All practice questions

Question 1521Question →

A researcher recorded the annual rainfall totals (in inches) for a specific region over a 20-year period, forming Dataset $P$ . Dataset $P$ has a range of $18$ inches and an interquartile range (IQR) of $8$ inches. A second dataset, Dataset $Q$ , is created by multiplying each rainfall total in Dataset $P$ by $1.2$ and then subtracting $3$ inches. What is the sum of the range and the interquartile range (IQR) of Dataset $Q$ ?

25.2

26.0

28.2

31.2

37.2

Show answer & explanation

Answer: $31.2$

Answer

31.2

For any linear transformation of the form

Y = aX + b

, measures of spread such as range and interquartile range (IQR) are multiplied by

|a|

, while the constant term

b

has no effect. Multiplying the original range (

18

) and IQR (

8

) by

1.2

yields a new range of

21.6

and a new IQR of

9.6

. Summing these values gives

21.6 + 9.6 = 31.2

.

Step-by-Step Solution

1

Determine the impact of a linear transformation

Y = aX + b

on measures of dispersion.

Measures of dispersion (Range, IQR, Standard Deviation) scale by

|a|

and are unaffected by the additive constant

b

.

Adding or subtracting a constant shifts all data points by the same amount, leaving the distance between points unchanged, whereas multiplying by a factor scales all distances between points.

2

Calculate the range of Dataset

Q

.

\text{Range}_Q = 1.2 \times \text{Range}_P = 1.2 \times 18 = 21.6

The range scales by the multiplier

1.2

and is not affected by subtracting

3

.

3

Calculate the interquartile range (IQR) of Dataset

Q

.

\text{IQR}_Q = 1.2 \times \text{IQR}_P = 1.2 \times 8 = 9.6

The IQR scales by the multiplier

1.2

and is not affected by subtracting

3

.

4

Compute the sum of the range and the IQR of Dataset

Q

.

\text{Sum} = 21.6 + 9.6 = 31.2

Adding the newly calculated Range and IQR gives the required total spread measure.

Key Concept

Linear Transformations on Dispersion Metrics

Estimated Time:1m 30s

Question 1522Question →

A $10$ -foot straight ladder leans against a vertical wall on flat horizontal ground. If the bottom of the ladder is $6$ feet away from the base of the wall, how high up the wall, in feet, does the top of the ladder reach?

Show answer & explanation

Answer: 8

Answer

The top of the ladder reaches

8

feet up the wall.

Applying the Pythagorean theorem

a^2 + b^2 = c^2

with hypotenuse

c = 10

and ground leg

a = 6

gives

6^2 + b^2 = 10^2

, which simplifies to

36 + b^2 = 100

, so

b^2 = 64

and

b = 8

feet.

Step-by-Step Solution

1

Set up the right triangle relationship using the Pythagorean theorem.

a^2 + b^2 = c^2

, where

a = 6

and

c = 10

.

The ladder forms the hypotenuse of a right triangle with the vertical wall and horizontal ground.

2

Substitute the known values into the formula and solve for

b

.

6^2 + b^2 = 10^2 \implies 36 + b^2 = 100 \implies b^2 = 64 \implies b = 8

.

Subtracting

36

from

100

gives

64

, and taking the positive square root yields the vertical height.

Key Concept

Applying the Pythagorean Theorem (

a^2 + b^2 = c^2

) to find an unknown leg length in a right triangle context.

Question 1523Question →

For all real numbers $a$ and $b$ , the custom operation $\star$ is defined by $a \star b = a(b + 2) - b$ . What is the value of $4 \star (-3)$ ?

-7

-4

-1

1

7

Show answer & explanation

Answer: $-1$

Answer

-1

Substituting

a = 4

and

b = -3

into the given operation gives

4((-3) + 2) - (-3) = 4(-1) + 3 = -4 + 3 = -1

. The value of the expression is

-1

.

Step-by-Step Solution

1

Substitute

a = 4

and

b = -3

into the custom operation formula

a \star b = a(b + 2) - b

.

4 \star (-3) = 4((-3) + 2) - (-3)

The definition specifies replacing variable

a

with

4

and variable

b

with

-3

.

2

Evaluate the expression inside the parentheses.

-3 + 2 = -1

, simplifying the expression to

4(-1) - (-3)

Perform operations inside grouping symbols first according to the standard order of operations.

3

Perform the multiplication and simplify the double negative.

4(-1) = -4

and

-(-3) = +3

, yielding

-4 + 3 = -1

Multiplying positive and negative yields negative, and subtracting a negative number is equivalent to addition.

Key Concept

Evaluating Custom Binary Operations

Estimated Time:45s

Question 1524Question →

For all real numbers

x

such that

x \neq 0

and

x \neq 1

, the function

f

satisfies the equation

f(x) + f\left(\frac{x - 1}{x}\right) = 1 + x

What is the value of

f(2)

?

Show answer & explanation

Answer: 0.75

Answer

The value of

f(2)

is

0.75

(or

\frac{3}{4}

).

Evaluating the given equation at

x = 2

,

x = 1/2

, and

x = -1

generates a system of three linear equations:

f(2) + f(1/2) = 3

,

f(1/2) + f(-1) = 1.5

, and

f(-1) + f(2) = 0

. Adding all three equations yields

2(f(2) + f(1/2) + f(-1)) = 4.5

, which simplifies to

f(2) + f(1/2) + f(-1) = 2.25

. Subtracting the equation

f(1/2) + f(-1) = 1.5

from

2.25

gives

f(2) = 0.75

.

Step-by-Step Solution

1

Determine the cyclic sequence of inputs starting at

x = 2

Evaluating the inner transformation

g(x) = \frac{x-1}{x}

yields

g(2) = \frac{1}{2}

,

g\left(\frac{1}{2}\right) = -1

, and

g(-1) = 2

. The input sequence cycles every 3 steps.

Identifying that

g(g(g(x))) = x

allows constructing a closed system of linear equations.

2

Set up a system of three linear equations using the cycle values

For

x = 2

:

f(2) + f\left(\frac{1}{2}\right) = 3

.
For

x = \frac{1}{2}

:

f\left(\frac{1}{2}\right) + f(-1) = \frac{3}{2}

.
For

x = -1

:

f(-1) + f(2) = 0

.

Substituting each element of the cycle into the given relation produces three linear equations in three unknown function values.

3

Sum the equations and isolate

f(2)

Summing all three equations gives

2\left[f(2) + f\left(\frac{1}{2}\right) + f(-1)\right] = \frac{9}{2}

, so

f(2) + f\left(\frac{1}{2}\right) + f(-1) = \frac{9}{4}

. Subtracting

f\left(\frac{1}{2}\right) + f(-1) = \frac{3}{2}

yields

f(2) = \frac{9}{4} - \frac{6}{4} = \frac{3}{4} = 0.75

.

Symmetric summation provides the total sum of the three function values, allowing direct elimination of the other two variables.

Key Concept

Solving cyclic functional equations by forming a system of linear equations

Question 1525Question →

A solid sphere of radius $r$ is placed inside a hollow right circular cylinder of base radius $r$ and height $3r$ such that the sphere rests on the bottom base of the cylinder. Water is then poured into the cylinder until the water level just covers the top of the sphere (a height of $2r$ ). What is the ratio of the volume of the empty space remaining in the cylinder above the water level to the volume of the water inside the cylinder?

\frac{1}{3}

\frac{1}{2}

\frac{3}{4}

\frac{3}{2}

2

Show answer & explanation

Answer: $\frac{3}{2}$

Answer

The ratio of the volume of the empty space remaining in the cylinder above the water level to the volume of the water inside the cylinder is

\frac{3}{2}

.

The empty cylinder portion at the top has height

r

, giving a volume of

\pi r^3

. The submerged section up to height

2r

has total volume

2\pi r^3

, of which the sphere accounts for

\frac{4}{3}\pi r^3

. Subtracting the sphere's volume leaves

\frac{2}{3}\pi r^3

of water. The ratio of empty space to water volume is

\frac{\pi r^3}{\frac{2}{3}\pi r^3} = \frac{3}{2}

.

Step-by-Step Solution

1

Calculate the volume of the empty cylindrical space above the water height.

The total height of the cylinder is

3r

and the water level reaches height

2r

, leaving an empty height of

3r - 2r = r

. The volume of empty space is

V_{\text{empty}} = \pi r^2 \cdot r = \pi r^3

.

The empty space forms a smaller cylinder of base radius

r

and height

r

at the top.

2

Calculate the total volume of the cylindrical region filled up to the water height of

2r

.

V_{\text{region}} = \pi r^2 \cdot (2r) = 2\pi r^3

.

This represents the combined volume of the water and the submerged sphere.

3

Calculate the volume of the solid sphere of radius

r

.

V_{\text{sphere}} = \frac{4}{3}\pi r^3

.

Using the standard formula for the volume of a sphere,

V = \frac{4}{3}\pi r^3

.

4

Find the volume of the water by subtracting the sphere's volume from the total filled region.

V_{\text{water}} = 2\pi r^3 - \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3

.

The space taken up by the water is the total volume up to height

2r

minus the displacement volume of the sphere.

5

Compute the required ratio of

V_{\text{empty}}

to

V_{\text{water}}

.

\frac{V_{\text{empty}}}{V_{\text{water}}} = \frac{\pi r^3}{\frac{2}{3}\pi r^3} = \frac{1}{\frac{2}{3}} = \frac{3}{2}

.

Dividing the empty space volume by the water volume simplifies to

\frac{3}{2}

.

Key Concept

Displacement and volume calculations combining 3D geometric figures (cylinders and spheres).

Estimated Time:2m 30s

Question 1526Question →

A non-degenerate triangle has integer side lengths $a$ , $b$ , and $c$ such that $a \le b \le c$ . If the perimeter of the triangle is $30$ and its area is an integer, what is the length of the longest side $c$ ?

Show answer & explanation

Answer: 13

Answer

13

Using Heron's formula with semi-perimeter

s = 15

, the area of the triangle is

\sqrt{15(15-a)(15-b)(15-c)}

. By defining

x = 15-a

,

y = 15-b

, and

z = 15-c

, we have

x+y+z = 15

with

x \ge y \ge z \ge 1

. For the area to be an integer,

15xyz

must be a perfect square. The AM-GM inequality bounds

xyz \le 125

, leaving

xyz = 60

as the only valid case (

k=2

). The system

x+y+z=15

and

xyz=60

has a unique positive integer solution

(10, 3, 2)

, giving side lengths

(5, 12, 13)

. Thus, the longest side length

c

is

13

.

Step-by-Step Solution

1

Find the semi-perimeter of the triangle.

s = \frac{a+b+c}{2} = \frac{30}{2} = 15

Heron's formula requires the semi-perimeter

s

to compute the area of a triangle.

2

Set up Heron's formula for the area.

\text{Area} = \sqrt{15(15-a)(15-b)(15-c)}

Heron's formula relates the side lengths of a triangle directly to its area.

3

Define variables for the differences from the semi-perimeter.

Let

x = 15-a

,

y = 15-b

, and

z = 15-c

. Then

x+y+z = 15

and

x \ge y \ge z \ge 1

.

Because

a \le b \le c

, their complements with respect to

15

satisfy

x \ge y \ge z

.

4

Determine the condition for the area to be an integer.

\text{Area} = \sqrt{15xyz}

must be an integer, so

15xyz = m^2

for some integer

m

, implying

xyz = 15k^2

for an integer

k \ge 1

.

For

\sqrt{15xyz}

to be an integer,

15xyz

must be a perfect square containing

15

as a factor.

5

Bound

xyz

using the AM-GM inequality.

xyz \le \left(\frac{x+y+z}{3}\right)^3 = 5^3 = 125 \implies 15k^2 \le 125 \implies k=1 \text{ or } k=2

The AM-GM inequality provides a maximum upper bound for the product of three positive numbers with a fixed sum.

6

Evaluate the possible values of

k

.

If

k=1

,

xyz = 15

, which has no integer solutions for

x+y+z=15

. If

k=2

,

xyz = 60

, which yields

(x, y, z) = (10, 3, 2)

.

Solving the system

x+y+z=15

and

xyz=60

under

x \ge y \ge z \ge 1

gives

z=2

,

y=3

,

x=10

uniquely.

7

Calculate the side lengths and identify the longest side

c

.

a = 5

,

b = 12

,

c = 13

, so the longest side length is

13

.

Since

c = 15 - z = 15 - 2 = 13

, the longest side of the right triangle

(5, 12, 13)

is

13

.

Key Concept

Heron's formula, integer partitioning, and the AM-GM inequality for triangle area optimization

Question 1527Question →

Consider the following system of three linear equations in variables

x

,

y

, and

z

, where

k

is a real constant:

\begin{aligned} 2x + 3y - z &= 11 \\ x - 2y + 4z &= -3 \\ 4x - y + 7z &= k \end{aligned}

For what value of

k

does the system have at least one solution

(x, y, z)

?

-5

5

8

17

23

Show answer & explanation

Answer: $5$

Answer

5

The left-hand side of the third equation is a linear combination of the first two equations:

1 \cdot (2x + 3y - z) + 2 \cdot (x - 2y + 4z) = 4x - y + 7z

. For the linear system to be consistent and possess at least one solution, the same linear combination must hold for the constant terms on the right-hand side:

1(11) + 2(-3) = 11 - 6 = 5

. Therefore, the value of

k

must be

5

.

Step-by-Step Solution

1

Analyze the variable coefficients across the three equations for linear dependence.

Notice that the coefficients of the third equation can be expressed as a linear combination of the first two equations.

If the left-hand side of the third equation is a linear combination of the first two equations, the system will only be consistent if the right-hand side constants satisfy the exact same linear combination.

2

Determine the multiplier needed to produce the third equation's left-hand side.

Multiply the second equation by

2

and add it to the first equation:

(2x + 3y - z) + 2(x - 2y + 4z) = 4x - y + 7z

.

This yields the exact expression

4x - y + 7z

present on the left-hand side of the third equation.

3

Apply the identical combination to the right-hand side constant terms.

The combined constant value is

11 + 2(-3) = 11 - 6 = 5

.

For the system to have at least one solution (i.e., to avoid contradiction and be consistent),

k

must equal this computed value of

5

.

Key Concept

Linear Dependence and Consistency in 3x3 Linear Systems

Estimated Time:2m 0s

Question 1528Question →

The function $f(x)$ is defined by $f(x) = |x - 3| - 2$ . The graph of $y = f(x)$ in the $xy$ -plane is reflected across the $y$ -axis to produce the graph of $y = g(x)$ . The graph of $y = g(x)$ is then translated $1$ unit to the left and $4$ units downward to produce the graph of $y = h(x)$ . Which of the following statements about the graph of $y = h(x)$ must be true? Select all that apply.

Select all that apply

The

y

-intercept of the graph of

y = h(x)

is

(0, -2)

.

The graph of

y = h(x)

is symmetric with respect to the line

x = -4

.

The area of the triangular region bounded by the graph of

y = h(x)

and the

x

-axis is

36

.

The minimum value of

h(x)

is

-2

.

The distance between the two

x

-intercepts of the graph of

y = h(x)

is

6

.

Show answer & explanation

Answer: The $y$ -intercept of the graph of $y = h(x)$ is $(0, -2)$ .; The graph of $y = h(x)$ is symmetric with respect to the line $x = -4$ .; The area of the triangular region bounded by the graph of $y = h(x)$ and the $x$ -axis is $36$ .

Answer

The statements establishing that the

y

-intercept is

(0, -2)

, that the line of symmetry is

x = -4

, and that the bounded region with the

x

-axis has an area of

36

are all correct.

Reflecting

f(x) = |x - 3| - 2

across the

y

-axis yields

g(x) = |x + 3| - 2

. Translating

g(x)

left by

1

unit and down by

4

units yields

h(x) = |x + 4| - 6

. From

h(x) = |x + 4| - 6

, evaluating

h(0) = -2

verifies the

y

-intercept of

(0, -2)

. The vertex at

(-4, -6)

defines the axis of symmetry at

x = -4

. Setting

h(x) = 0

gives

x

-intercepts at

-10

and

2

, producing a triangle bounded by the

x

-axis with base

12

and height

6

, which has an area of

\frac{1}{2} \times 12 \times 6 = 36

.

Step-by-Step Solution

1

Apply reflection across the

y

-axis to determine

g(x)

.

g(x) = f(-x) = |-x - 3| - 2 = |x + 3| - 2

Reflecting

y = f(x)

across the

y

-axis replaces every occurrence of

x

with

-x

.

2

Apply horizontal and vertical translations to determine

h(x)

.

h(x) = g(x + 1) - 4 = |(x + 1) + 3| - 2 - 4 = |x + 4| - 6

Translating a graph

1

unit left adds

1

to the input variable, and translating

4

units down subtracts

4

from the expression.

3

Find the

y

-intercept of

y = h(x)

.

h(0) = |0 + 4| - 6 = 4 - 6 = -2

, so the intercept point is

(0, -2)

.

The

y

-intercept is found by setting

x = 0

.

4

Determine the vertex, axis of symmetry, and minimum value.

Vertex is at

(-4, -6)

, line of symmetry is

x = -4

, and minimum value is

-6

.

For

y = a|x - h| + k

, the vertex is at

(h, k)

and the axis of symmetry is

x = h

.

5

Find the

x

-intercepts and compute the bounded triangular area.

|x + 4| - 6 = 0 \implies |x + 4| = 6 \implies x = 2

or

x = -10

. Base

= 2 - (-10) = 12

, height

= |-6| = 6

. Area

= \frac{1}{2} \times 12 \times 6 = 36

.

The region bounded by the V-shaped graph below the

x

-axis forms a triangle with base on the

x

-axis and height equal to the magnitude of the vertex's

y

-coordinate.

Key Concept

Coordinate transformations of algebraic functions including

y

-axis reflection, horizontal and vertical translations, and analyzing key geometric features of the resulting graph.

Question 1529Question →

In a meteorological study, the daily peak wind speeds (in kilometers per hour) recorded over 7 consecutive days were $14, 18, 21, 24, 27, 31,$ and $37$ . As part of a data calibration process, each recorded wind speed is increased by $15\%$ and then increased by an additional constant of $5\text{ km/h}$ . What is the range, in kilometers per hour, of the calibrated daily peak wind speeds?

Show answer & explanation

Answer: 26.45

Answer

The range of the calibrated daily peak wind speeds is 26.45 km/h.

For any dataset transformed linearly according to

y = a \cdot x + b

where

a > 0

, the range of the transformed dataset is given by

\text{Range}(Y) = a \cdot \text{Range}(X)

. Here, increasing each value by

15\%

corresponds to

a = 1 + 0.15 = 1.15

, and adding

5

corresponds to

b = 5

. The original maximum is

37

and the original minimum is

14

, giving an original range of

37 - 14 = 23

. Therefore, the new range is

1.15 \times 23 = 26.45\text{ km/h}

. The constant shift of

5\text{ km/h}

affects the center (mean, median) but has zero effect on the dispersion.

Step-by-Step Solution

1

Determine the range of the original set of wind speeds.

Original Range = 37 - 14 = 23 km/h.

The range is defined as the difference between the maximum and minimum values in a dataset.

2

Apply the properties of linear transformations to measures of dispersion.

The transformation is y = 1.15x + 5. The range is scaled by 1.15 and unaffected by the addition of 5.

Adding a constant shift to all data points shifts the entire distribution without changing the spread (dispersion), whereas multiplying all data points by a constant factor 'a' scales all measures of dispersion by |a|.

3

Compute the calibrated range.

Calibrated Range = 1.15 * 23 = 26.45 km/h.

Multiplying the original range of 23 by the scale factor of 1.15 gives the exact range of the transformed dataset.

Key Concept

Effect of linear transformations on measures of dispersion (range, standard deviation, IQR)

Estimated Time:1m 30s

Question 1530Question →

A retailer purchases items for $x$ dollars each and sets the regular selling price at a $40\%$ markup above the purchase price. During a clearance event, the retailer discounts the regular selling price by $15\%$ . Which of the following statements regarding the pricing of an item must be true? Select all such statements.

Select all that apply

The regular selling price of an item is

1.40x

dollars.

The clearance sale price of an item is

1.19x

dollars.

The clearance sale price represents a

25\%

markup above the original purchase price

x

.

The dollar profit earned on each item sold during the clearance event is

0.25x

dollars.

Show answer & explanation

Answer: The regular selling price of an item is $1.40x$ dollars.; The clearance sale price of an item is $1.19x$ dollars.

Answer

The statements confirming that the regular selling price is

1.40x

dollars and that the clearance sale price is

1.19x

dollars are correct.

The regular selling price is computed by increasing the cost

x

by 40%, yielding

1.40x

. The clearance price is found by reducing

1.40x

by 15%, which equals

1.40x \times 0.85 = 1.19x

. Therefore, both the statement that the regular price is

1.40x

dollars and the statement that the clearance price is

1.19x

dollars are correct.

Step-by-Step Solution

1

Model the regular selling price using the original purchase price

x

and the

40\%

markup.

Regular price

= x \times (1 + 0.40) = 1.40x

.

A 40% markup increases the base cost

x

by 0.40 times its original value.

2

Model the clearance sale price by applying a

15\%

discount to the regular selling price.

Clearance price

= 1.40x \times (1 - 0.15) = 1.40x \times 0.85 = 1.19x

.

A 15% discount reduces the regular price to 85% of its value.

3

Calculate the effective markup percentage and profit per item to evaluate all statements.

Profit

= 1.19x - x = 0.19x

, which is a

19\%

net markup over the purchase price

x

.

Comparing

1.19x

to the base cost

x

shows a net gain of

0.19x

per unit.

Key Concept

Applying successive percent changes in algebraic modeling by using correct base values.

Estimated Time:1m 0s

Question 1531Question →

Two water pumps, Pump $A$ and Pump $B$ , operate at constant rates to fill a $910$ -gallon storage tank. The rate at which Pump $A$ fills the tank is $15$ gallons per hour more than twice the rate of Pump $B$ . Both pumps begin working together to fill the empty tank. After $4$ hours, Pump $A$ shuts down, and Pump $B$ continues working alone for another $5$ hours to finish filling the tank completely. What is the pumping rate of Pump $A$ , in gallons per hour?

50

85

100

115

130

Show answer & explanation

Answer: 115

Answer

115 gallons per hour

Letting

r

represent the rate of Pump B in gallons per hour, the rate of Pump A is

2r + 15

. During the first 4 hours, both pumps operate together at a combined rate of

(3r + 15)

gallons per hour, contributing

4(3r + 15) = 12r + 60

gallons. In the remaining 5 hours, Pump B alone contributes

5r

gallons. Setting the total amount equal to the tank capacity gives

12r + 60 + 5r = 910

, which simplifies to

17r = 850

, so

r = 50

. Substituting

r = 50

into

2r + 15

yields

115

gallons per hour.

Step-by-Step Solution

1

Define the variable for the unknown rate.

Let

r

be the pumping rate of Pump

B

in gallons per hour. Then the rate of Pump

A

is

2r + 15

gallons per hour.

Expressing both rates in terms of a single variable

r

allows us to formulate a linear equation in one variable.

2

Determine the combined pumping rate and the work done during the first phase.

Combined rate

= r + (2r + 15) = 3r + 15

gallons per hour. Work completed in 4 hours

= 4(3r + 15) = 12r + 60

gallons.

When both pumps work together, their individual rates add together.

3

Formulate the total work equation for filling the 910-gallon tank.

(12r + 60) + 5r = 910 \implies 17r + 60 = 910

The sum of the work done in the first 4 hours by both pumps and the work done in the next 5 hours by Pump B alone equals the total capacity.

4

Solve the linear equation for

r

.

17r = 850 \implies r = 50

gallons per hour.

Subtracting 60 from both sides and dividing by 17 gives the rate of Pump B.

5

Calculate the pumping rate of Pump A.

Rate of Pump

A = 2(50) + 15 = 115

gallons per hour.

Substitute

r = 50

into the expression for Pump A's rate (

2r + 15

).

Key Concept

Linear equation formulation from work and rate relationships

Question 1532Question →

A regular polygon has interior angles that each measure $140^\circ$ . What is the total number of diagonals that can be drawn in this polygon?

9

14

27

36

54

Show answer & explanation

Answer: 27

Answer

The total number of diagonals that can be drawn in the polygon is 27.

To find the total number of diagonals in a regular polygon, first determine its number of sides

n

. Using the interior angle formula

\frac{(n-2) \times 180^\circ}{n} = 140^\circ

, we find

n = 9

. Next, applying the diagonal formula

D = \frac{n(n-3)}{2}

for

n = 9

gives

D = \frac{9 \times 6}{2} = 27

. Thus, the value 27 is correct.

Step-by-Step Solution

1

Find the number of sides

n

of the regular polygon using the interior angle measure.

The polygon has

n = 9

sides.

Each interior angle of a regular

n

-gon is given by

\frac{(n - 2) \times 180^\circ}{n} = 140^\circ

. Solving for

n

:

180n - 360 = 140n \implies 40n = 360 \implies n = 9

.

2

Calculate the total number of diagonals using the formula

D = \frac{n(n - 3)}{2}

.

The total number of diagonals is 27.

Substituting

n = 9

into the formula yields

D = \frac{9(9 - 3)}{2} = \frac{9 \times 6}{2} = 27

.

Key Concept

Interior Angle Measure and Diagonal Formula for Regular Polygons

Estimated Time:1m 0s

Question 1533Question →

Container X contains a liquid solution that is $80\%$ alcohol by volume, and Container Y contains a liquid solution that is $25\%$ alcohol by volume. A chemist removes a specific volume of solution from Container X and mixes it with a solution from Container Y to produce $100$ liters of a new mixture that is $58\%$ alcohol by volume. Following this removal, $20$ liters of pure alcohol are added to the liquid remaining in Container X. If Container X initially held $120$ liters of solution, what is the concentration of alcohol, by volume, in Container X after the pure alcohol is added?

56.67\%

68\%

70\%

80\%

85\%

Show answer & explanation

Answer: $85\%$

Answer

85%

To find the final concentration, we first determine the volume of solution removed from Container X. Using the weighted average for the 100-liter mixture:

0.80 V_x + 0.25 (100 - V_x) = 58

, which simplifies to

0.55 V_x = 33

, so

V_x = 60

liters. Container X originally held 120 liters, so removing 60 liters leaves 60 liters of solution containing

80\% \times 60 = 48

liters of alcohol. Adding 20 liters of pure alcohol increases the total alcohol to

48 + 20 = 68

liters and the total volume to

60 + 20 = 80

liters. The final concentration is

\frac{68}{80} = 85\%

.

Step-by-Step Solution

1

Find the volume of solution

V_x

removed from Container X to make the 100-liter mixture.

V_x = 60

liters.

Let

V_x

be the volume from X and

100 - V_x

be the volume from Y. Setting up the alcohol concentration equation:

0.80 V_x + 0.25 (100 - V_x) = 0.58(100) \implies 0.55 V_x + 25 = 58 \implies 0.55 V_x = 33 \implies V_x = 60

liters.

2

Determine the remaining solution volume and alcohol volume in Container X after removing 60 liters.

Remaining solution =

60

liters; Remaining alcohol =

48

liters.

Container X initially had

120

liters. Removing

60

liters leaves

120 - 60 = 60

liters. Since the mixture is homogeneous, the remaining liquid retains an

80\%

alcohol concentration, yielding

0.80 \times 60 = 48

liters of alcohol.

3

Calculate the total alcohol volume and total solution volume in Container X after adding 20 liters of pure alcohol.

New alcohol volume =

68

liters; New total solution volume =

80

liters.

Adding

20

liters of pure alcohol increases both the alcohol amount (

48 + 20 = 68

liters) and the total liquid volume (

60 + 20 = 80

liters).

4

Calculate the final concentration of alcohol in Container X.

Concentration =

\frac{68}{80} = 0.85 = 85\%

.

The final alcohol concentration is the ratio of final alcohol volume to final total liquid volume.

Key Concept

Algebraic Modeling of Multi-Step Mixture Solutions

Estimated Time:2m 30s

Question 1534Question →

If $x$ is a real number that satisfies both $|4 - x| < 3$ and $\frac{2x - 15}{-3} \le 1$ , which of the following inequalities expresses all possible values of $x$ ?

6 \le x < 7

1 < x \le 6

1 < x < 7

x \ge 6

6 < x < 7

Show answer & explanation

Answer: $6 \le x < 7$

Answer

The inequality expressing all possible values of

x

is

6 \le x < 7

.

The correct inequality is determined by finding the intersection of both given inequalities. The absolute value inequality

|4 - x| < 3

simplifies to

1 < x < 7

. The linear inequality

\frac{2x - 15}{-3} \le 1

simplifies to

x \ge 6

after reversing the inequality sign when multiplying by

-3

. Combining

1 < x < 7

and

x \ge 6

yields the compound inequality stating

x

is greater than or equal to

6

and strictly less than

7

.

Step-by-Step Solution

1

Solve the absolute value inequality

|4 - x| < 3

.

-3 < 4 - x < 3 \implies -7 < -x < -1 \implies 1 < x < 7

.

An absolute value inequality

|u| < k

translates to

-k < u < k

. Multiplying or dividing by

-1

flips the inequality signs.

2

Solve the linear inequality

\frac{2x - 15}{-3} \le 1

.

2x - 15 \ge -3 \implies 2x \ge 12 \implies x \ge 6

.

Multiplying both sides of an inequality by a negative number reverses the direction of the inequality sign.

3

Find the intersection of the two solution sets.

1 < x < 7

and

x \ge 6

combine to give

6 \le x < 7

.

Values of

x

must satisfy both inequalities simultaneously.

Key Concept

Solving systems of absolute value inequalities and linear inequalities, including sign reversal rules for negative multipliers.

Question 1535Question →

A high school basketball team scored the following points in five consecutive games: $4$ , $6$ , $6$ , $7$ , and $12$ . Which of the following statements regarding the measures of central tendency for these scores must be true? Select all such statements.

Select all that apply

The mean score is equal to

7

.

The median score is equal to the mode score.

The median score is equal to

7

.

The mode score is equal to

6

.

The mean score is less than the median score.

Show answer & explanation

Answer: The mean score is equal to $7$ .; The median score is equal to the mode score.; The mode score is equal to $6$ .

Answer

The statements asserting that the mean score is equal to 7, the median score is equal to the mode score, and the mode score is equal to 6 are all true.

For the dataset

4, 6, 6, 7, 12

, the mean is

\frac{35}{5} = 7

, the median (middle score) is

6

, and the mode (most frequent score) is

6

. Consequently, the mean equals

7

, the median and mode are equal (

6 = 6

), and the mode is

6

.

Step-by-Step Solution

1

Calculate the mean of the dataset.

Mean =

\frac{4 + 6 + 6 + 7 + 12}{5} = \frac{35}{5} = 7

.

The mean is calculated by dividing the sum of all data points by the total count.

2

Determine the median of the dataset.

Median =

6

.

The data is already arranged in ascending order:

4, 6, 6, 7, 12

. The 3rd value (middle position) is

6

.

3

Determine the mode of the dataset.

Mode =

6

.

The value

6

occurs twice, whereas all other values occur only once.

4

Evaluate each given statement against the calculated metrics.

Mean =

7

, Median =

6

, Mode =

6

. Thus: Mean (

7

) is true; Median = Mode (

6 = 6

) is true; Mode =

6

is true.

Direct comparison verifies which statements hold true.

Key Concept

Calculating and comparing mean, median, and mode for a discrete set of numerical data.

Estimated Time:1m 0s

Question 1536Question →

If $x$ is an integer that satisfies both $|3x - 12| \le 15$ and $|2x + 4| \ge 6$ , what is the sum of the maximum possible value of $x$ and the minimum possible value of $x$ ?

Show answer & explanation

Answer: 10

Answer

The correct answer is 10.

Solving the first inequality

|3x - 12| \le 15

yields

-1 \le x \le 9

. Solving the second inequality

|2x + 4| \ge 6

yields

x \ge 1

or

x \le -5

. The values of

x

that satisfy both constraints are those in the interval

[1, 9]

. Within this interval, the minimum integer value is

1

and the maximum integer value is

9

. Adding these values together gives

1 + 9 = 10

.

Step-by-Step Solution

1

Solve the bounded absolute value inequality

|3x - 12| \le 15

.

-15 \le 3x - 12 \le 15 \implies -3 \le 3x \le 27 \implies -1 \le x \le 9.

An inequality of the form

|A| \le B

(where

B \ge 0

) unwraps to the compound inequality

-B \le A \le B

.

2

Solve the boundary absolute value inequality

|2x + 4| \ge 6

.

2x + 4 \ge 6 \implies 2x \ge 2 \implies x \ge 1

, or

2x + 4 \le -6 \implies 2x \le -10 \implies x \le -5$.

An inequality of the form

|A| \ge B

(where

B > 0

) splits into two separate cases:

A \ge B

or

A \le -B

.

3

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

The intersection of

[-1, 9]

and

(-\infty, -5] \cup [1, \infty)

is

[1, 9]

.

To satisfy both inequalities simultaneously,

x

must belong to both set

[-1, 9]

and set

(-\infty, -5] \cup [1, \infty)

.

4

Identify the extreme integer values of

x

and compute their sum.

\text{Minimum integer } x = 1, \quad \text{Maximum integer } x = 9; \quad 1 + 9 = 10.

The range of valid integer values for

x

is all integers from 1 to 9 inclusive.

Key Concept

Linear Inequalities and Absolute Value

Estimated Time:2m 0s

Question 1537Question →

The side lengths of triangle $ABC$ , in units, are given by $AB = x$ , $BC = x + 2$ , and $CA = x + 4$ , where $x$ is a positive integer. If the perimeter of triangle $ABC$ is strictly less than $36$ , how many possible integer values of $x$ exist such that triangle $ABC$ is an acute triangle?

2

3

4

5

7

Show answer & explanation

Answer: 3

Answer

There are 3 possible integer values of

x

.

To form an acute triangle with side lengths

x

,

x+2

, and

x+4

, three conditions must be met: the perimeter bound (

3x+6 < 36 \implies x < 10

), the non-degeneracy condition (

x + x + 2 > x + 4 \implies x > 2

), and the acute angle condition (

(x+4)^2 < x^2 + (x+2)^2 \implies x^2 - 4x - 12 > 0 \implies x > 6

). Combining these bounds restricts

x

to integer values in the range

6 < x < 10

, giving exactly three valid integer values:

7, 8,

and

9

.

Step-by-Step Solution

1

Apply the perimeter constraint to find an upper bound for

x

.

The perimeter is

P = x + (x + 2) + (x + 4) = 3x + 6

. Given

3x + 6 < 36

, subtracting

6

yields

3x < 30

, so

x < 10

.

The total length around the triangle must be strictly less than

36

.

2

Apply the triangle inequality theorem to find a lower bound for non-degenerate triangles.

AB + BC > CA \implies x + (x + 2) > x + 4 \implies 2x + 2 > x + 4 \implies x > 2

.

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

3

Formulate and solve the acute triangle condition.

For an acute triangle with longest side

CA = x + 4

, we require

CA^2 < AB^2 + BC^2

.

(x + 4)^2 < x^2 + (x + 2)^2

x^2 + 8x + 16 < x^2 + x^2 + 4x + 4

x^2 + 8x + 16 < 2x^2 + 4x + 4

0 < x^2 - 4x - 12

(x - 6)(x + 2) > 0

Since

x

is a positive integer,

x + 2 > 0

, which implies

x - 6 > 0

, so

x > 6

.

In any acute triangle, the square of the longest side is strictly less than the sum of the squares of the remaining two sides.

4

Combine the bounds to determine the number of valid integer values for

x

.

Combining

x > 6

and

x < 10

gives integer values

x \in \{7, 8, 9\}

. There are

3

such values.

Only integer values in the open interval

(6, 10)

satisfy all conditions simultaneously.

Key Concept

Acute Triangle Criteria and Triangle Inequality

Question 1538Question →

In acute triangle $ABC$ , point $D$ lies on side $BC$ such that segment $AD$ is perpendicular to $BC$ . The length of side $BC$ is $15$ , and the ratio of the area of triangle $ABD$ to the area of triangle $ADC$ is $2 : 3$ . If the lengths of sides $AB$ and $AC$ are both integers, what is the perimeter of triangle $ABC$ ?

30

45

54

60

75

Show answer & explanation

Answer: $60$

Answer

The perimeter of triangle

ABC

is

60

.

The correct answer is derived by recognizing that the altitude splits the base into segments of lengths 6 and 9 based on the area ratio 2:3. Applying the Pythagorean theorem to both right triangles yields

AC^2 - AB^2 = 45

. Factoring 45 into positive integer pairs shows that the only valid side lengths yielding a real, positive height are

AB = 22

and

AC = 23

, giving a total perimeter of

15 + 22 + 23 = 60

.

Step-by-Step Solution

1

Determine the lengths of base segments

BD

and

DC

.

BD = 6

and

DC = 9

.

Triangles

ABD

and

ADC

share the common height

AD = h

. The ratio of their areas is equal to the ratio of their bases:

\frac{\text{Area}(ABD)}{\text{Area}(ADC)} = \frac{BD}{DC} = \frac{2}{3}

. Since

BD + DC = 15

, we have

BD = 6

and

DC = 9

.

2

Express the square of height

h^2

using the Pythagorean theorem in right triangles

ABD

and

ADC

.

h^2 = AB^2 - 36 = AC^2 - 81

.

In right triangle

ABD

,

AB^2 = BD^2 + h^2 = 36 + h^2

. In right triangle

ADC

,

AC^2 = DC^2 + h^2 = 81 + h^2

.

3

Set up a difference of squares equation for side lengths

AB

and

AC

.

(AC - AB)(AC + AB) = 45

.

Equating the two expressions for

h^2

gives

AC^2 - 81 = AB^2 - 36 \implies AC^2 - AB^2 = 45

.

4

Find positive integer solutions for

AB

and

AC

.

AB = 22

and

AC = 23

.

Since

AB

and

AC

are positive integers and

AC > AB

, we analyze factor pairs

(AC - AB, AC + AB)

of

45

with same parity (both odd):
- Pair

(1, 45)

:

AC - AB = 1

and

AC + AB = 45 \implies AC = 23, AB = 22

. Here

h^2 = 22^2 - 36 = 448 > 0

, giving a valid non-degenerate triangle.
- Pair

(3, 15)

:

AC - AB = 3

and

AC + AB = 15 \implies AC = 9, AB = 6

. Here

h^2 = 6^2 - 36 = 0

, which means

h = 0

(degenerate line segment, invalid).
- Pair

(5, 9)

:

AC - AB = 5

and

AC + AB = 9 \implies AC = 7, AB = 2

. Here

h^2 = 2^2 - 36 = -32 < 0

(impossible).

5

Calculate the total perimeter of triangle

ABC

.

Perimeter

= 15 + 22 + 23 = 60

.

Summing all three side lengths gives

BC + AB + AC = 15 + 22 + 23 = 60

.

Key Concept

Triangles: Properties, Perimeter, and Area

Question 1539Question →

A company has two sales divisions, Division X and Division Y. Division X has 12 representatives with an average monthly sales volume of $\$14,000$ . Division Y has 18 representatives with an average monthly sales volume of $\$24,000$ . If 2 representatives leave Division Y who together accounted for $\$82,000$ in monthly sales, what is the new combined average monthly sales volume, in dollars, for all remaining representatives in both divisions?

Show answer & explanation

Answer: 18500

Answer

The new combined average monthly sales volume for all remaining representatives is $18,500.

To find the combined mean, calculate the sum of all remaining values across both groups (

518,000

) and divide by the total number of remaining items (

28

), yielding

18,500

.

Step-by-Step Solution

1

Find total monthly sales for Division X

12 \times 14,000 = 168,000

The total value of a group is the product of its count and its arithmetic mean.

2

Find total monthly sales for Division Y before representatives left

18 \times 24,000 = 432,000

Multiply the number of representatives in Division Y by their average sales volume.

3

Determine the remaining total sales in Division Y

432,000 - 82,000 = 350,000

Subtract the sales volume of the 2 departing representatives from Division Y's original total.

4

Calculate the total combined sales for both divisions

168,000 + 350,000 = 518,000

Add the total sales of Division X and the remaining sales of Division Y.

5

Calculate total number of remaining representatives across both divisions

12 + (18 - 2) = 28

Division X retains 12 representatives, while Division Y retains 16 representatives.

6

Compute the combined mean sales volume

\frac{518,000}{28} = 18,500

Divide the total combined sales volume by the total number of remaining representatives.

Key Concept

Weighted Mean and Combining Group Data

Question 1540Question →

In quadrilateral $ABCD$ , diagonal $AC$ divides the quadrilateral into two right triangles, $\triangle ABC$ and $\triangle ADC$ . The measure of angle $ABC$ is $90^\circ$ , the measure of angle $ADC$ is $90^\circ$ , and the measure of angle $BAC$ is $30^\circ$ . If $AB = 12$ and $AD = CD$ , what is the area of quadrilateral $ABCD$ ?

48 + 24\sqrt{3}

36 + 18\sqrt{3}

192 + 24\sqrt{3}

96 + 48\sqrt{3}

24 + 48\sqrt{3}

Show answer & explanation

Answer: $48 + 24\sqrt{3}$

Answer

48 + 24\sqrt{3}

The total area of quadrilateral ABCD is the sum of the areas of right triangles ABC and ADC. Using 30-60-90 triangle properties on ABC gives

BC = 4\sqrt{3}

and hypotenuse

AC = 8\sqrt{3}

, resulting in

\text{Area}(\triangle ABC) = 24\sqrt{3}

. Triangle ADC is a 45-45-90 triangle with hypotenuse

8\sqrt{3}

, yielding leg lengths

4\sqrt{6}

and

\text{Area}(\triangle ADC) = 48

. Adding both areas gives

48 + 24\sqrt{3}

.

Step-by-Step Solution

1

Analyze right triangle ABC to find side lengths BC and hypotenuse AC.

BC = 4\sqrt{3}

and

AC = 8\sqrt{3}

In 30-60-90 triangle ABC,

\cos(30^\circ) = \frac{AB}{AC} \implies \frac{\sqrt{3}}{2} = \frac{12}{AC} \implies AC = 8\sqrt{3}

. Also,

\tan(30^\circ) = \frac{BC}{AB} \implies \frac{1}{\sqrt{3}} = \frac{BC}{12} \implies BC = 4\sqrt{3}

.

2

Calculate the area of right triangle ABC.

\text{Area}(\triangle ABC) = 24\sqrt{3}

\text{Area}(\triangle ABC) = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 12 \times 4\sqrt{3} = 24\sqrt{3}

.

3

Analyze isosceles right triangle ADC to find leg length

s = AD = CD

.

s = 4\sqrt{6}

In 45-45-90 triangle ADC, the hypotenuse

AC = s\sqrt{2} = 8\sqrt{3} \implies s = \frac{8\sqrt{3}}{\sqrt{2}} = 4\sqrt{6}

.

4

Calculate the area of right triangle ADC.

\text{Area}(\triangle ADC) = 48

\text{Area}(\triangle ADC) = \frac{1}{2} \times s^2 = \frac{1}{2} \times (4\sqrt{6})^2 = \frac{1}{2} \times 96 = 48

.

5

Sum the areas of the two triangles to get total area of quadrilateral ABCD.

\text{Total Area} = 48 + 24\sqrt{3}

\text{Area}(ABCD) = \text{Area}(\triangle ABC) + \text{Area}(\triangle ADC) = 24\sqrt{3} + 48 = 48 + 24\sqrt{3}

.

Key Concept

Special Right Triangle Ratios (30-60-90 and 45-45-90)

Alternative Method

Use trigonometric ratios directly:

\text{Area}(\triangle ABC) = \frac{1}{2} (12)^2 \tan(30^\circ) = 72 \times \frac{1}{\sqrt{3}} = 24\sqrt{3}

, and

\text{Area}(\triangle ADC) = \frac{1}{4} AC^2 = \frac{1}{4} (8\sqrt{3})^2 = 48

.

Estimated Time:2m 30s

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