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
For the elements in the periodic table having atomic numbers 1 though 18, the bar graph summarizes the number of elements by their number of valence electrons. How many of these elements have at most 4 valence electrons?
▶️Answer/Explanation
Ans:10
To find the elements with at most 4 valence electrons, we count the elements with 1, 2, 3, or 4 valence electrons:
- 1 valence electron: Hydrogen (H), Lithium (Li), Sodium (Na)
- 2 valence electrons: Helium (He), Beryllium (Be), Magnesium (Mg)
- 3 valence electrons: Boron (B), Aluminum (Al)
- 4 valence electrons: Carbon (C), Silicon (Si)
Summing these:
- Elements with 1 valence electron: 3 (H, Li, Na)
- Elements with 2 valence electrons: 3 (He, Be, Mg)
- Elements with 3 valence electrons: 2 (B, Al)
- Elements with 4 valence electrons: 2 (C, Si)
Adding these counts together, we get:
\[
3+3+2+2=10
\]
Therefore, 10 elements have at most 4 valence electrons.
Question medium
A state representative wants to increase the amount of driving practice time required before a student can earn a driver’s license. The representative surveyed a random sample of 50 from the 250 students taking the driver’s education class at Jefferson High School.
The survey reported that 62% of students taking the class agree that driving practice time should be increased, with an associated margin of error of 5%.
Which of the following populations can the results of the survey at Jefferson High School be extended to?
A. Students at any US high school
B. Students taking a driver’s education class at any US high school
C. Students at Jefferson High School
D. Students taking a driver’s education class at Jefferson High School
▶️Answer/Explanation
Ans: D
Given information:
- The survey was conducted on a random sample of 50 students from the population of 250 students taking the driver’s education class at Jefferson High School.
Based on this information, the results of the survey can only be reliably extended to the population from which the sample was drawn, which is:D.
The survey results cannot be generalized to other populations, such as:
A. Students at any US high school (because the sample was specific to Jefferson High School) B. Students taking a driver’s education class at any US high school (because the sample was specific to Jefferson High School) C. Students at Jefferson High School (because the sample was specific to the driver’s education class)
To extend the results to a larger population, such as students taking driver’s education classes at other high schools or all high school students, a different sampling method that ensures representativeness of the larger population would be required.
Question medium
In 1980, the world population was 4.44 billion people. A model created at the time predicted that this population would grow at a rate of 0.076 billion people each year after 1980. In 2015, the world population was 7.35 billion people. What is the positive difference, in billions of people, between the actual world population in 2015 and the model’s predicted world population in 2015?
▶️Answer/Explanation
Ans: 1/4, .25
To find the model’s predicted world population in 2015, we need to use the model that predicts the population growth rate.
Given:
World population in 1980: \(4.44\) billion people
Population growth rate: \(0.076\) billion people each year after 1980
calculate the predicted population in 2015 using the model:
\[ \text{Predicted population in 2015} = 4.44 + (2015 – 1980) \times 0.076 \]
\[ = 4.44 + 35 \times 0.076 \]
\[ = 4.44 + 2.66 \]
\[ = 7.10 \text{ billion people} \]
The actual world population in 2015 was \(7.35\) billion people.
Now, difference between the actual and predicted populations in 2015:
\[ \text{Difference} = \text{Actual population} – \text{Predicted population} \]
\[ = 7.35 – 7.10 \]
\[ = 0.25 \text{ billion people} \]
So, the positive difference between the actual world population in 2015 and the model’s predicted world population in 2015 is \(0.25\) billion people.
Question Medium
The dot plot shows the estimated market values for 15 houses in a neighborhood. What is the maximum estimated market value, in thousands of dollars, for the data set?
▶️Answer/Explanation
Ans: 178
The dot plot shows the estimated market values for 15 houses in a neighborhood. To determine the maximum estimated market value, we need to identify the highest value on the horizontal axis that has at least one dot above it.
From the provided dot plot, the horizontal axis is labeled with market values in thousands of dollars, ranging from 168 to 178. The maximum estimated market value is represented by the rightmost dot on the plot.
The highest value on the plot where dots are present is at 178 thousand dollars.
Therefore, the maximum estimated market value for the data set is 178 thousand dollars.
Question Medium
The bar graph shows the number of discovered near-Earth asteroids, by classification, as of April 2017 Of the near-Earth asteroids, how many more are classified as medium, large, or very large than are classified as very small or small?
▶️Answer/Explanation
Ans:1422
From the bar graph, the numbers are:
Very Small: 2911
Small: 4406
Medium: 4055
Large: 3808
Very Large: 876
First, sum the numbers for medium, large, and very large asteroids:
\[ 4055 + 3808 + 876 = 8739 \]
Next, sum the numbers for very small and small asteroids:
\[ 2911 + 4406 = 7317 \]
Now, find the difference between these two sums:
\[ 8739 – 7317 = 1422 \]
Therefore, there are 1422 more near-Earth asteroids classified as medium, large, or very large than are classified as very small or small.