Digital SAT Math Practice Questions

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

Oocytes are a type of cell that can be modeled as a sphere. The table shows the surface area, in square micrometers $\left(\mu \mathrm{m}^2\right)$, and volume, in cubic micrometers $\left(\mu \mathrm{m}^3\right)$, based on the average radius for oocytes at the same stage of development in four types of mammals.
(The surface area of a sphere with a radius of $r$ is $4 \pi r^2$, and the volume of a sphere with a radius of $r$ is equal to $\frac{4}{3} \pi r^3$.)

The volume of a mouse oocyte is approximately what percent of the volume of a human oocyte?
A) $4.29 \%$
B) $8.16 \%$
C) $12.25 \%$
D) $23.32 \%$

▶️Answer/Explanation

Ans:A

To determine the volume of a mouse oocyte as a percentage of the volume of a human oocyte, we use the given volumes for both:

Volume of mouse oocyte: $1,047.4 \, \mu \text{m}^3$
Volume of human oocyte: $24,429 \, \mu \text{m}^3$

We calculate the percentage as follows:

\[
\text{Percentage} = \left(\frac{\text{Volume of mouse oocyte}}{\text{Volume of human oocyte}}\right) \times 100
\]

Substitute the given volumes into the formula:

\[
\text{Percentage} = \left(\frac{1,047.4}{24,429}\right) \times 100 \approx 4.29\%
\]

[Calc] Question Medium

In 1976, there were approximately 1,000 gray
wolves in northern Minnesota. The number of gray wolves in northern Minnesota in 2008 was $190 \%$ greater than in 1976. Approximately how many gray wolves were in northern Minnesota in 2008?

A. 1,190
B. 1,900
C. 2,900
D. 19,000

▶️Answer/Explanation

Ans:C

To find the number of gray wolves in northern Minnesota in 2008, we need to calculate $190\%$ of the number of wolves in 1976 and then add that to the number in 1976.

Let $x$ be the number of gray wolves in 1976.

$190\%$ of $x$ is $1.9x$.

So, the total number of gray wolves in 2008 is $x + 1.9x = 2.9x$.

Given that there were approximately 1,000 gray wolves in northern Minnesota in 1976, we can set $x = 1000$.

$2.9 \times 1000 = 2900$

Therefore, approximately 2,900 gray wolves were in northern Minnesota in 2008. So, the correct answer is option C: 2,900.

[Calc] Question medium

Kai went on a bike ride. During the first mile of his bike ride, he rode at a constant speed, s. After the first mile of his ride, Kai increased his speed by 100%. Which of the following expressions represents the speed Kai rode his bike after the first mile, in terms of s ?
A. s
B. s + 1
C. 1.5s
D. 2s

▶️Answer/Explanation

Ans: D

After the first mile of his ride, Kai increased his speed by $100\%$, which means his speed doubled. Therefore, the expression representing the speed Kai rode his bike after the first mile, in terms of $s$, would be $2s$. So, the correct answer is option D, $2s$.

[Calc] Question Medium

What is the result of increasing 300 by $200 \%$ ?

▶️Answer/Explanation

Ans: 900

To increase 300 by $200\%$, we multiply 300 by 2 (which represents a $200\%$ increase), then add the result to the original value.

\[300 + 200\% \times 300 = 300 + 2 \times 300 = 300 + 600 = 900\]

So, increasing 300 by $200\%$ results in $\boxed{900}$.

[Calc] Question Medium

Joe was asked to memorize a list of 200 vocabulary words, and he was assessed on his memorization of the words over 3 days. On day 1, he remembered all 200 words. On each of the next two days, Joe remembered $10 \%$ fewer words than he did the preceding day. How many words did Joe remember on day 3 ?
A) 160
B) 162
C) 172
D) 180

▶️Answer/Explanation

Joe was asked to memorize a list of 200 vocabulary words. On day 1, he remembered all 200 words. On each of the next two days, Joe remembered $10\%$ fewer words than he did the preceding day.

On day 1:
\[ 200 \text{ words} \]

On day 2:
\[ 200 – 0.1 \times 200 = 200 \times 0.9 = 180 \text{ words} \]

On day 3:
\[ 180 – 0.1 \times 180 = 180 \times 0.9 = 162 \text{ words} \]

So the answer is:
\[ \boxed{B} \]

[Calc] Question Medium

In 2015, a certain country had an adult population of 250 million people, of which 160 million were internet users and 90 million were not internet users. Of the adult population that used the internet, 52.8 million people had accessed two or more social media websites.

The adult population of this country in 2015 was $77 \%$ of the total population. Which of the following was the approximate total population of this country in 2015 ?
A) 140 million
B) 190 million
C) 320 million
D) 440 million

▶️Answer/Explanation

Adult population in 2015: 250 million people.
Adult internet users: 160 million people.
Adult non-internet users: 90 million people.
Adult internet users accessing two or more social media websites: 52.8 million people.
Adult population is $77\%$ of the total population.

To find the total population of the country in 2015, we need to find $100\%$ of the adult population and then convert it back to the total population.

\[ \text{Adult population} = 250 \text{ million} \]

\[ \text{Total population} = \frac{\text{Adult population}}{0.77} = \frac{250}{0.77} \]

Calculating:
\[ \text{Total population} \approx \frac{250}{0.77} \approx 324.68 \]

Approximately, the total population was $ \approx 325 $ million people.

Therefore, the answer is:
\[ \boxed{C) \, 320 \text{ million}} \]

[Calc] Question Medium

Researchers estimated that $0.07 \%$, by mass, of a 12 gram sample of an orchid plant consists of the fatty acid eicosadienoic acid. Based on this estimate, what is the mass of eicosadienoic acid, in grams, in this orchid sample?
A) 0.0084
B) 0.084
C) 0.84
D) 8.4

▶️Answer/Explanation

$0.07\%$ of a 12 gram sample of an orchid plant consists of eicosadienoic acid.

To find the mass of eicosadienoic acid in the orchid sample, we need to calculate $0.07\%$ of 12 grams.

\[ \text{Mass of eicosadienoic acid} = 0.07\% \times 12 \, \text{grams} \]

First, let’s convert $0.07\%$ to a decimal:
\[ 0.07\% = \frac{0.07}{100} = 0.0007 \]

Now, multiply by 12 grams:
\[ \text{Mass of eicosadienoic acid} = 0.0007 \times 12 = 0.0084 \, \text{grams} \]

Therefore, the mass of eicosadienoic acid in the orchid sample is $ \boxed{0.0084} $ grams.

So the answer is:
\[ \boxed{A) \, 0.0084} \]

[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.

Sean also read d books that were written in the twentieth century by European authors other than those in the table. The number of books referred to by the table that were written by British authors in the twentieth century is approximately 39% of d. Of the following, which is closest to the value of d ?
A. 6
B. 15
C. 38
D. 46

▶️Answer/Explanation

Ans: C

Let’s denote $ d $ as the number of books written in the twentieth century by European authors other than those listed in the table.

According to the information given, the number of books written by British authors in the twentieth century is approximately $ 39\% $ of $ d $.

\[ \frac{39}{100}d \approx 15 \]

To find the value of $ d $, we can solve for $ d $:

\[ \frac{39}{100}d \approx 15 \]

\[ d \approx \frac{15 \times 100}{39} \]

\[ d \approx 38.46 \]

The closest integer to $ 38.46 $ is $ \boxed{\text{C. } 38} $.

[Calc] Question medium

The height of a certain tree in 2016 was 1.35 times the height of the tree in 2011. By what percentage did the height of the tree increase from 2011 to 2016?

A) 3.5%

B)35.2%

C) 35%

D) 135%

▶️Answer/Explanation

C) 35%

To find the percentage increase in the height of the tree from 2011 to 2016, we’ll use the ratio of the heights in 2016 and 2011.

Given that the height of the tree in 2016 was $1.35$ times the height in 2011, we can express this as:
\[ \text{Height in 2016} = 1.35 \times \text{Height in 2011} \]

Now, let’s calculate the percentage increase:

Calculate the difference in heights:
\[ \text{Difference} = \text{Height in 2016} – \text{Height in 2011} = 1.35 \times \text{Height in 2011} – \text{Height in 2011} \]
\[ \text{Difference} = 0.35 \times \text{Height in 2011} \]

Calculate the percentage increase:
\[ \text{Percentage Increase} = \frac{\text{Difference}}{\text{Height in 2011}} \times 100 \]
\[ \text{Percentage Increase} = \frac{0.35 \times \text{Height in 2011}}{\text{Height in 2011}} \times 100 \]
\[ \text{Percentage Increase} = 35\% \]

So, the height of the tree increased by $35\%$ from 2011 to 2016.

[Calc] Question Medium

A certificate of deposit (CD) is an investment account in which money is deposited for a specific amount of time, called the term. The investment earns a guaranteed yearly interest during the term. The table shows the annual percentage yields (APY) for CDs with a term of 18 months and the total interest earned on an initial deposit of $2,000 at four different
banks. Interest is calculated on the total balance of the account and added to the account after each day.

The total interest earned in the l8-month term at Bank B is what percentage greater than the total interest earned in the 18-month term at Bank C?
A) 1.4%
B) 10.7%
C) 26.1%
D) 35.3%

▶️Answer/Explanation

D) 35.3%

The total interest earned in the 18-month term at Bank B is \$40.91, and at Bank C it is \$30.23. We need to find what percentage greater the total interest earned at Bank B is compared to Bank C.

1. difference in total interest earned:
\[
\text{Difference} = \$40.91 – \$30.23 = \$10.68
\]

2. percentage increase:
\[
\text{Percentage increase} = \left( \frac{\text{Difference}}{\text{Interest at Bank C}} \right) \times 100 = \left( \frac{\$10.68}{\$30.23} \right) \times 100
\]
\[
\text{Percentage increase} \approx 35.3\%
\]

Thus, the total interest earned in the 18-month term at Bank B is approximately 35.3% greater than the total interest earned at Bank C.

The answer is:
\[ \boxed{35.3\%} \]

[Calc] Question medium

At the beginning of the day, there were 500 items for sale in a store. The number of items for sale at the end of the day was r% less than the number at the beginning of the day. Which expression represents the number of items for sale at the end of the day?

A) $(\frac{100-r}{100}) 500$

B) $(\frac{100+r}{100}) 500$

C) $(\frac{r}{100}) 500$

D) (100 − r)500

▶️Answer/Explanation

A) $(\frac{100-r}{100}) 500$

If the number of items for sale at the end of the day is $r\%$ less than the number at the beginning of the day, we can express the number of items for sale at the end of the day as a percentage of the initial number of items. Since it’s $r\%$ less, it means we’re left with $100 – r\%$ of the initial amount.

The expression representing the number of items for sale at the end of the day would be:

\[
\left(\frac{100-r}{100}\right) \times 500
\]

[Calc] Question medium

A plant’s height is 1.25 times its height from last week. What was the percentage increase in the plant’s height from last week?

A)1.25%

B)2.5%

C)12.5%

D)25%

▶️Answer/Explanation

D)25%

Let’s denote the plant’s height from last week as $h_{\text{last week}}$ and its current height as $h_{\text{current}}$.

Given that the current height is 1.25 times its height from last week, we can express this mathematically as:
\[ h_{\text{current}} = 1.25 \times h_{\text{last week}} \]

To find the percentage increase, we use the formula:
\[ \text{Percentage Increase} = \left( \frac{h_{\text{current}} – h_{\text{last week}}}{h_{\text{last week}}} \right) \times 100\]

Substituting the given relationship between current and last week’s height:
\[ \text{Percentage Increase} = \left( \frac{1.25h_{\text{last week}} – h_{\text{last week}}}{h_{\text{last week}}} \right) \times 100\]

\[ \text{Percentage Increase} = \left( \frac{0.25h_{\text{last week}}}{h_{\text{last week}}} \right) \times 100\]

\[ \text{Percentage Increase} = 25\% \]

So, the correct answer is D) $25 \%$.

[Calc] Questions Medium

A local restaurant gives teachers a $20 \%$ discount on all their meals. If a teacher pays $\$ 14.00$ for a meal after the discount was applied, what was the price of the meal before the discount?

A) $\$ 16.80$
B) $\$ 17.50$
C) $\$ 20.00$
D) $\$ 25.20$

▶️Answer/Explanation

Ans: B

To find the original price of the meal before the 20% discount was applied, let’s denote the original price by $ P $.

Given that the teacher pays $14.00 after the discount, we can set up the equation based on the discount formula. Since the discount is 20%, the teacher pays 80% of the original price.

Thus:
\[ 0.80P = 14.00 \]

To find $ P $, divide both sides by 0.80:
\[ P = \frac{14.00}{0.80} \]

Calculate the division:
\[ P = 17.50 \]

[Calc] Questions Medium

The number of books in a library increased by 30% from 2002 to 2014. There were x books in the library in 2002. Which expression represents the number of books in the library in 2014 in terms of x ?
A) 130x
B) 30x
C) 1.3x
D) 0.3x

▶️Answer/Explanation

Ans: C

If the number of books in the library increased by $30\%$ from 2002 to 2014, and there were $x$ books in 2002, then in 2014, there would be $x$ books plus $30\%$ of $x$ more books, which can be represented as $x + 0.3x$.

So, the expression representing the number of books in the library in 2014 in terms of $x$ is $x + 0.3x$, which simplifies to $1.3x$.

Therefore, the correct answer is option $\mathbf{C}$ – $1.3x$.

[Calc] Question Medium

$$
g(x)=-0.038 x+2.136
$$

The given linear function $g$ models the annual percentage increase in the population of India $x$ years after 1990 . What is the best interpretation of $g(20)=1.376$ in this context?
A) 1.376 years after 1990, the percentage increase in the population of India was $20 \%$ over the previous year.
B) 1.376 years after 1990, India’s population was approximately 20 times its population in 1990
C) 20 years after 1990, the percentage increase in the population of India was $1.376 \%$ over the previous year.
D) 20 years after 1990, India’s population was approximately 1.376 times its population in 1990.

▶️Answer/Explanation

[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.

In 2008 , what percentage of the total businesses in Iceland were tourism-related?
A) $52 \%$
B) $20 \%$
C) $5 \%$
D) $2 \%$

▶️Answer/Explanation

Questions 35 and 36 refer to the following information.

As of 2016, there were 118 known elements. These elements can be described with 10 classifications as summarized in the given table.

Question

The percentage of the 10 classifications that have 11 or more elements is $p$%. What is the value of $p$ ?

▶️Answer/Explanation

Question

As of 2016, there were 118 known elements. These elements can be described with 10 classifications as summarized in the given table.

An element is a nonmetal if it is classified as a halogen, noble gas, or other nonmetal. The ratio of all nonmetals to all elements is 10 to $k$. What is the value of $k$?

▶️Answer/Explanation

Questions 37 and 38 refer to the following information.
The table gives the age groups of the total population of women and the number of registered women voters in the United States in 2012, rounded to the nearest million.

Total population of women (in millions) Registered women voters(in millions)

18 to 24 years old 15 8

25 to 44 years old 41 25

45 to 64 years old 42 30

65 to 74 years old 13 10

75 years old and over 11 8

Total 122 81

Question

If a woman is selected at random from the total population of women ages 45 to 64 years old, what is the probability of selecting a registered woman voter, rounded to the nearest hundredth? (Express your answer as a decimal, not as a percent.)

▶️Answer/Explanation

.71, 71/100

Question

$(8-\sqrt{x})^2=(4+\sqrt{x})^2$

What is the solution to the equation above?
A. $x=2$
B. $x=4$
C. $x=8$
D. $x=16$

▶️Answer/Explanation

Ans: B

Question

Of 100 people who played a certain video game, 85 scored more than 0 but less than 10,000 points, 14 scored between 10,000 and 100,000 points, and the remaining player scored 5,350,000 points. Which of the following statements about the mean and median of the 100 scores is true?
A. The mean is greater than the median.
B. The median is greater than the mean.
C. The mean and the median are equal.
D. There is not enough information to determine whether the mean or the median is greater.

▶️Answer/Explanation

Ans: A

Question

In spring 2015, three separate studies on the fitness level of tenth graders were conducted in the city of Mistwick. In each study, every student in a group of tenth graders took the same fitness test and received a score on it. The possible scores on the fitness test are the whole numbers from 50 to 100, inclusive. The distribution of the scores for each of the studies is shown in the table below.

The participants for the studies were selected as follows.
• For Study I, 100 tenth graders were selected at random from all tenth graders in Mistwick.
• For Study II, 200 tenth graders were selected at random from all tenth graders in Mistwick.
• For Study III, 300 tenth graders from Mistwick volunteered to participate.
No tenth grader participated in more than one of the three studies.

Which of the following could be the median score in Study III?

59
68
70
82
▶️Answer/Explanation
Ans: B

Question

The results of which of the studies can appropriately be generalized to all tenth graders in Mistwick in spring 2015?

Study III only
Studies I and II only
Studies II and III only
Studies I, II, and III

▶️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.

The band was composed of freshmen, sophomores, juniors, and seniors. If 30% of the students in the band were juniors, how many students were in the band? 3.2

49
54
56
60

▶️Answer/Explanation

Ans: D

Questions

A magazine article on video game habits in the United States reported that in 2012 gamers spent an average of 5.6 hours per week playing games. The article also reported the average for 2013 to be 6.3 hours per week. Based on the article, how did the average number of hours that gamers spent playing games per week change from 2012 to 2013?
A. It decreased by $12.5 \%$.
B. It increased by $7.0 \%$.
C. It increased by $11.1 \%$.
D. It increased by $12.5 \%$.

▶️Answer/Explanation

Ans: D

Questions

The budget for a school band was $\$ 8,000$ in 2010 . The budget decreased by $15 \%$ from 2010 to 2011 and then increased by $22 \%$ from 2011 to 2012. Which of the following expressions represents the budget, in dollars, for the school band in 2012?
A. $(1.15)(1.22)(8,000)$
B. $(0.85)(1.22)(8,000)$
C. $(1.15)(0.78)(8,000)$
D. $(0.85)(0.78)(8,000)$

▶️Answer/Explanation

Ans: B

SAT MAth Practice questions – all topics

SAT MAth and English – full syllabus practice tests

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