All practice questions

Question 1501Question →

A water reservoir is filled by Pipe A and Pipe B operating simultaneously at their respective constant rates. Operating together at their original rates, the two pipes can fill the empty reservoir completely in $12$ hours. On a certain day, both pipes begin filling the empty reservoir together at their original rates. After $4$ hours, Pipe A's rate decreases by $25\%$ , while Pipe B's rate increases by $50\%$ . Operating at these new constant rates, the two pipes require an additional $7$ hours to fill the remainder of the reservoir. How many hours would it take Pipe A, operating alone at its original rate, to fill the entire reservoir?

Show answer & explanation

Answer: 25.2

Answer

It would take Pipe A 25.2 hours operating alone at its original rate to fill the entire reservoir.

By defining the original work rates

a

and

b

in reservoirs per hour, the initial condition yields

a + b = \frac{1}{12}

. In the first 4 hours,

\frac{1}{3}

of the job is completed, leaving

\frac{2}{3}

. Setting up the equation for the remaining job with modified rates

0.75a

and

1.5b

over 7 hours produces

7(0.75a + 1.5b) = \frac{2}{3}

. Solving this system of two linear equations yields

a = \frac{5}{126}

reservoirs per hour. Taking the reciprocal gives the time required for Pipe A alone to fill the reservoir, which is

25.2

hours.

Step-by-Step Solution

1

Set up equations for the original rates of Pipe A (

a

) and Pipe B (

b

).

The combined original rate is

a + b = \frac{1}{12}

reservoir per hour.

Together they complete

1

reservoir in

12

hours.

2

Determine the fraction of the reservoir filled in the first 4 hours and the remaining fraction.

Work completed =

4 \times \frac{1}{12} = \frac{1}{3}

; Remaining work =

\frac{2}{3}

.

The pipes worked at their original combined rate for 4 hours.

3

Set up an equation for the work done during the remaining 7 hours at the adjusted rates.

7 \left(0.75a + 1.5b\right) = \frac{2}{3} \implies 5.25a + 10.5b = \frac{2}{3} \implies 63a + 126b = 8

.

Pipe A's rate decreases by

25\%

to

0.75a

, and Pipe B's rate increases by

50\%

to

1.5b

.

4

Solve the system of linear equations for

a

.

a = \frac{5}{126}

reservoir per hour.

Multiplying

a + b = \frac{1}{12}

by

126

yields

126a + 126b = 10.5

. Subtracting

63a + 126b = 8

gives

63a = 2.5

, so

a = \frac{2.5}{63} = \frac{5}{126}

.

5

Calculate the time for Pipe A alone to fill the entire reservoir.

Time

= \frac{1}{a} = \frac{126}{5} = 25.2

hours.

Time equals total work divided by individual rate.

Key Concept

Algebraic modeling of combined work and rates with mid-process rate modifications

Question 1502Question →

If $x$ is a real number satisfying the radical equation $\sqrt{x + 7} - x = 1$ , what is the value of $x$ ?

-3

-2

2

3

6

Show answer & explanation

Answer: $2$

Answer

The value of

x

is

2

.

Isolating the square root yields

\sqrt{x + 7} = x + 1

. Squaring both sides produces

x + 7 = x^2 + 2x + 1

, which reduces to the quadratic equation

x^2 + x - 6 = 0

. Factoring yields

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

, giving potential roots of

x = -3

and

x = 2

. Testing

x = 2

in the original expression yields

\sqrt{2+7} - 2 = 3 - 2 = 1

, confirming it is correct.

Step-by-Step Solution

1

Isolate the radical expression on one side of the equation.

\sqrt{x + 7} = x + 1

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

2

Square both sides of the equation.

x + 7 = (x + 1)^2 = x^2 + 2x + 1

Squaring removes the square root on the left side.

3

Rearrange into standard quadratic form

ax^2 + bx + c = 0

.

x^2 + x - 6 = 0

Subtracting

x + 7

from both sides collects all terms on one side.

4

Factor the quadratic equation and solve for potential values of

x

.

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

Setting each factor to zero identifies potential solutions.

5

Substitute each potential solution back into the original equation

\sqrt{x + 7} - x = 1

to check for extraneous roots.

For

x = 2

:

\sqrt{2 + 7} - 2 = 3 - 2 = 1

(Valid). For

x = -3

:

\sqrt{-3 + 7} - (-3) = 2 + 3 = 5 \neq 1

(Extraneous).

Squaring an equation can introduce false solutions that must be eliminated.

Key Concept

Solving radical equations and checking for extraneous solutions

Estimated Time:1m 30s

Question 1503Question →

For all real numbers $x$ , the function $g$ is defined by $g(x) = 5 - 2x^2$ . What is the value of $g(g(2))$ ?

23

-13

11

3

9

Show answer & explanation

Answer: -13

Answer

-13

Evaluating the inner expression

g(2)

gives

5 - 2(2)^2 = 5 - 8 = -3

. Substituting

-3

back into the function

g(x)

gives

g(-3) = 5 - 2(-3)^2 = 5 - 2(9) = -13

. Thus, the correct value is

-13

.

Step-by-Step Solution

1

Evaluate the inner function

g(2)

g(2) = 5 - 2(2)^2 = 5 - 2(4) = 5 - 8 = -3

Substitute

x = 2

into the definition of

g(x)

and follow standard order of operations (exponents before multiplication).

2

Substitute the result into the outer function to evaluate

g(-3)

g(-3) = 5 - 2(-3)^2 = 5 - 2(9) = 5 - 18 = -13

Squaring a negative number yields a positive result, so

(-3)^2 = 9

, which is then multiplied by

2

.

Key Concept

Nested Function Evaluation and Order of Operations

Estimated Time:45s

Question 1504Question →

If

x

satisfies the linear equation

\frac{2(3x - 1)}{5} - \frac{4x - 3}{3} = \frac{x + 18}{15} - 2

what is the value of

3x - 5

?

7

16

-14

-12

40

Show answer & explanation

Answer: $16$

Answer

16

Multiplying the entire equation by the common denominator

15

clears all fractions, yielding

6(3x - 1) - 5(4x - 3) = (x + 18) - 30

. Expanding both sides produces

18x - 6 - 20x + 15 = x - 12

, which simplifies to

-2x + 9 = x - 12

. Rearranging terms yields

-3x = -21

, so

x = 7

. Substituting

x = 7

into

3x - 5

gives

3(7) - 5 = 16

.

Step-by-Step Solution

1

Clear the denominators by multiplying both sides of the equation by the least common multiple (LCM) of 5, 3, and 15, which is 15.

15 \cdot \left(\frac{2(3x - 1)}{5}\right) - 15 \cdot \left(\frac{4x - 3}{3}\right) = 15 \cdot \left(\frac{x + 18}{15}\right) - 15 \cdot 2

Eliminating fractions simplifies the linear equation into standard integer polynomial terms.

2

Simplify the products and distribute coefficients across parentheses.

3 \cdot 2(3x - 1) - 5(4x - 3) = (x + 18) - 30 \implies 6(3x - 1) - 5(4x - 3) = x - 12

Perform fractional reduction and simplify constants on the right side.

3

Expand both groupings and combine like terms on the left-hand side.

18x - 6 - 20x + 15 = x - 12 \implies -2x + 9 = x - 12

Ensure the negative sign is properly distributed to both terms inside

-5(4x - 3)

.

4

Isolate the variable

x

by subtracting

x

and

9

from both sides.

-3x = -21 \implies x = 7

Solve for the single variable

x

.

5

Substitute

x = 7

into the target expression

3x - 5

.

3(7) - 5 = 21 - 5 = 16

The question asks for the value of the algebraic expression

3x - 5

, not

x

alone.

Key Concept

Linear Equations in One Variable with Fractional Coefficients

Estimated Time:2m 0s

Question 1505Question →

If $x$ and $y$ are real numbers that satisfy the absolute value inequalities $|2x - 6| \le 4$ and $|y + 3| \le 5$ , which of the following values could be equal to the product $xy$ ? Indicate all such values.

Select all that apply

-45

-35

0

8

15

Show answer & explanation

Answer: $-35$ ; $0$ ; $8$

Answer

The possible values for the product

xy

are

-35

,

0

, and

8

.

Solving

|2x - 6| \le 4

yields

1 \le x \le 5

, and solving

|y + 3| \le 5

yields

-8 \le y \le 2

. The product

xy

attains its minimum at

5 \times (-8) = -40

and its maximum at

5 \times 2 = 10

. Since

xy

can take any value in the continuous interval

[-40, 10]

, the values

-35

,

0

, and

8

are all valid choices.

Step-by-Step Solution

1

Solve the inequality

|2x - 6| \le 4

for

x

.

-4 \le 2x - 6 \le 4 \implies 2 \le 2x \le 10 \implies 1 \le x \le 5

.

Unfold the absolute value into a compound inequality to determine the valid range for

x

.

2

Solve the inequality

|y + 3| \le 5

for

y

.

-5 \le y + 3 \le 5 \implies -8 \le y \le 2

.

Unfold the absolute value into a compound inequality to determine the valid range for

y

.

3

Determine the minimum and maximum possible values of the product

xy

.

Evaluating the extreme product combinations of endpoints:

1 \times (-8) = -8

,

1 \times 2 = 2

,

5 \times (-8) = -40

, and

5 \times 2 = 10

. Thus,

-40 \le xy \le 10

.

The continuous product of two real intervals

[a, b]

and

[c, d]

spans from the minimum endpoint product to the maximum endpoint product.

4

Select all options that fall within the interval

[-40, 10]

.

The values

-35

,

0

, and

8

lie within

[-40, 10]

, while

-45

and

15

fall outside.

Any real number within the closed interval

[-40, 10]

can be formed by valid choices of

x

and

y

.

Key Concept

Determining the range of a product from two independent absolute value inequalities.

Question 1506Question →

A logistics company recorded the daily delivery processing times (in minutes) for a warehouse over a given period, forming Dataset $X$ . Dataset $X$ has a range of $40$ minutes, an interquartile range ( $\text{IQR}$ ) of $15$ minutes, and a standard deviation of $8.5$ minutes. A new dataset, Dataset $Y$ , is created by transforming each processing time $x$ in Dataset $X$ according to the formula $y = 1.5x - 10$ . Which of the following statements regarding the measures of dispersion for Dataset $Y$ must be true? Select all such statements.

Select all that apply

The range of Dataset

Y

is

60

minutes.

The standard deviation of Dataset

Y

is

12.75

minutes.

The interquartile range of Dataset

Y

is

12.5

minutes.

The standard deviation of Dataset

Y

is

2.75

minutes.

The interquartile range of Dataset

Y

is equal to the interquartile range of Dataset

X

.

Show answer & explanation

Answer: The range of Dataset $Y$ is $60$ minutes.; The standard deviation of Dataset $Y$ is $12.75$ minutes.

Answer

The statements asserting that the range of Dataset Y is 60 minutes and that the standard deviation of Dataset Y is 12.75 minutes are both correct.

Under a transformation of the form

y = ax + b

(where

a > 0

), any measure of dispersion

D

transforms according to

D_Y = a \cdot D_X

. The constant

b

does not affect spread. Therefore, the range becomes

1.5 \times 40 = 60

minutes and the standard deviation becomes

1.5 \times 8.5 = 12.75

minutes.

Step-by-Step Solution

1

Recall the effect of a linear transformation

y = ax + b

on measures of dispersion.

Measures of dispersion (range, IQR, standard deviation) are scaled by

|a|

and are completely unaffected by the constant addition/subtraction

b

.

Adding or subtracting a constant shifts all data points by the exact same amount without altering the relative distances between data points.

2

Calculate the range for Dataset Y.

\text{Range}_Y = 1.5 \times \text{Range}_X = 1.5 \times 40 = 60

minutes.

The multiplicative factor is

a = 1.5

.

3

Calculate the standard deviation for Dataset Y.

S_Y = 1.5 \times S_X = 1.5 \times 8.5 = 12.75

minutes.

The standard deviation scales proportionally by

1.5

.

4

Calculate the interquartile range (IQR) for Dataset Y.

\text{IQR}_Y = 1.5 \times \text{IQR}_X = 1.5 \times 15 = 22.5

minutes.

The IQR also scales proportionally by

1.5

.

Key Concept

Linear Transformations of Dispersion Measures

Question 1507Question →

In $\triangle ABC$ , the length of side $AB$ is $10$ and the length of side $BC$ is $17$ . Point $D$ lies on the line containing segment $AC$ such that segment $BD$ is perpendicular to line $AC$ . If the length of altitude $BD$ and the length of side $AC$ are both integers, and the area of $\triangle ABC$ is strictly greater than $36$ , what is the perimeter of $\triangle ABC$ ?

36

42

48

52

56

Show answer & explanation

Answer: $48$

Answer

The perimeter of

\triangle ABC

is

48

.

Applying the Pythagorean theorem to both right triangles formed by altitude

BD = h

gives

100 - m^2 = h^2

and

289 - n^2 = h^2

, where

m = AD

and

n = CD

. Subtracting these equations yields

n^2 - m^2 = 189

, which factors as

(n - m)(n + m) = 189

. Testing integer factor pairs of

189

while enforcing

m \le 10

isolates a single non-degenerate solution:

m = 6

,

n = 15

, and

h = 8

. When point

D

lies between

A

and

C

,

AC = 6 + 15 = 21

. This gives an area of

\frac{1}{2} \times 21 \times 8 = 84

(which is strictly greater than

36

) and a total perimeter of

10 + 17 + 21 = 48

.

Step-by-Step Solution

1

Set up Pythagorean relationships for the right triangles formed by altitude

BD

.

Let

BD = h

,

AD = m

, and

CD = n

. In right

\triangle ABD

,

m^2 + h^2 = 10^2 = 100

. In right

\triangle CBD

,

n^2 + h^2 = 17^2 = 289

.

Altitude

BD

divides the figure into two right triangles sharing leg

h

.

2

Subtract the two equations to eliminate

h^2

and factor the difference of squares.

n^2 - m^2 = 289 - 100 = 189 \implies (n - m)(n + m) = 189.

Since

h

and

AC

are integers,

m

and

n

must also be integers for

AC = n \pm m

to be an integer.

3

Analyze integer factor pairs

(u, v)

of

189

where

u = n - m

and

v = n + m

.

Factor pairs

(u, v)

with

u \cdot v = 189

:
-

(1, 189) \implies m = 94

(invalid,

m \le 10

)
-

(3, 63) \implies m = 30

(invalid,

m \le 10

)
-

(7, 27) \implies m = 10, h = 0

(degenerate triangle)
-

(9, 21) \implies n = 15, m = 6, h = \sqrt{100 - 36} = 8

.

The leg

m = AD

cannot exceed the hypotenuse

AB = 10

.

4

Evaluate side

AC

, area, and perimeter for valid geometric configurations.

Case 1:

D

lies on segment

AC \implies AC = n + m = 15 + 6 = 21

.
Area =

\frac{1}{2} \times 21 \times 8 = 84 > 36

.
Perimeter =

10 + 17 + 21 = 48

.
Case 2:

D

lies outside segment

AC \implies AC = n - m = 15 - 6 = 9

.
Area =

\frac{1}{2} \times 9 \times 8 = 36

, which does not satisfy area

> 36

.

The problem specifies that the area must be strictly greater than

36

.

Key Concept

Properties of triangles, Pythagorean theorem system solver, and geometric area constraints

Question 1508Question →

For all non-zero real numbers $x$ , the function $f$ satisfies the relation $2f(x) + f\left(\frac{1}{x}\right) = 3x$ . What is the value of $f(2)$ ?

\frac{7}{2}

\frac{8}{3}

\frac{12}{5}

\frac{11}{4}

\frac{5}{2}

Show answer & explanation

Answer: $\frac{7}{2}$

Answer

The value of

f(2)

is

\frac{7}{2}

.

To solve for

f(2)

, set up a system of equations by evaluating the given relation

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

at

x = 2

and at

x = 1/2

. Substituting

x = 2

gives

2f(2) + f(1/2) = 6

, and substituting

x = 1/2

gives

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

. Expressing

f(1/2)

from the first equation as

6 - 2f(2)

and substituting it into the second yields

2(6 - 2f(2)) + f(2) = 3/2

, which simplifies to

-3f(2) = -21/2

, so

f(2) = 7/2

.

Step-by-Step Solution

1

Substitute

x = 2

into the functional relation.

2f(2) + f\left(\frac{1}{2}\right) = 3(2) = 6

This yields a linear relationship involving

f(2)

and

f\left(\frac{1}{2}\right)

.

2

Substitute

x = \frac{1}{2}

into the original functional relation.

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

This provides a second independent linear equation with the same two unknown quantities,

f(2)

and

f\left(\frac{1}{2}\right)

.

3

Multiply the second equation by

2

to eliminate

f\left(\frac{1}{2}\right)

.

4f\left(\frac{1}{2}\right) + 2f(2) = 3

Preparing to eliminate

f(2)

or

f\left(\frac{1}{2}\right)

using elimination.

4

Solve the system of equations for

f(2)

.

From

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

, we get

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

. Substituting this into the second equation

2(6 - 2f(2)) + f(2) = \frac{3}{2} \implies 12 - 4f(2) + f(2) = \frac{3}{2} \implies -3f(2) = \frac{3}{2} - 12 = -\frac{21}{2} \implies f(2) = \frac{7}{2}

.

Algebraic reduction isolates

f(2)

.

Key Concept

Functional Equations and System of Equations via Variable Substitution

Question 1509Question →

In triangle $ABC$ , point $D$ lies on side $AB$ such that $AD : DB = 3 : 1$ , and point $E$ lies on side $AC$ such that segment $DE$ is parallel to side $BC$ . If the area of triangle $ABC$ is $64$ , what is the area of triangle $ADE$ ?

16

27

36

48

52

Show answer & explanation

Answer: 36

Answer

36

Because segment

DE

is parallel to

BC

, triangle

ADE

is similar to triangle

ABC

. The ratio of side

AD

to side

AB

is

3 : (3 + 1) = 3 : 4

. For similar triangles, the ratio of their areas equals the square of the ratio of their corresponding side lengths, which is

(3/4)^2 = 9/16

. Multiplying the area of triangle

ABC

(

64

) by

9/16

yields an area of

36

for triangle

ADE

.

Step-by-Step Solution

1

Determine the scale factor between similar triangles

ADE

and

ABC

.

Since

DE \parallel BC

, triangle

ADE

is similar to triangle

ABC

. The side length ratio is

AD / AB = AD / (AD + DB) = 3 / (3 + 1) = 3/4

.

Parallel lines create corresponding equal angles, making triangles

ADE

and

ABC

similar.

2

Calculate the ratio of the areas of the similar triangles.

The area ratio is the square of the side scale factor:

(3/4)^2 = 9/16

.

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

3

Compute the area of triangle

ADE

.

\text{Area}(\triangle ADE) = \frac{9}{16} \times 64 = 36.

Multiply the total area of triangle

ABC

by the area ratio

9/16

.

Key Concept

Area Ratio of Similar Triangles

Question 1510Question →

In the $xy$ -plane, line $l_1$ intersects the positive $x$ -axis at $(a, 0)$ and the positive $y$ -axis at $(0, b)$ , where $a > b > 0$ . Line $l_2$ is perpendicular to line $l_1$ and passes through the point $(a, b)$ . If the perpendicular distance from the origin $(0, 0)$ to line $l_1$ is equal to the perpendicular distance from the origin to line $l_2$ , what is the value of $\frac{a}{b}$ ?

\frac{\sqrt{5} - 1}{2}

1

\frac{\sqrt{5} + 1}{2}

\sqrt{2}

2

Show answer & explanation

Answer: $\frac{\sqrt{5} + 1}{2}$

Answer

The ratio

\frac{a}{b}

is

\frac{\sqrt{5} + 1}{2}

.

The option stating

\frac{\sqrt{5} + 1}{2}

is correct because equating the distances of both lines from the origin leads to the equation

a^2 - ab - b^2 = 0

. Expressed in terms of the ratio

r = \frac{a}{b}

, this becomes

r^2 - r - 1 = 0

. Because

a > b > 0

,

r

must be greater than

1

, yielding

r = \frac{\sqrt{5} + 1}{2}

.

Step-by-Step Solution

1

Find the equation and distance from origin for line

l_1

.

The equation of

l_1

is

\frac{x}{a} + \frac{y}{b} = 1

, or

bx + ay - ab = 0

. The perpendicular distance from

(0,0)

to

l_1

is

d_1 = \frac{|-ab|}{\sqrt{a^2 + b^2}} = \frac{ab}{\sqrt{a^2 + b^2}}

.

Standard intercept form of a line and point-to-line distance formula.

2

Find the slope and equation of line

l_2

.

The slope of

l_1

is

-\frac{b}{a}

, so the slope of perpendicular line

l_2

is

\frac{a}{b}

. Since

l_2

passes through

(a, b)

, its equation is

y - b = \frac{a}{b}(x - a)

, which simplifies to

ax - by - (a^2 - b^2) = 0

.

Perpendicular lines have slopes that are negative reciprocals.

3

Calculate the distance from the origin to line

l_2

and set

d_1 = d_2

.

The distance

d_2 = \frac{|-(a^2 - b^2)|}{\sqrt{a^2 + b^2}} = \frac{a^2 - b^2}{\sqrt{a^2 + b^2}}

(since

a > b > 0

). Equating

d_1 = d_2

gives

\frac{ab}{\sqrt{a^2 + b^2}} = \frac{a^2 - b^2}{\sqrt{a^2 + b^2}}

, so

ab = a^2 - b^2

.

The distances from the origin to both lines are specified to be equal.

4

Solve for the ratio

r = \frac{a}{b}

.

Divide

a^2 - ab - b^2 = 0

by

b^2

to obtain

r^2 - r - 1 = 0

. Applying the quadratic formula gives

r = \frac{1 \pm \sqrt{5}}{2}

. Since

a > b > 0

,

r > 1

, so

r = \frac{\sqrt{5} + 1}{2}

.

Determines the unique positive value greater than 1 that satisfies the algebraic relationship.

Key Concept

Perpendicular line slopes, point-to-line distance formula, and algebraic substitution for ratio determination.

Question 1511Question →

For all positive real numbers $a$ and $b$ , which of the following expressions are equivalent to $\left( \frac{a^{-2} b^3}{\sqrt{a^4 b^{-2}}} \right)^{-1/2}$ ? Select all that apply.

Select all that apply

\frac{a^2}{b^2}

\left( \frac{b}{a} \right)^{-2}

\sqrt{\frac{a^4}{b^4}}

\frac{a^{-2}}{b^{-2}}

\frac{b^2}{a^2}

Show answer & explanation

Answer: $\frac{a^2}{b^2}$ ; $\left( \frac{b}{a} \right)^{-2}$ ; $\sqrt{\frac{a^4}{b^4}}$

Answer

The expressions equivalent to the given quantity are

\frac{a^2}{b^2}

,

\left( \frac{b}{a} \right)^{-2}

, and

\sqrt{\frac{a^4}{b^4}}

.

Simplifying the original expression step-by-step yields

\frac{a^2}{b^2}

. The expressions

\frac{a^2}{b^2}

,

\left( \frac{b}{a} \right)^{-2} = \left( \frac{a}{b} \right)^2 = \frac{a^2}{b^2}

, and

\sqrt{\frac{a^4}{b^4}} = \frac{a^2}{b^2}

are all identical to the simplified form.

Step-by-Step Solution

1

Simplify the radical in the denominator of the inner expression.

Since

a > 0

and

b > 0

,

\sqrt{a^4 b^{-2}} = \sqrt{a^4} \cdot \sqrt{b^{-2}} = a^2 b^{-1}

.

Applying the square root to each variable power individually.

2

Simplify the expression inside the outer parenthesis.

\frac{a^{-2} b^3}{a^2 b^{-1}} = a^{-2 - 2} b^{3 - (-1)} = a^{-4} b^4.

Subtracting exponents of like bases according to the quotient rule of exponents.

3

Apply the outer exponent of

-\frac{1}{2}

.

(a^{-4} b^4)^{-1/2} = (a^{-4})^{-1/2} (b^4)^{-1/2} = a^2 b^{-2} = \frac{a^2}{b^2}.

Multiplying internal exponents by

-\frac{1}{2}

according to the power rule of exponents.

4

Evaluate each choice against the simplified form

\frac{a^2}{b^2}

.

The expressions

\frac{a^2}{b^2}

,

\left(\frac{b}{a}\right)^{-2}

, and

\sqrt{\frac{a^4}{b^4}}

are all algebraically equivalent to

\frac{a^2}{b^2}

.

Testing algebraic equivalence using standard exponent and radical laws.

Key Concept

Properties of exponents and radicals, including power of a quotient, quotient rule, negative exponents, and square roots of powers.

Estimated Time:1m 30s

Question 1512Question →

A total initial capital of $\$100,000$ is split between Fund X and Fund Y. Fund X yields an annual simple interest rate of $r\%$ , while Fund Y yields an annual simple interest rate of $(r + 2)\%$ . Under the original capital allocation, the ratio of the annual interest earned from Fund X to the annual interest earned from Fund Y is $15 : 14$ . If the initial allocation to Fund X had been increased by $25\%$ by transferring funds directly from Fund Y, the total annual interest earned from both funds combined would have been $\$5,500$ . What was the original amount allocated to Fund X?

\$40,000

\$50,000

\$60,000

\$64,000

\$75,000

Show answer & explanation

Answer: $\$60,000$

Answer

The original amount allocated to Fund X was

\$60,000

.

The correct option is

\$60,000

. Substituting

X = 60,000

gives Fund Y allocation

Y = 40,000

. From the modified interest condition,

100,000r - 2.5(60,000) = 350,000

, giving

r = 5\%

. Fund X yields

60,000 \times 0.05 = \$3,000

and Fund Y yields

40,000 \times 0.07 = \$2,800

, which satisfies the ratio

3,000 : 2,800 = 15 : 14

. Under the modified allocation (

75,000

in X and

25,000

in Y), interest is

75,000(0.05) + 25,000(0.07) = 3,750 + 1,750 = \$5,500

.

Step-by-Step Solution

1

Formulate variables and express initial conditions.

Let

X

be the initial capital in Fund X and

Y = 100,000 - X

be the initial capital in Fund Y. The interest rates are

\frac{r}{100}

and

\frac{r+2}{100}

respectively.

Establish a single-variable representation for the fund allocations.

2

Set up the interest ratio equation.

\frac{X \cdot r}{(100,000 - X)(r+2)} = \frac{15}{14} \implies 14 X r = 15(100,000 - X)(r+2).$

Relate the original interest outputs according to the

15:14

ratio.

3

Model the modified allocation scenario.

Fund X becomes

1.25X

and Fund Y becomes

100,000 - 1.25X

. Total interest equation:

(1.25X)\left(\frac{r}{100}\right) + (100,000 - 1.25X)\left(\frac{r+2}{100}\right) = 5,500.

Express total combined interest under the hypothetical

25\%

transfer.

4

Simplify the total interest equation to express

r

in terms of

X

.

1.25Xr + 100,000r + 200,000 - 1.25Xr - 2.5X = 550,000 \implies 100,000r - 2.5X = 350,000 \implies r = 3.5 + 0.000025X.$

Eliminate the

Xr

product term to solve for

r

linearly.

5

Substitute

r

into the ratio equation and solve for

X

.

Expanding

29Xr + 30X = 1,500,000r + 3,000,000

with

r = 3.5 + 0.000025X

yields

0.000725X^2 + 94X - 8,250,000 = 0

, which factors to give

X = 60,000

.

Determine the exact value for the initial allocation to Fund X.

Key Concept

Algebraic modeling of multi-variable financial rate and allocation systems.

Estimated Time:3m 0s

Question 1513Question →

Two automated data processing algorithms, Algorithm X and Algorithm Y, operate at their respective constant rates. Algorithm X takes $x$ hours to process a full dataset when working alone. Algorithm Y operates at a constant rate that is $50\%$ faster than Algorithm X.

In Scenario 1, Algorithm X processes the dataset alone for $2$ hours, after which Algorithm Y joins, and both algorithms work together for an additional $t$ hours to complete the dataset.

In Scenario 2, Algorithm Y processes the dataset alone for $3$ hours, after which Algorithm X joins, and both algorithms complete the remaining work together in $\frac{4}{5}t$ hours.

Which of the following statements must be true? Indicate all such statements.

Select all that apply

Algorithm X operating alone requires 14 hours and 30 minutes to process one full dataset.

Algorithm Y operating alone requires 10 hours and 15 minutes to process one full dataset.

If both algorithms operate together from the beginning, they will complete one full dataset in 5 hours and 48 minutes.

The total time spent by Algorithm X working across Scenario 1 is 5 hours.

Algorithm Y processes exactly 60 percent of the dataset in Scenario 2.

Show answer & explanation

Answer: Algorithm X operating alone requires 14 hours and 30 minutes to process one full dataset.; If both algorithms operate together from the beginning, they will complete one full dataset in 5 hours and 48 minutes.

Answer

The true statements are that Algorithm X operating alone requires 14 hours and 30 minutes to process one full dataset, and that operating together from the start, both algorithms complete one full dataset in 5 hours and 48 minutes.

The statement specifying that Algorithm X takes 14 hours and 30 minutes alone is correct because solving the system yields

x = 14.5

hours. The statement regarding the combined time of 5 hours and 48 minutes is correct because

\frac{1}{R_X + R_Y} = 5.8

hours, which equals 5 hours and 48 minutes.

Step-by-Step Solution

1

Define individual and combined work rates in terms of

x

.

Rate of Algorithm X is

R_X = \frac{1}{x}

datasets/hr. Since Algorithm Y is

50\%

faster, its rate is

R_Y = 1.5 R_X = \frac{3}{2x}

datasets/hr. Their combined rate is

R_{X+Y} = \frac{1}{x} + \frac{3}{2x} = \frac{5}{2x}

datasets/hr.

Establishing accurate rate expressions is essential for modeling combined work scenarios.

2

Formulate equations for both scenarios and solve for

t

and

x

.

From Scenario 1:

2 \cdot \frac{1}{x} + t \cdot \frac{5}{2x} = 1 \implies 2 + 2.5t = x

.
From Scenario 2:

3 \cdot \frac{3}{2x} + \frac{4}{5}t \cdot \frac{5}{2x} = 1 \implies 4.5 + 2t = x

.
Equating the two expressions:

2 + 2.5t = 4.5 + 2t \implies 0.5t = 2.5 \implies t = 5

hours.
Substituting

t = 5

yields

x = 2 + 2.5(5) = 14.5

hours (14 hours 30 minutes).

Setting work completed equal to 1 dataset in both scenarios yields a solvable system of equations.

3

Evaluate the given statement choices using the derived parameters.

1. Algorithm X alone time:

x = 14.5

hours = 14 hours 30 minutes (True).
2. Algorithm Y alone time:

\frac{1}{R_Y} = \frac{2x}{3} = \frac{29}{3} = 9

hours 40 minutes (False).
3. Combined time from start:

\frac{1}{R_{X+Y}} = \frac{2x}{5} = \frac{29}{5} = 5.8

hours = 5 hours 48 minutes (True).
4. Total time Algorithm X works in Scenario 1:

2 + t = 2 + 5 = 7

hours (False).
5. Work fraction by Algorithm Y in Scenario 2:

(3 + \frac{4}{5}(5)) \cdot \frac{3}{2x} = 7 \cdot \frac{3}{29} = \frac{21}{29} \approx 72.4\%

(False).

Direct numerical verification reveals which statements hold true.

Key Concept

Combined work rate equations with variable initial delays and relative rate multipliers.

Question 1514Question →

In circle $O$ , line segments $AB$ and $CD$ are perpendicular diameters, each of length $12$ . An arc of a second circle, centered at point $A$ with radius $AC$ , is drawn from point $C$ to point $D$ through the interior of circle $O$ . What is the area of the crescent-shaped region bounded by the semicircle $CBD$ of circle $O$ and arc $CD$ of the second circle?

Show answer & explanation

Answer: 36

Answer

36

Circle

O

has radius

r = 6

, giving semicircle

CBD

an area of

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

. The distance

AC = \sqrt{6^2 + 6^2} = 6\sqrt{2}

is the radius of circle

A

. Because

\angle CAD = 90^\circ

, sector

ACD

has area

\frac{90^\circ}{360^\circ}\pi(6\sqrt{2})^2 = 18\pi

. Subtracting the area of triangle

ACD

(

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

) yields a segment area of

18\pi - 36

. Subtracting this segment area from the semicircle area yields

18\pi - (18\pi - 36) = 36

.

Step-by-Step Solution

1

Find the radii of circle

O

and circle

A

.

Radius of circle

O

is

r = 6

. In right triangle

AOC

,

OA = OC = 6

, so radius

AC = \sqrt{6^2 + 6^2} = 6\sqrt{2}

.

Perpendicular diameters

AB

and

CD

intersect at center

O

, dividing each diameter into radii of length

6

.

2

Calculate the area of sector

ACD

of circle

A

and triangle

ACD

.

Sector area

= \frac{90^\circ}{360^\circ} \pi (6\sqrt{2})^2 = 18\pi

. Triangle area

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

.

Angle

\angle CAD = 90^\circ

because

\triangle ACD

is a right isosceles triangle with hypotenuse

CD = 12

.

3

Find the area of the circular segment bounded by chord

CD

and arc

CD

of circle

A

.

Segment Area

= 18\pi - 36

.

The area of a circular segment is equal to the sector area minus the triangle area.

4

Subtract the segment area from the area of semicircle

CBD

of circle

O

.

Region Area

= 18\pi - (18\pi - 36) = 36

.

Semicircle

CBD

has radius

6

and area

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

. Subtracting the segment area leaves the crescent region.

Key Concept

Area of circular sectors, segments, and compound regions (Lune of Hippocrates)

Question 1515Question →

A rectangular park measures $8$ meters in width and $15$ meters in length. A straight walking path extends diagonally across the park from one corner to the opposite corner. What is the length, in meters, of the path?

23

17

289

161

\sqrt{23}

Show answer & explanation

Answer: $17$

Answer

17 meters

The diagonal of a rectangle forms a right triangle where the width (

8

m) and length (

15

m) are legs, and the diagonal path is the hypotenuse. Applying the Pythagorean theorem:

d = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17

meters.

Step-by-Step Solution

1

Identify the geometric model and given dimensions

The park is a rectangle with perpendicular sides of length

8

meters and

15

meters. The diagonal forms a right-angled triangle with these two sides as legs.

Adjacent sides of a rectangle meet at right angles (

90^\circ

).

2

Apply the Pythagorean theorem to calculate the diagonal length

d

d^2 = 8^2 + 15^2 = 64 + 225 = 289

In any right triangle, the square of the hypotenuse equals the sum of the squares of the legs (

a^2 + b^2 = c^2

).

3

Solve for

d

by taking the square root

d = \sqrt{289} = 17

The distance

d

must be positive.

Key Concept

Pythagorean Theorem (

a^2 + b^2 = c^2

) and the

8-15-17

Pythagorean Triple

Question 1516Question →

A customer service representative resolved the following number of support tickets over five consecutive days: $18$ , $24$ , $15$ , $31$ , and $22$ . What is the arithmetic mean of the number of tickets resolved per day by the representative?

Show answer & explanation

Answer: 22

Answer

The arithmetic mean of the number of tickets resolved per day is

22

.

The mean of a data set is calculated by taking the sum of all values and dividing by the total number of items. The sum of the numbers is

18 + 24 + 15 + 31 + 22 = 110

. Dividing

110

by

5

yields

22

.

Step-by-Step Solution

1

Sum all data values in the set

18 + 24 + 15 + 31 + 22 = 110

To find the mean, the first step is to calculate the total sum of all observations.

2

Divide the total sum by the total number of values

\frac{110}{5} = 22

The arithmetic mean is defined as the sum of the values divided by the count of the values.

Key Concept

Arithmetic Mean

Question 1517Question →

An electronics manufacturer produces three types of circuit boards: Alpha, Beta, and Gamma. Production requires processing across three specialized workstations: Solder, Component Placement, and Inspection.

- Each Alpha board requires 2 hours of Solder, 3 hours of Component Placement, and 1 hour of Inspection.
- Each Beta board requires 1 hour of Solder, 4 hours of Component Placement, and 2 hours of Inspection.
- Each Gamma board requires 3 hours of Solder, 2 hours of Component Placement, and 4 hours of Inspection.

During a given production cycle, the Solder station was operated for 55 hours, the Component Placement station for 85 hours, and the Inspection station for 65 hours. If all three workstations were operated at full capacity with no downtime, what was the total number of circuit boards produced?

Show answer & explanation

Answer: 28

Answer

The total number of circuit boards produced is 28.

Setting up equations for total machine hours yields

2x + y + 3z = 55

,

3x + 4y + 2z = 85

, and

x + 2y + 4z = 65

, where

x

,

y

, and

z

represent the quantities of Alpha, Beta, and Gamma boards produced, respectively. Subtracting the third equation from the first equation gives

(2x + y + 3z) - (x + 2y + 4z) = 55 - 65

, which simplifies to

x - y - z = -10

, or

x = y + z - 10

. Substituting

x = y + z - 10

into the second and third equations produces the 2x2 system

7y + 5z = 115

and

3y + 5z = 75

. Subtracting these two equations eliminates

z

, giving

4y = 40

, so

y = 10

. Substituting

y = 10

into

3y + 5z = 75

gives

30 + 5z = 75

, so

z = 9

. Substituting

y = 10

and

z = 9

into

x = y + z - 10

gives

x = 9

. The total number of circuit boards produced is

x + y + z = 9 + 10 + 9 = 28

.

Step-by-Step Solution

1

Set up the 3x3 system of linear equations based on workstation hours.

2x + y + 3z = 55

,

3x + 4y + 2z = 85

, and

x + 2y + 4z = 65

Each equation models the total operational hours used across the three product types.

2

Subtract the third equation from the first equation to isolate

x

in terms of

y

and

z

.

x = y + z - 10

Eliminating terms directly reduces coefficient complexity.

3

Substitute

x = y + z - 10

into the second and third equations to construct a 2x2 system.

3y + 5z = 75

and

7y + 5z = 115

Reducing to a two-variable system allows direct elimination.

4

Subtract the two reduced equations to solve for

y

and

z

.

y = 10

and

z = 9

The

5z

terms cancel out upon subtraction.

5

Determine

x

and compute the total sum

x + y + z

.

x = 9

, total

= 9 + 10 + 9 = 28

The question asks for the total quantity of circuit boards produced.

Key Concept

Systems of Linear Equations

Question 1518Question →

In a circle centered at point $O$ , the radius is $12$ units and central angle $\angle AOB$ measures $150^\circ$ . Which of the following statements regarding sector $AOB$ and minor arc $AB$ are correct? Select all such statements.

Select all that apply

The length of minor arc

AB

is

10\pi

units.

The area of sector

AOB

is

60\pi

square units.

The total perimeter of sector

AOB

is

24 + 10\pi

units.

The length of minor arc

AB

is

24\pi

units.

The perimeter of sector

AOB

is equal to

10\pi

units.

Show answer & explanation

Answer: The length of minor arc $AB$ is $10\pi$ units.; The area of sector $AOB$ is $60\pi$ square units.; The total perimeter of sector $AOB$ is $24 + 10\pi$ units.

Answer

The correct statements are those indicating that the minor arc length is

10\pi

units, the sector area is

60\pi

square units, and the perimeter of the sector is

24 + 10\pi

units.

The central angle fraction is

\frac{150^\circ}{360^\circ} = \frac{5}{12}

. Multiplying the full circumference (

24\pi

) by

\frac{5}{12}

gives an arc length of

10\pi

. Multiplying the full circle area (

144\pi

) by

\frac{5}{12}

gives a sector area of

60\pi

. Adding the two radii (

2 \times 12 = 24

) to the arc length (

10\pi

) gives the total sector perimeter of

24 + 10\pi

.

Step-by-Step Solution

1

Determine the central angle fraction of the circle.

The fraction is

\frac{150^\circ}{360^\circ} = \frac{5}{12}

.

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

360^\circ

.

2

Calculate the circumference of the circle and the arc length of minor arc

AB

.

Circumference =

2\pi(12) = 24\pi

. Arc length =

\frac{5}{12} \times 24\pi = 10\pi

units.

Arc length equals full circumference multiplied by the central angle fraction.

3

Calculate the total area of the circle and the area of sector

AOB

.

Circle Area =

\pi(12^2) = 144\pi

. Sector Area =

\frac{5}{12} \times 144\pi = 60\pi

square units.

Sector area equals total circle area multiplied by the central angle fraction.

4

Calculate the total perimeter of sector

AOB

.

Perimeter =

\text{arc length} + 2r = 10\pi + 2(12) = 24 + 10\pi

units.

The perimeter of a sector consists of the bounding arc plus the two radii.

Key Concept

Circles, Arc Lengths, and Sector Areas

Question 1519Question →

If $x$ is an integer that satisfies both $|2x - 3| \le 9$ and $\frac{5 - 3x}{-2} > 4$ , how many possible values of $x$ exist?

2

3

5

8

10

Show answer & explanation

Answer: 2

Answer

There are 2 possible integer values of

x

(specifically

x = 5

and

x = 6

).

Solving

|2x - 3| \le 9

yields

-3 \le x \le 6

. Solving

\frac{5 - 3x}{-2} > 4

requires reversing the inequality sign twice (first when multiplying by

-2

, giving

5 - 3x < -8

, and second when dividing by

-3

, giving

x > \frac{13}{3} \approx 4.33

). The integers satisfying both

-3 \le x \le 6

and

x > 4.33

are

5

and

6

, which gives exactly 2 possible values.

Step-by-Step Solution

1

Solve the absolute value inequality

|2x - 3| \le 9

.

-9 \le 2x - 3 \le 9 \implies -6 \le 2x \le 12 \implies -3 \le x \le 6

.

An absolute value inequality of the form

|u| \le k

unfolds into the compound inequality

-k \le u \le k

.

2

Solve the linear inequality

\frac{5 - 3x}{-2} > 4

.

5 - 3x < -8 \implies -3x < -13 \implies x > \frac{13}{3} \approx 4.33

.

Multiplying or dividing an inequality by a negative number requires reversing the direction of the inequality sign at each step.

3

Find the intersection of the two solution sets for integer values of

x

.

The range of valid values is

4.33 < x \le 6

. The integers in this interval are

x = 5

and

x = 6

.

Combining

-3 \le x \le 6

and

x > 4.33

restricts

x

to integer values strictly greater than

4.33

and less than or equal to

6

.

4

Count the valid integer values.

There are 2 integers (

5

and

6

).

Direct counting of the valid integer set

\{5, 6\}

.

Key Concept

Solving absolute value inequalities and compound linear inequalities with negative multipliers

Question 1520Question →

A catering service charges a fixed base setup fee of $\$150$ plus $\$25$ per guest for a private dinner event. If a client was billed a total of $\$900$ for an event, how many guests attended?

24

30

34

36

42

Show answer & explanation

Answer: $30$

Answer

30 guests attended the event.

The total billing model is represented by the equation

150 + 25g = 900

, where

g

is the number of guests. Subtracting the fixed setup fee of

\$150

gives

25g = 750

. Dividing

750

by

25

yields

g = 30

, showing that exactly

30

guests attended.

Step-by-Step Solution

1

Formulate a linear modeling equation for total cost

\text{Total Cost} = \text{Base Fee} + (\text{Rate per Guest} \times g)

, where

g

represents the number of guests. Thus,

150 + 25g = 900

.

The total charge consists of a fixed constant component plus a variable component that depends on the number of guests.

2

Isolate the variable term by subtracting the fixed setup fee from both sides

25g = 900 - 150 \implies 25g = 750

.

Subtracting the fixed fee isolates the portion of the total cost spent strictly on guests.

3

Solve for

g

by dividing by the per-guest rate

g = \frac{750}{25} = 30

.

Dividing the variable total cost by the rate per guest yields the exact number of attending guests.

Key Concept

Linear Modeling and Single-Variable Word Problems

Estimated Time:1m 0s

Answer

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