All practice questions

Question 181Question →

At the beginning of Year 1, a commercial real estate portfolio consisted of two properties, Property A and Property B, where the value of Property A was $50\%$ greater than the value of Property B. During Year 1, the value of Property A increased by $20\%$ while the value of Property B decreased by $20\%$ . During Year 2, the value of Property A decreased by $x\%$ while the value of Property B increased by $x\%$ . If the combined value of the two properties at the end of Year 2 was equal to the combined value of the two properties at the beginning of Year 1, what is the value of $x$ ?

Show answer & explanation

Answer: 10

Answer

10

Tracking base value changes across both properties yields a Year 1 ending total of

1.8B + 0.8B = 2.6B

. Applying Year 2 percentage adjustments gives a total value of

1.8B(1 - 0.01x) + 0.8B(1 + 0.01x) = 2.6B - 0.01Bx

. Setting this equal to the initial portfolio value of

2.5B

gives

0.1B = 0.01Bx

, which solves to

x = 10

.

Step-by-Step Solution

1

Define initial values for both properties relative to a single variable at the beginning of Year 1

Property B value =

B

, Property A value =

1.5B

, and total portfolio value =

2.5B

Setting Property B as

B

allows all subsequent values and combined sums to be expressed cleanly in terms of

B

.

2

Calculate individual property values at the end of Year 1 after applying respective

+20\%

and

-20\%

changes

Property A value =

1.5B \times 1.20 = 1.8B

; Property B value =

B \times 0.80 = 0.8B

Year 1 percentage changes must be calculated using each property's initial starting base.

3

Formulate algebraic expressions for the property values at the end of Year 2

Total Year 2 end value =

1.8B \left(1 - \frac{x}{100}\right) + 0.8B \left(1 + \frac{x}{100}\right) = 2.6B - 0.01Bx

Year 2 percentage shifts apply to the new updated base values (

1.8B

and

0.8B

) obtained at the end of Year 1.

4

Set the Year 2 end total value equal to the Year 1 start total value and solve for

x

2.6B - 0.01Bx = 2.5B \implies 0.1B = 0.01Bx \implies x = 10

Equating the two expressions eliminates

B

from both sides, leaving a linear equation in

x

.

Key Concept

Successive Percentage Changes and Base Shifts across Multiple Portfolio Assets

Question 182Question →

Three automated production lines—Line A, Line B, and Line C—run continuous maintenance cycles every $45$ minutes, $60$ minutes, and $75$ minutes, respectively. All three lines completed a maintenance cycle simultaneously at 8:00 a.m. on Monday. A supervisory inspection is triggered whenever at least two of the three lines complete a maintenance cycle at the exact same time. Between 8:01 a.m. on Monday and 8:00 a.m. on Tuesday, inclusive, how many supervisory inspections will be triggered?

Show answer & explanation

Answer: 16

Answer

16

To find the number of times at least two lines finish maintenance simultaneously, we determine the LCM for each pair of intervals (180, 225, and 300 minutes) and calculate how many multiples of these values occur within 1,440 minutes. Within this duration, there are 8 multiples of 180, 6 multiples of 225, and 4 multiples of 300. The single time all three coincide (at 900 minutes) is included in all three pairwise counts. Subtracting 2 instances of this overlap yields 8 + 6 + 4 - 2 = 16 total supervisory inspections.

Step-by-Step Solution

1

Prime factorize each interval length

45 = 3^2 × 5; 60 = 2^2 × 3 × 5; 75 = 3 × 5^2

Prime factorization allows systematic calculation of pairwise and triple LCMs.

2

Compute pairwise and triple LCMs

lcm(45, 60) = 180; lcm(45, 75) = 225; lcm(60, 75) = 300; lcm(45, 60, 75) = 900

The LCM determines the interval between simultaneous completions.

3

Count occurrences within the 1,440-minute time window

A & B: 8 times; A & C: 6 times; B & C: 4 times; All three: 1 time

Dividing the total time window by each LCM yields the frequency of simultaneous maintenance events.

4

Apply Principle of Inclusion-Exclusion for sets

8 + 6 + 4 - 2(1) = 16 distinct inspection events

Events where all three lines coincide (at t = 900) are counted in all three pair counts and must be adjusted so that the time point is counted exactly once.

Key Concept

Least Common Multiple (LCM) and Inclusion-Exclusion Principle

Question 183Question →

Line $Q$ passes through the coordinates $(-1, 7)$ and $(7, 1)$ in the Cartesian plane. Line $P$ is perpendicular to line $Q$ and intersects the $x$ -axis at $(-6, 0)$ . What is the $y$ -intercept of line $P$ ?

Show answer & explanation

Answer: 8

Answer

The

y

-intercept of line

P

is 8.

The slope of line

Q

is calculated as

\frac{1 - 7}{7 - (-1)} = -\frac{3}{4}

. Since line

P

is perpendicular to line

Q

, its slope is the negative reciprocal,

\frac{4}{3}

. Using the

x

-intercept point

(-6, 0)

, the line equation is

y - 0 = \frac{4}{3}(x + 6)

, which simplifies to

y = \frac{4}{3}x + 8

. Setting

x = 0

yields the

y

-intercept of

8

.

Step-by-Step Solution

1

Calculate the slope of line

Q

Slope of line

Q

is

-\frac{3}{4}

Using the slope formula

m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 7}{7 - (-1)} = \frac{-6}{8} = -\frac{3}{4}

.

2

Determine the slope of perpendicular line

P

Slope of line

P

is

\frac{4}{3}

Perpendicular lines have slopes that are negative reciprocals of each other:

m_P = -\frac{1}{m_Q} = -\frac{1}{-3/4} = \frac{4}{3}

.

3

Find the equation and

y

-intercept of line

P

The

y

-intercept is

8

Line

P

passes through

(-6, 0)

. Using point-slope form:

y - 0 = \frac{4}{3}(x - (-6)) \implies y = \frac{4}{3}x + 8

. Substituting

x = 0

gives

y = 8

.

Key Concept

Perpendicular lines and line equations in coordinate geometry

Question 184Question →

When the expression $\frac{3x^2 - 12}{x - 2}$ is completely simplified, it can be written in the form $ax + b$ for all $x \neq 2$ , where $a$ and $b$ are constants. What is the value of $a + b$ ?

Show answer & explanation

Answer: 9

Answer

The value of

a + b

is 9.

Factoring the numerator gives

3x^2 - 12 = 3(x^2 - 4) = 3(x - 2)(x + 2)

. Canceling the common factor

(x - 2)

in the denominator yields

3(x + 2)

, which expands to

3x + 6

. Matching this expression to

ax + b

reveals

a = 3

and

b = 6

. Adding these constants gives

a + b = 9

.

Step-by-Step Solution

1

Factor the numerator of the rational expression.

3x^2 - 12 = 3(x^2 - 4) = 3(x - 2)(x + 2)

Factoring out the common numerical factor 3 and using the difference of squares identity

(u^2 - v^2 = (u-v)(u+v))

completely factors the numerator.

2

Simplify the expression by canceling common factors.

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

Because

x \neq 2

, the factor

(x - 2)

is non-zero and can be canceled from both numerator and denominator.

3

Identify the values of

a

and

b

and compute

a + b

.

a = 3

,

b = 6

, so

a + b = 3 + 6 = 9

Comparing

3x + 6

to

ax + b

gives

a = 3

and

b = 6

.

Key Concept

Factoring algebraic expressions using common monomial factors and the difference of squares to simplify rational expressions.

Question 185Question →

In Month 1, a data center's two facilities, Facility A and Facility B, consumed a combined total of $120,000\text{ kWh}$ of electricity. In Month 2, Facility A decreased its electricity consumption by $15\%$ , while Facility B increased its electricity consumption by $20\%$ . If the combined electricity consumption of the two facilities increased by $6\%$ overall in Month 2, how many kilowatt-hours ( $\text{kWh}$ ) of electricity did Facility A consume in Month 1?

Show answer & explanation

Answer: 48000

Answer

Facility A consumed 48,000 kWh of electricity in Month 1.

Let

A

represent the electricity consumed by Facility A in Month 1. The remaining electricity consumed by Facility B in Month 1 is

120,000 - A

. In Month 2, the decrease in Facility A's consumption is

15\%

of

A

, or

0.15A

, and the increase in Facility B's consumption is

20\%

of

(120,000 - A)

, or

0.20(120,000 - A)

. The net increase overall is

6\%

of

120,000

, which is

7,200\text{ kWh}

. Setting up the change equation

-0.15A + 0.20(120,000 - A) = 7,200

leads to

-0.35A + 24,000 = 7,200

, which simplifies to

0.35A = 16,800

. Solving for

A

yields

A = 48,000\text{ kWh}

.

Step-by-Step Solution

1

Define variables for the initial Month 1 consumption of each facility.

Let

A

= Month 1 consumption of Facility A. Month 1 consumption of Facility B =

120,000 - A

.

Expressing one unknown in terms of the total reduces the system to a single variable equation.

2

Determine the net kilowatt-hour change for Month 2.

Net change =

0.06 \times 120,000 = 7,200\text{ kWh}

.

An overall increase of 6% applies to the total starting value of 120,000 kWh.

3

Express the individual percentage changes in terms of

A

and equate to the net change.

-0.15A + 0.20(120,000 - A) = 7,200

.

Facility A's 15% decrease contributes

-0.15A

and Facility B's 20% increase contributes

+0.20(120,000 - A)

.

4

Solve the linear equation for

A

.

-0.35A + 24,000 = 7,200 \Rightarrow 0.35A = 16,800 \Rightarrow A = 48,000

.

Dividing

16,800

by

0.35

yields the original consumption of Facility A.

Key Concept

Weighted Percent Change and Systems of Percent Equations

Question 186Question →

An automated micro-dispenser delivers liquid reagents in precise doses. Each dose has a volume of $3.6 \times 10^{-5}$ liters. If a reservoir containing $0.0162$ liters of reagent is emptied completely by delivering these equal doses, how many doses were delivered?

Show answer & explanation

Answer: 450

Answer

450

Dividing the total volume of

1.62 \times 10^{-2}

liters by the single dose volume of

3.6 \times 10^{-5}

liters yields

\frac{1.62}{3.6} \times 10^{-2 - (-5)} = 0.45 \times 10^3 = 450

doses.

Step-by-Step Solution

1

Express the total reservoir volume in scientific notation

0.0162 = 1.62 \times 10^{-2}

liters

Converting decimals into standard scientific notation simplifies multiplication and division operations.

2

Set up the division for the number of doses

\text{Number of doses} = \frac{1.62 \times 10^{-2}}{3.6 \times 10^{-5}}

The total volume divided by the volume per single dose yields the total dose count.

3

Compute the division of coefficients and exponent terms independently

\frac{1.62}{3.6} = 0.45

and

\frac{10^{-2}}{10^{-5}} = 10^{-2 - (-5)} = 10^3

Applying the exponent quotient rule

10^a / 10^b = 10^{a-b}

yields

-2 - (-5) = 3

.

4

Convert from scientific notation to a standard integer

0.45 \times 10^3 = 450

Multiplying

0.45

by

1,000

shifts the decimal point 3 places to the right.

Key Concept

Division of numbers in scientific notation and place value manipulation

Estimated Time:1m 30s

Question 187Question →

A specialty coffee shop prepares two custom coffee bean blends using Arabica and Robusta beans. The first 10-pound blend consists of 4 pounds of Arabica beans and 6 pounds of Robusta beans and costs $\$52$ . The second 10-pound blend consists of 7 pounds of Arabica beans and 3 pounds of Robusta beans and costs $\$61$ . What is the cost, in dollars, of 1 pound of Arabica coffee beans?

Show answer & explanation

Answer: 7

Answer

The cost of 1 pound of Arabica coffee beans is 7 dollars.

By setting up the system

4A + 6R = 52

and

7A + 3R = 61

, we can simplify the first equation to

2A + 3R = 26

. Subtracting this simplified equation from the second equation eliminates

R

, giving

5A = 35

, which simplifies directly to

A = 7

.

Step-by-Step Solution

1

Define variables and set up the system of linear equations.

Let

A

be the price per pound of Arabica beans and

R

be the price per pound of Robusta beans.
Equation 1:

4A + 6R = 52

Equation 2:

7A + 3R = 61

Translating the quantitative relationship given in the problem statement into algebraic equations.

2

Simplify Equation 1 and eliminate variable

R

by subtraction.

Dividing Equation 1 by 2 gives

2A + 3R = 26

. Subtracting this from Equation 2 yields

(7A + 3R) - (2A + 3R) = 61 - 26

, which reduces to

5A = 35

.

Matching coefficients of

R

allows for straightforward elimination of

R

.

3

Solve for variable

A

.

A = 7

Dividing 35 by 5 yields the price per pound of Arabica beans.

Key Concept

Solving a system of 2x2 linear equations using substitution or elimination.

Question 188Question →

What is the smallest positive integer that is a multiple of 18, 24, and 30, and is also a perfect square?

Show answer & explanation

Answer: 3600

Answer

The smallest positive integer that is a multiple of 18, 24, and 30, and is also a perfect square is 3600.

The least common multiple of 18, 24, and 30 is 360, which factors into

2^3 \cdot 3^2 \cdot 5^1

. For an integer to be a perfect square, all exponents in its prime factorization must be even. Multiplying 360 by

2 \cdot 5 = 10

completes the odd exponents to even values (

2^4 \cdot 3^2 \cdot 5^2

), giving 3600, which is the smallest perfect square divisible by 18, 24, and 30.

Step-by-Step Solution

1

Find the prime factorization of 18, 24, and 30

18 = 2^1 \cdot 3^2

,

24 = 2^3 \cdot 3^1

,

30 = 2^1 \cdot 3^1 \cdot 5^1

Decomposing numbers into prime factors allows calculation of the LCM and analysis of perfect square conditions.

2

Calculate the LCM of 18, 24, and 30

\text{LCM}(18, 24, 30) = 2^3 \cdot 3^2 \cdot 5^1 = 360

Any common multiple must be a multiple of the LCM of these three numbers.

3

Determine the smallest factor required to make the prime exponents even

Multiply 360 by

2^1 \cdot 5^1 = 10

A perfect square requires all prime exponents to be even; 2 has power 3 and 5 has power 1 in 360, so one more factor of 2 and one more factor of 5 are required.

4

Compute the final result

360 \times 10 = 3600

3600 = 2^4 \cdot 3^2 \cdot 5^2 = 60^2

, which is a perfect square.

Key Concept

Prime Factorization, LCM, and Exponent Properties of Perfect Squares

Question 189Question →

A craft brewery blends two batches of cider. Batch A contains a $12\%$ sugar solution by volume, and Batch B contains a $20\%$ sugar solution by volume. The brewer mixes $15$ liters of Batch A with $x$ liters of Batch B to create a resulting mixture that is $17\%$ sugar by volume. What is the value of $x$ ?

Show answer & explanation

Answer: 25

Answer

The value of

x

is

25

.

Equating the total amount of pure sugar before and after mixing gives

0.12(15) + 0.20x = 0.17(15 + x)

. Simplifying yields

1.8 + 0.20x = 2.55 + 0.17x

, which reduces to

0.03x = 0.75

, giving

x = 25

.

Step-by-Step Solution

1

Calculate the volume of pure sugar contributed by Batch A

0.12 \times 15 = 1.8

liters of pure sugar

Batch A is

12\%

sugar by volume and has a total volume of

15

liters.

2

Express the total volume of pure sugar in terms of

x

Total sugar volume =

1.8 + 0.20x

liters

Batch B adds

x

liters of a

20\%

sugar solution.

3

Set up the mixture equation

1.8 + 0.20x = 0.17(15 + x)

The final mixture has a total volume of

(15 + x)

liters with a concentration of

17\%

sugar.

4

Expand and solve the linear equation for

x

1.8 + 0.20x = 2.55 + 0.17x \implies 0.03x = 0.75 \implies x = 25

Subtract

0.17x

and

1.8

from both sides, then divide by

0.03

.

Key Concept

Linear Equations in One Variable (Mixture Word Problems)

Estimated Time:1m 30s

Question 190Question →

On the real number line, points $P$ , $Q$ , and $R$ have coordinates $x$ , $y$ , and $z$ , respectively, such that $|x - 2| = 5$ , $|y + 4| = 3$ , and $z$ is the midpoint of segment $PQ$ . If $x < y$ , what is the value of $z$ ?

Show answer & explanation

Answer: -2

Answer

The coordinate of point

R

(the value of

z

) is

-2

.

Solving the two absolute value equations yields

x \in \{-3, 7\}

and

y \in \{-7, -1\}

. Testing the given constraint

x < y

across all four possible ordered pairs reveals that only

x = -3

and

y = -1

satisfy the condition, since

-3 < -1

is true while all other pairs fail. The midpoint of points with coordinates

-3

and

-1

is

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

.

Step-by-Step Solution

1

Solve the absolute value equation

|x - 2| = 5

.

x = 7

or

x = -3

The equation

|x - 2| = 5

splits into

x - 2 = 5 \implies x = 7

and

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

.

2

Solve the absolute value equation

|y + 4| = 3

.

y = -1

or

y = -7

The equation

|y + 4| = 3

splits into

y + 4 = 3 \implies y = -1

and

y + 4 = -3 \implies y = -7

.

3

Evaluate all candidate pairs

(x, y)

under the inequality constraint

x < y

.

The only valid pair is

x = -3

and

y = -1

.

Comparing all four combinations:

7 < -1

(false),

7 < -7

(false),

-3 < -7

(false), and

-3 < -1

(true).

4

Calculate the midpoint

z

of the segment

PQ

.

z = -2

The midpoint of coordinates

-3

and

-1

on a number line is given by their average:

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

.

Key Concept

Absolute Value Equations and Midpoint on the Real Number Line

Question 191Question →

A venture capital fund allocated an initial sum of money between Portfolio Alpha and Portfolio Beta, with Portfolio Alpha receiving $60\%$ of the total sum and Portfolio Beta receiving the remaining $40\%$ . During the first year, the value of Portfolio Alpha increased by $20\%$ , while the value of Portfolio Beta decreased by $15\%$ . During the second year, the value of Portfolio Alpha decreased by $10\%$ relative to its value at the end of the first year, while the value of Portfolio Beta increased by $25\%$ relative to its value at the end of the first year. By what percent did the total combined value of the two portfolios increase from the initial allocation to the end of the second year?

Show answer & explanation

Answer: 7.3

Answer

The total combined value of the two portfolios increased by 7.3%.

To find the net percent change over the two-year period, track each portfolio's value year by year. Assuming an initial combined total of

100, Portfolio Alpha begins at

60 and Portfolio Beta at

40. By the end of Year 1, Alpha increases by 20% to

72 (

60 \times 1.20

), while Beta decreases by 15% to

34 (

40 \times 0.85

). In Year 2, Alpha decreases by 10% relative to its Year 1 ending value, becoming

64.80 (

72 \times 0.90

), while Beta increases by 25% relative to its Year 1 ending value, becoming

42.50 (

34 \times 1.25

). The total combined value at the end of Year 2 is

64.80 +

42.50 =

107.30. Relative to the initial $100 allocation, this represents a net increase of 7.3%.

Step-by-Step Solution

1

Assign a convenient base value for the initial total investment.

Assume an initial total allocation of

100. Portfolio Alpha starts with

60 and Portfolio Beta starts with $40.

Percent changes are proportional and scale-invariant, making $100 a simple base value for calculation.

2

Compute the value of each portfolio at the end of Year 1.

Portfolio Alpha =

60 \times (1 + 0.20) =

72. Portfolio Beta =

40 \times (1 - 0.15) =

34.

Apply Year 1 growth (+20%) and loss (-15%) to their respective starting funds.

3

Compute the value of each portfolio at the end of Year 2.

Portfolio Alpha =

72 \times (1 - 0.10) =

64.80. Portfolio Beta =

34 \times (1 + 0.25) =

42.50.

Apply Year 2 changes (-10% and +25%) to the Year 1 ending values rather than the original principal.

4

Find the final total combined value and calculate the overall net percentage increase.

Combined Year 2 value =

64.80 +

42.50 = $107.30. Net percentage increase = \frac{107.30 - 100}{100} \times 100\% = 7.3\%.

Compare the final combined sum of

107.30 to the initial combined sum of

100.

Key Concept

Successive percent changes with shifting base values across multiple assets

Question 192Question →

Three water pumps, $P_1$ , $P_2$ , and $P_3$ , operate at constant individual rates to fill a large reservoir. The ratio of the pumping rate of $P_1$ to that of $P_2$ is $3 : 4$ , and the ratio of the pumping rate of $P_2$ to that of $P_3$ is $3 : 5$ .

At 8:00 AM, all three pumps begin filling an empty reservoir together. At 10:00 AM, pump $P_1$ shuts down, while $P_2$ and $P_3$ continue operating at their original rates. At 11:00 AM, the operating rate of $P_2$ is decreased by $25\%$ , and the operating rate of $P_3$ is increased by $25\%$ . The two remaining pumps continue at these adjusted rates until the reservoir is completely full at 1:00 PM.

If pump $P_3$ were to fill the empty reservoir working alone at its original constant rate, how many hours would it take?

Show answer & explanation

Answer: 9.1

Answer

It would take pump

P_3

exactly

9.1

hours (or

9.1

when entered numerically) to fill the empty reservoir alone at its original constant rate.

By unifying the given ratios

r_1 : r_2 = 3 : 4

and

r_2 : r_3 = 3 : 5

, we obtain the relative rates

r_1 = 9k

,

r_2 = 12k

, and

r_3 = 20k

. Summing the work done across the three intervals (8–10 AM at rate

41k

, 10–11 AM at rate

32k

, and 11 AM–1 PM at rate

34k

) yields a total capacity of

182k

. Dividing total work

182k

by

P_3

's original rate of

20k

gives

9.1

hours.

Step-by-Step Solution

1

Unify the individual pumping rate ratios into a single ratio

r_1 : r_2 : r_3

.

r_1 = 9k

,

r_2 = 12k

,

r_3 = 20k

Aligning the ratio of

P_1:P_2 = 3:4 = 9:12

and

P_2:P_3 = 3:5 = 12:20

establishes a common scale factor

k

.

2

Calculate the work completed in each of the three time intervals.

W_1 = 82k

,

W_2 = 32k

,

W_3 = 68k

Multiply the duration of each interval by the sum of active rates during that period.

3

Sum the total work completed to get total capacity and divide by

P_3

's original rate.

Total capacity

= 182k

; Time

= \frac{182k}{20k} = 9.1

hours

The total work equals the sum of work done across all three phases, and time is work divided by rate.

Key Concept

Multi-stage work rates, compound ratio unification, and percentage rate adjustments

Estimated Time:3m 0s

Question 193Question →

An isosceles trapezoid has parallel base lengths of $10$ and $26$ , and an altitude of $15$ . A line segment connects the midpoints of the two non-parallel legs, dividing the figure into two smaller trapezoids. What is the area of the larger of these two smaller trapezoids?

Show answer & explanation

Answer: 165

Answer

165

The midsegment of a trapezoid connects the midpoints of the non-parallel legs, and its length is the average of the two parallel bases:

\frac{10 + 26}{2} = 18

. Because the line connects midpoints, it also bisects the altitude, making the height of each smaller trapezoid equal to

\frac{15}{2} = 7.5

. The larger of the two resulting trapezoids has bases of lengths

18

and

26

. Using the trapezoid area formula

\text{Area} = \frac{b_1 + b_2}{2} \times h

, we obtain

\frac{18 + 26}{2} \times 7.5 = 22 \times 7.5 = 165

.

Step-by-Step Solution

1

Calculate the length of the midsegment connecting the midpoints of the non-parallel legs.

The midsegment length is

18

.

The midsegment of a trapezoid is parallel to the bases and its length equals the average of the two base lengths:

\frac{10 + 26}{2} = 18

.

2

Determine the height of the smaller subtrapezoid.

The height of the subtrapezoid is

7.5

.

The segment connecting the midpoints of the legs bisects the overall altitude of

15

, giving a height of

\frac{15}{2} = 7.5

for each subtrapezoid.

3

Compute the area of the larger subtrapezoid.

The area is

165

.

The larger subtrapezoid is bounded by the midsegment (length

18

) and the bottom base (length

26

). Applying the trapezoid area formula yields

\text{Area} = \frac{18 + 26}{2} \times 7.5 = 22 \times 7.5 = 165

.

Key Concept

Trapezoid Midsegment Theorem and Subdivided Area Calculation

Question 194Question →

A manufacturing facility produces precision metal components. Each component has a mass of $8.4 \times 10^{-4}$ kilograms. If a shipment container holds a batch of these components with a total mass of $1.05$ kilograms, how many components are in the container?

Show answer & explanation

Answer: 1250

Answer

The container holds 1,250 components.

To find the number of components, divide the total batch mass by the mass per component:

\frac{1.05 \text{ kg}}{8.4 \times 10^{-4} \text{ kg}}

. Expressing

\frac{1}{10^{-4}}

as

10^4

transforms the expression into

\frac{1.05}{8.4} \times 10^4

. Dividing

1.05

by

8.4

yields

0.125

. Finally,

0.125 \times 10^4 = 1250

.

Step-by-Step Solution

1

Set up the ratio of total mass to single component mass

Number of components =

\frac{1.05}{8.4 \times 10^{-4}}

Dividing total mass by individual component mass gives the total count.

2

Apply exponent rules to move the power of 10 to the numerator

\frac{1.05}{8.4} \times 10^4

Since

\frac{1}{10^{-4}} = 10^4

, shifting the negative exponent to the numerator changes its sign.

3

Divide the decimal coefficients

\frac{1.05}{8.4} = 0.125

Simplifying

\frac{105}{840}

reduces to

\frac{1}{8} = 0.125

.

4

Evaluate the product with the place value shift

0.125 \times 10,000 = 1250

Multiplying by

10^4

shifts the decimal point 4 places to the right.

Key Concept

Division with scientific notation and decimal place value adjustment

Estimated Time:1m 30s

Question 195Question →

A commercial property was purchased at an initial value of $V$ dollars. In its first year, the value of the property increased by $25\%$ . In its second year, the property's value decreased by $20\%$ relative to its value at the end of the first year. In its third year, the value of the property increased by $15\%$ relative to its value at the end of the second year. If the value of the property at the end of the third year was $\$276,000$ , what was the initial purchase value $V$ , in dollars, of the property?

Show answer & explanation

Answer: 240000

Answer

240000

Each percent change applies to the value at the end of the previous period. A

25\%

increase scales the value by

1.25

, a

20\%

decrease scales it by

0.80

, and a

15\%

increase scales it by

1.15

. The net multiplier is

1.25 \times 0.80 \times 1.15 = 1.15

. Solving

1.15V = 276,000

gives

V = 240,000

.

Step-by-Step Solution

1

Convert percent changes to growth and decay multiplier factors

Year 1 factor = 1.25, Year 2 factor = 0.80, Year 3 factor = 1.15

Percent changes are compounded by multiplying successive growth/decay factors.

2

Multiply the factors sequentially to express the final value in terms of initial value V

Final Value = V × 1.25 × 0.80 × 1.15 = 1.15V

Combining successive multipliers simplifies the relationship between original and final amounts.

3

Set the algebraic expression equal to $276,000 and solve for V

V = 276,000 / 1.15 = 240,000

Dividing the final value by the overall multiplier determines the original base value.

Key Concept

Successive Percent Changes and Base Shift

Estimated Time:1m 30s

Question 196Question →

A technical consultant charges a one-time setup fee of $\$150$ plus a standard rate of $\$45$ per hour for the first $20$ hours of work on a project. For any hours worked beyond $20$ hours, the consultant charges an increased hourly rate that is $20\%$ higher than the standard rate. If the total bill for a project was $\$1,320$ , how many total hours did the consultant work on the project?

Show answer & explanation

Answer: 25

Answer

The consultant worked a total of 25 hours on the project.

To find the total hours worked, first subtract the setup fee (

150) and the cost of the first 20 hours (20 *

45 =

900) from the total bill of

1,320. This leaves

1,320 -

1,050 =

270. The hourly rate after 20 hours increases by 20% to

45 * 1.20 =

54 per hour. Dividing the remaining

270 by $54 gives 5 additional hours. Adding these 5 hours to the initial 20 hours gives a total of 25 hours.

Step-by-Step Solution

1

Determine the base cost for the setup fee and the initial 20 hours.

Base cost =

150 + (20 *

45) = $1,050.

The initial cost tier applies up to 20 hours of work.

2

Calculate the higher hourly rate applied to additional hours worked beyond 20.

Increased rate =

45 * 1.20 =

54 per hour.

The rate increases by 20% over the standard rate of $45 per hour.

3

Formulate and solve a linear equation for total hours

h

.

1050 + 54(h - 20) = 1320 \implies 54(h - 20) = 270 \implies h - 20 = 5 \implies h = 25

.

Subtracting the base cost leaves

270 for overtime hours, which divides by

54 per hour to yield 5 additional hours, for 25 hours total.

Key Concept

Linear Equations in One Variable

Question 197Question →

Three decorative light signals flash at regular intervals of $15$ seconds, $20$ seconds, and $36$ seconds, respectively. If all three signals flash simultaneously at 12:00 PM, how many times will all three signals flash simultaneously between 12:01 PM and 1:00 PM, inclusive?

Show answer & explanation

Answer: 20

Answer

20

The three signals flash together every 180 seconds (3 minutes), which is the least common multiple of 15, 20, and 36. Within the 60-minute interval from 12:01 PM to 1:00 PM inclusive, simultaneous flashes occur at minute 3, 6, 9, ..., up to minute 60, resulting in exactly 20 simultaneous flashes.

Step-by-Step Solution

1

Find the prime factorization of each interval in seconds

15 = 3 \cdot 5

,

20 = 2^2 \cdot 5

,

36 = 2^2 \cdot 3^2

Prime factorization allows systematic calculation of the least common multiple.

2

Calculate the LCM of 15, 20, and 36

\text{LCM} = 2^2 \cdot 3^2 \cdot 5 = 180

seconds

The LCM gives the minimum period of time after which all three signals flash together.

3

Convert the period into minutes and find the frequency in 60 minutes

180\text{ seconds} = 3\text{ minutes}

;

60 \div 3 = 20

flashes

The timeframe between 12:01 PM and 1:00 PM inclusive contains 60 minutes, yielding 20 multiples of 3 minutes.

Key Concept

Least Common Multiple (LCM) for periodic events

Question 198Question →

During a quality assurance test of a dual-sensor monitoring device, Sensor $A$ operates independently of Sensor $B$ . The probability that Sensor $A$ detects a target signal during a test trial is $0.80$ , and the probability that Sensor $B$ detects the target signal during the same trial is $0.75$ . What is the probability that exactly one of the two sensors detects the target signal during a test trial?

Show answer & explanation

Answer: 0.35

Answer

0.35

To find the probability that exactly one sensor detects the signal, we must evaluate two disjoint scenarios: Sensor

A

detects while Sensor

B

fails (

0.80 \times 0.25 = 0.20

), or Sensor

A

fails while Sensor

B

detects (

0.20 \times 0.75 = 0.15

). Summing these two probabilities gives

0.20 + 0.15 = 0.35

. Alternatively, subtracting the probability that both sensors detect (

0.80 \times 0.75 = 0.60

) from the probability that at least one detects (

1 - 0.20 \times 0.25 = 0.95

) yields

0.95 - 0.60 = 0.35

.

Step-by-Step Solution

1

Determine the probabilities of individual event non-occurrences

P(A') = 0.20

and

P(B') = 0.25

The probability of an event not occurring is

1

minus the probability that it occurs.

2

Calculate joint probability for each mutually exclusive event outcome

P(A \text{ and } B') = 0.80 \times 0.25 = 0.20

and

P(A' \text{ and } B) = 0.20 \times 0.75 = 0.15

Since the sensors operate independently,

P(X \text{ and } Y) = P(X) \times P(Y)

.

3

Combine the independent outcomes that satisfy the condition

0.20 + 0.15 = 0.35

The scenarios (only

A

detects, or only

B

detects) are mutually exclusive, so their probabilities are added.

Key Concept

Probability of Independent and Mutually Exclusive Events

Question 199Question →

Let $x$ and $y$ be integers such that $x < 0 < y$ . If $x$ and $y$ satisfy all of the following conditions:

1. $(-1)^{x y + x} = -1$
2. $(-1)^{x^2 y + y} = 1$
3. $y^2 - x^2 = 19$

What is the value of $x + y$ ?

Show answer & explanation

Answer: 1

Answer

The value of

x + y

is 1.

Condition 1 dictates that

(-1)^{x y + x} = -1

, meaning

x y + x = x(y + 1)

is odd. For the product

x(y+1)

to be odd, both

x

and

y+1

must be odd, which means

x

is odd and

y

is even. Condition 3 factors as

(y - x)(y + x) = 19

. Given

x < 0 < y

, we know

y - x > y + x

. Since 19 is prime, its unique positive factor pair requires

y - x = 19

and

y + x = 1

. Solving this system yields

y = 10

and

x = -9

, which satisfies all sign and parity constraints. Thus,

x + y = -9 + 10 = 1

.

Step-by-Step Solution

1

Analyze parity requirements from Condition 1

x

is odd and

y

is even

Since

(-1)^{x(y+1)} = -1

, the exponent

x(y+1)

must be odd, requiring both

x

and

y+1

to be odd.

2

Check consistency with Condition 2

Condition 2 is satisfied

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

is always even when

y

is even, making

(-1)^{x^2 y + y} = 1

true.

3

Factor difference of squares and set up system using sign rules

y - x = 19

and

y + x = 1

Since 19 is prime and

x < 0 < y

,

y - x > y + x > 0

, forcing the factor pair to be 19 and 1.

4

Solve for

x

and

y

and sum them

x = -9

,

y = 10

, giving

x + y = 1

Adding the system yields

2y = 20 \implies y = 10

, and substituting into

y + x = 1

yields

x = -9

.

Key Concept

Even-Odd Exponent Rules and Sign Properties of Integers

Question 200Question →

If

x

is a real number such that

\sqrt{\frac{25^{x+1} - 25^x}{5^{x+2} + 5^{x+1}}} = 250

what is the value of

x

?

Show answer & explanation

Answer: 7

Answer

The value of

x

is 7.

Factoring out common powers in the numerator and denominator yields

25^x(25-1) = 24 \cdot 5^{2x}

and

5^{x+1}(5+1) = 6 \cdot 5^{x+1}

. Simplifying their ratio inside the square root gives

\frac{24 \cdot 5^{2x}}{6 \cdot 5^{x+1}} = 4 \cdot 5^{x-1}

. Taking the square root gives

2 \cdot 5^{(x-1)/2}

. Setting this equal to 250 yields

5^{(x-1)/2} = 125 = 5^3

. Equating the exponents gives

(x-1)/2 = 3

, which solves to

x = 7

.

Step-by-Step Solution

1

Factor the numerator and express terms with a common base of 5

25^{x+1} - 25^x = 25^x(25 - 1) = 24 \cdot (5^2)^x = 24 \cdot 5^{2x}

Factoring out

25^x

simplifies the difference into a single term with base 5.

2

Factor the denominator using base 5

5^{x+2} + 5^{x+1} = 5^{x+1}(5 + 1) = 6 \cdot 5^{x+1}

Factoring out the common power

5^{x+1}

simplifies the sum into a single term.

3

Simplify the fraction inside the square root

\frac{24 \cdot 5^{2x}}{6 \cdot 5^{x+1}} = 4 \cdot 5^{2x - (x+1)} = 4 \cdot 5^{x-1}

Dividing coefficients (24/6 = 4) and applying exponent rules for division (

a^m / a^n = a^{m-n}

).

4

Take the square root of the simplified expression

\sqrt{4 \cdot 5^{x-1}} = \sqrt{4} \cdot \sqrt{5^{x-1}} = 2 \cdot 5^{\frac{x-1}{2}}

Using radical rules

\sqrt{ab} = \sqrt{a}\sqrt{b}

and

\sqrt{a^k} = a^{k/2}

.

5

Set the simplified radical expression equal to 250 and solve for x

2 \cdot 5^{\frac{x-1}{2}} = 250 \implies 5^{\frac{x-1}{2}} = 125 \implies 5^{\frac{x-1}{2}} = 5^3 \implies \frac{x-1}{2} = 3 \implies x = 7

Dividing both sides by 2 gives

5^{(x-1)/2} = 125 = 5^3

. Equating exponents yields

(x-1)/2 = 3

, so

x = 7

.

Key Concept

Exponent rules, base conversion, factoring exponential terms, and radical simplification

Estimated Time:2m 0s

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