## 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**** **medium

A circle has been divided into three nonoverlapping regions: I, II, and III. The area of region I is \(4\pi \) square centimeters (\(cm^{2}\)), the area of region II is \(12\pi \) \(cm^{2}\), and the area of region III is \(16\pi \) \(cm^{2}\). If a point in the circle is selected at random, what is the probability of selecting a point that does not lie in region II? (Express your answer as a decimal or fraction, not as a percent.)

**▶️Answer/Explanation**

Ans: 5/8, .625

To find the probability of selecting a point that does not lie in region II, we need to find the total area of the circle and subtract the area of region II, then divide by the total area of the circle.

Given:

Area of region I = \(4\pi \text{ cm}^2\)

Area of region II = \(12\pi \text{ cm}^2\)

Area of region III = \(16\pi \text{ cm}^2\)

The total area of the circle is the sum of the areas of all three regions:

\[4\pi + 12\pi + 16\pi = 32\pi \text{ cm}^2\]

So, the probability of selecting a point that does not lie in region II is:

\[\frac{4\pi + 16\pi}{32\pi} = \frac{20\pi}{32\pi} = \frac{5}{8}\]

Therefore, the probability is \(\frac{5}{8}\).

**[Calc]**** ****Question**** Medium**

As a literature major in college, Sean has read books written by a variety of European authors. The table above shows the numbers of books written by British, French, and German authors that Sean has read, categorized by the century in which the books were written.

If a book referred to by the table that was written in the twentieth century is to be selected at random, the probability that the book was written by a British author is 15/n Which of the following best describes n in this context?

A. The total number of books referred to by the table

B. The number of books referred to by the table that were written in the twentieth century

C. The number of books referred to by the table that were written by British authors

D. The number of books referred to by the table that were written by either French authors or German authors

**▶️Answer/Explanation**

Ans: B

In this context, \( n \) represents the number of books referred to by the table that were written in the twentieth century.

The probability of selecting a book written by a British author from those written in the twentieth century is given as \( \frac{15}{n} \).

So, the correct answer is:

\(\boxed{\text{B) The number of books referred to by the table that were written in the twentieth century}}\).

**[Calc]**** ****Question** ** **medium

A forest contains different species of trees. Let *t *represent the total number of trees in the forest, let *h *represent the number of hickory trees, and let *k *represent the number of oak trees. If a tree is selected at random from the forest,which expression represents the probability of selecting a tree that is neither hickory nor oak?

A) \(\frac{h+k}{t}\)

B) \(\frac{t-h-k}{t}\)

C) \(\frac{h+k-t}{t}\)

D) \(\frac{t+h+k}{t}\)

**▶️Answer/Explanation**

**B) \(\frac{t-h-k}{t}\)**

We’re looking for the probability of selecting a tree that is neither hickory nor oak. This means we want the total number of trees that are neither hickory nor oak divided by the total number of trees.

The total number of trees that are neither hickory nor oak is \(t – (h + k)\), as \(h\) represents the number of hickory trees and \(k\) represents the number of oak trees. The total number of trees is \(t\).

So, the probability expression is:

\[

\frac{t – (h + k)}{t}

\]

Simplifying:

\[

\frac{t – h – k}{t}

\]

Thus, the correct answer is B) \(\frac{t – h – k}{t}\).

**[Calc]**** ***Questions ***Medium**

In a survey of 240 television viewers, 3/5 indicated that they like comedies, some indicated that they do not like comedies, and the rest did not respond. If one of the 240 viewers is selected at random, the probability is 1/15 that the viewer selected did not respond. How many of the 240 viewers indicated that they do not like comedies?

**▶️Answer/Explanation**

Ans: 80

In the survey, \(\frac{3}{5}\) of the 240 television viewers indicated that they like comedies. So, the number of viewers who like comedies is \(\frac{3}{5} \times 240 = 144\).

The probability that the viewer selected did not respond is \(\frac{1}{15}\) of the total viewers, which is 240.

The number of viewers who indicated that they do not like comedies can be calculated by subtracting the number of viewers who like comedies and the number of viewers who did not respond from the total number of viewers:

\[240 – 144 – \frac{1}{15} \times 240 = 240 – 144 – 16 = 80\]

Therefore, \(80\) of the \(240\) viewers indicated that they do not like comedies.

**[Calc]**** ****Question**** ****Medium**

The figure shown is divided into 100 squares of equal area, where 60 squares are shaded.

If one of these squares is selected at random how much greater is the probability of selecting a shaded square than the probability of selecting a square that is not shaded ?

A) 0.20

8) 0.40

C) 0,60

D) 0.80

**▶️Answer/Explanation**

A) 0.20

Probability of selecting a shaded square = $\frac{\text{Number of shaded squares }}{ \text{Total number of squares Probability of selecting a shaded square}}= \frac{60}{ 100 }= 0.6$

Probability of selecting an unshaded square =$\frac{\text{ Number of unshaded squares} }{\text{ Total number of squares Probability of selecting an unshaded square}}= \frac{40}{ 100 }= 0.4$

Difference between the probabilities $\text{= Probability of selecting a shaded square – Probability of selecting an unshaded square Difference between the probabilities }= 0.6 – 0.4 = 0.2$

Therefore, the probability of selecting a shaded square is 0.2 greater than the probability of selecting an unshaded square.

The correct answer is A) 0.20.

**[Calc]**** ****Question**** **** Medium**

The table shows the results of a poll that was used to determine support for a county proposal. The results are categorized by county and opinion. If one person who responded to the poll is selected at random, which of the following statements results in the greatest value?

A) The probability that the person is undecided, given that the person is from County 1

B) The probability that the person is undecided, given that the person is from County 2

C) The probability that the person is from County 1, given that the person is undecided

D) The probability that the person is from County 2, given that the person is undecided

**▶️Answer/Explanation**

D

**[Calc]**** ****Question**** **** Medium**

A county school board in a certain state is proposing a start time of 7:30 a.m. for all high schools in the county. A sample of 100 high school students was selected at random from all high school students in the county. The selected students were asked whether they approved of the proposed change in the school start time, and 70 students responded that they did not approve of the proposed change. Which of the following is the largest population to which the results of the survey can be generalized?

A) The 70 students who responded that they did not approve of the proposed change

B) The 100 students who were surveyed

C) All high school students in the county

D) All high school students in the state

**▶️Answer/Explanation**

C

**[Calc]**** ****Question**** **** Medium**

The bar graph shown summarizes the total number of businesses, in thousands, and the total number of tourismrelated businesses, in thousands, in Iceland for each of 7 years.

Tourism-related businesses employed 13.5 thousand people in Iceland in 2009. Which of the following is closest to the mean number of employees per tourism-related business in Iceland in 2009?

A) 5

B) 14

C) 23

D) 58

**▶️Answer/Explanation**

C

**[Calc]**** ****Question**** Medium**

The two-way table categorizes the change in value in July and August for 50 stocks. If one of the stocks that increased in value in August is chosen at random, what is the probability that the stock also increased in value in July?

A) 0.42

B) 0.60

C) 0.70

D) 0.84

**▶️Answer/Explanation**

C

*Question*

A one-digit positive integer will be chosen at random. What is the probability that the chosen integer will be greater than 7? (Express your answer as a fraction or decimal, not as a percent.)

**▶️Answer/Explanation**

2/9, .222

*Question*

According to the 2010 Census, the adult population aged 18 or greater of the United States in 2010 was 234,564,071. In 2010, a survey was conducted among a randomly chosen sample of adults aged 18 or greater in the United States about their preference to live in a warm climate or a cool climate. The table below displays a summary of the survey results.

Based on the data, which of the following is closest to the probability that a randomly selected adult who is 18-35 years old prefers to live in a cool climate?

- 0.11
- 0.30
- 0.33
- 0.49
**▶️Answer/Explanation**Ans: C

*Question*

According to the 2010 Census, the adult population aged 18 or greater of the United States in 2010 was 234,564,071. In 2010, a survey was conducted among a randomly chosen sample of adults aged 18 or greater in the United States about their preference to live in a warm climate or a cool climate. The table below displays a summary of the survey results.

Which of the following is closest to the difference between the percentage of adults aged 18-50 years who responded “warm” and the percentage of adults aged 51 years or greater who responded “warm”?

- 4%
- 5%
- 10%
- 18%
**▶️Answer/Explanation**Ans: A

*Questions *

A park ranger asked a random sample of visitors how far they hiked during their visit. Based on the responses, the estimated mean was found to be 4.5 miles, with an associated margin of error of 0.5 miles. Which of the following is the best conclusion from these data?

A. It is likely that all visitors hiked between 4 and 5 miles.

B. It is likely that most visitors hiked exactly 4.5 miles.

C. It is not possible that any visitor hiked less than 3 miles.

D. It is plausible that the mean distance hiked for all visitors is between 4 and 5 miles.

**▶️Answer/Explanation**

Ans: D

*Questions *

Sierra recorded the gender and eye color of all the students in her biology class. The results are shown in the table above. If a male student is selected at random from Sierra’s biology class, what is the probability that he will have brown eyes?

- \(\frac{2}{3}\)
- \(\frac{2}{5}\)
- \(\frac{3}{7}\)
- \(\frac{3}{13}\)
**▶️Answer/Explanation**Ans: B

*Questions *

Of all juniors and seniors who attended a particular high school during the 2014-2015 school year, 149 participated in the clubs listed in the table above. Each of the 149 students participated in only one of the four school clubs listed. The table shows the distribution of the 149 students by class and club participation.

Of the number of juniors and seniors in the drama 1 club, the 25% who walked to school represents \(\frac{1}{8}\) of the total number of juniors and seniors who walked to school. How many juniors and seniors walked to school? 3.2

- 96
- 60
- 24
- 12

**▶️Answer/Explanation**Ans: A