All practice questions

Question 1401Question →

In the geometric plane, line $k$ is parallel to line $m$ . A transversal line $t$ intersects line $k$ at point $P$ and line $m$ at point $Q$ . Ray $PR$ extends along line $k$ to the right of $P$ , and ray $QS$ extends along line $m$ to the right of $Q$ . The measure of interior angle $\angle RPQ$ is represented by $(3x + 20)^\circ$ and the measure of interior angle $\angle PQS$ is represented by $(2x + 10)^\circ$ . Which of the following statements must be true? Select all such statements.

Select all that apply

The value of

x

is

30

.

The measure of angle

\angle RPQ

is

110^\circ

.

The acute angle formed by line

t

and line

k

at point

P

measures

70^\circ

.

The value of

x

is

12

.

Angles

\angle RPQ

and

\angle PQS

are complementary angles.

Show answer & explanation

Answer: The value of $x$ is $30$ .; The measure of angle $\angle RPQ$ is $110^\circ$ .; The acute angle formed by line $t$ and line $k$ at point $P$ measures $70^\circ$ .

Answer

The true statements are that

x = 30

, the measure of angle

\angle RPQ

is

110^\circ

, and the acute angle formed by line

t

and line

k

at point

P

measures

70^\circ

.

Lines

k

and

m

are parallel, so interior angles on the same side of transversal

t

(angles

\angle RPQ

and

\angle PQS

) are supplementary. Solving

(3x + 20) + (2x + 10) = 180

gives

x = 30

. Substituting

x = 30

yields

\angle RPQ = 110^\circ

and

\angle PQS = 70^\circ

. The adjacent angle to

\angle RPQ

at point

P

is

180^\circ - 110^\circ = 70^\circ

, which is acute.

Step-by-Step Solution

1

Set up the geometric equation using the parallel line angle relationship.

(3x + 20) + (2x + 10) = 180

When two parallel lines are cut by a transversal, consecutive interior angles on the same side of the transversal are supplementary (their sum is

180^\circ

).

2

Solve the linear equation for

x

.

5x + 30 = 180 \implies 5x = 150 \implies x = 30

Combine like terms and isolate

x

using basic algebra.

3

Calculate the specific angle measures.

m

\angle RPQ = 3(30) + 20 = 110^\circ

and m

\angle PQS = 2(30) + 10 = 70^\circ

Substitute

x = 30

into the given algebraic angle expressions.

4

Determine the supplementary acute angle at point

P

.

180^\circ - 110^\circ = 70^\circ

Angles along a straight line form a linear pair and sum to

180^\circ

.

Key Concept

Parallel Lines and Consecutive Interior Angles

Question 1402Question →

Circle $K$ has a radius of $10$ . Points $P$ and $Q$ lie on circle $K$ such that central angle $\angle POQ$ measures $72^\circ$ , where point $O$ is the center of circle $K$ . Which of the following statements are true? Select all that apply.

Select all that apply

The length of minor arc

PQ

is

4\pi

.

The area of sector

POQ

is

20\pi

.

The perimeter of sector

POQ

is

4\pi

.

The area of sector

POQ

is

40\pi

.

The ratio of the area of sector

POQ

to the total area of circle

K

is

1

to

4

.

Show answer & explanation

Answer: The length of minor arc $PQ$ is $4\pi$ .; The area of sector $POQ$ is $20\pi$ .

Answer

The correct statements are that the length of minor arc

PQ

is

4\pi

, and the area of sector

POQ

is

20\pi

.

The central angle of

72^\circ

corresponds to

\frac{72^\circ}{360^\circ} = \frac{1}{5}

of the entire circle. The circle has a circumference of

2\pi(10) = 20\pi

and an area of

\pi(10)^2 = 100\pi

. Taking

\frac{1}{5}

of the circumference gives an arc length of

4\pi

, and taking

\frac{1}{5}

of the total area gives a sector area of

20\pi

. Both of these statements are mathematically accurate.

Step-by-Step Solution

1

Find the central angle fraction relative to the full circle

\text{Fraction} = \frac{72^\circ}{360^\circ} = \frac{1}{5}

A complete circle subtends

360^\circ

, so the arc and sector comprise one-fifth of the circle.

2

Calculate the circumference and length of minor arc

PQ

\text{Circumference} = 2\pi(10) = 20\pi, \quad \text{Arc Length } PQ = \frac{1}{5} \times 20\pi = 4\pi

Arc length is the central angle fraction multiplied by the total circumference.

3

Calculate the total circle area and area of sector

POQ

\text{Total Area} = \pi(10)^2 = 100\pi, \quad \text{Sector Area } POQ = \frac{1}{5} \times 100\pi = 20\pi

Sector area is the central angle fraction multiplied by the total area of the circle.

4

Evaluate sector perimeter and area ratio for remaining options

\text{Perimeter} = 4\pi + 2(10) = 4\pi + 20; \quad \text{Ratio} = \frac{20\pi}{100\pi} = \frac{1}{5}

Sector perimeter requires adding the two straight radii to the arc length, and the ratio of sector area to total area is

1:5

.

Key Concept

Arc Length and Sector Area of a Circle

Estimated Time:1m 30s

Question 1403Question →

For a constant

a

, consider the following linear equation in one variable

x

:

\frac{a(x - 2)}{3} - \frac{2x + 1}{4} = \frac{(a - 3)x + 5}{12}

Which of the following statements are true? Select all such statements.

Select all that apply

If

a = 1

, the equation has no solution for

x

.

If

a = 5

, then

x = 3

.

If

a = -1

, then

x = 0

.

If

x = 4

, then

a = 5

.

If

a = 0

, the equation has a positive solution for

x

.

Show answer & explanation

Answer: If $a = 1$ , the equation has no solution for $x$ .; If $a = -1$ , then $x = 0$ .; If $x = 4$ , then $a = 5$ .

Answer

The correct statements are the ones asserting that

a = 1

leaves the equation with no solution,

a = -1

leads to

x = 0

, and

x = 4

requires

a = 5

.

Multiplying the equation by 12 and simplifying yields

3(a - 1)x = 8(a + 1)

. When

a = 1

, the left side becomes 0 while the right side becomes 16, creating an inconsistent equation

0 = 16

with no solution. When

a = -1

, the equation becomes

-6x = 0

, which gives

x = 0

. When

x = 4

, substituting into the simplified equation gives

12(a - 1) = 8(a + 1)

, which simplifies to

4a = 20 \implies a = 5

.

Step-by-Step Solution

1

Clear denominators by multiplying every term of the equation by the least common multiple, 12.

4a(x - 2) - 3(2x + 1) = (a - 3)x + 5

Eliminating fractional coefficients simplifies the distribution and collection of variable terms.

2

Expand products on both sides of the equation.

4ax - 8a - 6x - 3 = ax - 3x + 5

Applying the distributive property isolates individual algebraic terms.

3

Rearrange terms to collect all

x

-terms on the left side and constant terms on the right side.

(4a - 6 - a + 3)x = 8a + 8 \implies 3(a - 1)x = 8(a + 1)

Factoring out

x

provides the canonical linear form

A x = B

.

4

Evaluate the given conditions for

a

and

x

against the canonical form

3(a - 1)x = 8(a + 1)

.

For

a = 1

:

0 = 16

(no solution). For

a = 5

:

12x = 48 \implies x = 4

. For

a = -1

:

-6x = 0 \implies x = 0

. For

x = 4

:

12(a - 1) = 8(a + 1) \implies a = 5

. For

a = 0

:

-3x = 8 \implies x = -\frac{8}{3}

(negative).

Direct substitution verifies which algebraic relationships hold true.

Key Concept

Linear equations in one variable containing symbolic parameters can be analyzed for existence of solutions, zero-roots, and explicit values by reducing to the form

A x = B

.

Estimated Time:2m 30s

Question 1404Question →

Which of the following values of $x$ satisfy the inequality $|2x - 1| \le 5 - x$ ? Select all that apply.

Select all that apply

-4

-1

1

3

-5

Show answer & explanation

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

Answer

The values of

x

that satisfy the inequality are

-4

,

-1

, and

1

.

Solving the compound inequality

-(5 - x) \le 2x - 1 \le 5 - x

yields the solution range

-4 \le x \le 2

. The candidate values

-4

,

-1

, and

1

all lie within this range, making them valid solutions.

Step-by-Step Solution

1

Determine the non-negativity constraint for the right-hand side expression.

Since an absolute value

|2x - 1|

must be non-negative, we must have

5 - x \ge 0

, which simplifies to

x \le 5

.

An absolute value cannot be less than a negative number.

2

Rewrite the absolute value inequality

|2x - 1| \le 5 - x

as a compound linear inequality.

-(5 - x) \le 2x - 1 \le 5 - x

, which expands to

-5 + x \le 2x - 1 \le 5 - x

.

For any non-negative expression

B

,

|A| \le B

is logically equivalent to

-B \le A \le B

.

3

Solve the left-hand inequality

-5 + x \le 2x - 1

.

Subtracting

x

from both sides gives

-5 \le x - 1

. Adding

1

to both sides yields

x \ge -4

.

Isolating

x

establishes the lower bound of the solution set.

4

Solve the right-hand inequality

2x - 1 \le 5 - x

.

Adding

x

to both sides gives

3x - 1 \le 5

. Adding

1

yields

3x \le 6

, so

x \le 2

.

Isolating

x

establishes the upper bound of the solution set.

5

Intersect all constraints to find the valid domain for

x

.

Combining

x \ge -4

,

x \le 2

, and

x \le 5

gives the closed interval

[-4, 2]

.

A valid value of

x

must satisfy all component inequalities simultaneously.

6

Evaluate the candidate choices against the solution interval

[-4, 2]

.

The values

-4

,

-1

, and

1

fall within

[-4, 2]

, whereas

3

and

-5

fall outside this interval.

Only numbers inside

[-4, 2]

satisfy the original inequality.

Key Concept

Solving Absolute Value Inequalities with Variable Expressions

Question 1405Question →

If $x$ is a real number that satisfies both of the inequalities $|3x + 4| \ge 11$ and $|x - 1| < 6$ , which of the following could be the value of $x$ ? Select all that apply.

Select all that apply

-5

-3

3

5

7

Show answer & explanation

Answer: $3$ ; $5$

Answer

The values 3 and 5 satisfy both inequalities.

Solving the first inequality

|3x + 4| \ge 11

yields

x \le -5

or

x \ge \frac{7}{3}

. Solving the second inequality

|x - 1| < 6

yields

-5 < x < 7

. Intersecting these two regions, the interval

x \le -5

does not overlap with

-5 < x < 7

because

-5

is excluded from the second inequality. The overlap occurs only for

\frac{7}{3} \le x < 7

. Among the choices, 3 and 5 fall within this valid interval.

Step-by-Step Solution

1

Solve the absolute value inequality

|3x + 4| \ge 11

.

x \le -5

or

x \ge \frac{7}{3}

.

An absolute value inequality of the form

|u| \ge k

(where

k > 0

) splits into two separate inequalities:

u \ge k

or

u \le -k

. Solving

3x + 4 \ge 11

gives

3x \ge 7 \implies x \ge \frac{7}{3}

. Solving

3x + 4 \le -11

gives

3x \le -15 \implies x \le -5

.

2

Solve the absolute value inequality

|x - 1| < 6

.

-5 < x < 7

.

An absolute value inequality of the form

|u| < k

is equivalent to the compound inequality

-k < u < k

. Thus,

-6 < x - 1 < 6

. Adding 1 to all parts yields

-5 < x < 7

.

3

Determine the intersection of the two solution sets.

\frac{7}{3} \le x < 7

.

The portion

x \le -5

has no overlap with

-5 < x < 7

because

-5

is excluded by the strict inequality in the second condition. The portion

x \ge \frac{7}{3}

overlaps with

-5 < x < 7

to yield the interval

[\frac{7}{3}, 7)

.

4

Evaluate which of the given options fall inside

[\frac{7}{3}, 7)

.

The numbers 3 and 5 belong to the interval, while

-5

,

-3

, and

7

do not.

Since

\frac{7}{3} \approx 2.33

, the values 3 and 5 fall strictly between 2.33 and 7.

Key Concept

Solving systems of linear absolute value inequalities by finding the intersection of compound solution intervals

Question 1406Question →

If $x$ is a real number that satisfies both $|3 - 2x| \ge 5$ and $\frac{7 - 3x}{-2} > -1$ , which of the following expresses all possible values of $x$ ?

x \ge 4

x \le -1

\frac{5}{3} < x \le 4

-1 \le x \le 4

x > \frac{5}{3}

Show answer & explanation

Answer: $x \ge 4$

Answer

The condition is satisfied by all values of

x

such that

x \ge 4

.

Solving

|3 - 2x| \ge 5

yields two separate intervals:

x \le -1

or

x \ge 4

. Solving

\frac{7 - 3x}{-2} > -1

involves multiplying by

-2

and dividing by

-3

, both of which flip the inequality sign, leading to

x > \frac{5}{3}

. The values of

x

that satisfy both constraints are those in the overlap of

x \in (-\infty, -1] \cup [4, \infty)

and

x > \frac{5}{3}

, which simplifies directly to

x \ge 4

.

Step-by-Step Solution

1

Solve the absolute value inequality

|3 - 2x| \ge 5

.

Splitting into two cases:

3 - 2x \ge 5 \implies -2x \ge 2 \implies x \le -1

, or

3 - 2x \le -5 \implies -2x \le -8 \implies x \ge 4

. So

x \in (-\infty, -1] \cup [4, \infty)

.

An absolute value inequality

|u| \ge k

(for

k > 0

) decouples into

u \ge k

or

u \le -k

.

2

Solve the linear inequality

\frac{7 - 3x}{-2} > -1

.

Multiply both sides by

-2

(reversing the inequality):

7 - 3x < 2

. Subtract 7:

-3x < -5

. Divide by

-3

(reversing the inequality again):

x > \frac{5}{3}

.

Multiplying or dividing an inequality by a negative quantity reverses the direction of the inequality sign.

3

Find the intersection of the two solution sets.

We require

x \in ((-\infty, -1] \cup [4, \infty)) \cap (\frac{5}{3}, \infty)

. Since

(-\infty, -1]

has no overlap with

(\frac{5}{3}, \infty)

, the intersection is

[4, \infty)

, or

x \ge 4

.

A real number must satisfy both inequalities simultaneously.

Key Concept

Solving systems of absolute value and linear inequalities with negative multipliers

Question 1407Question →

A store manager recorded the number of orders processed online each day during a one-week period: $14$ , $7$ , $25$ , $11$ , $18$ , $2$ , and $16$ . What is the median number of daily online orders processed during this week?

11

14

16

18

93

Show answer & explanation

Answer: $14$

Answer

The median number of daily online orders is

14

.

To find the median, the seven numbers must first be placed in ascending order:

2, 7, 11, 14, 16, 18, 25

. Since there is an odd number of data points (

7

), the median is the single middle value located at the 4th position, which is

14

.

Step-by-Step Solution

1

Arrange the given dataset of seven numbers in ascending order.

The sorted dataset is

2, 7, 11, 14, 16, 18, 25

.

To determine the median of any dataset, the numerical values must first be ordered from smallest to largest.

2

Find the position of the middle element for a dataset with

n = 7

values.

The median position is

\frac{7 + 1}{2} = 4

.

When a dataset contains an odd number of elements

n

, the median is located at the

\frac{n+1}{2}

-th position.

3

Identify the 4th element in the sorted list.

The 4th element is

14

.

In the ordered sequence

2, 7, 11, 14, 16, 18, 25

, the 4th entry is

14

.

Key Concept

The median is the middle value in a set of numerical data that has been arranged in ascending or descending order.

Question 1408Question →

For all non-zero real numbers

x

, the function

f

satisfies the functional equation

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

Which of the following statements must be true for all $x \neq 0$ ? Select all such statements.

Select all that apply

f(-1) = 3

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

for all

x \neq 0

f(x) \leq -2\sqrt{2}

for all

x > 0

f(x) + f(-x) = 6x

for all

x \neq 0

f(2) = 3

Show answer & explanation

Answer: $f(-1) = 3$ ; $f(x) \leq -2\sqrt{2}$ for all $x > 0$

Answer

The statements stating that

f(-1) = 3

and that

f(x) \leq -2\sqrt{2}

for all

x > 0

are the correct choices.

Solving the functional equation yields

f(x) = -x - \frac{2}{x}

. Substituting

x = -1

gives

f(-1) = 3

, making the statement asserting

f(-1) = 3

correct. Furthermore, for any positive real number

x

, the sum

x + \frac{2}{x} \geq 2\sqrt{2}

by the AM-GM inequality, so

f(x) = -\left(x + \frac{2}{x}\right) \leq -2\sqrt{2}

, making the statement asserting

f(x) \leq -2\sqrt{2}

for all

x > 0

correct.

Step-by-Step Solution

1

Set up a system of functional equations by substituting

x \to -\frac{1}{x}

.

Replacing

x

with

-\frac{1}{x}

in the original equation

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

gives

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

.

This creates a system of two linear algebraic equations with two unknowns:

f(x)

and

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

.

2

Solve the system of equations for

f(x)

.

Multiply the second equation by

2

to obtain

4f(x) + 2f\left(-\frac{1}{x}\right) = -\frac{6}{x}

. Subtracting the first equation

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

from this gives

3f(x) = -3x - \frac{6}{x}

, so

f(x) = -x - \frac{2}{x}

.

Eliminating the composite term

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

yields an explicit expression for

f(x)

.

3

Evaluate

f(-1)

.

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

.

This confirms that the statement asserting

f(-1) = 3

is correct.

4

Analyze the maximum value of

f(x)

for

x > 0

using the AM-GM inequality.

For

x > 0

,

x + \frac{2}{x} \geq 2\sqrt{x \cdot \frac{2}{x}} = 2\sqrt{2}

. Therefore,

f(x) = -\left(x + \frac{2}{x}\right) \leq -2\sqrt{2}

.

This confirms that the statement asserting

f(x) \leq -2\sqrt{2}

for all

x > 0

is correct.

Key Concept

Solving functional equations via variable substitution and analyzing function bounds via the AM-GM inequality.

Estimated Time:2m 30s

Question 1409Question →

In a geometric plane, straight lines $L_1$ and $L_2$ intersect at point $P$ . The measure of the obtuse angle formed by the intersection of $L_1$ and $L_2$ is $(5x - 10)^\circ$ , and the measure of an adjacent acute angle is $(2x + 15)^\circ$ . Ray $PQ$ originates from point $P$ , is perpendicular to line $L_1$ , and lies entirely within the interior of the $(5x - 10)^\circ$ obtuse angle. What is the measure, in degrees, of the angle formed between ray $PQ$ and line $L_2$ ?

25^\circ

35^\circ

50^\circ

65^\circ

155^\circ

Show answer & explanation

Answer: $25^\circ$

Answer

25^\circ

Two intersecting lines form adjacent angles that sum to

180^\circ

. Solving

(5x - 10) + (2x + 15) = 180

gives

7x + 5 = 180

, so

x = 25

. The obtuse angle measure is

5(25) - 10 = 115^\circ

. Since ray

PQ

is perpendicular to line

L_1

, it forms a

90^\circ

angle with line

L_1

. The remaining angle between ray

PQ

and line

L_2

within the obtuse angle region is

115^\circ - 90^\circ = 25^\circ

.

Step-by-Step Solution

1

Set up an algebraic equation using the supplementary angle relationship.

(5x - 10) + (2x + 15) = 180

Adjacent angles formed by two intersecting straight lines lie on a straight line and are supplementary, summing to

180^\circ

.

2

Solve for the variable

x

.

7x + 5 = 180 \implies 7x = 175 \implies x = 25

Combining like terms simplifies the linear equation.

3

Calculate the degree measure of the obtuse angle.

Obtuse angle =

5(25) - 10 = 125 - 10 = 115^\circ

Substitute

x = 25

back into the expression

(5x - 10)^\circ

.

4

Calculate the angle between ray

PQ

and line

L_2

.

Angle =

115^\circ - 90^\circ = 25^\circ

Ray

PQ

is perpendicular to line

L_1

(

90^\circ

) and lies inside the

115^\circ

angle, dividing the obtuse angle into a

90^\circ

portion and the remaining angle adjacent to line

L_2

.

Key Concept

Supplementary angles on a straight line and angle subtraction with perpendicular rays

Estimated Time:1m 30s

Question 1410Question →

If $x$ is an integer that satisfies both $|4x - 5| \le 13$ and $\frac{2x - 1}{-3} \le -1$ , what is the sum of all possible values of $x$ ?

-7

0

7

9

14

Show answer & explanation

Answer: 9

Answer

9

Solving the absolute value inequality

|4x - 5| \le 13

yields

-13 \le 4x - 5 \le 13

, which simplifies to

-2 \le x \le 4.5

. Next, solving

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

requires multiplying both sides by

-3

and reversing the inequality sign, giving

2x - 1 \ge 3

, which simplifies to

x \ge 2

. Combining

-2 \le x \le 4.5

and

x \ge 2

for integer

x

gives the set

\{2, 3, 4\}

. The sum of these values is

2 + 3 + 4 = 9

.

Step-by-Step Solution

1

Solve the absolute value inequality

|4x - 5| \le 13

-13 \le 4x - 5 \le 13 \implies -8 \le 4x \le 18 \implies -2 \le x \le 4.5

An absolute value inequality

|u| \le k

unfolds into the compound inequality

-k \le u \le k

.

2

Solve the linear inequality

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

2x - 1 \ge 3 \implies 2x \ge 4 \implies x \ge 2

Multiplying or dividing 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 for integer

x

x \in \{2, 3, 4\}

The integers satisfying both

-2 \le x \le 4.5

and

x \ge 2

are 2, 3, and 4.

4

Calculate the sum of the possible integer values

2 + 3 + 4 = 9

Summing the valid integer solutions yields the final requested answer.

Key Concept

Linear Inequalities and Absolute Value

Estimated Time:2m 0s

Question 1411Question →

A sector of a circle has an area of $15\pi$ square units and a perimeter of $10 + 6\pi$ units. What is the measure of the central angle of the sector, in degrees?

108^\circ

135^\circ

144^\circ

216^\circ

270^\circ

Show answer & explanation

Answer: $216^\circ$

Answer

The correct answer is

216^\circ

, corresponding to the option stating

216^\circ

.

The perimeter of a sector is defined as

2r + s = 10 + 6\pi

, which yields a radius of

r = 5

and an arc length

s = 6\pi

. Checking with the sector area formula

\frac{1}{2}rs = \frac{1}{2}(5)(6\pi) = 15\pi

confirms these measurements. The total area of the circle is

\pi (5)^2 = 25\pi

. The sector thus constitutes

\frac{15\pi}{25\pi} = \frac{3}{5}

of the circle. Multiplying this fraction by

360^\circ

gives a central angle of

216^\circ

.

Step-by-Step Solution

1

Express the sector perimeter formula and solve for radius and arc length.

Radius

r = 5

and arc length

s = 6\pi

.

The perimeter of a sector equals two radii plus its arc length:

\text{Perimeter} = 2r + s = 10 + 6\pi

. Equating standard and

\pi

-termed components gives

2r = 10 \implies r = 5

and arc length

s = 6\pi

.

2

Verify consistency using the sector area formula.

\text{Area} = 15\pi

, matching the given information.

The area of a sector can also be calculated as

\frac{1}{2} r s = \frac{1}{2}(5)(6\pi) = 15\pi

.

3

Find the total area of the circle and the fraction of the circle occupied by the sector.

Total area

= 25\pi

, area fraction

= \frac{3}{5}

.

The full circle area is

\pi r^2 = \pi (5^2) = 25\pi

. The sector represents

\frac{15\pi}{25\pi} = \frac{3}{5}

of the entire circle.

4

Calculate the central angle

\theta

in degrees.

\theta = 216^\circ

.

Multiply the fraction by

360^\circ

:

\theta = \frac{3}{5} \times 360^\circ = 216^\circ

.

Key Concept

Perimeter, Arc Length, and Area of a Circle Sector

Question 1412Question →

If $x$ and $y$ are real numbers greater than $1$ such that $x^{\sqrt{y}} = y^{\sqrt{x}}$ and $x^3 = y^2$ , what is the value of $x$ ?

\frac{81}{16}

\frac{9}{4}

\frac{27}{8}

\frac{729}{64}

\frac{243}{32}

Show answer & explanation

Answer: $\frac{81}{16}$

Answer

The correct answer is

\frac{81}{16}

.

Expressing

y

as

x^{3/2}

and substituting it into

x^{\sqrt{y}} = y^{\sqrt{x}}

transforms the equation into

x^{x^{3/4}} = x^{\frac{3}{2}x^{1/2}}

. Equating exponents yields

x^{3/4} = \frac{3}{2}x^{1/2}

, which simplifies to

x^{1/4} = \frac{3}{2}

. Raising both sides to the fourth power gives

x = \frac{81}{16}

.

Step-by-Step Solution

1

Express

y

in terms of

x

using the second given equation.

Since

x > 1

and

y > 1

, taking the square root of both sides of

y^2 = x^3

gives

y = x^{3/2}

.

Converting

y

to an exponential expression of

x

allows single-variable substitution into the first equation.

2

Substitute

y = x^{3/2}

into the first equation

x^{\sqrt{y}} = y^{\sqrt{x}}

.

Note that

\sqrt{y} = \sqrt{x^{3/2}} = (x^{3/2})^{1/2} = x^{3/4}

. Thus, the left side becomes

x^{x^{3/4}}

, and the right side becomes

(x^{3/2})^{\sqrt{x}} = x^{\frac{3}{2}x^{1/2}}

.

Applying exponent rules

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

simplifies both sides to base

x

expressions.

3

Equate the exponents since the bases are equal and greater than 1.

x^{3/4} = \frac{3}{2}x^{1/2}

.

If

x^a = x^b

for

x > 1

, then

a = b

.

4

Divide both sides by

x^{1/2}

to isolate the power of

x

.

\frac{x^{3/4}}{x^{1/2}} = \frac{3}{2} \implies x^{3/4 - 1/2} = \frac{3}{2} \implies x^{1/4} = \frac{3}{2}

.

Using the quotient rule for exponents,

\frac{x^a}{x^b} = x^{a-b}

where

3/4 - 1/2 = 1/4

.

5

Raise both sides to the 4th power to solve for

x

.

x = \left(\frac{3}{2}\right)^4 = \frac{3^4}{2^4} = \frac{81}{16}

.

Raising

(x^{1/4})^4

eliminates the fractional exponent to give

x

.

Key Concept

Solving systems of exponential equations using fractional exponent rules and base equality properties.

Question 1413Question →

A container originally holds a liquid mixture consisting of substance A and substance B, where substance A constitutes $\frac{2}{5}$ of the total volume. After $15$ liters of substance B are added to the container and $3$ liters of substance A evaporate, the volume of substance A in the container becomes $\frac{1}{4}$ of the new total liquid volume. What was the original total volume, in liters, of the liquid mixture in the container?

32

40

45

50

52

Show answer & explanation

Answer: 40

Answer

40 liters

Letting

V

represent the original volume of the liquid mixture in liters, the initial amount of substance A is

\frac{2}{5}V

. After adding 15 liters of substance B and losing 3 liters of substance A to evaporation, the updated volume of substance A is

\frac{2}{5}V - 3

, and the updated total volume is

V + 15 - 3 = V + 12

. Setting up the relationship

\frac{2}{5}V - 3 = \frac{1}{4}(V + 12)

and expanding the right side gives

\frac{2}{5}V - 3 = \frac{1}{4}V + 3

. Subtracting

\frac{1}{4}V

from both sides yields

\frac{3}{20}V = 6

, which solves to

V = 40

liters.

Step-by-Step Solution

1

Define the unknown variable and express initial quantities algebraically.

Let

V

be the original total volume of the liquid mixture in liters. The original volume of substance A is

\frac{2}{5}V

.

Establishing a variable for the initial total volume allows all changes to be modeled in terms of one variable.

2

Express the modified quantities after additions and evaporation.

New volume of substance A

= \frac{2}{5}V - 3

. New total volume

= V + 15 - 3 = V + 12

.

Adding 15 liters of substance B increases the total volume by 15, and losing 3 liters of substance A decreases both substance A and the total volume by 3.

3

Set up the linear equation based on the given ratio condition.

\frac{2}{5}V - 3 = \frac{1}{4}(V + 12)

Substance A forms one-fourth of the updated total liquid volume.

4

Expand and solve the linear equation for

V

.

\frac{2}{5}V - 3 = \frac{1}{4}V + 3 \implies \frac{2}{5}V - \frac{1}{4}V = 6 \implies \frac{8 - 5}{20}V = 6 \implies \frac{3}{20}V = 6 \implies V = 40.

Clearing terms and subtracting

\frac{1}{4}V

from

\frac{2}{5}V

gives

\frac{3}{20}V = 6

, which yields

V = 40

.

Key Concept

Linear Equations in One Variable

Question 1414Question →

If $x$ and $y$ are positive real numbers such that $\left(\frac{x^{-1}y^{\frac{2}{3}}}{\sqrt{x^3 y^{-2}}}\right)^{-\frac{3}{4}} = x^a y^b$ , what is the value of $a + b$ ?

\frac{5}{8}

-\frac{5}{8}

\frac{17}{8}

-\frac{1}{8}

-\frac{13}{8}

Show answer & explanation

Answer: $\frac{5}{8}$

Answer

The value of

a + b

is

\frac{5}{8}

.

Simplifying the fraction inside the parentheses gives

x^{-\frac{5}{2}} y^{\frac{5}{3}}

. Raising this to the power of

-\frac{3}{4}

produces

x^{\frac{15}{8}} y^{-\frac{5}{4}}

. Equating exponents yields

a = \frac{15}{8}

and

b = -\frac{5}{4} = -\frac{10}{8}

. Adding these together gives

a + b = \frac{5}{8}

.

Step-by-Step Solution

1

Rewrite the radical expression in the denominator using fractional exponents.

\sqrt{x^3 y^{-2}} = (x^3 y^{-2})^{\frac{1}{2}} = x^{\frac{3}{2}} y^{-1}

Applying the power rule

(u^v)^w = u^{v \cdot w}

to radical expressions converts square roots to fractional exponents of

\frac{1}{2}

.

2

Simplify the expression inside the main parentheses by combining like base exponents.

\frac{x^{-1}y^{\frac{2}{3}}}{x^{\frac{3}{2}} y^{-1}} = x^{-1 - \frac{3}{2}} y^{\frac{2}{3} - (-1)} = x^{-\frac{5}{2}} y^{\frac{5}{3}}

When dividing exponential expressions with the same base, subtract the denominator exponent from the numerator exponent.

3

Apply the outer exponent

-\frac{3}{4}

to each factor inside the parentheses.

(x^{-\frac{5}{2}} y^{\frac{5}{3}})^{-\frac{3}{4}} = x^{(-\frac{5}{2})(-\frac{3}{4})} y^{(\frac{5}{3})(-\frac{3}{4})} = x^{\frac{15}{8}} y^{-\frac{5}{4}}

According to exponent rules,

(u \cdot v)^p = u^p \cdot v^p

and

(u^p)^q = u^{p \cdot q}

.

4

Identify

a

and

b

and calculate their sum

a + b

.

a = \frac{15}{8}, \quad b = -\frac{5}{4} = -\frac{10}{8} \implies a + b = \frac{15}{8} - \frac{10}{8} = \frac{5}{8}

Match the simplified expression with

x^a y^b

and add the resulting fractional exponents using a common denominator.

Key Concept

Simplifying nested algebraic expressions with negative and rational exponents using fundamental laws of exponents.

Estimated Time:2m 0s

Question 1415Question →

In a circle centered at point $O$ with radius $R$ , radii $OA$ and $OB$ bound a sector $AOB$ with a central angle of $120^\circ$ . Point $P$ lies on minor arc $AB$ such that the ratio of the length of arc $AP$ to the length of arc $PB$ is $1:3$ . Segment $PQ$ is drawn perpendicular to radius $OA$ , intersecting $OA$ at point $Q$ . What is the ratio of the area of the region bounded by line segment $PQ$ , line segment $AQ$ , and minor arc $AP$ to the area of sector $AOB$ ?

\frac{4\pi - 3\sqrt{3}}{16\pi}

\frac{4\pi - 3\sqrt{3}}{4\pi}

\frac{4\pi - 3\sqrt{3}}{8\pi}

\frac{\pi - 3}{4\pi}

\frac{1}{4}

Show answer & explanation

Answer: $\frac{4\pi - 3\sqrt{3}}{16\pi}$

Answer

\frac{4\pi - 3\sqrt{3}}{16\pi}

The correct answer

\frac{4\pi - 3\sqrt{3}}{16\pi}

is obtained by finding the central angle of sector

AOP

(

30^\circ

), subtracting the area of right triangle

OQP

(

\frac{\sqrt{3}}{16}R^2

) from the area of sector

AOP

(

\frac{\pi}{12}R^2

), and dividing this bounded area by the total sector

AOB

area (

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

).

Step-by-Step Solution

1

Determine the central angle

\angle AOP

of sector

AOP

\angle AOP = \frac{1}{1+3} \times 120^\circ = 30^\circ

Arc length is directly proportional to central angle. A

1:3

ratio means sector

AOP

represents

\frac{1}{4}

of the

120^\circ

central angle.

2

Calculate the area of sector

AOP

and the total area of sector

AOB

\text{Area(Sector } AOP) = \frac{30^\circ}{360^\circ} \pi R^2 = \frac{\pi}{12} R^2

, and

\text{Area(Sector } AOB) = \frac{120^\circ}{360^\circ} \pi R^2 = \frac{\pi}{3} R^2

The area of a sector with central angle

\theta

is

\frac{\theta}{360^\circ} \pi R^2

.

3

Calculate the area of right triangle

OQP

\text{Area}(\triangle OQP) = \frac{\sqrt{3}}{16} R^2

In

\triangle OQP

,

\angle QOP = 30^\circ

and hypotenuse

OP = R

. Thus

OQ = R \cos(30^\circ) = \frac{R\sqrt{3}}{2}

and

PQ = R \sin(30^\circ) = \frac{R}{2}

.

\text{Area} = \frac{1}{2} \times OQ \times PQ = \frac{1}{2} \left(\frac{R\sqrt{3}}{2}\right)\left(\frac{R}{2}\right) = \frac{\sqrt{3}}{16} R^2

.

4

Calculate the area of the bounded region and take the ratio to the area of sector

AOB

\text{Ratio} = \frac{\frac{\pi}{12}R^2 - \frac{\sqrt{3}}{16}R^2}{\frac{\pi}{3}R^2} = \frac{4\pi - 3\sqrt{3}}{16\pi}

Subtract the triangle area from sector

AOP

area to get the bounded region area, then divide by sector

AOB

area.

Key Concept

Sector Area, Arc Length Proportions, and Geometric Region Subdivision

Estimated Time:3m 0s

Question 1416Question →

What is the least integer value of $x$ that satisfies the inequality $|4x - 5| \le 11$ ?

Show answer & explanation

Answer: -1

Answer

The least integer value of

x

that satisfies the inequality is

-1

.

Expanding the absolute value inequality

|4x - 5| \le 11

produces

-11 \le 4x - 5 \le 11

. Adding 5 across all sections yields

-6 \le 4x \le 16

, and dividing by 4 results in

-1.5 \le x \le 4

. The set of integers within this range is

\{-1, 0, 1, 2, 3, 4\}

, making

-1

the least integer value.

Step-by-Step Solution

1

Rewrite the absolute value inequality as a double inequality.

-11 \le 4x - 5 \le 11

An absolute value inequality of the form

|u| \le a

for

a \ge 0

is equivalent to

-a \le u \le a

.

2

Add 5 to all parts of the compound inequality.

-6 \le 4x \le 16

To isolate

x

, first add 5 to eliminate the constant term

-5

.

3

Divide all parts by 4.

-1.5 \le x \le 4

Dividing by a positive constant preserves the direction of the inequality signs.

4

Select the minimum integer contained within the solution interval

[-1.5, 4]

.

-1

The integer values satisfying

-1.5 \le x \le 4

are

-1, 0, 1, 2, 3, 4

. The smallest among these is

-1

.

Key Concept

Linear inequalities involving absolute value

Question 1417Question →

What is the sum of all real solutions to the equation $\sqrt{3x + 10} - x = 2$ ?

2

-1

-3

5

3

Show answer & explanation

Answer: $2$

Answer

The sum of all real solutions is

2

.

To solve

\sqrt{3x + 10} - x = 2

, isolate the radical to get

\sqrt{3x + 10} = x + 2

. Squaring both sides yields

3x + 10 = x^2 + 4x + 4

. Rearranging into standard quadratic form gives

x^2 + x - 6 = 0

, which factors as

(x + 3)(x - 2) = 0

, yielding candidates

x = 2

and

x = -3

. Substituting

x = 2

into the original equation yields

\sqrt{16} - 2 = 2

, which is true. Substituting

x = -3

yields

\sqrt{1} - (-3) = 4 \neq 2

, so

x = -3

is extraneous. The only valid solution is

x = 2

, so the sum of all valid solutions is

2

.

Step-by-Step Solution

1

Isolate the radical expression on one side of the equation

\sqrt{3x + 10} = x + 2

Isolating the radical allows squaring both sides cleanly to eliminate the radical sign.

2

Square both sides of the equation

3x + 10 = (x + 2)^2 = x^2 + 4x + 4

Squaring eliminates the square root, converting the equation into a polynomial form.

3

Rearrange terms into standard quadratic form

ax^2 + bx + c = 0

x^2 + x - 6 = 0

Grouping all terms on one side sets up the equation for factoring.

4

Factor the quadratic equation to find candidate solutions

(x + 3)(x - 2) = 0 \implies x = -3 \text{ or } x = 2

Factoring determines the values of

x

that solve the algebraic polynomial.

5

Test candidate solutions in the original radical equation to filter extraneous roots

For

x = 2

:

\sqrt{3(2) + 10} - 2 = \sqrt{16} - 2 = 4 - 2 = 2

(Valid). For

x = -3

:

\sqrt{3(-3) + 10} - (-3) = \sqrt{1} + 3 = 4 \neq 2

(Extraneous). Thus,

x = 2

is the only valid solution.

Squaring an equation can introduce extraneous roots that do not satisfy the original principal root definition.

Key Concept

Solving Radical Equations and Filtering Extraneous Roots

Estimated Time:1m 30s

Question 1418Question →

A coffee shop owner creates a $30$ -pound blend of coffee by mixing Bean X, which costs $\$8$ per pound, with Bean Y, which costs $\$14$ per pound. If the final mixture costs $\$10$ per pound, how many pounds of Bean X are in the mixture?

Show answer & explanation

Answer: 20

Answer

The mixture contains 20 pounds of Bean X.

The quantity of Bean X required is 20 pounds. Setting up the cost equation

8x + 14(30 - x) = 300

simplifies to

-6x = -120

, giving

x = 20

.

Step-by-Step Solution

1

Define the unknown variables representing the quantities of each component.

Let

x

be the weight of Bean X in pounds. The weight of Bean Y is

(30 - x)

pounds.

Since the total weight of the mixture is 30 pounds, expressing Bean Y in terms of

x

creates a single-variable system.

2

Set up a linear equation based on total financial value.

8x + 14(30 - x) = 10 \times 30

The total cost of Bean X plus the total cost of Bean Y equals the total cost of the combined 30-pound mixture.

3

Solve the algebraic equation for

x

.

8x + 420 - 14x = 300 \implies -6x = -120 \implies x = 20

Combining like terms isolates the variable

x

to yield the required quantity of Bean X.

Key Concept

Linear mixture modeling and single-variable algebraic modeling

Question 1419Question →

If $x$ is a real number that satisfies both $|5 - 2x| \le 11$ and $-3x + 7 < 1$ , what is the least possible integer value of $x$ ?

-3

2

3

-8

8

Show answer & explanation

Answer: 3

Answer

The least possible integer value of

x

is 3.

Solving

|5 - 2x| \le 11

leads to

-11 \le 5 - 2x \le 11

. Subtracting 5 gives

-16 \le -2x \le 6

, and dividing by

-2

(flipping the inequalities) yields

-3 \le x \le 8

. Next, solving

-3x + 7 < 1

gives

-3x < -6

, which upon dividing by

-3

(and flipping the inequality sign) gives

x > 2

. Combining these two requirements yields

2 < x \le 8

. The integer values satisfying this inequality are 3, 4, 5, 6, 7, and 8. The least possible integer value among these is 3.

Step-by-Step Solution

1

Solve the absolute value inequality

|5 - 2x| \le 11

.

-11 \le 5 - 2x \le 11 \implies -16 \le -2x \le 6 \implies -3 \le x \le 8

.

An absolute value inequality

|u| \le k

expands to

-k \le u \le k

. Dividing by

-2

flips the inequality direction.

2

Solve the linear inequality

-3x + 7 < 1

.

-3x < -6 \implies x > 2

.

Subtract 7 from both sides, then divide by

-3

, remembering to reverse the inequality sign.

3

Determine the intersection of both solution sets.

2 < x \le 8

.

x

must be strictly greater than 2 and less than or equal to 8.

4

Identify the smallest integer within the range

2 < x \le 8

.

3

Since

x > 2

is strict, 2 is excluded, making 3 the smallest integer in the range.

Key Concept

Solving systems of linear inequalities involving absolute values and correctly applying sign-flipping rules when multiplying or dividing by negative quantities.

Alternative Method

Test integer candidates directly: for

x = 2

,

-3(2) + 7 = 1

, which is not strictly less than 1. For

x = 3

,

-3(3) + 7 = -2 < 1

(valid) and

|5 - 2(3)| = |-1| = 1 \le 11

(valid), confirming 3 is the smallest integer solution.

Estimated Time:1m 30s

Question 1420Question →

If $-5x + 3 \le 18$ , which of the following inequalities represents all possible real values of $x$ ?

x \ge -3

x \le -3

x \ge 3

x \le 3

x \ge -\frac{21}{5}

Show answer & explanation

Answer: $x \ge -3$

Answer

x \ge -3

Subtracting 3 from both sides gives

-5x \le 15

. Dividing both sides by

-5

requires flipping the inequality sign from

\le

to

\ge

, resulting in

x \ge -3

.

Step-by-Step Solution

1

Subtract 3 from both sides of the inequality

-5x + 3 \le 18

.

-5x \le 15

Isolate the variable term on the left side.

2

Divide both sides by

-5

and reverse the inequality sign from

\le

to

\ge

.

x \ge -3

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

Key Concept

Linear Inequalities and Sign Reversal

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

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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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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

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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

Alternative Method

Answer

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