Key Concept

Relationship between arc length, radius, central angle, and sector area

Estimated Time:2m 30s

Question 1424Question →

If $x$ and $y$ are real numbers that satisfy the inequalities $|3 - 2x| \le 9$ and $|4y + 3| \le 13$ , what is the maximum possible value of the expression $|3x - 4y|$ ?

Show answer & explanation

Answer: 34

Answer

The maximum possible value of

|3x - 4y|

is

34

.

To find the maximum possible value of

|3x - 4y|

, we first solve for the ranges of

x

and

y

from their respective absolute value inequalities. From

|3 - 2x| \le 9

, we get

-9 \le 3 - 2x \le 9 \implies -3 \le x \le 6

. From

|4y + 3| \le 13

, we get

-13 \le 4y + 3 \le 13 \implies -4 \le y \le 2.5

. To maximize

3x - 4y

, we take the maximum value of

3x

(

3 \times 6 = 18

) and the maximum value of

-4y

(

-4 \times -4 = 16

), giving

18 + 16 = 34

. To minimize

3x - 4y

, we take the minimum value of

3x

(

3 \times -3 = -9

) and the minimum value of

-4y

(

-4 \times 2.5 = -10

), giving

-9 - 10 = -19

. The expression

3x - 4y

ranges from

-19

to

34

, so the maximum possible magnitude

|3x - 4y|

is

\max(|-19|, |34|) = 34

.

Step-by-Step Solution

1

Solve the inequality

|3 - 2x| \le 9

for

x

.

-3 \le x \le 6

Unpacking the absolute value yields

-9 \le 3 - 2x \le 9

. Subtracting

3

gives

-12 \le -2x \le 6

. Dividing by

-2

and reversing the inequality signs produces

-3 \le x \le 6

.

2

Solve the inequality

|4y + 3| \le 13

for

y

.

-4 \le y \le 2.5

Unpacking the absolute value yields

-13 \le 4y + 3 \le 13

. Subtracting

3

gives

-16 \le 4y \le 10

. Dividing by

4

gives

-4 \le y \le 2.5

.

3

Find the range of possible values for

3x

and

-4y

.

-9 \le 3x \le 18

and

-10 \le -4y \le 16

Multiplying

-3 \le x \le 6

by

3

gives

-9 \le 3x \le 18

. Multiplying

-4 \le y \le 2.5

by

-4

and flipping the signs gives

-10 \le -4y \le 16

.

4

Combine the bounds for

3x

and

-4y

to find the range for

3x - 4y

.

-19 \le 3x - 4y \le 34

The minimum value of

3x - 4y

is

(-9) + (-10) = -19

. The maximum value of

3x - 4y

is

18 + 16 = 34

.

5

Determine the maximum absolute value

|3x - 4y|

over the interval

[-19, 34]

.

34

The absolute value of any number in the interval

[-19, 34]

ranges from

0

to

\max(|-19|, |34|) = 34

.

Key Concept

Absolute Value Inequalities and Expression Bounding

Question 1425Question →

In triangle $ABC$ , the measure of angle $A$ is $40^\circ$ and the measure of angle $B$ is $70^\circ$ . If the perimeter of triangle $ABC$ is $22$ centimeters and the length of side $AB$ is $8$ centimeters, what is the length, in centimeters, of side $BC$ ?

4

6

7

8

14

Show answer & explanation

Answer: $6$

Answer

6 centimeters

First, find the measure of angle

C

:

180^\circ - 40^\circ - 70^\circ = 70^\circ

. Because angle

B

and angle

C

both measure

70^\circ

, triangle

ABC

is isosceles with sides

AC

and

AB

of equal length (

AC = AB = 8

cm). Since the perimeter is the sum of all three sides (

AB + AC + BC = 22

), substituting the known side lengths gives

8 + 8 + BC = 22

, which yields

BC = 6

cm.

Step-by-Step Solution

1

Calculate the measure of the third angle, angle

C

.

Angle

C = 180^\circ - (40^\circ + 70^\circ) = 70^\circ

.

The sum of interior angles in any triangle is

180^\circ

.

2

Determine the relationship between the side lengths using the angle measures.

Side

AC = AB = 8

centimeters.

Since angle

B = 70^\circ

and angle

C = 70^\circ

, triangle

ABC

is isosceles. Sides opposite equal angles are equal in length, so the side opposite angle

C

(which is

AB

) equals the side opposite angle

B

(which is

AC

).

3

Use the perimeter formula to find the length of side

BC

.

Length of side

BC = 22 - (8 + 8) = 6

centimeters.

Perimeter is the total boundary distance (

AB + AC + BC = 22

). Substituting known values gives

8 + 8 + BC = 22

, so

BC = 6

.

Key Concept

Isosceles Triangle Side-Angle Properties and Perimeter

Estimated Time:45s

Question 1426Question →

Dataset $A$ consists of $n$ numerical values with a standard deviation of $\sigma_A$ and an interquartile range of $\text{IQR}_A$ . Dataset $B$ is constructed by multiplying each value in Dataset $A$ by $4$ and then subtracting $10$ from each resulting value. Which of the following correctly expresses the standard deviation $\sigma_B$ and the interquartile range $\text{IQR}_B$ of Dataset $B$ in terms of $\sigma_A$ and $\text{IQR}_A$ ?

\sigma_B = 4\sigma_A - 10

and

\text{IQR}_B = 4\text{IQR}_A - 10

\sigma_B = 4\sigma_A

and

\text{IQR}_B = 4\text{IQR}_A

\sigma_B = 16\sigma_A

and

\text{IQR}_B = 4\text{IQR}_A

\sigma_B = 4\sigma_A

and

\text{IQR}_B = \text{IQR}_A

\sigma_B = 4\sigma_A - 10

and

\text{IQR}_B = 4\text{IQR}_A

Show answer & explanation

Answer: $\sigma_B = 4\sigma_A$ and $\text{IQR}_B = 4\text{IQR}_A$

Answer

\sigma_B = 4\sigma_A

and

\text{IQR}_B = 4\text{IQR}_A

When a dataset undergoes a linear transformation of the form

Y = cX + k

, any measure of dispersion

D

(such as standard deviation or interquartile range) transforms according to

D_Y = |c| \cdot D_X

. The constant shift

k

shifts all points equally and does not affect spread. Therefore, multiplying by

4

multiplies both

\sigma_A

and

\text{IQR}_A

by

4

, and subtracting

10

has no effect on either measure.

Step-by-Step Solution

1

Analyze the effect of multiplying each value in a dataset by a positive constant

c = 4

.

Multiplying all data values by

c

scales all measures of spread (including standard deviation, range, and interquartile range) by

|c| = 4

. Thus, initial scaled values give

4\sigma_A

and

4\text{IQR}_A

.

Measures of dispersion reflect distances between data points, which scale linearly when all values are multiplied by a constant.

2

Analyze the effect of subtracting a constant

k = 10

from each data point.

Subtracting a constant shifts the entire distribution along the number line without altering the relative distances between any data points.

Because measures of dispersion measure relative spread rather than absolute location, adding or subtracting a constant has zero effect on standard deviation or interquartile range.

3

Combine the transformation effects to state

\sigma_B

and

\text{IQR}_B

.

\sigma_B = 4\sigma_A

and

\text{IQR}_B = 4\text{IQR}_A

.

The multiplicative factor scales both metrics by 4, and the additive shift of

-10

leaves both metrics unchanged.

Key Concept

Linear Transformations on Measures of Dispersion

Estimated Time:1m 15s

Question 1427Question →

Lines $L_1$ and $L_2$ intersect at point $P$ to form an acute angle of $54^\circ$ . Line $M$ passes through point $P$ and is perpendicular to line $L_1$ . Ray $R$ originates at point $P$ and lies in the interior of one of the obtuse angles formed by $L_1$ and $L_2$ . If Ray $R$ bisects the angle formed between line $M$ and line $L_2$ , what is the measure, in degrees, of the acute angle formed by Ray $R$ and line $L_1$ ?

Show answer & explanation

Answer: 72

Answer

The measure of the acute angle formed by Ray

R

and line

L_1

is

72^\circ

.

Lines

L_1

and

L_2

form an obtuse angle of

180^\circ - 54^\circ = 126^\circ

. Perpendicular line

M

consumes

90^\circ

of this angle relative to

L_1

, leaving

36^\circ

between line

M

and line

L_2

. Bisecting this remaining angle gives

18^\circ

. Ray

R

therefore lies

18^\circ

away from perpendicular line

M

, making an acute angle of

90^\circ - 18^\circ = 72^\circ

with line

L_1

.

Step-by-Step Solution

1

Determine the measure of the obtuse angle between lines

L_1

and

L_2

.

The obtuse angle measures

180^\circ - 54^\circ = 126^\circ

.

Adjacent angles along intersecting straight lines are supplementary and sum to

180^\circ

.

2

Find the angle between perpendicular line

M

and line

L_2

.

The angle between line

M

and line

L_2

within the obtuse region is

126^\circ - 90^\circ = 36^\circ

.

Line

M

is perpendicular to line

L_1

, taking up

90^\circ

of the

126^\circ

obtuse angle.

3

Determine the angle formed by Ray

R

after bisecting the

36^\circ

angle.

The angle between Ray

R

and line

M

is

36^\circ / 2 = 18^\circ

.

An angle bisector divides an angle into two equal parts.

4

Calculate the acute angle between Ray

R

and line

L_1

.

The acute angle formed between Ray

R

and line

L_1

is

90^\circ - 18^\circ = 72^\circ

.

Line

M

forms a

90^\circ

angle with line

L_1

. Subtracting the

18^\circ

offset created by Ray

R

yields the acute angle of

72^\circ

.

Key Concept

Supplementary angles, perpendicular lines, and angle bisectors

Estimated Time:2m 0s

Question 1428Question →

If $x$ is a real number that satisfies both $|4 - 3x| > 7$ and $\frac{1 - 2x}{3} \ge -3$ , which of the following represents the complete set of all possible values of $x$ ?

x < -1

or

\frac{11}{3} < x \le 5

-1 < x \le 5

x \ge 5

-1 < x < \frac{11}{3}

x < -1

or

x > \frac{11}{3}

Show answer & explanation

Answer: $x < -1$ or $\frac{11}{3} < x \le 5$

Answer

x < -1

or

\frac{11}{3} < x \le 5

The correct option correctly solves

|4 - 3x| > 7

to yield

x < -1

or

x > \frac{11}{3}

, solves

\frac{1 - 2x}{3} \ge -3

to yield

x \le 5

, and takes their intersection to produce

x < -1

or

\frac{11}{3} < x \le 5

.

Step-by-Step Solution

1

Solve the absolute value inequality

|4 - 3x| > 7

.

4 - 3x > 7

or

4 - 3x < -7

. Solving

4 - 3x > 7

gives

-3x > 3 \implies x < -1

. Solving

4 - 3x < -7

gives

-3x < -11 \implies x > \frac{11}{3}

. Thus,

x \in (-\infty, -1) \cup (\frac{11}{3}, \infty)

.

An absolute value inequality of the form

|u| > c

splits into

u > c

or

u < -c

. Dividing by a negative number reverses the inequality direction.

2

Solve the linear inequality

\frac{1 - 2x}{3} \ge -3

.

Multiply both sides by

3

:

1 - 2x \ge -9

. Subtract

1

:

-2x \ge -10

. Divide by

-2

and flip the inequality sign:

x \le 5

.

Isolating the variable

x

requires reversing the inequality sign when dividing by the negative constant

-2

.

3

Find the intersection of the solution sets from Step 1 and Step 2.

We require

x

to satisfy

(x < -1 \text{ or } x > \frac{11}{3})

AND

x \le 5

. Case 1:

x < -1

automatically satisfies

x \le 5

, giving

x < -1

. Case 2:

x > \frac{11}{3}

combined with

x \le 5

gives

\frac{11}{3} < x \le 5

. Combining both cases yields

x < -1 \text{ or } \frac{11}{3} < x \le 5

.

The word 'both' in the stem indicates a logical AND (intersection) between the two conditions.

Key Concept

Solving systems involving absolute value inequalities and linear inequalities requires handling disjunctions (OR) for absolute values greater than a positive constant, reversing inequality signs when multiplying or dividing by negative numbers, and taking the intersection (AND) of all valid regions.

Estimated Time:2m 15s

Question 1429Question →

A right circular cylinder has a base radius of $3$ units and a height of $4$ units. Which of the following statements regarding this cylinder are true? Select all such statements.

Select all that apply

The volume of the cylinder is

36\pi

cubic units.

The total surface area of the cylinder is

42\pi

square units.

The volume of the cylinder is

24\pi

cubic units.

The total surface area of the cylinder is

36\pi

square units.

The ratio of the area of one circular base to the lateral surface area is

1

to

2

.

Show answer & explanation

Answer: The volume of the cylinder is $36\pi$ cubic units.; The total surface area of the cylinder is $42\pi$ square units.

Answer

The correct statements are that the volume of the cylinder is

36\pi

cubic units and the total surface area of the cylinder is

42\pi

square units.

The volume of a cylinder with radius

3

and height

4

is

V = \pi r^2 h = \pi(3)^2(4) = 36\pi

. The total surface area is

A = 2\pi r h + 2\pi r^2 = 2\pi(3)(4) + 2\pi(3)^2 = 24\pi + 18\pi = 42\pi

. Therefore, both statements asserting these exact values are correct.

Step-by-Step Solution

1

Calculate the volume of the right circular cylinder

Volume

V = \pi r^2 h = \pi (3^2)(4) = 36\pi

The formula for the volume of a right circular cylinder is

V = \pi r^2 h

.

2

Calculate the lateral surface area and total surface area of the cylinder

Lateral Area =

2\pi r h = 24\pi

; Total Area =

24\pi + 2(\pi \cdot 3^2) = 42\pi

Total surface area is the sum of the lateral surface area (

2\pi r h

) and the areas of the two circular bases (

2\pi r^2

).

3

Evaluate the base-to-lateral surface area ratio

Ratio =

\frac{9\pi}{24\pi} = \frac{3}{8}

Comparing the base area (

9\pi

) to the lateral area (

24\pi

) simplifies to

3:8

.

Key Concept

Volume and surface area formulas for right circular cylinders

Question 1430Question →

If $x$ is a real number that satisfies the inequality $-2|3x - 4| + 5 \ge -7$ , which of the following inequalities represents all possible values of $x$ ?

-\frac{2}{3} \le x \le \frac{10}{3}

x \le -\frac{2}{3}

or

x \ge \frac{10}{3}

-\frac{10}{3} \le x \le \frac{2}{3}

x \le \frac{10}{3}

-\frac{1}{3} \le x \le 3

Show answer & explanation

Answer: $-\frac{2}{3} \le x \le \frac{10}{3}$

Answer

-\frac{2}{3} \le x \le \frac{10}{3}

Isolating

|3x - 4|

requires dividing

-2|3x - 4| \ge -12

by

-2

, which reverses the inequality to

|3x - 4| \le 6

. Expanding this into the compound inequality

-6 \le 3x - 4 \le 6

and solving for

x

yields the interval

-\frac{2}{3} \le x \le \frac{10}{3}

.

Step-by-Step Solution

1

Isolate the absolute value expression by subtracting 5 from both sides of the inequality.

-2|3x - 4| \ge -12

Before removing absolute value bars, the term

|3x - 4|

must be isolated on one side.

2

Divide both sides by

-2

and reverse the inequality sign.

|3x - 4| \le 6

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

3

Express the absolute value inequality

|u| \le k

as a double inequality

-k \le u \le k

.

-6 \le 3x - 4 \le 6

The distance of

3x - 4

from 0 on the number line must be at most 6 units.

4

Add 4 to all three parts of the inequality and divide by 3.

-2 \le 3x \le 10 \implies -\frac{2}{3} \le x \le \frac{10}{3}

Solving for

x

isolates the variable in the center of the double inequality.

Key Concept

Solving linear absolute value inequalities requires reversing the inequality sign when multiplying or dividing by negative numbers, and expressing

|A| \le B

as

-B \le A \le B

.

Question 1431Question →

In the geometric plane, line $k$ is parallel to line $m$ ( $k \parallel m$ ). Transversal line $t$ intersects line $k$ at point $A$ and line $m$ at point $B$ . Point $C$ lies on line $k$ to the left of $A$ , such that interior acute angle $\angle CAB = (3x - 10)^\circ$ . Ray $AD$ bisects $\angle CAB$ . Ray $BF$ is drawn into the region between lines $k$ and $m$ making an angle $\angle ABF = (x + 35)^\circ$ with transversal segment $AB$ . Ray $AD$ and ray $BF$ intersect at point $P$ inside the parallel region. Line $PB$ is extended past $P$ to intersect line $k$ at point $Q$ . If ray $AD$ is perpendicular to ray $BF$ , what is the measure of the obtuse angle formed at the intersection of line $QB$ and line $k$ ?

121^\circ

59^\circ

118^\circ

128^\circ

149^\circ

Show answer & explanation

Answer: $121^\circ$

Answer

The measure of the obtuse angle formed at the intersection of line

QB

and line

k

is

121^\circ

.

The measure

121^\circ

is correct because ray

AD

bisects

\angle CAB

, giving

\angle PAB = (1.5x - 5)^\circ

. Since

AD \perp BF

,

\triangle APB

is a right triangle where

(1.5x - 5) + (x + 35) = 90

, yielding

x = 24

. Consequently,

\angle QAB = 62^\circ

and

\angle ABQ = 59^\circ

. In

\triangle ABQ

, the acute angle

\angle AQB = 180^\circ - (62^\circ + 59^\circ) = 59^\circ

. The supplementary obtuse angle along line

k

is

180^\circ - 59^\circ = 121^\circ

.

Step-by-Step Solution

1

Express the angle

\angle PAB

in terms of

x

.

Since ray

AD

bisects

\angle CAB = (3x - 10)^\circ

,

\angle PAB = \frac{3x - 10}{2} = (1.5x - 5)^\circ

.

An angle bisector divides an angle into two equal halves.

2

Set up an equation using triangle

\triangle APB

.

In

\triangle APB

,

\angle APB = 90^\circ

because ray

AD \perp

ray

BF

. Therefore,

\angle PAB + \angle ABP = 90^\circ \implies (1.5x - 5) + (x + 35) = 90 \implies 2.5x + 30 = 90 \implies 2.5x = 60 \implies x = 24

.

The acute angles in a right triangle sum to

90^\circ

.

3

Calculate the measures of

\angle CAB

and

\angle ABQ

.

\angle CAB = 3(24) - 10 = 62^\circ

, and

\angle ABQ = \angle ABF = 24 + 35 = 59^\circ

.

Substitute

x = 24

back into the original angle expressions.

4

Determine the acute angle

\angle AQB

in triangle

\triangle ABQ

.

In

\triangle ABQ

, points

Q, C, A

lie on line

k

, so

\angle QAB = \angle CAB = 62^\circ

. Sum of angles in

\triangle ABQ

:

\angle AQB = 180^\circ - (62^\circ + 59^\circ) = 180^\circ - 121^\circ = 59^\circ

.

The interior angles of any triangle sum to

180^\circ

.

5

Find the supplementary obtuse angle at intersection point

Q

.

Obtuse angle

= 180^\circ - 59^\circ = 121^\circ

.

Angles forming a linear pair on a straight line are supplementary.

Key Concept

Parallel Lines, Transversals, Angle Bisectors, and Triangle Angle Sum Theorem

Question 1432Question →

If $|x - 3| < 5$ , which of the following values could be the value of $x$ ? Select all such values.

Select all that apply

-3

-1

0

4

8

Show answer & explanation

Answer: $-1$ ; $0$ ; $4$

Answer

The possible values of

x

are

-1

,

0

, and

4

.

The absolute value inequality

|x - 3| < 5

represents all real numbers

x

whose distance from

3

on the number line is strictly less than

5

. Expressed as a compound inequality, this means

-5 < x - 3 < 5

. Adding

3

across the entire inequality yields

-2 < x < 8

. The values

-1

,

0

, and

4

are the only options that fall strictly within the interval

(-2, 8)

.

Step-by-Step Solution

1

Rewrite the absolute value inequality as a compound inequality.

-5 < x - 3 < 5

An absolute value inequality of the form

|u| < k

(where

k > 0

) is equivalent to

-k < u < k

.

2

Solve for

x

by adding

3

to all parts of the inequality.

-5 + 3 < x < 5 + 3 \implies -2 < x < 8

Adding a positive constant to all parts of an inequality isolates

x

while preserving inequality directions.

3

Test each given option against the range

-2 < x < 8

.

The values

-1

,

0

, and

4

lie within

(-2, 8)

, whereas

-3

and

8

lie outside.

-3

is less than or equal to

-2

, and

8

is not strictly less than

8

.

Key Concept

Solving linear absolute value inequalities

Estimated Time:45s

Question 1433Question →

If

x

satisfies the linear equation

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

what is the value of

2x - 5

?

43

24

123

-1

51

Show answer & explanation

Answer: 43

Answer

43

The correct answer is obtained by clearing denominators with the least common denominator 12, yielding

9(x - 2) - 4(x - 5) = 2(2x + 1) + 24

. Expanding and simplifying gives

5x + 2 = 4x + 26

, which isolates

x = 24

. Substituting

x = 24

into

2x - 5

results in

2(24) - 5 = 43

.

Step-by-Step Solution

1

Clear the denominators by multiplying every term in the equation by the least common multiple of 4, 3, and 6, which is 12.

12 \cdot \frac{3(x - 2)}{4} - 12 \cdot \frac{x - 5}{3} = 12 \cdot \frac{2x + 1}{6} + 12 \cdot 2 \implies 9(x - 2) - 4(x - 5) = 2(2x + 1) + 24

Eliminating fractions simplifies the linear equation into standard polynomial form.

2

Distribute the constants through the parentheses, taking careful note of negative signs.

9x - 18 - 4x + 20 = 4x + 2 + 24

Parentheses must be removed to collect like variable and constant terms.

3

Combine like terms on both sides of the equation.

5x + 2 = 4x + 26

Consolidating terms on each side allows for isolating the variable.

4

Isolate

x

on one side of the equation.

5x - 4x = 26 - 2 \implies x = 24

Subtracting

4x

and 2 from both sides solves directly for

x

.

5

Evaluate the target expression

2x - 5

using

x = 24

.

2(24) - 5 = 48 - 5 = 43

The question requests the value of

2x - 5

, not

x

alone.

Key Concept

Solving multi-step linear equations in one variable with fractional terms and evaluating algebraic expressions.

Estimated Time:2m 0s

Question 1434Question →

A square is inscribed in a circle with center $O$ . The perimeter of the square is $16\sqrt{2}$ units. A sector of this circle has an area equal to the total area of the region inside the circle that lies outside the square. What is the arc length of this sector?

4\pi - 8

8\pi - 16

16\pi - 32

16\pi - 8

8\pi - 32

Show answer & explanation

Answer: $8\pi - 16$

Answer

8\pi - 16

The correct answer is

8\pi - 16

. The inscribed square has side length

4\sqrt{2}

and diagonal

8

. Since the diagonal of an inscribed square is the circle's diameter, the circle has radius

r = 4

and total area

16\pi

. The square has area

32

, so the region outside the square has area

16\pi - 32

. For any sector of radius

r

, the arc length

L

and sector area

A

satisfy

L = \frac{2A}{r}

. Substituting

A = 16\pi - 32

and

r = 4

yields

L = \frac{2(16\pi - 32)}{4} = 8\pi - 16

.

Step-by-Step Solution

1

Find the side length and diagonal of the inscribed square.

Side length

s = \frac{16\sqrt{2}}{4} = 4\sqrt{2}

. Diagonal

d = s\sqrt{2} = (4\sqrt{2})\sqrt{2} = 8

.

The perimeter of a square is

4s

, and the diagonal of a square with side

s

is

s\sqrt{2}

.

2

Determine the radius and total area of the circle.

Diameter equals diagonal

d = 8

, so radius

r = 4

. Area of circle

A_{\text{circle}} = \pi r^2 = 16\pi

.

A square inscribed in a circle has its diagonal aligned with the diameter of the circle.

3

Calculate the area of the region inside the circle but outside the square.

Area of square

A_{\text{square}} = (4\sqrt{2})^2 = 32

. Area outside square

A_{\text{outside}} = 16\pi - 32

.

Subtract the area of the inscribed square from the total area of the circle.

4

Relate sector area to arc length using

L = \frac{2 A_{\text{sector}}}{r}

.

L = \frac{2(16\pi - 32)}{4} = \frac{32\pi - 64}{4} = 8\pi - 16

.

Since sector area

A_{\text{sector}} = \frac{\theta}{360^\circ}\pi r^2

and arc length

L = \frac{\theta}{360^\circ}(2\pi r)

, we have

L = \frac{2A_{\text{sector}}}{r}

.

Key Concept

Relationship between circle area, inscribed figures, sector area, and arc length

Estimated Time:1m 30s

Question 1435Question →

Working alone at a constant rate, Pump A can drain a full water reservoir in $6$ hours, while Pump B working alone at a constant rate can drain the same full reservoir in $4$ hours. Which of the following statements must be true? Select all such statements.

Select all that apply

Working together, Pump A and Pump B can drain the entire reservoir in

2.4

hours.

In

1

hour, Pump A and Pump B working together drain

\frac{5}{12}

of the total reservoir volume.

The combined rate of Pump A and Pump B working together is

\frac{1}{10}

of the reservoir per hour.

Pump A's individual work rate accounts for

\frac{3}{5}

of the combined hourly work rate of both pumps.

Pump B's hourly rate is

33.\overline{3}\%

greater than Pump A's hourly rate.

Show answer & explanation

Answer: Working together, Pump A and Pump B can drain the entire reservoir in $2.4$ hours.; In $1$ hour, Pump A and Pump B working together drain $\frac{5}{12}$ of the total reservoir volume.

Answer

The statements confirming that the two pumps together drain the reservoir in 2.4 hours and that they drain 5/12 of the reservoir in 1 hour are correct.

The rate for Pump A is

\frac{1}{6}

job/hr and for Pump B is

\frac{1}{4}

job/hr. Their combined rate is

\frac{1}{6} + \frac{1}{4} = \frac{5}{12}

of the reservoir per hour, which directly validates the statement regarding 1 hour of combined work. Dividing

1

full reservoir by

\frac{5}{12}

yields

2.4

hours, which validates the total elapsed time statement.

Step-by-Step Solution

1

Calculate individual work rates

Pump A rate =

\frac{1}{6}

reservoir/hr, Pump B rate =

\frac{1}{4}

reservoir/hr

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

2

Calculate combined work rate

Combined rate =

\frac{1}{6} + \frac{1}{4} = \frac{2}{12} + \frac{3}{12} = \frac{5}{12}

reservoir/hr

Rates add when workers/pumps perform simultaneously.

3

Calculate time needed for full task

Total time =

\frac{1}{\text{Combined Rate}} = \frac{12}{5} = 2.4

hours

Total work (

1

) divided by combined rate gives total elapsed time.

Key Concept

Combined Work Rates and Inverse Time Relationships

Estimated Time:1m 0s

Question 1436Question →

Four rays, $\vec{OA}$ , $\vec{OB}$ , $\vec{OC}$ , and $\vec{OD}$ , radiate from a common point $O$ in consecutive clockwise order such that $\vec{OA} \perp \vec{OC}$ and $\vec{OB} \perp \vec{OD}$ . If the measure of angle $\angle AOD$ is $3.5$ times the measure of angle $\angle BOC$ , what is the measure, in degrees, of angle $\angle AOB$ ?

36^\circ

40^\circ

50^\circ

65^\circ

140^\circ

Show answer & explanation

Answer: $50^\circ$

Answer

50^\circ

Because

\vec{OA} \perp \vec{OC}

and

\vec{OB} \perp \vec{OD}

, we know

\angle AOB + \angle BOC = 90^\circ

and

\angle BOC + \angle COD = 90^\circ

, which implies

\angle AOB = \angle COD

. The total angle

\angle AOD = \angle AOB + \angle BOD = \angle AOB + 90^\circ

. Using the given condition

\angle AOD = 3.5 \times \angle BOC

, we substitute

\angle BOC = 90^\circ - \angle AOB

to get

\angle AOB + 90^\circ = 3.5(90^\circ - \angle AOB)

, which solves to

\angle AOB = 50^\circ

.

Step-by-Step Solution

1

Set up angle variable definitions and perpendicular relationships.

Let

\angle AOB = x

,

\angle BOC = y

, and

\angle COD = z

. Since

\vec{OA} \perp \vec{OC}

, we have

x + y = 90^\circ

. Since

\vec{OB} \perp \vec{OD}

, we have

y + z = 90^\circ

.

Perpendicular rays form right angles measuring

90^\circ

.

2

Deduce the relationship between

x

,

y

, and

z

, and express

\angle AOD

in terms of

x

.

Subtracting

y

from both equations gives

x = 90^\circ - y

and

z = 90^\circ - y

, so

x = z

. Thus,

\angle AOD = x + y + z = x + 90^\circ

.

Adjacent angles sharing a vertex add up to form the overall combined angle.

3

Formulate and solve the equation based on the given ratio.

We are given

\angle AOD = 3.5 \times \angle BOC

, so

x + 90^\circ = 3.5y

. Substituting

y = 90^\circ - x

yields

x + 90^\circ = 3.5(90^\circ - x) \implies x + 90^\circ = 315^\circ - 3.5x \implies 4.5x = 225^\circ \implies x = 50^\circ

.

Substitution creates a single linear equation in terms of

x = \angle AOB

.

Key Concept

Perpendicular Ray Systems and Angle Addition

Estimated Time:1m 30s

Question 1437Question →

Let $P$ be the product of all positive real numbers $x$ that satisfy the exponential equation $x^{\sqrt{x}} = \left(x^2\sqrt{x}\right)^x$ . What is the value of $P$ ?

\frac{4}{25}

\frac{2}{5}

\frac{25}{4}

\frac{16}{625}

1

Show answer & explanation

Answer: $\frac{4}{25}$

Answer

\frac{4}{25}

Simplifying the right-hand side yields

x^{\sqrt{x}} = x^{\frac{5}{2}x}

. For

x = 1

, both sides equal 1, making

x = 1

a valid solution. For

x > 0

and

x \neq 1

, equating the exponents gives

\sqrt{x} = \frac{5}{2}x

, which reduces to

\sqrt{x} = \frac{2}{5}

, so

x = \frac{4}{25}

. Multiplying all valid solutions together yields

1 \cdot \frac{4}{25} = \frac{4}{25}

.

Step-by-Step Solution

1

Simplify the right-hand side of the equation using fractional exponent rules.

Since

x^2\sqrt{x} = x^2 \cdot x^{1/2} = x^{5/2}

, the equation becomes

x^{\sqrt{x}} = \left(x^{5/2}\right)^x = x^{\frac{5}{2}x}

.

Combining terms with the same base into a single exponent simplifies comparison between both sides.

2

Check for the base root

x = 1

.

Substituting

x = 1

gives

1^{\sqrt{1}} = 1^1 = 1

and

(1^2\sqrt{1})^1 = 1^1 = 1

. Thus,

x = 1

is a valid solution.

For any exponential equation of the form

x^{f(x)} = x^{g(x)}

,

x = 1

is always a candidate solution because

1^a = 1^b = 1

for all real exponents.

3

Equate the exponents for positive real solutions where

x \neq 1

.

Setting the exponents equal gives

\sqrt{x} = \frac{5}{2}x

.

When the base

x > 0

and

x \neq 1

,

x^{f(x)} = x^{g(x)}

implies

f(x) = g(x)

.

4

Solve the resulting radical equation for

x

.

Divide both sides by

\sqrt{x}

(since

x > 0

):

1 = \frac{5}{2}\sqrt{x} \implies \sqrt{x} = \frac{2}{5} \implies x = \left(\frac{2}{5}\right)^2 = \frac{4}{25}

.

Isolating

\sqrt{x}

and squaring both sides gives the non-trivial solution.

5

Compute the product

P

of all positive real solutions.

P = 1 \cdot \frac{4}{25} = \frac{4}{25}

.

The question requests the product of all positive real values of

x

satisfying the original equation.

Key Concept

Solving exponential equations with variable bases and radical powers

Question 1438Question →

A research team measured the height, in centimeters, of 15 plants in a greenhouse. The dataset has a range of $24\text{ cm}$ , an interquartile range (IQR) of $10\text{ cm}$ , and a standard deviation of $5.2\text{ cm}$ . If the height of every plant is increased by exactly $8\text{ cm}$ , which of the following statements regarding the modified dataset must be true? Select all such statements.

Select all that apply

The range of the modified dataset is

32\text{ cm}

.

The range of the modified dataset is

24\text{ cm}

.

The interquartile range of the modified dataset is

18\text{ cm}

.

The standard deviation of the modified dataset is

5.2\text{ cm}

.

The standard deviation of the modified dataset is

13.2\text{ cm}

.

Show answer & explanation

Answer: The range of the modified dataset is $24\text{ cm}$ .; The standard deviation of the modified dataset is $5.2\text{ cm}$ .

Answer

The statement that the range of the modified dataset is

24\text{ cm}

and the statement that the standard deviation of the modified dataset is

5.2\text{ cm}

are both true.

When a constant is added to every value in a dataset, the entire distribution shifts by that amount. Consequently, the relative positions and distances between data points do not change, leaving measures of dispersion (range, IQR, standard deviation) unchanged. Therefore, the range remains

24\text{ cm}

and the standard deviation remains

5.2\text{ cm}

.

Step-by-Step Solution

1

Analyze the effect of adding a constant to a dataset on measures of dispersion.

Adding a constant

k

to every data value shifts the center (mean, median) by

k

, but all measures of spread (range, IQR, standard deviation) remain completely unchanged.

Measures of dispersion quantify the spread or distance between data values, which is invariant under a uniform translation.

2

Evaluate each statement against the unchanged measures of dispersion.

The range remains

24\text{ cm}

, the IQR remains

10\text{ cm}

, and the standard deviation remains

5.2\text{ cm}

.

Comparing the calculated original measures to the statements shows that the range of

24\text{ cm}

and standard deviation of

5.2\text{ cm}

are correct.

Key Concept

Effect of Linear Transformations (Addition of a Constant) on Measures of Dispersion

Question 1439Question →

In circle $O$ , minor arc $AB$ is subtended by a central angle of $120^\circ$ , and the radius of circle $O$ is $6$ . A smaller circle, circle $C$ , is constructed such that it is tangent to chord $AB$ at its midpoint and tangent to minor arc $AB$ at its midpoint. What is the area of a sector of circle $C$ subtended by a central angle of $90^\circ$ ?

\frac{9\pi}{16}

\frac{9\pi}{8}

\frac{9\pi}{4}

\frac{27\pi}{16}

9\pi

Show answer & explanation

Answer: $\frac{9\pi}{16}$

Answer

\frac{9\pi}{16}

The distance from the center of circle

O

to chord

AB

is

6 \cos(60^\circ) = 3

, leaving a height of

6 - 3 = 3

for the minor segment. Circle

C

is tangent to the chord and arc midpoints, making its diameter equal to

3

and its radius

\frac{3}{2}

. The area of a

90^\circ

sector of circle

C

is

\frac{90^\circ}{360^\circ} \pi \left(\frac{3}{2}\right)^2 = \frac{9\pi}{16}

.

Step-by-Step Solution

1

Find the perpendicular distance from center

O

to chord

AB

.

In isosceles triangle

OAB

with

OA = OB = 6

and central angle

\angle AOB = 120^\circ

, the perpendicular bisector from

O

to

AB

meets

AB

at midpoint

M

.

\angle AOM = 60^\circ

, so

OM = OA \cos(60^\circ) = 6 \times \frac{1}{2} = 3

.

Determining

OM

allows us to find the sagitta (height) of the minor segment bounded by chord

AB

and minor arc

AB

.

2

Calculate the height of the minor segment (diameter of circle

C

).

The height of the minor segment along the radius passing through

M

to arc midpoint

N

is

MN = ON - OM = 6 - 3 = 3

. Since circle

C

is tangent to

AB

at

M

and to the arc at

N

, segment

MN

is a diameter of circle

C

, giving a diameter of

3

.

The space between the chord midpoint and arc midpoint bounds circle

C

, defining its diameter.

3

Find the radius of circle

C

.

Radius

r_C = \frac{\text{diameter}}{2} = \frac{3}{2}

.

The radius of circle

C

is needed to calculate its sector area.

4

Compute the area of the

90^\circ

sector of circle

C

.

\text{Area} = \frac{\theta}{360^\circ} \pi r_C^2 = \frac{90^\circ}{360^\circ} \pi \left(\frac{3}{2}\right)^2 = \frac{1}{4} \pi \left(\frac{9}{4}\right) = \frac{9\pi}{16}.

Multiply the fraction of the full circle represented by the central angle by the total area of circle

C

.

Key Concept

Calculating sector area of an inscribed circle within a circle segment

Estimated Time:2m 0s

Question 1440Question →

Dataset $S$ consists of 10 distinct numerical values. Dataset $T$ is created by replacing the maximum value of Dataset $S$ with a number that is strictly greater than that maximum value, while all other 9 values remain unchanged. Which of the following statements comparing Dataset $S$ and Dataset $T$ must be true? Select all that apply.

Select all that apply

The range of Dataset

T

is greater than the range of Dataset

S

.

The interquartile range (IQR) of Dataset

T

is equal to the interquartile range of Dataset

S

.

The standard deviation of Dataset

T

is equal to the standard deviation of Dataset

S

.

The median of Dataset

T

is strictly greater than the median of Dataset

S

.

The 75th percentile of Dataset

T

is strictly greater than the 75th percentile of Dataset

S

.

Show answer & explanation

Answer: The range of Dataset $T$ is greater than the range of Dataset $S$ .; The interquartile range (IQR) of Dataset $T$ is equal to the interquartile range of Dataset $S$ .

Answer

The statements asserting that the range of Dataset

T

is greater than the range of Dataset

S

and that the interquartile range (IQR) of Dataset

T

is equal to the interquartile range of Dataset

S

are correct.

The statement regarding the range is correct because increasing the maximum value increases the difference between the maximum and minimum values. The statement regarding the interquartile range is correct because the 25th percentile (

Q_1

) and 75th percentile (

Q_3

) depend only on the first 8 ordered elements of a 10-element dataset, so changing the 10th element leaves

Q_1

and

Q_3

identical.

Step-by-Step Solution

1

Analyze the impact on Range

\text{Range}(T) > \text{Range}(S)

Range is defined as

\text{Maximum} - \text{Minimum}

. Since the minimum is unchanged and the maximum increases, the range must increase.

2

Analyze the impact on Interquartile Range (IQR)

\text{IQR}(T) = \text{IQR}(S)

IQR is

Q_3 - Q_1

. For 10 ordered elements,

Q_1

and

Q_3

depend on the positions of the bottom 75% of the data. Modifying only the 10th (largest) element leaves

Q_1

and

Q_3

unchanged.

3

Analyze the impact on Median and Standard Deviation

Median is unchanged; Standard deviation increases.

The median depends on the 5th and 6th values, which are unaffected. Moving an extreme value further out increases distance from the mean, strictly increasing standard deviation.

Key Concept

Effect of outlier modification on measures of position (

Q_1, Q_3

, Median) versus measures of dispersion (Range, Standard Deviation, IQR).

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

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Key Concept

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

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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

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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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