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 Hard
A fitness membership costs \(\$ 45\) per month. All new members receive a discount of \(\$ 20\) off the cost of their first month of membership. Which function \(c\) gives the total cost \(c(t)\), in dollars, that a new member pays after \(t\) months of membership?
A) \(c(t)=20+45 t\)
B) \(c(t)=25+45 t\)
C) \(c(t)=20+45(t-1)\)
D) \(c(t)=25+45(t-1)\)
▶️Answer/Explanation
Ans:D
To determine the correct function \( c(t) \) that represents the total cost for a new member after \( t \) months of membership, let’s analyze the given information:
1. The regular monthly cost of membership is \(\$45\) .
2. New members receive a discount of \(\$20\) off their first month.
Let’s break down the costs:
The first month costs \(\$45 – \$20 = \$25\).
Each subsequent month costs \(\$45\).
The function \( c(t) \) needs to account for the special pricing in the first month and the regular pricing for the subsequent months. Let’s define \( t \) as the number of months:
For \( t = 1 \):
\[ c(1) = 25 \]
For \( t > 1 \):
\[ c(t) = 25 + 45(t – 1) \]
This formula can be derived from:
\(\$25\) for the first month.
\(\$45\) for each of the remaining \( t – 1 \) months.
Thus, the correct function is:
\[ c(t) = 25 + 45(t – 1) \]
So, the answer is:
D) \( c(t) = 25 + 45(t – 1) \)
Question Hard
The scatterplot shows the number of housing units in 2010 and 2018 for each of 11 US states. A line of best fit with equation 𝑦 = 𝑎x, where a is a constant, is also shown. A point lies above this line if and only if it represents a state with an increase in housing units from 2010 to 2018 greater than 7%. What is the value of a?
▶️Answer/Explanation
Ans: 107/100, 1.07
To find the value of the constant a in the equation \(y=a x\) representing the line of best fit, \(I\) will use the given information that a point lies above the line if and only if it represents a state with an increase in housing units from 2010 to 2018 greater than \(7 \%\).
Let’s consider a sample point on the line, say (400,000, 400,000a).
For this point to lie on the line \(y=a x\), we must have:
\[
\begin{aligned}
& 400,000 a=a x \\
& a=1
\end{aligned}
\]
So the equation of the line is \(\mathrm{y}=\mathrm{x}\).
Now, for a point \((\mathrm{x}, \mathrm{y})\) to lie above this line, we must have \(\mathrm{y}>\mathrm{x}\).
Rearranging, we get:
\[
\begin{aligned}
& y / x>1 \\
& (y-x) / x>0 \\
& (y-x) / x=(y / x)-1>0.07 \text { (since increase is greater than } 7 \%) \\
& y / x>1.07
\end{aligned}
\]
Therefore, the value of a that satisfies the given condition is \(\mathrm{a}=1.07\).
Question Hard
Kiara uses her propane grill for an average of 11 hours each week. Her grill can run an average of
18 hours per 20-pound tank. Kiara would like to reduce her weekly expenditure on propane by \($\)5.
Assuming propane costs \($\)16 per 20-pound tank, which equation can Kiara use to determine how
many fewer average hours, h, she should use her grill each week ?
A) \(\frac{18}{16}h=6\)
B) \(\frac{18}{16}h=5\)
C) \(\frac{16}{18}h=6\)
D) \(\frac{16}{18}h=5\)
▶️Answer/Explanation
D) \(\frac{16}{18}h=5\)
Kiara uses her propane grill for an average of 11 hours each week. Her grill can run an average of 18 hours per 20-pound tank, and propane costs \($\)16 per 20-pound tank. She wants to reduce her weekly expenditure on propane by \($\)5. We need to find the equation to determine how many fewer average hours, \( h \), she should use her grill each week.
1. Determine the current weekly propane usage:
Kiara uses 11 hours per week.
The grill runs for 18 hours per 20-pound tank.
The number of tanks used per week:
\[
\frac{11}{18} \text{ tanks/week}
\]
2. Calculate the current weekly cost:
Propane costs $16 per tank.
\[
\text{Current weekly cost} = \left(\frac{11}{18}\right) \times 16
\]
3. Determine the desired weekly cost reduction:
Kiara wants to reduce her weekly expenditure by $5.
\[
\left(\frac{11 – h}{18}\right) \times 16 = \left(\frac{11}{18}\right) \times 16 – 5
\]
4. Formulate the equation:
To find \(h\), the number of fewer hours she should use her grill:
\[
\frac{16}{18} h = 5
\]
Thus, the correct equation is:
\[ \boxed{\frac{16}{18} h = 5} \]
Question Hard
A company offers its salespeople two different weekly compensation plans. Salespeople on Plan X
earn \($\)1,000 plus a 10% commission on their sales each week. Salespeople on Plan Y earn \($\)500 plus a 20% commission on their sales each week. Which inequality models the amount in sales each week, d dollars, for which salespeople on Plan X earn more than salespeople on Plan Y?
A) d < 5,000
B) d > 5,000
C) d < 1,500
D) d > 1,500
▶️Answer/Explanation
A) d < 5,000
To determine the inequality that models the sales amount \(d\) dollars for which salespeople on Plan X earn more than those on Plan Y, :
Write the weekly earnings for Plan X:
\[
\text{Earnings for Plan X} = 1000 + 0.10d
\]
Write the weekly earnings for Plan Y:
\[
\text{Earnings for Plan Y} = 500 + 0.20d
\]
Set up the inequality where Plan X earnings are greater than Plan Y earnings:
\[
1000 + 0.10d > 500 + 0.20d
\]
Subtract \(500\) from both sides:
\[
500 + 0.10d > 0.20d
\]
Subtract \(0.10d\) from both sides:
\[
500 > 0.10d
\]
Divide by \(0.10\):
\[
d < 5000
\]
Thus, salespeople on Plan X earn more than those on Plan Y when \(d < 5000\).