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
[Calc] Question Hard
The population of a city in 2000 was 2.6 times its population in 1999. The population of this city increased by \(p \%\) from 1999 to 2000 . What is the value of \(p\) ?
▶️Answer/Explanation
Ans:160
Let’s denote the population of the city in 1999 as \(P_{1999}\) and the population in 2000 as \(P_{2000}\).
According to the given information, the population of the city in 2000 was 2.6 times its population in 1999. Mathematically, we can express this as:
\[ P_{2000} = 2.6 \times P_{1999} \]
We are also given that the population of the city increased by \(p \%\) from 1999 to 2000. This means the increase in population from 1999 to 2000 is \(p \%\) of the population in 1999, which can be represented as:
\[ \text{Increase} = \frac{p}{100} \times P_{1999} \]
So, the population in 2000 is the sum of the population in 1999 and the increase:
\[ P_{2000} = P_{1999} + \text{Increase} \]
Substituting the given expressions for \(P_{2000}\) and the increase, we get:
\[ 2.6 \times P_{1999} = P_{1999} + \frac{p}{100} \times P_{1999} \]
Now, let’s solve this equation for \(p\):
\[ 2.6 = 1 + \frac{p}{100} \]
Subtracting \(1\) from both sides:
\[ 2.6 – 1 = \frac{p}{100} \]
\[ 1.6 = \frac{p}{100} \]
Multiplying both sides by \(100\):
\[ 160 = p \]
So, the value of \(p\) is \( \boxed{160} \).
[Calc] Question Hard
The box plots summarize data set \(X\) and data set \(Y\). Each of the data sets consists of a total of 100 integers. Which of the following statements must be true?
1) The mean of data set \(X\) is less than the mean of data set \(Y\).
2) The median of data set \(X\) is less than the median of data set \(Y\).
A. I only
B. II only
C. I and II
D. Neither I nor II
▶️Answer/Explanation
Ans:B
To determine which statements must be true based on the given box plots for data set \( X \) and data set \( Y \), let’s analyze the box plots carefully.
Analysis of the Box Plots:
1. Median:
The median is represented by the line inside the box of the box plot.
For data set \( X \), the median appears to be around 5.
For data set \( Y \), the median appears to be around 6.
2. Mean:
The mean is not directly visible from the box plot, but we can make inferences based on the symmetry of the box plot.
For data set \( X \), the box plot looks fairly symmetric around the median.
For data set \( Y \), the box plot is slightly skewed to the left (since the left whisker is shorter than the right whisker), indicating the mean might be slightly less than the median but not necessarily conclusive from the plot alone.
Evaluating Statements:
1. The mean of data set \( X \) is less than the mean of data set \( Y \):
This statement cannot be conclusively determined from the box plots alone as we need the exact data values to calculate the means.
2. The median of data set \( X \) is less than the median of data set \( Y \):
This statement is true based on the medians shown in the box plots. The median for data set \( X \) is around 5, while the median for data set \( Y \) is around 6.
Conclusion:
The second statement about the medians is true based on the box plots.
The first statement about the means cannot be conclusively determined from the box plots.
Therefore, the correct answer is:
B. II only
[Calc] Question Hard
The table gives the typical adult weight ranges and life spans for African and Asian elephants in the wild.
Based on the table, what is the typical minimum weight of an adult African elephant in the wild, in pounds? (1 ton = 2000 pounds)
▶️Answer/Explanation
Ans: 5000
To find the typical minimum weight of an adult African elephant in the wild in pounds, we’ll convert the weight from tons to pounds. Given that 1 ton is equal to 2000 pounds, we can multiply the minimum weight of an African elephant (2.50 tons) by 2000:
\[2.50 \text{ tons} \times 2000 \text{ pounds/ton} = 5000 \text{ pounds}\]
So, the typical minimum weight of an adult African elephant in the wild is \(\mathbf{5000}\) pounds
[Calc] Question Hard
The graph shows the power capacity for the nine countries that had the greatest geothermal power capacity in 2010. What was the capacity, in megawatts, of the country that had the median capacity of the nine countries?
▶️Answer/Explanation
Ans: 863
First geothermal power capacity of the nine countries, list these values in ascending order:
- El Salvador: 204 MW
- Japan: 536 MW
- Iceland: 575 MW
- New Zealand: 628 MW
- Italy: 863 MW
- Mexico: 958 MW
- Indonesia: 1197 MW
- Philippines: 1904 MW
- United States: 3087 MW
With nine data points, the median is the fifth value in the ordered list:
- Median capacity = Italy: 863 MW
Thus, the geothermal power capacity of the country with the median capacity is 863 megawatts.
Questions
The line graph above shows the population, in thousands, of people living in Alaska every 10 years from 1900 to 2000.
What was the population of Alaska, in thousands, in 1990?
▶️Answer/Explanation
Ans: 550
Questions
An ecologist selected a random sample of 30 prairie dogs from a colony and found that the mean mass of the prairie dogs in the sample was 0.94 kilograms (kg) with an associated margin of error of $0.12 \mathrm{~kg}$. Which of the following is the best interpretation of the ecologist’s findings?
A. All prairie dogs in the sample have a mass between $0.82 \mathrm{~kg}$ and $1.06 \mathrm{~kg}$.
B. Most prairie dogs in the colony have a mass between $0.82 \mathrm{~kg}$ and $1.06 \mathrm{~kg}$.
C. Any mass between $0.82 \mathrm{~kg}$ and $1.06 \mathrm{~kg}$ is a plausible value for the mean mass of the prairie dogs in the sample.
D. Any mass between $0.82 \mathrm{~kg}$ and $1.06 \mathrm{~kg}$ is a plausible value for the mean mass of the prairie dogs in the colony.
▶️Answer/Explanation
Ans: D
Question
The table above shows the distribution of United States presidents according to the century and the region of the country in which they were born. Based on the information in the table, what fraction of presidents who were not born in the nineteenth century were born in the South?
▶️Answer/Explanation
Ans: 1/2, .5