Problem-Solving and Data Analysis Practice Questions — SAT — Page 4

Question 61Question →

An administrator installs a new security update on a computer server. The update is designed to identify and quarantine system errors such that the total number of errors on the server is reduced by $12\%$ each hour. Which of the following best describes the relationship between the time, in hours, since the update was installed and the number of system errors remaining on the server?

Decreasing exponential, because the number of errors decreases by a constant percentage each hour.

Decreasing linear, because the number of errors decreases by a constant quantity each hour.

Increasing exponential, because the number of errors increases by

12\%

each hour.

Decreasing linear, because the number of errors decreases by exactly

12

errors each hour.

Show answer & explanation

Answer: Decreasing exponential, because the number of errors decreases by a constant percentage each hour.

Answer

Decreasing exponential, because the number of errors decreases by a constant percentage each hour.

The correct answer is the option stating that the relationship is decreasing exponential because the number of errors decreases by a constant percentage each hour. In mathematical modeling, when a quantity changes by a fixed percentage over equal intervals of time, it is modeled by an exponential function. Since the quantity is being reduced, it is a decreasing exponential relationship.

Step-by-Step Solution

1

Analyze the rate of change described in the problem.

The number of system errors is reduced by

12\%

each hour.

Identifying how the quantity changes per unit of time is key to determining the model type.

2

Distinguish between linear and exponential relationships based on the rate of change.

Since the change is a constant percent decrease (

12\%

) rather than a constant numerical amount, the relationship is exponential rather than linear.

Exponential models represent situations with constant percentage changes, while linear models represent constant absolute changes.

3

Determine the direction of the change (growth or decay).

Because the number of errors is reduced, the quantity is decreasing, representing exponential decay (decreasing exponential).

A reduction in quantity indicates decay (decreasing) rather than growth (increasing).

Key Concept

Linear and Exponential Growth

Question 62Question →

A solar power station consists of solar panels that generate electricity at a constant rate of $15\text{ watts}$ per square foot of panel surface area (where $1\text{ watt} = 1\text{ joule per second}$ ). The panels have a total surface area of $12,000\text{ square yards}$ . How many megajoules (MJ) of energy do the panels generate in $2\text{ hours}$ ? (Given that $1\text{ yard} = 3\text{ feet}$ and $1\text{ MJ} = 1,000,000\text{ joules}$ )

144

194.4

3,888

11,664

Show answer & explanation

Answer: $11,664$

Answer

The solar panels generate

11,664

megajoules of energy in

2\text{ hours}

.

To find the energy generated, first convert the area from square yards to square feet. Since

1\text{ yard} = 3\text{ feet}

,

1\text{ square yard} = (3\text{ feet})^2 = 9\text{ square feet}

. Multiplying

12,000\text{ square yards}

by

9

gives

108,000\text{ square feet}

. The power generated is

108,000\text{ square feet} \times 15\text{ watts/square foot} = 1,620,000\text{ watts}

, which is

1,620,000\text{ joules per second}

. In

2\text{ hours}

, which is

2 \times 3,600 = 7,200\text{ seconds}

, the total energy generated is

1,620,000\text{ joules/second} \times 7,200\text{ seconds} = 11,664,000,000\text{ joules}

. Since

1\text{ MJ} = 1,000,000\text{ joules}

, this is equivalent to

11,664\text{ MJ}

.

Step-by-Step Solution

1

Convert the surface area of the solar panels from square yards to square feet.

Since

1\text{ yard} = 3\text{ feet}

,

1\text{ square yard} = 3^2 = 9\text{ square feet}

. Therefore, the total area in square feet is

12,000 \times 9 = 108,000\text{ square feet}

.

To apply the generation rate given in watts per square foot, the area must be in square feet.

2

Determine the total rate of energy generation in joules per second.

The total power generated is

108,000\text{ square feet} \times 15\text{ watts per square foot} = 1,620,000\text{ watts}

. Since

1\text{ watt} = 1\text{ joule per second}

, this equals

1,620,000\text{ joules per second}

.

This calculates the total energy produced by the entire solar farm every second.

3

Convert the total time from hours to seconds.

2\text{ hours} \times 60\text{ minutes/hour} \times 60\text{ seconds/minute} = 7,200\text{ seconds}

.

To find the total energy generated, the rate (joules per second) must be multiplied by the duration in seconds.

4

Calculate the total energy generated in joules, then convert it to megajoules (MJ).

Total energy in joules is

1,620,000\text{ joules/second} \times 7,200\text{ seconds} = 11,664,000,000\text{ joules}

. Converting to megajoules:

11,664,000,000\text{ joules} \div 1,000,000\text{ joules/MJ} = 11,664\text{ MJ}

.

This yields the final quantity of energy in the requested units.

Key Concept

Multi-step unit conversions involving area scaling, compound rates, and time conversions.

Estimated Time:2m 30s

Question 63Question →

An express delivery truck travels from Warehouse A to Warehouse B at a constant speed, and then returns from Warehouse B to Warehouse A along the same route at a different constant speed. On the first trip, the ratio of the outbound travel time to the inbound travel time is $3$ to $5$ . On a second trip along the same route, the truck's outbound speed is $50\%$ greater than its outbound speed on the first trip, and its inbound speed is $25\%$ less than its inbound speed on the first trip. If the total round-trip travel time for the second trip is $13$ hours, what is the total round-trip travel time, in hours, for the first trip?

Show answer & explanation

Answer: 12

Answer

The total round-trip travel time for the first trip is 12 hours.

The correct answer is

12

. By setting the outbound and inbound travel times of the first trip as

t_1

and

t_2

with

t_1 = 0.6t_2

, we apply the inverse relationships of speed and time to express the second trip's times as

t_1' = \frac{2}{3}t_1

and

t_2' = \frac{4}{3}t_2

. Substituting these into the second trip's total time equation

\frac{2}{3}t_1 + \frac{4}{3}t_2 = 13

yields

t_2 = 7.5

and

t_1 = 4.5

, summing to

12

hours.

Step-by-Step Solution

1

Set up the ratio of outbound to inbound travel times for the first trip.

t_1 = 0.6t_2

(or

t_1 = \frac{3}{5}t_2

)

The ratio of the outbound travel time (

t_1

) to the inbound travel time (

t_2

) is given as

3

to

5

.

2

Express the new travel times for the second trip (

t_1'

and

t_2'

) in terms of the original travel times.

t_1' = \frac{2}{3}t_1

and

t_2' = \frac{4}{3}t_2

For a constant distance, travel time is inversely proportional to speed. A

50\%

speed increase multiplies time by

\frac{1}{1.5} = \frac{2}{3}

, and a

25\%

speed decrease multiplies time by

\frac{1}{0.75} = \frac{4}{3}

.

3

Create an equation for the total travel time of the second trip.

\frac{2}{3}t_1 + \frac{4}{3}t_2 = 13

The sum of the outbound and inbound travel times on the second trip is given as

13

hours.

4

Substitute the first trip's relationship into the equation and solve for

t_2

.

t_2 = 7.5

Substituting

t_1 = \frac{3}{5}t_2

gives

\frac{2}{3}\left(\frac{3}{5}t_2\right) + \frac{4}{3}t_2 = 13 \implies \frac{26}{15}t_2 = 13 \implies t_2 = 13 \times \frac{15}{26} = 7.5

.

5

Determine

t_1

and calculate the total travel time for the first trip.

t_1 + t_2 = 12

Since

t_2 = 7.5

, then

t_1 = 0.6(7.5) = 4.5

. The total round-trip time is

4.5 + 7.5 = 12

.

Key Concept

Inverse proportionality of speed and time, and solving linear systems involving ratios.

Question 64Question →

A customer service representative recorded the call durations, in minutes, for $5$ phone calls on a Tuesday morning. The call durations are $1$ , $2$ , $3$ , $5$ , and $9$ minutes. What is the mean duration, in minutes, of these $5$ phone calls?

3

4

5

8

Show answer & explanation

Answer: $4$

Answer

The correct answer is

4

minutes, which represents the mean duration of the phone calls.

The mean duration of the calls is found by dividing the sum of the durations by the total number of calls. The sum of the durations is

1 + 2 + 3 + 5 + 9 = 20

minutes. Dividing this sum by the

5

calls yields

20 \div 5 = 4

minutes.

Step-by-Step Solution

1

Calculate the sum of all the call durations.

Sum =

1 + 2 + 3 + 5 + 9 = 20

To find the mean, the sum of all values in the dataset must first be calculated.

2

Divide the sum by the total number of call durations.

20 \div 5 = 4

The mean is found by dividing the sum of the values by the number of values in the dataset, which is

5

.

Key Concept

Calculating the mean (average) of a data set by dividing the sum of all values by the count of values.

Question 65Question →

An environmental study monitored the water volume of two reservoirs, Reservoir A and Reservoir B. In June, the water volume of Reservoir A was $20\%$ greater than the water volume of Reservoir B. By August, the water volume of Reservoir A had decreased by $15\%$ , while the water volume of Reservoir B had increased by $10\%$ . If the water volume of Reservoir A in August was $10.2$ million gallons, what was the water volume of Reservoir B in June, in millions of gallons?

8.5

9.6

10.0

14.4

Show answer & explanation

Answer: 10.0

Answer

The water volume of Reservoir B in June was 10.0 million gallons.

The correct answer is the value that represents Reservoir B's volume in June. Let

B

represent Reservoir B's water volume in June. Since Reservoir A's June volume was

20\%

greater than Reservoir B's, it is represented as

1.20B

. Between June and August, Reservoir A's volume decreased by

15\%

, so its August volume is

85\%

of its June volume, which is

0.85 \times 1.20B = 1.02B

. Setting this equal to the given August volume of

10.2

million gallons yields

1.02B = 10.2

. Solving for

B

gives

B = 10.0

million gallons.

Step-by-Step Solution

1

Define a variable for the unknown quantity and express Reservoir A's volume in June.

Let

B

represent the water volume of Reservoir B in June. Since Reservoir A's June volume was

20\%

greater than Reservoir B's, the volume of Reservoir A in June is

1.20B

.

This establishes the relationship between the initial volumes of the two reservoirs.

2

Determine Reservoir A's volume in August in terms of the variable

B

.

Since Reservoir A's volume decreased by

15\%

by August, its new volume is

85\%

of its June volume. Thus, the volume in August is

0.85 \times 1.20B = 1.02B

.

This represents the two-stage percentage change for Reservoir A.

3

Solve for the variable

B

using the given August volume.

Set the expression for Reservoir A's August volume equal to

10.2

million gallons:

1.02B = 10.2

. Dividing both sides by

1.02

gives

B = 10.0

million gallons.

This finds the initial volume of Reservoir B.

Key Concept

Multi-step percentage change and setting up equations to solve for initial values.

Estimated Time:1m 30s

Question 66Question →

A manufacturing plant uses a machine lubricant at a constant rate of $0.08$ fluid ounces per second of operation. The plant operates the machinery for $8$ hours per day, $5$ days a week. The lubricant is purchased in drums, where each drum contains $15$ gallons of lubricant. How many drums of lubricant does the plant use in a period of $5$ weeks? (Note: $1\text{ gallon} = 128\text{ fluid ounces}$ )

Show answer & explanation

Answer: 30

Answer

The plant uses 30 drums of lubricant in 5 weeks.

To find the number of drums, calculate the total seconds of operation: 5 weeks * 5 days/week * 8 hours/day * 3,600 seconds/hour = 720,000 seconds. Multiply this by the consumption rate of 0.08 fluid ounces/second to get 57,600 fluid ounces. Convert this to gallons by dividing by 128 fluid ounces/gallon to get 450 gallons. Finally, divide by 15 gallons/drum to find that 30 drums are used.

Step-by-Step Solution

1

Calculate the total number of operational seconds in the 5-week period.

720{}000\text{ seconds}

Multiply 5 weeks by 5 days per week, 8 hours per day, and 3,600 seconds per hour:

5 \times 5 \times 8 \times 3{}600 = 720{}000

.

2

Calculate the total volume of lubricant used in fluid ounces.

57{}600\text{ fluid ounces}

Multiply the rate of

0.08\text{ fluid ounces per second}

by the total operational seconds:

720{}000 \times 0.08 = 57{}600

.

3

Convert the total fluid ounces to gallons.

450\text{ gallons}

Divide the total fluid ounces by the conversion factor of

128\text{ fluid ounces per gallon}

:

57{}600 / 128 = 450

.

4

Calculate the total number of drums needed.

30\text{ drums}

Divide the total gallons by the capacity of a single drum (

15\text{ gallons}

):

450 / 15 = 30

.

Key Concept

Multi-step dimensional analysis and unit conversion under compound rates

Question 67Question →

Two environmental cleanup projects, Project A and Project B, begin treating separate, identical bodies of water that each contain $V_0$ gallons of a certain chemical. The volume of the chemical remaining in the water treated by Project A is modeled by the exponential decay function $A(t) = V_0(k)^t$ , where $t$ is the number of days since treatment began and $k$ is a constant. The volume of the chemical remaining in the water treated by Project B is modeled by the linear decay function $B(t) = V_0 - c \cdot t$ , where $c$ is a positive constant. After $1$ day of treatment, the volume of the chemical remaining in both bodies of water is the same. After $2$ days of treatment, the volume of the chemical remaining in the water treated by Project A is exactly $\frac{16}{15}$ times the volume of the chemical remaining in the water treated by Project B. If the treatment for Project B continues at this constant rate, after how many days will the chemical in Project B's water be completely removed?

Show answer & explanation

Answer: 5

Answer

5

Equating both models at

t = 1

gives

\frac{c}{V_0} = 1 - k

. At

t = 2

, Project A's remaining volume is

V_0 k^2

and Project B's is

V_0(2k - 1)

. Setting

k^2 = \frac{16}{15}(2k - 1)

results in the quadratic equation

15k^2 - 32k + 16 = 0

. Factoring this yields

k = 0.8

(discarding

1.33

because the scenario represents decay). The chemical in Project B's water is completely removed when

B(t) = 0 \implies t = \frac{V_0}{c} = \frac{1}{1-k} = \frac{1}{0.2} = 5

days.

Step-by-Step Solution

1

Relate parameters

k

and

c

using the condition

A(1) = B(1)

.

\frac{c}{V_0} = 1 - k

Establishing a link between the rates of linear decay and exponential decay.

2

Express the remaining volumes at

t = 2

in terms of

k

.

A(2) = V_0 k^2

and

B(2) = V_0(2k - 1)

Preparing equations to apply the ratio condition at

t = 2

using a single variable.

3

Apply the condition

A(2) = \frac{16}{15}B(2)

to solve for

k

.

k = 0.8

Setting up the quadratic equation

15k^2 - 32k + 16 = 0

, factoring it to find roots

0.8

and

1.33

, and selecting

0.8

because

k < 1

for a decay process.

4

Calculate the time

t

when

B(t) = 0

.

t = 5

Solving

V_0 - c \cdot t = 0

yields

t = \frac{V_0}{c} = \frac{1}{1-k} = \frac{1}{0.2} = 5

days.

Key Concept

Linear and exponential decay modeling

Question 68Question →

A bicycle factory produces road bikes and mountain bikes at a constant ratio of $2$ to $3$ . If the factory produces $120$ road bikes in one day, how many mountain bikes does the factory produce in the same day?

80

180

300

360

Show answer & explanation

Answer: 180

Answer

180

The ratio of road bikes to mountain bikes is

2

to

3

. Since the factory produces

120

road bikes, we can set up the proportion

\frac{2}{3} = \frac{120}{x}

, where

x

represents the number of mountain bikes. Solving for

x

gives

2x = 360

, which simplifies to

x = 180

.

Step-by-Step Solution

1

Identify the given ratio and set up a proportion comparing road bikes to mountain bikes.

The ratio of road bikes to mountain bikes is

\frac{2}{3}

, so the proportion is

\frac{\text{Road Bikes}}{\text{Mountain Bikes}} = \frac{2}{3}

.

To find the unknown number of mountain bikes using the constant ratio.

2

Substitute

120

for the number of road bikes and solve for the number of mountain bikes,

x

.

\frac{120}{x} = \frac{2}{3} \implies 2x = 360 \implies x = 180

.

To calculate the exact number of mountain bikes produced.

Key Concept

Using ratios and proportions to solve word problems

Question 69Question →

A scientist is monitoring the population of a bacterial culture in a petri dish. The population of the culture grows exponentially, doubling every $3$ hours. If the population of the culture after $9$ hours is $1,600$ , what was the initial population of the culture?

Show answer & explanation

Answer: 200

Answer

The initial population of the culture was

200

.

Since the bacterial culture doubles every

3

hours, the population undergoes

9 \div 3 = 3

doubling periods over a span of

9

hours. An initial population

P_0

that doubles

3

times will grow to

P_0 \times 2^3 = 8P_0

. Setting this expression equal to the final population of

1,600

gives

8P_0 = 1,600

. Dividing both sides of this equation by

8

yields

P_0 = 200

. Therefore, the initial population of the culture was

200

.

Step-by-Step Solution

1

Determine the number of doubling periods that occurred in

9

hours.

3

doubling periods

Since the population doubles every

3

hours, dividing the total time of

9

hours by the doubling time of

3

hours gives

9 \div 3 = 3

periods.

2

Set up an equation representing the exponential growth.

P_0 \times 2^3 = 1,600

An initial population

P_0

doubling

3

times grows by a factor of

2^3

, which is equal to

8

times the initial amount.

3

Solve the equation

8P_0 = 1,600

for the initial population

P_0

.

P_0 = 200

Dividing both sides of the equation by

8

isolates

P_0

and gives the initial value.

Key Concept

Exponential growth models and solving for the initial value.

Question 70Question →

A cargo ship travels at a constant speed of $24\text{ knots}$ . The ship's engine consumes fuel at a constant rate of $23\text{ liters per minute}$ . The fuel has a density of $0.80\text{ grams per milliliter}$ . Given that $1\text{ knot} = 1.15\text{ miles per hour}$ and $1\text{ metric ton} = 1,000\text{ kilograms}$ , what is the ship's fuel consumption rate in metric tons per mile?

0.04

0.053

0.0625

0.40

Show answer & explanation

Answer: 0.04

Answer

0.04

The correct answer is obtained by first converting the ship's speed to miles per hour (

24 \times 1.15 = 27.6\text{ mph}

). Next, the fuel usage is converted to liters per hour (

23 \times 60 = 1,380\text{ L/hr}

). Using the density conversion where

0.80\text{ g/mL}

is equal to

0.80\text{ kg/L}

, the fuel consumption is converted to mass (

1,380 \times 0.80 = 1,104\text{ kg/hr}

), which is equivalent to

1.104

metric tons per hour. Finally, dividing the fuel rate by the speed gives the rate per mile:

1.104 / 27.6 = 0.04

metric tons per mile.

Step-by-Step Solution

1

Convert the speed of the cargo ship from knots to miles per hour.

24\text{ knots} \times 1.15\text{ miles per hour per knot} = 27.6\text{ miles per hour}

To match the final unit of miles, we convert knots to miles per hour.

2

Convert the fuel consumption rate from liters per minute to liters per hour.

23\text{ liters/minute} \times 60\text{ minutes/hour} = 1,380\text{ liters/hour}

Since speed is in miles per hour, we convert the fuel consumption to a per-hour rate.

3

Convert the volume consumption rate to a mass consumption rate using the density.

1,380\text{ liters/hour} \times 0.80\text{ kilograms/liter} = 1,104\text{ kilograms/hour}

Density of

0.80\text{ g/mL}

is equivalent to

0.80\text{ kg/L}

because there are

1,000

grams in a kilogram and

1,000

milliliters in a liter.

4

Convert the mass rate from kilograms per hour to metric tons per hour.

\frac{1,104\text{ kilograms/hour}}{1,000\text{ kilograms/metric ton}} = 1.104\text{ metric tons/hour}

To match the final unit of metric tons, we divide the kilograms by

1,000

.

5

Calculate the fuel consumption rate in metric tons per mile by dividing the hourly fuel consumption by the hourly speed.

\frac{1.104\text{ metric tons/hour}}{27.6\text{ miles/hour}} = 0.04\text{ metric tons/mile}

Dividing the rate of fuel consumed per hour by the distance traveled per hour gives the fuel consumed per mile.

Key Concept

Unit conversions with compound rates and density scaling

Estimated Time:2m 30s

Question 71Question →

A quality control analyst inspected a batch of 180 microchips from two production lines, Line A and Line B. The microchips were classified as Defective, Acceptable, or Premium. Some of the data from the inspection are shown in the table below.

Production Line	Defective	Acceptable	Premium
Line A	10	50	$x$
Line B	15	$y$	35
Total	25	55	180

If a microchip is selected at random from those classified as Defective or Premium, what is the probability that the chip was produced by Line A?

Show answer & explanation

Answer: 0.375

Answer

The correct answer is 3/8 (or 0.375).

To find the probability that a randomly selected microchip was produced by Line A given that it is classified as Defective or Premium, we restrict the sample space to only the Defective and Premium microchips. From the table, the total number of Defective microchips is 25, and the total number of Premium microchips is 55, giving a combined group size of 25 + 55 = 80 microchips. Next, we determine how many of these 80 microchips were produced by Line A. Line A produced 10 Defective microchips and x Premium microchips. Since the total number of Premium microchips is 55 and Line B produced 35, Line A produced x = 55 - 35 = 20 Premium microchips. Thus, the number of microchips produced by Line A that are Defective or Premium is 10 + 20 = 30. The conditional probability is the number of favorable outcomes divided by the restricted total outcomes, which is 30/80 = 3/8 (or 0.375).

Step-by-Step Solution

1

Find the value of

x

(Line A Premium chips) using the total number of Premium chips.

x = 20

Since the total number of Premium chips is 55 and Line B produced 35, Line A must have produced

55 - 35 = 20

Premium chips.

2

Calculate the total number of chips produced by Line A.

Line A Total = 80

Sum the Defective, Acceptable, and Premium chips produced by Line A:

10 + 50 + 20 = 80

.

3

Calculate the total number of chips produced by Line B.

Line B Total = 100

Subtract the total number of Line A chips from the grand total of 180 chips:

180 - 80 = 100

.

4

Find the value of

y

(Line B Acceptable chips).

y = 50

Subtract the Defective (15) and Premium (35) chips of Line B from its total (100):

100 - 15 - 35 = 50

.

5

Identify the total number of chips in the conditioning category 'Defective or Premium'.

Total Defective or Premium = 80

Sum the total number of Defective chips (25) and Premium chips (55):

25 + 55 = 80

.

6

Identify the number of chips produced by Line A that are either Defective or Premium.

Favorable chips = 30

Sum the Defective chips from Line A (10) and the Premium chips from Line A (

x = 20

):

10 + 20 = 30

.

7

Calculate the probability by dividing the favorable outcomes by the total outcomes of the conditioning category.

30/80 = 3/8 = 0.375

The probability of selecting a Line A chip from the Defective or Premium group is the ratio of favorable chips to total chips in that group.

Key Concept

Conditional Probability from Two-Way Tables

Estimated Time:2m 30s

Question 72Question →

A manufacturing company uses two types of machines, Type P and Type Q, to complete a production run of custom components. The rate of production of a Type P machine is to that of a Type Q machine in the ratio $3:2$ . A team consisting of $4$ Type P machines and $3$ Type Q machines can complete the entire production run in $10$ hours when working together. The company starts the run with this team, but after $4$ hours, $2$ of the Type P machines break down and are removed from service. To complete the remainder of the production run on schedule, additional Type Q machines are added to the remaining machines. What is the total number of Type Q machines that must be added?

2

3

5

6

Show answer & explanation

Answer: 3

Answer

3 additional Type Q machines must be added.

To complete the production run on schedule, we express the rate of Type P and Type Q machines as

3r

and

2r

respectively. The initial team of

4

Type P and

3

Type Q machines produces at a combined rate of

4(3r) + 3(2r) = 18r

components per hour. The total work for the

10

-hour project is

18r \times 10 = 180r

components. In the first

4

hours, the team completes

18r \times 4 = 72r

components, leaving

180r - 72r = 108r

components to be completed in the remaining

6

hours. This requires a combined rate of

108r / 6 = 18r

components per hour. After

2

Type P machines break down, the remaining machines (

2

Type P and

3

Type Q) provide a rate of

2(3r) + 3(2r) = 12r

components per hour. To reach the required rate of

18r

, an additional rate of

18r - 12r = 6r

is needed. Since each Type Q machine has a rate of

2r

, the number of additional Type Q machines needed is

6r / 2r = 3

.

Step-by-Step Solution

1

Define the individual production rates of Type P and Type Q machines using a common variable

r

.

Let the rate of a Type P machine be

3r

components per hour, and the rate of a Type Q machine be

2r

components per hour.

This represents the given

3:2

ratio of their production rates.

2

Calculate the combined rate of the initial team and the total work required for the production run.

Combined initial rate

= 4(3r) + 3(2r) = 18r

components per hour. Total work

W = 18r \times 10 \text{ hours} = 180r

components.

The team consists of

4

Type P and

3

Type Q machines working for

10

hours.

3

Determine the work completed in the first

4

hours and the remaining work.

Completed work

= 18r \times 4 = 72r

components. Remaining work

= 180r - 72r = 108r

components.

The team worked for

4

hours before the breakdown occurred.

4

Find the required rate to complete the remaining work in the remaining time.

Remaining time

= 10 - 4 = 6

hours. Required rate

= 108r / 6 = 18r

components per hour.

The production run must be completed on the original schedule of

10

total hours.

5

Calculate the rate of the remaining machines and the rate deficit.

Current rate of remaining machines (

2

Type P and

3

Type Q)

= 2(3r) + 3(2r) = 12r

components per hour. Rate deficit

= 18r - 12r = 6r

components per hour.

Two Type P machines broke down, leaving

2

Type P and the original

3

Type Q machines.

6

Find the number of additional Type Q machines needed to cover the rate deficit.

Number of additional Type Q machines

= 6r / 2r = 3

.

Each Type Q machine contributes a rate of

2r

components per hour.

Key Concept

Work-rate problems with varying machine efficiencies and changing team compositions over time.

Question 73Question →

In a certain school district, the number of students enrolled in high school chemistry was $20\%$ of the total number of high school students. After a new curriculum was introduced, the number of chemistry students increased by $35\%$ , while the total number of high school students in the district remained the same. What percent of the total high school student population was enrolled in chemistry after the curriculum change?

7\%

27\%

48\%

55\%

Show answer & explanation

Answer: $27\%$

Answer

27\%

To find the new percentage of students enrolled in chemistry, we can express the initial enrollment in terms of the total student population,

T

. Initially, the number of chemistry students is

0.20T

. An increase of

35\%

on this enrollment is calculated by multiplying the initial enrollment by

1.35

. This gives the new enrollment as

0.20T \times 1.35 = 0.27T

. Since the total number of students

T

remains unchanged, the new enrollment represents

27\%

of the total population.

Step-by-Step Solution

1

Define variables for the initial values.

Let the total number of high school students in the district be

T

. The initial number of chemistry students is

0.20T

.

Establishing algebraic representations helps organize the relation between the variables.

2

Calculate the new number of chemistry students after the

35\%

increase.

The new number of chemistry students is

0.20T \times (1 + 0.35) = 0.20T \times 1.35 = 0.27T

.

An increase of

35\%

means multiplying the initial value by

1.35

.

3

Calculate the new percentage of the total student population.

Since the total student population

T

remained the same, the new fraction of chemistry students is

\frac{0.27T}{T} = 0.27

, which is

27\%

.

Dividing the new number of chemistry students by the total student population gives the final percentage.

Key Concept

Calculating percentage changes on a subgroup relative to a constant total population.

Question 74Question →

The table below summarizes the results of a survey conducted by a market research firm. The survey asked $160$ randomly selected adults about their age group and their primary source of news.

Age group	Digital/Social Media	Print/Television	Total
Under 40	$68$	$12$	$80$
40 and older	$32$	$48$	$80$
Total	$100$	$60$	$160$

Given that a randomly selected person from the survey has digital or social media as their primary news source, what is the probability that the person is under 40 years old?

\frac{17}{40}

\frac{17}{25}

\frac{17}{20}

\frac{5}{8}

Show answer & explanation

Answer: $\frac{17}{25}$

Answer

The correct answer is

\frac{17}{25}

(representing the ratio of individuals under 40 years old among those who use digital/social media as their primary news source).

The correct answer is

\frac{17}{25}

. To find the conditional probability that a randomly selected person is under 40 years old given that their primary news source is digital or social media, the denominator must represent the total number of people who satisfy the given condition (

100

). The numerator must represent the number of people who satisfy both conditions (are under 40 years old and have digital/social media as their primary news source, which is

68

). Dividing these values yields

\frac{68}{100}

, which simplifies to

\frac{17}{25}

.

Step-by-Step Solution

1

Identify the size of the restricted sample space based on the conditional statement in the question.

The total number of surveyed individuals whose primary news source is digital or social media is

100

.

The phrase 'Given that a randomly selected person... has digital or social media as their primary news source' restricts the denominator to the column total for digital/social media.

2

Identify the number of favorable outcomes within the restricted sample space.

Within the group of

100

individuals who use digital/social media, the number of individuals who are under 40 years old is

68

.

This value serves as the numerator for the conditional probability.

3

Divide the number of favorable outcomes by the size of the restricted sample space and simplify the resulting fraction.

The probability is

\frac{68}{100} = \frac{17}{25}

.

Both numbers are divided by their greatest common divisor,

4

, to simplify the fraction to its lowest terms.

Key Concept

Conditional Probability in Two-Way Tables

Question 75Question →

A community library starts a digital collection with $5,000$ e-books. The library's goal is to increase the number of e-books in the collection by $8\%$ each year. Which of the following functions best models the number of e-books, $E(t)$ , in the collection $t$ years after the collection was started?

E(t) = 400t + 5,000

E(t) = 5,000(0.08)^t

E(t) = 5,000(1.08)^t

E(t) = 1.08t + 5,000

Show answer & explanation

Answer: $E(t) = 5,000(1.08)^t$

Answer

E(t) = 5,000(1.08)^t

The correct option is the function that represents exponential growth, where the quantity increases by a constant percentage each year. The general formula for exponential growth is

E(t) = a(1 + r)^t

, where

a

is the initial value,

r

is the growth rate as a decimal, and

t

is the number of time periods. Here, the initial number of e-books is

a = 5,000

and the annual growth rate is

r = 8\% = 0.08

. Therefore, the growth factor is

1 + r = 1.08

. Substituting these values into the formula yields the correct model:

E(t) = 5,000(1.08)^t

.

Step-by-Step Solution

1

Determine if the growth is linear or exponential by analyzing the type of increase described.

The problem states that the collection increases by

8\%

each year. Since the increase is a constant percentage rather than a constant quantity, this represents exponential growth.

Identifying the growth type determines whether to use a linear equation or an exponential equation.

2

State the general form of an exponential growth function.

The general formula is

E(t) = a(1 + r)^t

, where

a

is the initial amount,

r

is the percent growth rate expressed as a decimal, and

1 + r

is the growth factor.

This formula provides the template for constructing the model.

3

Identify the values of

a

and

r

from the given context and substitute them into the template.

The initial value

a = 5,000

. The growth rate

r = 8\% = 0.08

, so the growth factor is

1 + 0.08 = 1.08

. Substituting these gives

E(t) = 5,000(1.08)^t

.

This completes the model construction to find the correct expression.

Key Concept

Linear and Exponential Growth

Question 76Question →

An investment firm analyzes two retirement portfolios, Portfolio P1 and Portfolio P2, each established with an initial deposit of $D$ dollars at time $t = 0$ years. The value of Portfolio P1 increases linearly by a constant amount each year. The value of Portfolio P2 increases exponentially at a constant annual rate. At $t = 10$ years, the value of both portfolios is equal. At $t = 20$ years, the value of Portfolio P2 is exactly $\frac{9}{8}$ times the value of Portfolio P1. Which of the following expressions represents the value of Portfolio P2, in dollars, at $t = 30$ years in terms of $D$ ?

\frac{5}{2}D

\frac{9}{4}D

\frac{27}{8}D

\frac{9}{2}D

Show answer & explanation

Answer: $\frac{27}{8}D$

Answer

The expression that represents the value of Portfolio P2 at

t = 30

years is

\frac{27}{8}D

.

To find the value of Portfolio P2 at

t = 30

, we model the linear portfolio as

V_1(t) = D + ct

and the exponential portfolio as

V_2(t) = D \cdot g^t

. At

t = 10

,

D + 10c = D \cdot g^{10}

, which gives

c = \frac{D}{10}(g^{10}-1)

. At

t = 20

,

V_2(20) = \frac{9}{8}V_1(20) \implies D \cdot g^{20} = \frac{9}{8}(D + 20c)

. Substituting

c

into the equation and simplifying yields the quadratic equation

8(g^{10})^2 - 18(g^{10}) + 9 = 0

. Solving for

g^{10}

and selecting the value greater than 1 (since the portfolio is growing) gives

g^{10} = 1.5

. Thus, at

t = 30

, the value of Portfolio P2 is

D \cdot g^{30} = D \cdot (g^{10})^3 = D \cdot (1.5)^3 = \frac{27}{8}D

.

Step-by-Step Solution

1

Write the models for both portfolios in terms of the initial deposit

D

.

For Portfolio P1 (linear):

V_1(t) = D + c \cdot t

, where

c

is the constant annual increase.
For Portfolio P2 (exponential):

V_2(t) = D \cdot g^t

, where

g

is the annual growth factor.

Establishing the mathematical definitions of linear and exponential growth models is necessary to set up equations representing the given relationships.

2

Use the condition at

t = 10

to express the linear growth rate

c

in terms of

D

and

g^{10}

.

V_1(10) = V_2(10) \implies D + 10c = D \cdot g^{10} \implies 10c = D(g^{10} - 1) \implies c = \frac{D}{10}(g^{10} - 1)

.

This allows us to eliminate the linear growth rate variable

c

and express all equations in terms of the exponential growth term

g^{10}

.

3

Set up the equation for

t = 20

using the given relationship and substitute

c

.

V_2(20) = \frac{9}{8} V_1(20) \implies D \cdot g^{20} = \frac{9}{8} (D + 20c)

. Substituting

c

gives:

D \cdot g^{20} = \frac{9}{8} \left( D + 20 \cdot \frac{D}{10}(g^{10} - 1) \right) \implies D \cdot g^{20} = \frac{9}{8} D \left( 1 + 2(g^{10} - 1) \right)

. Since

D > 0

, we divide both sides by

D

to get:

g^{20} = \frac{9}{8} (2g^{10} - 1)

.

This sets up a solvable quadratic equation for the variable

g^{10}

.

4

Solve the quadratic equation for

g^{10}

.

Let

x = g^{10}

. The equation is

x^2 = \frac{9}{8}(2x - 1) \implies 8x^2 - 18x + 9 = 0

. Factoring the quadratic yields

(2x - 3)(4x - 3) = 0

, giving

x = 1.5

or

x = 0.75

. Since the portfolios are growing, the growth factor

g > 1

, so

g^{10} > 1

. Therefore, we choose

g^{10} = 1.5

.

Determining the value of the 10-year growth factor

g^{10}

is the key step to finding the value at future time intervals.

5

Calculate the value of Portfolio P2 at

t = 30

years.

V_2(30) = D \cdot g^{30} = D \cdot (g^{10})^3 = D \cdot (1.5)^3 = D \cdot \left(\frac{3}{2}\right)^3 = \frac{27}{8}D

.

Using the properties of exponents, we express the value at 30 years in terms of the 10-year growth factor cubed.

Key Concept

Linear vs. Exponential Growth Models

Question 77Question →

A community organization wanted to estimate the proportion of residents in a town who support the construction of a new public library. The organization surveyed a random sample of 200 residents who were visiting the current public library on a Saturday afternoon. Of those surveyed, 150 supported the construction of a new library. Which of the following is the most appropriate conclusion?

The results cannot be generalized to all residents of the town because the sample of people surveyed at the current library is likely biased and not representative of the entire town's population.

Approximately 75% of all residents in the town support the construction of a new library.

The results can be generalized to all residents of the town because a random sample of 200 people is large enough to represent the entire population.

Because 150 out of 200 is 50%, approximately 50% of the people who visit the library on a Saturday support the construction of a new library.

Show answer & explanation

Answer: The results cannot be generalized to all residents of the town because the sample of people surveyed at the current library is likely biased and not representative of the entire town's population.

Answer

The results cannot be generalized to all residents of the town because the sample of people surveyed at the current library is likely biased and not representative of the entire town's population.

The correct option is the one stating that the results cannot be generalized to all town residents because library visitors are likely to support library-related initiatives more than the general public, making the sample non-representative.

Step-by-Step Solution

1

Analyze the population and the sample collection method.

The target population is all residents of the town, but the sample consists of residents visiting the current public library on a Saturday.

To determine if results can be generalized, the sample must be representative of the target population.

2

Evaluate the representativeness of the sample.

People visiting a library are more likely to support a new library than the general public. Thus, the sample has selection bias and is not representative of all town residents.

Bias in the sampling method prevents generalization to the broader population.

Key Concept

Generalizing results from a sample to a population requires that the sample be representative and randomly selected from that population.

Question 78Question →

A conservationist monitors the populations of two plant species, Species A and Species B, in a forest reserve. The table below shows the population of each species, $P(t)$ and $Q(t)$ respectively, $t$ years after the monitoring began.

$t$ (years)	Species A population, $P(t)$	Species B population, $Q(t)$
$0$	$100$	$100$
$2$	$150$	$144$
$4$	$200$	$207.36$

The population of Species A can be modeled by a linear function, where $r_P$ represents the constant increase in the number of plants per year. The population of Species B can be modeled by an exponential function, where $r_Q\%$ represents the constant percent increase in the number of plants per year. What is the value of $r_P - r_Q$ ?

5

6

19

30

Show answer & explanation

Answer: $5$

Answer

5

To find

r_P

, we calculate the constant amount by which the linear population

P(t)

increases each year. Over the 2-year interval from

t = 0

to

t = 2

, the population increases by

150 - 100 = 50

. Since the growth is linear, the annual increase is

r_P = \frac{50}{2} = 25

plants per year. To find

r_Q

, we write the exponential model in the form

Q(t) = 100(b)^t

, where

b

is the annual growth factor. Using the value at

t = 2

, we have

144 = 100(b)^2

, which simplifies to

b^2 = 1.44

. Taking the square root gives the annual growth factor

b = 1.2

. A growth factor of

1.2

corresponds to an annual growth rate of

20\%

, so

r_Q = 20

. The difference between the two values is

r_P - r_Q = 25 - 20 = 5

. This matches the option with value 5.

Step-by-Step Solution

1

Calculate the annual rate of change (

r_P

) for the linear population model

P(t)

.

r_P = 25

Since Species A's population grows linearly, its rate of change is constant. The population increases from 100 to 150 over 2 years, which is an increase of

150 - 100 = 50

plants. Dividing this by the 2-year interval gives an annual growth rate of

\frac{50}{2} = 25

plants per year.

2

Determine the annual growth factor and percentage growth rate (

r_Q

) for the exponential population model

Q(t)

.

r_Q = 20

Since Species B's population grows exponentially, its population after

t

years can be written in the form

Q(t) = a(b)^t

, where

a

is the initial population and

b

is the annual growth factor. Using the data in the table:

Q(0) = a = 100

and

Q(2) = 100(b)^2 = 144

. Solving for

b

gives

b^2 = 1.44

, which means

b = \sqrt{1.44} = 1.2

. An annual growth factor of

1.2

corresponds to a constant percent increase of

(1.2 - 1) \times 100\% = 20\%

per year. Thus,

r_Q = 20

.

3

Calculate the difference between the two rates (

r_P - r_Q

).

5

Subtracting

r_Q

from

r_P

yields

25 - 20 = 5

.

Key Concept

Identifying and interpreting linear and exponential growth models from data tables, and converting rate of change across different time intervals.

Estimated Time:2m 0s

Question 79Question →

The table below shows the distribution of the number of public library cards owned by a group of families in a certain neighborhood.

Number of library cards	Number of families
1	4
2	5
3	2
4	1

What is the median number of public library cards owned by these families?

Show answer & explanation

Answer: 2

Answer

The median number of public library cards is 2.

To find the median, first determine the total number of families by summing the frequencies:

4 + 5 + 2 + 1 = 12

. Since the number of data points is even, the median is the average of the two middle values (

6^{\text{th}}

and

7^{\text{th}}

values). Sorting the data, the first 4 values are 1, and the next 5 values (positions 5 through 9) are 2. Since both the

6^{\text{th}}

and

7^{\text{th}}

values are 2, the median is 2.

Step-by-Step Solution

1

Find the total number of families by adding the frequencies.

4 + 5 + 2 + 1 = 12

To determine the position of the median in the dataset, we must first find the total number of data points.

2

Identify the positions of the middle values.

The

6^{\text{th}}

and

7^{\text{th}}

values.

For an even number of data points (

n = 12

), the median is the average of the values at positions

\frac{n}{2}

and

\frac{n}{2} + 1

.

3

Find the values at the

6^{\text{th}}

and

7^{\text{th}}

positions using the frequency table.

Both the

6^{\text{th}}

and

7^{\text{th}}

values are 2.

Accumulating the frequencies from top to bottom: the first 4 values are 1, and the next 5 values (positions 5 to 9) are 2.

4

Calculate the average of the two middle values.

\frac{2 + 2}{2} = 2

The median of an even-sized dataset is the arithmetic mean of its two middle values.

Key Concept

Calculating the median from a frequency table

Question 80Question →

Two runners, Estela and Desmond, start at the same point on a circular track and run in opposite directions at constant speeds. The ratio of Estela's speed to Desmond's speed is $5$ to $4$ . If they start at the same time, they meet for the first time after $40$ seconds. If Desmond increases his speed by $25\%$ immediately after their first meeting, how many seconds after their first meeting will they meet for the second time?

Show answer & explanation

Answer: 36

Answer

36

To find the time between the first and second meetings, we determine the track length in terms of a constant

k

. Given the speed ratio of

5

to

4

, the speeds are

5k

and

4k

. The relative speed in opposite directions is the sum of these speeds, or

9k

. In

40

seconds, they cover the track length

L = 9k \times 40 = 360k

. When Desmond increases his speed by

25\%

, his new speed is

4k \times 1.25 = 5k

, making the new relative speed

5k + 5k = 10k

. The time to meet again is the track length divided by the new relative speed:

360k / 10k = 36

seconds.

Step-by-Step Solution

1

Represent the speeds of both runners using the given ratio of

5

to

4

.

Estela's speed is

5k

and Desmond's speed is

4k

, where

k

is a positive constant.

Using a common variable allows calculation of relative rates without needing their absolute speeds.

2

Calculate the initial relative speed of approach.

Relative speed =

5k + 4k = 9k

.

Since they run in opposite directions around the circular track, their rates of travel add up to determine their rate of approach.

3

Determine the track circumference

L

in terms of

k

.

Track length

L = 9k \times 40 = 360k

.

They meet for the first time when their combined distance traveled is exactly equal to one full lap of the track.

4

Calculate Desmond's updated speed after his speed increase.

Desmond's new speed =

4k \times (1 + 0.25) = 5k

.

His speed increases by

25\%

, which corresponds to multiplying his initial speed of

4k

by

1.25

.

5

Find the new relative speed of approach after the first meeting.

New relative speed =

5k + 5k = 10k

.

Estela's speed remains

5k

and Desmond's speed is now

5k

, so their sum gives the new combined rate.

6

Solve for the time elapsed between the first and second meetings.

Time

t = \frac{360k}{10k} = 36

seconds.

To meet a second time, they must cover the track length of

360k

again from their first meeting point at the new relative speed.

Key Concept

Solving multi-step rate-time-distance problems on circular tracks using ratios and percentage change.

Problem-Solving and Data Analysis

Answer

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