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
