SAT MAth Practice questions – all topics
- Algebra Weightage: 35% Questions: 13-15
- Linear equations in one variable
- Linear equations in two variables
- Linear functions
- Systems of two linear equations in two variables
- Linear inequalities in one or two variables
SAT MAth and English – full syllabus practice tests
Question Hard
Some values of \(x\) and their corresponding values of \(f(x)\) for the linear function \(f\) are shown in the table. What is the value of \(f(6)\) ?
A) 7
B) 8
C) 9
D) 10
▶️Answer/Explanation
Ans:A
To find the value of \(f(6)\), we can first identify the slope of the linear function \(f\). We’ll use the points \((-2, -5)\) and \((4, 4)\) from the table.
The slope \(m\) of a linear function is given by:
\[ m = \frac{{\Delta y}}{{\Delta x}} \]
Using the given points:
\[ m = \frac{{4 – (-5)}}{{4 – (-2)}} = \frac{{4 + 5}}{{4 + 2}} = \frac{9}{6} = \frac{3}{2} \]
Now that we have the slope, we can use the point-slope form of a linear equation to find the equation of the line. Let’s use the point \((4, 4)\):
\[ y – y_1 = m(x – x_1) \]
\[ y – 4 = \frac{3}{2}(x – 4) \]
Now, let’s find the value of \(f(6)\) using this equation:
\[ f(6) = \frac{3}{2}(6 – 4) + 4 = \frac{3}{2}(2) + 4 = 3 + 4 = 7 \]
So, the value of \(f(6)\) is \(7\).
Therefore, the correct answer is:
\[ \boxed{\text{A) 7}} \]
Question Hard
Line \(k\) has a slope of \(-\frac{4}{5}\) and an \(x\)-intercept of \(\left(\frac{r}{2}, 0\right)\), where \(r\) is a constant. What is the \(y\)-coordinate of the \(y\)-intercept of line \(k\) in terms of \(r\) ?
A) \(-\frac{2 r}{5}\)
B) \(\frac{2 r}{5}\)
C) \(-\frac{5 r}{8}\)
D) \(\frac{5 r}{8}\)
▶️Answer/Explanation
Ans: B
Line \(k\) has a slope of \(-\frac{4}{5}\) and an \(x\)-intercept of \(\left(\frac{r}{2}, 0\right)\). We need to find the \(y\)-coordinate of the \(y\)-intercept in terms of \(r\).
The equation of a line in slope-intercept form is:
\[
y = mx + b
\]
where \(m\) is the slope and \(b\) is the \(y\)-intercept.
Given the slope \(m = -\frac{4}{5}\) and the \(x\)-intercept \(\left(\frac{r}{2}, 0\right)\), we can use the point-slope form to find the \(y\)-intercept \(b\).
Using the point-slope form equation:
\[
y – y_1 = m(x – x_1)
\]
where \((x_1, y_1) = \left(\frac{r}{2}, 0\right)\), we have:
\[
y – 0 = -\frac{4}{5}\left(x – \frac{r}{2}\right)
\]
\[
y = -\frac{4}{5}x + \frac{4}{5} \cdot \frac{r}{2}
\]
\[
y = -\frac{4}{5}x + \frac{2r}{5}
\]
Therefore, the \(y\)-intercept \(b\) is:
\[
b = \frac{2r}{5}
\]
So, the correct answer is:
\[
\boxed{\frac{2r}{5}}
\]
Question Hard
Line k in the xy-plane has slope \(\frac{-2p}{5}\)and y-intercept (0,p), where p is a positive constant. What is the x-coordinate of the x-intercept of line k ?
▶️Answer/Explanation
Ans: 5/2, 2.5
The equation of line \(k\) is given as \(y = -\frac{2p}{5}x + p\). To find the \(x\)-intercept, we set \(y = 0\) and solve for \(x\):
\[0 = -\frac{2p}{5}x + p\]
\[ \frac{2p}{5}x = p \]
\[ x = \frac{5}{2} \]
So, the \(x\) coordinate of the \(x\)-intercept of line \(k\) is \(\frac{5}{2}\).
Question Hard
The table shows the list price, discount, and installation fee for tires from four different car repair stores. Assume there is no sales tax and the information in the table is for tires of the same brand and size
Store W’s total expenses for selling and installing 4 tires is $100. Which function represents the profit p(a), in dollars, from selling and installing 4 tires to which the store’s discount is applied? (profit = total amount of money received – expenses)
A) p(a) = 3a + 50
B) p(a) = 3a − 50
C) p(a) = 4a + 50
D) p(a) = 4a − 50
▶️Answer/Explanation
B) p(a) = 3a − 50
For Store W, the list price per tire is \( \$a \). The store offers a “Buy 3 tires at list price and get the 4th tire free” discount. The installation fee for all 4 tires is \( \$50 \).
Calculate the total amount received from selling the tires:
\[
\text{Total amount received} = 3 \times \$a + 0 \times \$a = 3a
\]
Add the installation fee:
\[
\text{Total amount received with installation} = 3a + 50
\]
The total expenses for selling and installing 4 tires is \( \$100 \). Therefore, the profit \( p(a) \) is given by:
\[
p(a) = (\text{Total amount received with installation}) – \text{Expenses}
\]
\[
p(a) = (3a + 50) – 100
\]
\[
p(a) = 3a – 50
\]
Question Hard
A company spent a total of \(\$ 9000\) on digital and print ads. The ratio of the money spent on digital ads to the money spent on print ads was 1 to 3 . How much money, in dollars, did the company spend on digital ads? (Disregard the \$ sign when entering your answer. For example, if your answer is \(\$ 4.97\), enter 4.97)
▶️Answer/Explanation
2250
Let’s represent the money spent on digital ads as \(d\) dollars and the money spent on print ads as \(3d\) dollars, based on the given ratio.
Given that the total amount spent is $9000, we can set up the equation:
\[ d + 3d = 9000 \]
Solve for \(d\):
\[ 4d = 9000 \]
\[ d = \frac{9000}{4} \]
\[ d = 2250 \]
Therefore, the company spent \(\$2250\) on digital ads.