All practice questions

If

x

and

y

are positive real numbers satisfying the system of exponential radical equations

\sqrt{x\sqrt{y}} = 3

\sqrt{y\sqrt{x}} = 9

which of the following statements must be true? Select all that apply.

Select all that apply

xy = 81

x^{1/2} + y^{1/2} = 10

\sqrt{xy} = 27

x

has two valid real solutions,

x = 1

and

x = -1

y^x - x^y = 80

Show answer & explanation

Answer: $xy = 81$ ; $x^{1/2} + y^{1/2} = 10$ ; $y^x - x^y = 80$

Answer

The statements asserting that

xy = 81

,

x^{1/2} + y^{1/2} = 10

, and

y^x - x^y = 80

are true.

Solving the system of radical equations gives the unique positive solution pair

x = 1

and

y = 81

. Substituting these values into the statements shows that

xy = 1 \cdot 81 = 81

,

x^{1/2} + y^{1/2} = \sqrt{1} + \sqrt{81} = 10

, and

y^x - x^y = 81^1 - 1^{81} = 80

are all mathematically valid.

Step-by-Step Solution

1

Square both sides of each equation to remove the outer radicals

x\sqrt{y} = 9 \implies x^2 y = 81

and

y\sqrt{x} = 81 \implies y^2 x = 6561

Squaring

\sqrt{A} = B

yields

A = B^2

, and rewriting radicals as rational powers gives

x y^{1/2} = 9

and

y x^{1/2} = 81

.

2

Express

y

in terms of

x

from the first equation and substitute into the second

y = \frac{81}{x^2} \implies \left(\frac{81}{x^2}\right)^2 x = 6561 \implies \frac{6561}{x^3} = 6561

Substituting

y

eliminates the variable

y

, leaving a single equation in terms of

x

.

3

Solve for

x

and

y

x^3 = 1 \implies x = 1

, which gives

y = \frac{81}{1^2} = 81

Since

x > 0

is a positive real number,

x = 1

is the unique real solution, yielding

y = 81

.

4

Evaluate each given statement using

x = 1

and

y = 81

xy = 81

(True),

x^{1/2} + y^{1/2} = 1 + 9 = 10

(True),

\sqrt{xy} = 9 \neq 27

(False),

x = -1

is invalid (False),

81^1 - 1^{81} = 80

(True)

Direct calculation confirms which individual statements hold.

Key Concept

Simplifying nested radical equations using fractional exponent rules

Estimated Time:2m 0s

Question 282Question →

If $x$ is a real number such that $2^{x+1} + 2^{x-1} = 40$ , what is the value of the expression $(x+1)^{x-1}$ ?

65

64

125

243

15

Show answer & explanation

Answer: 125

Answer

125

Factoring out

2^x

from the given equation yields

2^x(2 + 0.5) = 40

, which gives

2.5 \cdot 2^x = 40

and

2^x = 16

. This determines that

x = 4

. Substituting

x = 4

into

(x+1)^{x-1}

gives

(4+1)^{3} = 5^3 = 125

.

Step-by-Step Solution

1

Rewrite the given exponential terms with a common power of 2.

2^{x+1} = 2^x \cdot 2^1

and

2^{x-1} = 2^x \cdot 2^{-1} = \frac{2^x}{2}

.

Applying product rule for exponents allows factoring out

2^x

.

2

Factor out

2^x

and solve for

x

.

2^x \left(2 + \frac{1}{2}\right) = 40 \implies 2^x \cdot \frac{5}{2} = 40 \implies 2^x = 16 \implies x = 4

.

Combining fractional coefficients isolates the exponential term

2^x

.

3

Substitute

x = 4

into the target expression

(x+1)^{x-1}

.

(4+1)^{4-1} = 5^3 = 125

.

Simplifying the base and exponent yields the final numerical value.

Key Concept

Solving exponential equations using power distribution rules and factoring.

Question 283Question →

What value of $x$ satisfies the exponential equation $25^x - 4 \cdot 5^{x+1} = 125$ ?

Show answer & explanation

Answer: 2

Answer

The correct answer is

2

.

Rewriting

25^x

as

(5^x)^2

and

4 \cdot 5^{x+1}

as

20 \cdot 5^x

transforms the equation into

(5^x)^2 - 20(5^x) - 125 = 0

. Substituting

u = 5^x

produces

u^2 - 20u - 125 = 0

, which factors into

(u - 25)(u + 5) = 0

. Because

5^x

must be greater than zero for all real values of

x

,

u = -5

yields no valid real solution. Thus,

5^x = 25 = 5^2

, giving

x = 2

.

Step-by-Step Solution

1

Convert exponential expressions to a common base of 5.

25^x = (5^2)^x = (5^x)^2

and

4 \cdot 5^{x+1} = 4 \cdot 5 \cdot 5^x = 20 \cdot 5^x

.

Applying exponent laws

a^{m+n} = a^m \cdot a^n

and

(a^m)^n = a^{mn}

expresses terms in quadratic form with respect to

5^x

.

2

Formulate and factor the quadratic equation in terms of

u = 5^x

.

u^2 - 20u - 125 = 0 \implies (u - 25)(u + 5) = 0

, yielding

u = 25

or

u = -5

.

The equation reduces to standard quadratic form, which factors easily.

3

Solve for

x

while rejecting non-viable real roots.

5^x = 25 = 5^2 \implies x = 2

.

5^x = -5

has no real solution.

An exponential function with a positive base produces strictly positive output values for all real domain inputs.

Key Concept

Solving exponential equations reducible to quadratic form using exponent laws

Question 284Question →

At a charity fundraising event, standard tickets were sold for $\$45$ each and VIP tickets were sold for $\$80$ each. The number of standard tickets sold was $12$ more than twice the number of VIP tickets sold. If the total revenue generated from standard tickets exceeded the total revenue from VIP tickets by $\$1,830$ , how many VIP tickets were sold?

Show answer & explanation

Answer: 129

Answer

129 VIP tickets were sold.

Letting

v

represent the number of VIP tickets sold, the number of standard tickets sold is

2v + 12

. Expressing the revenue condition yields the linear equation

45(2v + 12) - 80v = 1830

. Distributing 45 gives

90v + 540 - 80v = 1830

. Combining like terms results in

10v + 540 = 1830

. Subtracting 540 from both sides gives

10v = 1290

, which yields

v = 129

.

Step-by-Step Solution

1

Define the unknown variable

Let

v

be the number of VIP tickets sold.

The question asks for the number of VIP tickets, making

v

a direct choice for the variable.

2

Translate the relationship between ticket quantities into an algebraic expression

Standard tickets sold

= 2v + 12

'12 more than twice the number of VIP tickets' translates directly to

2v + 12

.

3

Formulate total revenue expressions and construct the single-variable linear equation

45(2v + 12) - 80v = 1830

Total standard revenue minus total VIP revenue equals the given excess of

\$1,830

.

4

Distribute and combine like terms to solve for

v

90v + 540 - 80v = 1830 \implies 10v + 540 = 1830 \implies 10v = 1290 \implies v = 129

Applying standard algebraic operations isolates

v

on one side of the equation.

Key Concept

Formulating and solving a linear equation in one variable from a multi-step word problem context

Estimated Time:2m 0s

Question 285Question →

A reservoir initially contains $4,800$ gallons of water. A drainage pump operates at a constant rate of $120$ gallons per minute to empty the reservoir. How many minutes of continuous pumping will it take for the amount of water remaining in the reservoir to be reduced to $1,200$ gallons?

10

25

30

40

50

Show answer & explanation

Answer: $30$

Answer

30 minutes

To find the time needed for

1,200

gallons to remain, subtract

1,200

from the initial

4,800

gallons to determine that

3,600

gallons of water must be removed. Dividing the

3,600

gallons by the drainage rate of

120

gallons per minute gives exactly

30

minutes.

Step-by-Step Solution

1

Calculate the total volume of water that must be drained from the reservoir.

4,800 \text{ gallons} - 1,200 \text{ gallons} = 3,600 \text{ gallons}

The question asks for the time until

1,200

gallons remain, so the amount removed is the difference between the starting volume and ending volume.

2

Set up the linear rate equation to solve for elapsed time

t

in minutes.

\text{Rate} \times t = \text{Volume Drained} \implies 120t = 3,600

The rate of drainage is constant at

120

gallons per minute.

3

Solve for

t

.

t = \frac{3,600}{120} = 30 \text{ minutes}

Dividing total gallons to drain by the rate gives the required time.

Key Concept

Linear Modeling and Distance-Rate-Time / Work-Rate Relationships

Estimated Time:1m 0s

Question 286Question →

Based on the structural contrast and elaboration signals in the passage below, fill in the blank with the word that best completes the text.

Fill in the blanks below

In early twentieth-century ethnomusicology, field recordings captured on wax cylinders were initially celebrated as objective preservation tools that recorded acoustic reality without distortion. Far from capturing an unmediated performance, however, these early acoustic recordings inadvertently the very traditions they sought to document; indeed, the intrusive presence of recording apparatuses frequently compelled musicians to truncate improvisational structures and temper tonal nuances, thereby fundamentally altering authentic performance practices.

Show answer & explanation

Answer

compromised

The contrast transition 'Far from... however' signals a turn away from the idea that field recordings were purely objective and non-distorting. The continuation signal 'indeed' after the semicolon elaborates on this pivot, explaining that the presence of recording equipment forced musicians to alter their performance habits. Consequently, the blank requires a word conveying that the integrity or authenticity of the traditions was weakened or altered (such as 'compromised').

Step-by-Step Solution

1

Analyze the opening contrast transition

The initial clause describes wax cylinders as celebrated 'objective preservation tools' capturing music 'without distortion'. The structural signal 'Far from..., however' indicates a complete reversal of this initial assumption.

Contrast markers signal that the target word must express an outcome opposite to unmediated preservation.

2

Examine the elaboration signal following the semicolon

The continuation signal 'indeed' introduces explicit evidence: the presence of recording equipment forced performers to 'truncate improvisational structures' and 'temper tonal nuances,' resulting in altered performance practices.

Elaboration markers reinforce and clarify the negative impact established by the structural contrast.

3

Select the vocabulary word that synthesizes both structural clues

The term 'compromised' (or acceptable synonyms such as 'vitiated' or 'altered') accurately reflects how the recording process impaired or undermined the authenticity of the musical traditions.

It reconciles the structural reversal ('however') with the explanatory details of performance modification ('indeed').

Key Concept

Deciphering Meaning via Structural Clues and Contrast/Continuation Signals

Question 287Question →

Let $m$ and $n$ be negative integers such that $m < n$ and $m - n$ is an odd integer. Which of the following expressions must be a positive even integer?

(m - n)^2 + 1

m + n + 1

(-1)^n (m - n)

m^2 + n^2

(m - n)^3

Show answer & explanation

Answer: $(m - n)^2 + 1$

Answer

(m - n)^2 + 1

must be a positive even integer.

Because

m - n

is given as an odd integer and

m < n

,

m - n

is a negative odd integer. Squaring any negative odd integer gives a positive odd integer. Adding

1

to a positive odd integer yields a positive even integer, which guarantees the result is always positive and even.

Step-by-Step Solution

1

Analyze the parity and sign of

m - n

m - n

is a negative odd integer.

Since

m < n

,

m - n < 0

. The problem specifies that

m - n

is odd.

2

Evaluate the expression

(m - n)^2

(m - n)^2

is a positive odd integer.

Squaring any non-zero real number yields a positive result. Squaring an odd integer always yields an odd integer.

3

Add 1 to

(m - n)^2

(m - n)^2 + 1

is an even integer greater than or equal to 2.

Adding 1 to a positive odd integer converts it into a positive even integer.

Key Concept

Parity rules under arithmetic operations and exponents with signed integers

Estimated Time:1m 15s

Question 288Question →

A logistics warehouse uses two sizes of shipping crates: small crates and large crates. A shipment containing $3$ small crates and $5$ large crates has a total weight of $47$ pounds. A second shipment containing $6$ small crates and $2$ large crates has a total weight of $38$ pounds. What is the total weight, in pounds, of a shipment containing $2$ small crates and $3$ large crates?

11

22

26

29

33

Show answer & explanation

Answer: 29

Answer

29 pounds

By setting up the linear system

3s + 5l = 47

and

6s + 2l = 38

, using elimination gives

s = 4

pounds for a small crate and

l = 7

pounds for a large crate. Substituting these values into the target expression

2s + 3l

yields

2(4) + 3(7) = 29

pounds.

Step-by-Step Solution

1

Set up a system of linear equations from the given context.

Let

s

be the weight of a small crate and

l

be the weight of a large crate. The equations are: (1)

3s + 5l = 47

and (2)

6s + 2l = 38

.

Translate the physical constraints of each shipment into algebraic equations.

2

Solve for the variable

l

by eliminating

s

.

Multiply Equation (1) by

2

to get

6s + 10l = 94

. Subtract Equation (2) (

6s + 2l = 38

) from this new equation:

(6s + 10l) - (6s + 2l) = 94 - 38 \implies 8l = 56 \implies l = 7

.

Align the coefficients of

s

so that elimination via subtraction yields a single-variable linear equation for

l

.

3

Substitute

l = 7

back into Equation (2) to solve for

s

.

6s + 2(7) = 38 \implies 6s + 14 = 38 \implies 6s = 24 \implies s = 4

.

Determine the individual weight of a small crate.

4

Calculate the requested total weight for

2

small crates and

3

large crates.

2s + 3l = 2(4) + 3(7) = 8 + 21 = 29

.

Substitute the individual values of

s

and

l

into the target expression.

Key Concept

Solving 2x2 Systems of Linear Equations by Elimination and Linear Combination Evaluation

Estimated Time:1m 30s

Question 289Question →

Let $x$ and $y$ be non-zero integers that satisfy all of the following conditions:
I. $(-x)^{y + 1} < 0$
II. $x^2 y + x$ is odd
III. $x < y$

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

Select all that apply

y^x

is an even integer

x^y - y

is an odd integer

-x < -y

(x + y)^2

is an odd integer

(-x)^y < 0

Show answer & explanation

Answer: $y^x$ is an even integer; $x^y - y$ is an odd integer; $(x + y)^2$ is an odd integer

Answer

The statements that must be true are '

y^x

is an even integer', '

x^y - y

is an odd integer', and '

(x + y)^2

is an odd integer'.

From the given conditions, we deduce that

x

is a positive odd integer and

y

is a positive even integer with

x < y

. Therefore:
- Raising the positive even integer

y

to the positive power

x

yields an even integer (

y^x

is even).
- Raising the odd integer

x

to the positive power

y

yields an odd integer, and subtracting the even integer

y

leaves an odd integer (

x^y - y

is odd).
- The sum of odd

x

and even

y

is odd, and squaring an odd integer yields an odd integer (

(x + y)^2

is odd).

Step-by-Step Solution

1

Analyze Condition I:

(-x)^{y + 1} < 0

x > 0

(a positive integer) and

y

is an even integer.

For a number raised to an integer power to be strictly negative, the base must be negative and the exponent must be odd. Hence,

-x < 0 \implies x > 0

, and

y + 1

is odd

\implies y

is even.

2

Analyze Condition II:

x^2 y + x

is odd

x

is an odd integer.

Factor the expression as

x(xy + 1)

. A product of two integers is odd if and only if both factors are odd. Thus,

x

must be odd. (Additionally,

xy + 1

must be odd

\implies xy

is even, which holds since

y

is even).

3

Analyze Condition III:

x < y

y

is a positive even integer.

Since

x

is a positive odd integer (

x \ge 1

) and

x < y

,

y

must also be a positive integer (

y \ge 2

).

4

Evaluate the statements based on the derived properties (

x

is positive odd,

y

is positive even,

x < y

)

The statements '

y^x

is an even integer', '

x^y - y

is an odd integer', and '

(x + y)^2

is an odd integer' are guaranteed to be true.

1)

\text{even}^{\text{positive odd}} = \text{even}

. 2)

\text{odd}^{\text{positive even}} - \text{even} = \text{odd} - \text{even} = \text{odd}

. 3)

(\text{odd} + \text{even})^2 = \text{odd}^2 = \text{odd}

.

Key Concept

Deducing parity and signs of variables using exponent rules and arithmetic properties of even and odd numbers.

Estimated Time:2m 0s

Question 290Question →

Two water reservoirs, Reservoir A and Reservoir B, contain a combined total of $360$ liters of water. After $30$ liters of water are transferred from Reservoir A to Reservoir B, the volume of water in Reservoir B is equal to $\frac{2}{3}$ of the volume of water remaining in Reservoir A. How many liters of water were originally in Reservoir A?

114

210

216

246

228

Show answer & explanation

Answer: $246$

Answer

246 liters were originally in Reservoir A.

Let

x

represent the original volume of Reservoir A. Since the combined volume is

360

liters, Reservoir B initially contains

360 - x

liters. After transferring

30

liters from A to B, Reservoir A contains

x - 30

liters and Reservoir B contains

(360 - x) + 30 = 390 - x

liters. Setting up the equation

390 - x = \frac{2}{3}(x - 30)

and multiplying by 3 gives

1170 - 3x = 2x - 60

. Combining like terms yields

5x = 1230

, so

x = 246

liters. Thus,

246

is the correct original volume.

Step-by-Step Solution

1

Define the variable for the unknown quantity.

Let

x

be the original volume of water in Reservoir A (in liters).

Choosing a single variable simplifies setting up a linear equation.

2

Express the original volume of Reservoir B in terms of

x

.

Original volume in Reservoir B is

360 - x

.

The total volume across both reservoirs is given as

360

liters.

3

Write expressions for the volumes in each reservoir after the transfer of 30 liters.

Volume in Reservoir A after transfer:

x - 30

.
Volume in Reservoir B after transfer:

(360 - x) + 30 = 390 - x

.

Transferring 30 liters removes 30 liters from A and adds 30 liters to B.

4

Set up the linear equation based on the given relationship.

390 - x = \frac{2}{3}(x - 30)

The problem specifies that the new volume in Reservoir B is

\frac{2}{3}

of the new volume in Reservoir A.

5

Solve the equation for

x

.

3(390 - x) = 2(x - 30) \implies 1170 - 3x = 2x - 60 \implies 1230 = 5x \implies x = 246

.

Multiply both sides by 3 to eliminate the fraction, then collect like terms.

Key Concept

Formulating and solving a linear equation in one variable from a real-world conservation/transfer scenario.

Question 291Question →

If $x$ is a real number such that $|2x - 5| = |x + 4|$ , which of the following could be the value of $x^2$ ? Select all such values.

Select all that apply

\frac{1}{9}

\frac{1}{3}

1

9

81

Show answer & explanation

Answer: $\frac{1}{9}$ ; $81$

Answer

The possible values of

x^2

are

\frac{1}{9}

and

81

.

To solve

|2x - 5| = |x + 4|

, set

2x - 5

equal to both

x + 4

and

-(x + 4)

. Solving

2x - 5 = x + 4

gives

x = 9

, which leads to

x^2 = 81

. Solving

2x - 5 = -x - 4

gives

3x = 1

, so

x = \frac{1}{3}

, which leads to

x^2 = \frac{1}{9}

. Therefore, the options representing

\frac{1}{9}

and

81

are correct.

Step-by-Step Solution

1

Set up equations to remove the absolute value bars

Two linear equations:

2x - 5 = x + 4

or

2x - 5 = -(x + 4)

.

For real numbers

A

and

B

,

|A| = |B|

implies

A = B

or

A = -B

.

2

Solve the first equation

2x - 5 = x + 4

x = 9

.

Subtract

x

and add

5

to both sides.

3

Solve the second equation

2x - 5 = -x - 4

3x = 1 \implies x = \frac{1}{3}

.

Add

x

and add

5

to both sides.

4

Compute

x^2

for each possible value of

x

If

x = 9

, then

x^2 = 81

. If

x = \frac{1}{3}

, then

x^2 = \frac{1}{9}

.

The question asks for the values of

x^2

, not

x

.

Key Concept

Solving equations with absolute values on both sides requires considering positive and negative case equivalences.

Question 292Question →

If $x$ and $y$ are real numbers such that $|x - 4| \le 3$ and $|y + 2| \le 5$ , what is the maximum possible value of $|x - y|$ ?

Show answer & explanation

Answer: 14

Answer

The maximum possible value of

|x - y|

is 14.

Solving

|x - 4| \le 3

gives the closed interval

[1, 7]

for

x

. Solving

|y + 2| \le 5

gives the closed interval

[-7, 3]

for

y

. The maximum possible value of

|x - y|

is the maximum distance between a point in

[1, 7]

and a point in

[-7, 3]

, which is

7 - (-7) = 14

.

Step-by-Step Solution

1

Determine the range of possible values for

x

.

1 \le x \le 7

The inequality

|x - 4| \le 3

represents all numbers within distance 3 of 4 on the number line.

2

Determine the range of possible values for

y

.

-7 \le y \le 3

The inequality

|y + 2| \le 5

represents all numbers within distance 5 of -2 on the number line.

3

Find the maximum distance between any point

x

in

[1, 7]

and any point

y

in

[-7, 3]

.

14

The maximum absolute difference

|x - y|

occurs between the upper endpoint of the

x

-interval (

7

) and the lower endpoint of the

y

-interval (

-7

).

Key Concept

Absolute value inequalities as distance intervals on the real number line and maximizing differences between bounded variables

Question 293Question →

If $n$ is a positive integer such that $2^{n-2} + 2^{n-1} + 2^n + 2^{n+1} = 480$ , what is the value of $n$ ?

Show answer & explanation

Answer: 7

Answer

The value of

n

is 7.

Factoring out

2^{n-2}

converts the sum into

2^{n-2}(1 + 2 + 4 + 8) = 15 \cdot 2^{n-2} = 480

. Dividing 480 by 15 gives

2^{n-2} = 32

. Since

32 = 2^5

, setting

n - 2 = 5

yields

n = 7

.

Step-by-Step Solution

1

Factor out the lowest power of 2, which is

2^{n-2}

, from all terms on the left side of the equation.

2^{n-2}(1 + 2 + 4 + 8) = 480

Factoring out a common exponential term simplifies the addition of powers into a product of a single exponential term and a constant.

2

Evaluate the constant factor inside the parentheses and solve for the exponential expression

2^{n-2}

.

15 \cdot 2^{n-2} = 480 \implies 2^{n-2} = 32

Summing

1 + 2 + 4 + 8

yields 15. Dividing both sides by 15 isolates the base-2 term.

3

Express 32 as a power with base 2 and equate the exponents.

2^{n-2} = 2^5 \implies n - 2 = 5 \implies n = 7

Since

32 = 2^5

and the bases are identical, the exponents must be equal.

Key Concept

Factoring sum of exponential terms with common bases

Question 294Question →

Two concentric circles centered at point $O$ have radii of $6$ units and $10$ units. A central angle $\theta$ defines a sector that intersects the region between the concentric circles, creating a region bounded by an outer arc, an inner arc, and two straight line segments. If the area of this bounded region is $16\pi$ square units, what is the perimeter of the region?

8\pi

4 + 8\pi

8 + 8\pi

8 + 16\pi

20 + 8\pi

Show answer & explanation

Answer: $8 + 8\pi$

Answer

The perimeter of the region is

8 + 8\pi

units.

The region between the concentric circles has an area equal to the fraction of the central angle times the area of the ring:

\frac{\theta}{360^\circ} \cdot \pi (10^2 - 6^2) = 64\pi \left(\frac{\theta}{360^\circ}\right)

. Setting this equal to

16\pi

gives

\frac{\theta}{360^\circ} = \frac{1}{4}

. The outer arc length is

\frac{1}{4}(2\pi \cdot 10) = 5\pi

, and the inner arc length is

\frac{1}{4}(2\pi \cdot 6) = 3\pi

. The region is bounded on the sides by two segments of length

10 - 6 = 4

units each. The total perimeter is

5\pi + 3\pi + 4 + 4 = 8 + 8\pi

.

Step-by-Step Solution

1

Express the area of the bounded region in terms of the central angle fraction

f = \frac{\theta}{360^\circ}

.

\text{Area} = f \cdot \pi(R^2 - r^2) = f \cdot \pi(10^2 - 6^2) = 64\pi f

.

The area of the region between two concentric sector arcs is the difference between the outer sector area and the inner sector area.

2

Solve for the fraction

f

using the given area of

16\pi

.

64\pi f = 16\pi \implies f = \frac{16\pi}{64\pi} = \frac{1}{4}

.

Equating the algebraic expression for the region's area to

16\pi

allows finding the proportion of the circle represented by the central angle.

3

Calculate the lengths of the outer arc, inner arc, and straight boundary segments.

\text{Outer Arc} = \frac{1}{4}(2\pi \cdot 10) = 5\pi

,

\text{Inner Arc} = \frac{1}{4}(2\pi \cdot 6) = 3\pi

,

\text{Segment Length} = 10 - 6 = 4

.

Arc lengths use the formula

f \cdot 2\pi r

, and each straight boundary segment is the radial distance between the two circles.

4

Sum all boundary components to find the total perimeter.

\text{Perimeter} = 5\pi + 3\pi + 4 + 4 = 8 + 8\pi

.

The complete perimeter of the bounded region consists of the outer arc, inner arc, and two radial segments.

Key Concept

Annular Sector Area and Perimeter

Estimated Time:1m 30s

Question 295Question →

If $x$ is a real number satisfying the exponential equation $4^{x+1} - 4^{x-1} = 120$ , what is the value of $2^{2x + 1}$ ?

16

32

64

128

-64

Show answer & explanation

Answer: 64

Answer

The value of

2^{2x + 1}

is 64.

Factoring

4^{x-1}

from

4^{x+1} - 4^{x-1}

gives

4^{x-1}(16 - 1) = 15 \cdot 4^{x-1} = 120

. Dividing by 15 yields

4^{x-1} = 8

. Rewriting with base 2 gives

(2^2)^{x-1} = 2^{2x-2} = 2^3

, so

2x - 2 = 3

, meaning

2x = 5

. Substituting

2x = 5

into

2^{2x+1}

gives

2^{5+1} = 2^6 = 64

.

Step-by-Step Solution

1

Factor out

4^{x-1}

from the left side of the equation

4^{x+1} - 4^{x-1} = 120

.

4^{x-1}(4^2 - 1) = 120

, which simplifies to

4^{x-1}(15) = 120

.

Factoring out the lowest power of 4 allows simplification of the terms on the left side.

2

Divide both sides by 15 to isolate

4^{x-1}

.

4^{x-1} = \frac{120}{15} = 8

.

Isolating the exponential expression is necessary to solve for

x

.

3

Express both sides with a common base of 2.

(2^2)^{x-1} = 2^3 \implies 2^{2(x-1)} = 2^3 \implies 2^{2x - 2} = 2^3

.

Converting to a common prime base allows equating the exponents.

4

Equate exponents to solve for

2x

.

2x - 2 = 3 \implies 2x = 5

.

Since the bases are equal and positive, their exponents must be equal.

5

Substitute

2x = 5

into the target expression

2^{2x + 1}

.

2^{5 + 1} = 2^6 = 64

.

Evaluating the exact expression requested in the stem.

Key Concept

Factoring exponential expressions with variable exponents and converting bases

Estimated Time:1m 30s

Question 296Question →

In triangle $ABC$ , the length of side $AB$ is $7$ and the length of side $BC$ is $15$ . If the area of triangle $ABC$ is $42$ , which of the following could be the length of side $AC$ ? Select all such lengths.

Select all that apply

2\sqrt{37}

16

20

22

\sqrt{274}

Show answer & explanation

Answer: $2\sqrt{37}$ ; $20$

Answer

The possible lengths of side

AC

are

2\sqrt{37}

and

20

.

Using the triangle area formula

\text{Area} = \frac{1}{2} \cdot a \cdot b \cdot \sin B

, we find

\sin B = \frac{4}{5}

. Because

\sin B

is positive in both Quadrant I and Quadrant II, angle

B

can be either acute or obtuse. If angle

B

is acute,

\cos B = \frac{3}{5}

, giving

AC = \sqrt{7^2 + 15^2 - 2(7)(15)(0.6)} = 2\sqrt{37}

. If angle

B

is obtuse,

\cos B = -\frac{3}{5}

, giving

AC = \sqrt{7^2 + 15^2 - 2(7)(15)(-0.6)} = 20

. Both values represent valid triangle configurations.

Step-by-Step Solution

1

Determine the sine of angle

B

using the area formula.

\text{Area} = \frac{1}{2} \cdot AB \cdot BC \cdot \sin B \implies 42 = \frac{1}{2} \cdot 7 \cdot 15 \cdot \sin B \implies \sin B = \frac{84}{105} = \frac{4}{5}

.

The area of a triangle with two given sides and an included angle is

\frac{1}{2} a b \sin \theta

.

2

Find the possible values for

\cos B

.

Since

\sin B = \frac{4}{5}

,

\cos B

can be either

\frac{3}{5}

(if angle

B

is acute) or

-\frac{3}{5}

(if angle

B

is obtuse).

Sine is positive in both the first and second quadrants, permitting both acute and obtuse angles for triangle

ABC

.

3

Calculate the length of side

AC

when angle

B

is acute.

AC^2 = 7^2 + 15^2 - 2(7)(15)\left(\frac{3}{5}\right) = 49 + 225 - 126 = 148 \implies AC = \sqrt{148} = 2\sqrt{37}

.

Apply the Law of Cosines:

AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos B

with

\cos B = 0.6

.

4

Calculate the length of side

AC

when angle

B

is obtuse.

AC^2 = 7^2 + 15^2 - 2(7)(15)\left(-\frac{3}{5}\right) = 49 + 225 + 126 = 400 \implies AC = \sqrt{400} = 20

.

Apply the Law of Cosines with

\cos B = -0.6

.

Key Concept

Triangle Area via Included Angle and Dual Solutions in Non-Right Triangles

Question 297Question →

In triangle $PQR$ , the measure of angle $PQR$ is $90^\circ$ . Point $S$ lies on segment $PR$ such that line segment $QS$ is perpendicular to $PR$ . If $PS = 4$ and $SR = 9$ , what is the area of triangle $PQR$ ?

18

39

78

117

234

Show answer & explanation

Answer: $39$

Answer

The area of triangle

PQR

is 39.

By the geometric mean theorem (right triangle altitude theorem), the altitude

QS

satisfies

QS^2 = PS \cdot SR = 4 \cdot 9 = 36

, so

QS = 6

. The hypotenuse

PR = PS + SR = 4 + 9 = 13

. Substituting base

13

and height

6

into the triangle area formula

\frac{1}{2}bh

gives

\frac{1}{2} \times 13 \times 6 = 39

.

Step-by-Step Solution

1

Find the length of hypotenuse

PR

PR = PS + SR = 4 + 9 = 13

Point

S

lies on segment

PR

, so the total length of the hypotenuse is the sum of its two segments.

2

Calculate altitude

QS

using the geometric mean theorem for right triangles

QS = \sqrt{PS \times SR} = \sqrt{4 \times 9} = \sqrt{36} = 6

In a right triangle, the altitude to the hypotenuse is the geometric mean of the two segments into which the hypotenuse is divided.

3

Calculate the area of triangle

PQR

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 13 \times 6 = 39

The area of any triangle is half the product of its base and corresponding height.

Key Concept

Altitude to the hypotenuse in right triangles and triangle area calculation

Alternative Method

Alternatively, use similar triangles

\triangle PQS \sim \triangle QRS

. The ratio of corresponding sides gives

\frac{PQ}{PS} = \frac{PR}{PQ} \implies PQ^2 = PS \cdot PR = 4 \cdot 13 = 52

, and

QR^2 = SR \cdot PR = 9 \cdot 13 = 117

. Since

\triangle PQR

is a right triangle at

Q

, its area is

\frac{1}{2} \cdot PQ \cdot QR = \frac{1}{2} \sqrt{52 \cdot 117} = \frac{1}{2} \sqrt{6084} = \frac{1}{2} \cdot 78 = 39

.

Estimated Time:1m 30s

Question 298Question →

On the real number line, $x$ is a negative real number such that $|5 - 2x| = 11$ , and $y$ is a positive real number such that $|3y + 1| = 13$ . What is the distance on the real number line between $x$ and $y$ ?

Show answer & explanation

Answer: 7

Answer

The distance between

x

and

y

on the real number line is 7.

Solving

|5 - 2x| = 11

with

x < 0

yields

x = -3

. Solving

|3y + 1| = 13

with

y > 0

yields

y = 4

. The distance between

-3

and

4

on the number line is

|4 - (-3)| = 7

.

Step-by-Step Solution

1

Solve for the negative real number

x

using the equation

|5 - 2x| = 11

.

x = -3

The equation splits into

5 - 2x = 11

(yielding

x = -3

) and

5 - 2x = -11

(yielding

x = 8

). Since

x

must be negative,

x = -3

is selected.

2

Solve for the positive real number

y

using the equation

|3y + 1| = 13

.

y = 4

The equation splits into

3y + 1 = 13

(yielding

y = 4

) and

3y + 1 = -13

(yielding

y = -\frac{14}{3}

). Since

y

must be positive,

y = 4

is selected.

3

Compute the distance between

x

and

y

on the number line.

7

The distance between two points on the number line is given by

|y - x| = |4 - (-3)| = 7

.

Key Concept

Distance on the real number line between two points

a

and

b

is given by

|a - b|

, solved by evaluating absolute value equations under given sign constraints.

Alternative Method

Plot the candidate solutions for

x

(

x = -3

and

x = 8

) and

y

(

y = 4

and

y = -14/3

) on a number line, then directly count units between the valid points

x = -3

and

y = 4

.

Estimated Time:1m 15s

Question 299Question →

A manufacturing company's monthly total production cost

C

, in dollars, is related to the number of units produced,

x

, by the linear equation:

\frac{3(C - 1,200)}{5} = 2x + 180

If the total production cost in a given month was between

\$4,000

and

\$5,000

, inclusive, which of the following could be the number of units produced in that month? Select all such numbers.

Select all that apply

700

800

950

1,000

1,100

Show answer & explanation

Answer: 800; 950; 1,000

Answer

The numbers of units that could be produced are 800, 950, and 1,000.

Solving the linear equation for

x

gives

x = \frac{3C - 4,500}{10}

. Substituting the boundary cost values

C = 4,000

and

C = 5,000

yields

x = 750

and

x = 1,050

, respectively. Therefore, any valid number of units must lie in the range

[750, 1,050]

. The values 800, 950, and 1,000 fall strictly within this interval.

Step-by-Step Solution

1

Express x in terms of C by isolating the variable x in the linear equation.

3(C - 1,200) = 5(2x + 180) \implies 3C - 3,600 = 10x + 900 \implies 10x = 3C - 4,500 \implies x = \frac{3C - 4,500}{10}

Isolating

x

allows direct evaluation of the unit limits based on the cost boundary values.

2

Calculate the minimum value of x corresponding to the minimum cost C = 4,000.

x_{\min} = \frac{3(4,000) - 4,500}{10} = \frac{12,000 - 4,500}{10} = \frac{7,500}{10} = 750

Determines the lower bound of units produced.

3

Calculate the maximum value of x corresponding to the maximum cost C = 5,000.

x_{\max} = \frac{3(5,000) - 4,500}{10} = \frac{15,000 - 4,500}{10} = \frac{10,500}{10} = 1,050

Determines the upper bound of units produced.

4

Filter the given choices to find all values within the interval [750, 1,050].

The valid values are 800, 950, and 1,000.

Only values between 750 and 1,050 inclusive satisfy the cost constraint.

Key Concept

Linear Equations in One Variable

Question 300Question →

In a school club of $40$ students, the ratio of the number of boys to the number of girls is $3 : 5$ . Which of the following statements must be true? Select all such statements.

Select all that apply

The number of boys in the club is

15

.

The ratio of the number of boys to the total number of students is

3 : 8

.

The number of girls in the club exceeds the number of boys by

10

.

The ratio of the number of boys to the total number of students is

3 : 5

.

The number of girls in the club is

20

.

Show answer & explanation

Answer: The number of boys in the club is $15$ .; The ratio of the number of boys to the total number of students is $3 : 8$ .; The number of girls in the club exceeds the number of boys by $10$ .

Answer

The correct statements are those indicating that there are 15 boys in the club, that the ratio of boys to total students is 3:8, and that there are 10 more girls than boys in the club.

The total number of ratio parts is

3 + 5 = 8

, representing the entire group of

40

students. Dividing

40

by

8

gives

5

students per part. Therefore, the number of boys is

3 \times 5 = 15

, the number of girls is

5 \times 5 = 25

, and the ratio of boys to total students is

15:40 = 3:8

. Subtracting the number of boys from girls (

25 - 15 = 10

) shows there are

10

more girls than boys.

Step-by-Step Solution

1

Calculate total ratio parts and determine the value of one part.

Total ratio parts =

3 + 5 = 8

. Each part corresponds to

\frac{40}{8} = 5

students.

Converting ratio terms into parts allows calculation of actual quantities from the total group size.

2

Compute the total number of boys and girls.

Number of boys =

3 \times 5 = 15

. Number of girls =

5 \times 5 = 25

.

Multiply each component's ratio share by the value of one part.

3

Evaluate the given statements against calculated quantities.

There are

15

boys; the ratio of boys to total students is

15:40 = 3:8

; and the difference between girls and boys is

25 - 15 = 10

.

Determines which options represent true statements about the group.

Key Concept

Part-to-Part vs. Part-to-Whole Ratios

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer