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
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:
\[
\text{Percentage} = \left(\frac{1,047.4}{24,429}\right) \times 100 \approx 4.29\%
\]
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.
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, \(2s\).
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 \(\boxed{900}\).
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
B
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:
\[ \boxed{B} \]