SAT MAth Practice questions – all topics
- Advanced Math Weightage: 35% Questions: 13-15
- Equivalent expressions
- Nonlinear equations in one variable and systems of equations in two variables
- Nonlinear functions
SAT MAth and English – full syllabus practice tests
Question medium
The graph models the mass \(y\), in nanograms, of cobalt-60 (Co-60) remaining in a sample after \(x\) halflives. The half-life of Co-60 is 5.27 years.
What is the mass, in nanograms, of Co-60 remaining in the sample after 10.54 years?
A. 0.47
B. 1.25
C. 2.00
D. 2.64
▶️Answer/Explanation
Ans:B
The relationship between the remaining mass of cobalt-60 (Co-60) and the number of half-lives that have passed. The half-life of Co-60 is 5.27 years, and the mass of Co-60 decreases by half every 5.27 years.
Initial mass \( y_0 = 5 \) nanograms
Half-life \( t_{1/2} = 5.27 \) years
We need to find the mass remaining after 10.54 years. First, we determine how many half-lives have passed in 10.54 years:
\[ \text{Number of half-lives} = \frac{10.54 \text{ years}}{5.27 \text{ years/half-life}} = 2 \]
Since 10.54 years corresponds to 2 half-lives, the mass of Co-60 remaining after 2 half-lives can be calculated as follows:
\[ y = y_0 \times \left(\frac{1}{2}\right)^{\text{Number of half-lives}} \]
Substitute the values into the equation:
\[ y = 5 \times \left(\frac{1}{2}\right)^2 \]
\[ y = 5 \times \frac{1}{4} \]
\[ y = 5 \times 0.25 \]
\[ y = 1.25 \]
So, the mass of Co-60 remaining in the sample after 10.54 years is \( 1.25 \) nanograms.
Alternative method:
Which is only in option B (because left mass is greater then 1 and less then 2)
Question medium
The graph of y =f(x)-1 is shown. Which equation could define function f ?
A.\(f(x)=2^{x}\)
B. \(f(x)=2^{x}-1\)
c. \(f(x)=2^{x}+1\)
D. \(f(x)=2^{x}+2\)
▶️Answer/Explanation
Ans: D
For $x= 0$ , the value of $f(x)-1$ is 2 .
equating , $2=f(0)-1\Rightarrow f(0)=3$
If \(f(x) – 1\) equals \(2\) when \(x = 0\), then \(f(0) = 3\)
Given the options, let’s evaluate each one for \(x = 0\):
A. \(f(x) = 2^x\)
\(f(0) = 2^0 = 1\)
B. \(f(x) = 2^x – 1\)
\(f(0) = 2^0 – 1 = 1 – 1 = 0\)
C. \(f(x) = 2^x + 1\)
\(f(0) = 2^0 + 1 = 1 + 1 = 2\)
D. \(f(x) = 2^x + 2\)
\(f(0) = 2^0 + 2 = 1 + 2 = 3\)
So, among the options, only option D yields \(f(0) = 3\), consistent with what we’ve determined. Hence, the correct answer is option D) \(f(x) = 2^x + 2\).
Question Medium
Which table could represent values of x and their corresponding values of f(x) for a decreasing exponential function f ?
▶️Answer/Explanation
Ans: B
An exponential function is characterized by its growth or decay rate. For a decreasing exponential function, as \(x\) increases, \(f(x)\) decreases. This means that as the input \(x\) increases, the output \(f(x)\) decreases at an increasing rate.
Looking at the tables:
A) The values of \(f(x)\) increase as \(x\) increases. This does not represent a decreasing exponential function.
B) The values of \(f(x)\) decrease as \(x\) increases. This represents a decreasing exponential function.
C) The values of \(f(x)\) increase as \(x\) increases. This does not represent a decreasing exponential function.
D) The values of \(f(x)\) decrease as \(x\) increases. This represents a decreasing exponential function.
So, the table that could represent values of \(x\) and their corresponding values of \(f(x)\) for a decreasing exponential function is:B
Question Medium
The graph of the exponential function \(\mathrm{f}\) is shown. For what value of \(x\) is \(f(x)=0\) ?
A) -4
B) -3
C) -2
D) -1
▶️Answer/Explanation
Ans:C
The value of y at x=0 is -3.
The value of x at y=0 is -2.
Question medium
The graph of y = f(x) is shown. What is the graph of y = f(x) − 2 ?
▶️Answer/Explanation
C
To find the graph of $y = f(x) – 2$, We need to shift the original graph $y = f(x)$ downwards by 2 units.
The graph provided shows a curved increasing function passing through the points (2, 0), and continuing to increase.
To get y = f(x) – 2, I will shift this entire curve downwards by 2 units on the y-axis:
The new curve will pass through the point (0, 0) instead of (2, 0). Which is option- C
Question Medium
The half-life of the radioactive isotope iodine-131 is approximately 8 days, which means that at the end of each 8 -day time interval only half of the mass of the isotope that was present at the beginning of the time interval remains. Which of the following best describes how the amount of iodine-131 changes over time?
A) It increases linearly.
B) It decreases linearly.
C) It increases exponentially.
D) It decreases exponentially.
▶️Answer/Explanation
Ans:D
The half-life of iodine-131 indicates that the amount of iodine-131 halves every 8 days. This kind of change over time is best described by an exponential decay.
The correct description is:
\[
\boxed{\text{It decreases exponentially.}}
\]
The equation of the circle is:
\[
\boxed{x^2 + y^2 = 4}
\]
Questions Medium
An advertising agency guarantees that its services will increase website traffic by 3.5% compared to each previous week. Which type of function best models the weekly guaranteed website traffic as the number of weeks increases?
A) Increasing exponential
B) Decreasing exponential
C) Increasing linear
D) Decreasing linear
▶️Answer/Explanation
Ans: A
The type of function that best models the weekly guaranteed website traffic as the number of weeks increases is an \(\mathbf{increasing\ exponential}\). This is because the guaranteed increase in website traffic each week (3.5%) indicates exponential growth over time. So, the correct answer is \(\mathbf{A}\).
Question Medium
What is the $y$-intercept of the graph of $y=4^x$ in the $x y$-plane?
A) $(1,4)$
B) $(1,0)$
C) $(0,1)$
D) $(4,1)$
▶️Answer/Explanation
C