Extrapolation and Trend Prediction Practice Questions — ACT

Question 1Question →

A chemist investigated the reaction rate of a reactant, Substance Y, at various initial concentrations. The initial rate of reaction, $R$ , in millimoles per liter per second ( $\text{mmol}\cdot\text{L}^{-1}\cdot\text{s}^{-1}$ ), was recorded for each concentration, $[Y]$ , in millimoles per liter ( $\text{mmol/L}$ ), at a constant temperature of $298\text{ K}$ . The results are presented in the table below:

Initial Concentration $[Y]$ ( $\text{mmol/L}$ )	Initial Rate of Reaction $R$ ( $\text{mmol}\cdot\text{L}^{-1}\cdot\text{s}^{-1}$ )
$1.5$	$4.5$
$3.0$	$18.0$
$4.5$	$40.5$
$6.0$	$72.0$
$7.5$	$112.5$

Based on the trend shown in the table, what would be the expected initial rate of reaction, in $\text{mmol}\cdot\text{L}^{-1}\cdot\text{s}^{-1}$ , if the initial concentration of Substance Y is increased to $12.0\text{ mmol/L}$ ?

Show answer & explanation

Answer: 288

Answer

The expected initial rate of reaction at a concentration of 12.0 mmol/L is 288.0 mmol*L^-1*s^-1.

The rate of reaction scales quadratically with concentration. Calculating the ratio of the rate to the concentration for each data point reveals that the ratio is equal to 2.0 times the concentration, yielding the equation R = 2.0 * [Y]^2. Substituting the target concentration of 12.0 mmol/L gives R = 2.0 * (12.0)^2 = 288.0 mmol*L^-1*s^-1. Alternatively, using the method of finite differences, the second difference between successive values is constant at 9.0, and continuing this sequence to 12.0 mmol/L also results in 288.0.

Step-by-Step Solution

1

Calculate the ratio of the rate R to the concentration [Y] for each data point.

The ratios are 3.0, 6.0, 9.0, 12.0, and 15.0.

To determine whether a direct proportional or higher-order relationship exists.

2

Formulate the mathematical model that represents this trend.

The ratio R/[Y] increases by 3.0 for every 1.5 mmol/L increase in [Y], which corresponds to R/[Y] = 2.0 * [Y], or R = 2.0 * [Y]^2.

To establish the quadratic relationship governing the dataset.

3

Substitute the target concentration value of 12.0 mmol/L into the derived quadratic equation.

R = 2.0 * (12.0)^2 = 288.0.

To calculate the extrapolated reaction rate.

Key Concept

Extrapolation of Quadratic Trends

Question 2Question →

A student investigated the electrical properties of a negative temperature coefficient (NTC) thermistor. The thermistor was placed in a temperature-controlled water bath, and its electrical resistance, $R$ (in kilohms, $\text{k}\Omega$ ), was measured at various temperatures, $T$ (in degrees Celsius, $^\circ\text{C}$ ). The data from this experiment are presented in the table below.

Temperature ( $T$ , $^\circ\text{C}$ )	Resistance ( $R$ , $\text{k}\Omega$ )
10	48.0
30	24.0
50	12.0
70	6.0

Based on the trend shown in the table, what is the predicted electrical resistance, in $\text{k}\Omega$ , of the thermistor at a temperature of $110^\circ\text{C}$ ?

Show answer & explanation

Answer: 1.5

Answer

The predicted electrical resistance of the thermistor at

110^\circ\text{C}

is

1.5\text{ k}\Omega

.

The correct calculation identifies that the resistance decreases by a factor of 2 for every

20^\circ\text{C}

increase in temperature. Following this exponential trend, the resistance at

90^\circ\text{C}

is

3.0\text{ k}\Omega

, and at

110^\circ\text{C}

it is half of that, which equals

1.5\text{ k}\Omega

.

Step-by-Step Solution

1

Analyze the pattern of temperature changes in the table.

The temperature increments are constant at

\Delta T = 20^\circ\text{C}

(e.g.,

30 - 10 = 20^\circ\text{C}

,

50 - 30 = 20^\circ\text{C}

,

70 - 50 = 20^\circ\text{C}

).

Establishing a constant independent variable interval simplifies trend extrapolation.

2

Analyze the corresponding ratio of resistance values at each temperature step.

At each interval, the resistance value is divided by 2:

\frac{24.0}{48.0} = 0.5

,

\frac{12.0}{24.0} = 0.5

, and

\frac{6.0}{12.0} = 0.5

. This indicates a non-linear, exponential decay trend.

Determining the mathematical relationship allows for precise calculation of values outside the dataset range.

3

Extrapolate the trend to

90^\circ\text{C}

by applying the factor of

0.5

to the resistance at

70^\circ\text{C}

.

R(90^\circ\text{C}) = R(70^\circ\text{C}) \times 0.5 = 6.0 \times 0.5 = 3.0\text{ k}\Omega

.

Since

90^\circ\text{C}

is exactly

20^\circ\text{C}

above

70^\circ\text{C}

, the pattern dictates that the resistance halves.

4

Extrapolate the trend further to

110^\circ\text{C}

by applying the factor of

0.5

to the resistance at

90^\circ\text{C}

.

R(110^\circ\text{C}) = R(90^\circ\text{C}) \times 0.5 = 3.0 \times 0.5 = 1.5\text{ k}\Omega

.

Since

110^\circ\text{C}

is exactly

20^\circ\text{C}

above

90^\circ\text{C}

, the resistance halves once more.

Key Concept

Extrapolating non-linear (exponential) relationships by identifying constant ratios over equal intervals of the independent variable.

Question 3Question →

A student measured the speed of sound in a chamber filled with pure carbon dioxide ( $CO_2$ ) gas at various temperatures. The measured speed of sound, in meters per second ( $\text{m/s}$ ), at each temperature, in degrees Celsius ( $^\circ\text{C}$ ), is shown in the table below:

Temperature ( $^\circ\text{C}$ )	Speed of Sound ( $\text{m/s}$ )
$0$	$259$
$20$	$268$
$40$	$277$
$60$	$286$

Assuming the speed of sound continues to change at a constant rate with respect to temperature, what is the predicted speed of sound in $CO_2$ gas, in meters per second ( $\text{m/s}$ ), at a temperature of $100^\circ\text{C}$ ?

Show answer & explanation

Answer: 304

Answer

The predicted speed of sound in carbon dioxide gas at

100^\circ\text{C}

is

304\text{ m/s}

.

The speed of sound increases linearly by

9\text{ m/s}

for every

20^\circ\text{C}

increase in temperature, which is a rate of

0.45\text{ m/s}

per

1^\circ\text{C}

. The target temperature of

100^\circ\text{C}

is

40^\circ\text{C}

higher than the highest data point in the table (

60^\circ\text{C}

). The speed of sound will therefore increase by

40 \times 0.45 = 18\text{ m/s}

beyond the

60^\circ\text{C}

speed. Adding this to

286\text{ m/s}

yields

304\text{ m/s}

.

Step-by-Step Solution

1

Determine the constant rate of change of the speed of sound per

1^\circ\text{C}

temperature increase.

The speed of sound increases at a rate of

0.45\text{ m/s}

per

1^\circ\text{C}

.

This establishes the linear trend shown in the experimental data.

2

Find the temperature interval between the highest measured data point and the target temperature.

The difference is

40^\circ\text{C}

(from

60^\circ\text{C}

to

100^\circ\text{C}

).

This determines how far outside the measured data range the extrapolation must extend.

3

Multiply the temperature interval by the rate of change and add it to the speed of sound at the highest measured temperature.

286\text{ m/s} + (40^\circ\text{C} \times 0.45\text{ m/s/}^\circ\text{C}) = 304\text{ m/s}

.

This completes the linear extrapolation to predict the final value.

Key Concept

Linear extrapolation relies on determining a constant rate of change from the given data points and applying it to a target value outside the experimental range.

Question 4Question →

Aerodynamic drag force ( $F_d$ ) acts on vehicles as they move through the air. A group of students measured the drag force, in newtons ( $\text{N}$ ), acting on a scale model of a sports car in a wind tunnel at various wind velocities ( $v$ ), in meters per second ( $\text{m/s}$ ). The data from their trials are recorded in the table below:

Velocity ( $v$ , $\text{m/s}$ )	Drag Force ( $F_d$ , $\text{N}$ )
$10$	$12$
$20$	$48$
$30$	$108$
$40$	$192$

Based on the trend shown in the table, what is the expected aerodynamic drag force, in newtons ( $\text{N}$ ), acting on the scale model when the wind velocity is $50\text{ m/s}$ ?

Show answer & explanation

Answer: 300

Answer

The expected aerodynamic drag force is 300 N.

The correct answer is 300 N because the drag force scales quadratically with velocity according to the relation

F_d = 0.12 v^2

. Plugging in

v = 50\text{ m/s}

yields

0.12 \times 2500 = 300\text{ N}

.

Step-by-Step Solution

1

Calculate the ratio of drag force to the square of the velocity for the given data points.

For all data points,

F_d / v^2 = 0.12

. This establishes the quadratic trend

F_d = 0.12 v^2

.

Identifying the mathematical relationship between the variables is necessary to accurately extrapolate beyond the measured data range.

2

Substitute the target velocity of

50\text{ m/s}

into the identified quadratic formula.

F_d = 0.12 \times (50)^2 = 300\text{ N}

.

Applying the mathematical trend allows for the calculation of the drag force at the extrapolated velocity.

Key Concept

Extrapolation of a quadratic relationship between velocity and aerodynamic drag force.

Estimated Time:1m 30s

Question 5Question →

Geologists drilled a deep borehole into the Earth's crust at a research site and measured the rock temperature at various depths. The recorded temperatures are shown in the table below:

Depth ( $m$ )	Temperature ( $^\circ\text{C}$ )
$0$	$12.0$
$250$	$19.5$
$500$	$27.0$
$750$	$34.5$
$1,000$	$42.0$

Assuming the temperature continues to increase linearly with depth at the same rate observed between $0\text{ m}$ and $1,000\text{ m}$ , what will the rock temperature, in degrees Celsius ( $^\circ\text{C}$ ), be at a depth of $1,800\text{ m}$ ?

Show answer & explanation

Answer: 66

Answer

The projected rock temperature at a depth of 1,800 m is 66.0°C.

The correct calculation determines that the temperature increases by 7.5°C for every 250 m (a rate of 0.03°C/m). Extrapolating linearly to 1,800 m, the temperature increases by 54.0°C (0.03°C/m * 1,800 m) from the surface baseline of 12.0°C, yielding a final temperature of 66.0°C.

Step-by-Step Solution

1

Calculate the rate of temperature change per meter of depth.

0.03^\circ\text{C/m}

Using the interval from

0\text{ m}

to

250\text{ m}

, the temperature increases by

19.5^\circ\text{C} - 12.0^\circ\text{C} = 7.5^\circ\text{C}

. The rate of change is

\frac{7.5^\circ\text{C}}{250\text{ m}} = 0.03^\circ\text{C/m}

.

2

Calculate the total temperature change over the depth interval of 1,800 m.

54.0^\circ\text{C}

Multiplying the constant rate of temperature change (

0.03^\circ\text{C/m}

) by the target depth (

1,800\text{ m}

) yields the total increase in temperature from the surface:

0.03 \times 1,800 = 54.0^\circ\text{C}

.

3

Determine the final temperature by adding the increase to the baseline surface temperature.

66.0^\circ\text{C}

Adding the

54.0^\circ\text{C}

increase to the baseline temperature at the surface (

12.0^\circ\text{C}

) gives the projected temperature at depth:

12.0^\circ\text{C} + 54.0^\circ\text{C} = 66.0^\circ\text{C}

.

Key Concept

Extrapolation of linear data trends

Estimated Time:1m 30s

Extrapolation and Trend Prediction

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Common Mistakes

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Common Mistakes