SAT MAth Practice questions – all topics
- Problem-solving and Data Analysis Weightage: 15% Questions: 5-7
- Ratios, rates, proportional relationships, and units
- Percentages
- One-variable data: distributions and measures of centre and spread
- Two-variable data: models and scatterplots
- Probability and conditional probability
- Inference from sample statistics and margin of error
- Evaluating statistical claims: observational studies and Experiments
SAT MAth and English – full syllabus practice tests
Question Medium
The number of southern white rhinos was 3,800 in 1984 . Due to conservation methods over time, the number of southern white rhinos increased to 20,405 by 2012 . The scatterplot shows the relationship between time, in number of years since 1984, and the number of southern white rhinos. A line of best fit for the data is also shown.
Which value is closest to the slope of the line of best fit shown?
A) \(-2,500\)
B) -650
C) 650
D) 2,500
▶️Answer/Explanation
Ans:C
The line of best fit is represented in the graph, and we will calculate its slope using two points on the line.
At \(x = 0\) (the year 1984), the number of rhinos is approximately \(3,800\).
At \(x = 28\) (28 years after 1984, which is 2012), the number of rhinos is approximately \(20,405\).
The slope \(m\) of the line is calculated as follows:
\[
m = \frac{\Delta y}{\Delta x} = \frac{y_2 – y_1}{x_2 – x_1}
\]
Using the chosen points \((0, 3800)\) and \((28, 20,405)\):
\[
m = \frac{20,405 – 3800}{28 – 0} = \frac{16200}{30} = 593
\]
Therefore, the slope is closest to:C) 650
Question medium
The given function \(C\) models the annual soybean use in China, in millions of metric tons, between 1995 and 2014, where \(x\) is the number of years after 1995.
$
C(x)=4.3 x+19
$
According to the model, what is the best interpretation of 4.3 in this context?
A. Each year between 1995 and 2014, China used 4.3 million metric tons of soybeans.
B. Each year between 1995 and 2014, China’s annual use of soybeans increased by 4.3 million metric tons.
C. China used 4.3 million metric tons of soybeans in 1995.
D. China used a total of 4.3 million metric tons of soybeans between 1995 and 2014.
▶️Answer/Explanation
Ans:B
The given function \( C(x) = 4.3x + 19 \) models the annual soybean use in China, in millions of metric tons, where \( x \) is the number of years after 1995. To interpret the coefficient \( 4.3 \), let’s consider the structure of the linear function:
\[ C(x) = 4.3x + 19 \]
Here:
\( C(x) \) represents the annual soybean use in millions of metric tons.
\( x \) represents the number of years after 1995.
In a linear function of the form \( y = mx + b \):
\( m \) is the slope of the line.
\( b \) is the y-intercept.
The slope \( 4.3 \) indicates the rate of change of the annual soybean use per year. Specifically, it means that for each additional year after 1995, the annual soybean use in China increases by \( 4.3 \) million metric tons.
Therefore, the best interpretation of the coefficient \( 4.3 \) in this context is:
The annual soybean use in China increases by 4.3 million metric tons each year after 1995.
Question medium
𝑓(𝑡) = 0.17𝑡 + 2.54
The given function f models the annual worldwide production of avocados, in millions of metric tons, t years after 2000. According to the function, by how many millions of metric tons did the annual worldwide production of avocados increase from 2010 to 2011?
A. 0.17
B. 2.54
C. 2.71
D. 4.24
▶️Answer/Explanation
Ans: A
We need to find the difference in avocado production between 2010 and 2011, which corresponds to a difference of 1 year in \( t \). We’ll evaluate the function \( f(t) \) at \( t = 11 \) and \( t = 10 \), then find the difference.
\( f(11) = 0.17(11) + 2.54 = 1.87 + 2.54 = 4.41 \) million metric tons (2011)
\( f(10) = 0.17(10) + 2.54 = 1.7 + 2.54 = 4.24 \) million metric tons (2010)
Therefore, the increase from 2010 to 2011 is \( 4.41 – 4.24 = 0.17 \) million metric tons. So, the answer is A. \(0.17\).
Question Medium
Each data point on the scatterplot gives the height x, in inches. and weighty, in pounds, for a llama in a sample of 10 llamas. A line of best fit is also shown.
Which of the following best approximates the equation for the line of best flt shown ?
A) y = -706 + 15.3x
B) y = -15.3 + 706x
C) y = 15.3 – 706x
D) y = 706 – 15.3x
▶️Answer/Explanation
A) y = -706 + 15.3x
From the scatter plot, we can observe that the line has a positive slope, indicating a direct relationship between height (x) and weight (y).
To approximate the slope, We can choose any two distinct points on the line and calculate the rise over the run (change in y over change in x).
Let’s use the points (60, around 200) and (70, around 360).
Rise = Change in $y = 360 – 200 = 160$ , Run = Change in $x = 70 – 60 = 10$
$Slope =\frac{ Rise}{Run} = \frac{160}{10} ≈ 16$
Using the point $(60, 200)$ and the calculated slope of $16$:
$200 ≈ 16(60) + b$
$200 ≈ 960 + b$
$b ≈ -760$
Therefore, the equation of the line can be approximated as:
$y ≈ 16x – 760$
Question Medium
A procedure allows a researcher to determine the concentration of glucose \(y\), in micrograms per milliliter \((\mu \mathrm{g} / \mathrm{mL})\), in a soil sample by measuring the absorbance, \(x\), at a specific wavelength of light. The scatterplot shows this relationship for 5 soil samples.
Which equation is the most appropriate linear model for the data?
A) \(y=1.5+90 x\)
B) \(y=1.5+10 x\)
C) \(y=10+1.5 x\)
D) \(y=90+1.5 x\)
▶️Answer/Explanation
A
The slope \(m\) is given by:
\[
m=\frac{y_2-y_1}{x_2-x_1}
\]
Substitute the given points \(\left(x_1, y_1\right)=(0.09,9)\) and \(\left(x_2, y_2\right)=(0.19,18)\) :
\[
m=\frac{18-9}{0.19-0.09}=\frac{9}{0.10}=90
\]
Using the slope-intercept form \(y=m x+b\), we can substitute one of the points to solve for \(b\). Let’s use \((0.09,9)\) :
\[
\begin{aligned}
& 9=90 \cdot 0.09+b \\
& 9=8.1+b \\
& b=9-8.1 \\
& b=0.9
\end{aligned}
\]
The equation of the line is:
\[
y=90 x+0.9
\]