Q1: A recipe calls for 3 cups of flour to make 12 cookies. If you want to make 36 cookies, how many cups of flour do you need? (a) 6 cups (b) 9 cups (c) 12 cups (d) 15 cups
Solution:
Ans: (b) Explanation: This is a proportional relationship. Since 36 cookies is 3 times as many as 12 cookies, you need \(3 \times 3 = 9\) cups of flour. The ratio \(\frac{3}{12} = \frac{9}{36}\) shows the proportion is maintained.
Q2: Which table represents a proportional relationship between \(x\) and \(y\)? (a) \(x: 1, 2, 3\) and \(y: 3, 5, 7\) (b) \(x: 2, 4, 6\) and \(y: 6, 12, 18\) (c) \(x: 1, 2, 3\) and \(y: 2, 4, 8\) (d) \(x: 3, 6, 9\) and \(y: 9, 15, 21\)
Solution:
Ans: (b) Explanation: In a proportional relationship, the ratio \(\frac{y}{x}\) must be constant. For option (b): \(\frac{6}{2} = 3\), \(\frac{12}{4} = 3\), and \(\frac{18}{6} = 3\). All ratios equal 3, so this is proportional. Options (a), (c), and (d) do not maintain a constant ratio.
Q3: The graph of a proportional relationship always passes through which point? (a) \((1, 1)\) (b) \((0, 0)\) (c) \((1, 0)\) (d) \((0, 1)\)
Solution:
Ans: (b) Explanation: A proportional relationship is represented by the equation \(y = kx\), where \(k\) is the constant of proportionality. When \(x = 0\), \(y = 0\), so the graph always passes through the origin \((0, 0)\).
Q4: A car travels 150 miles in 3 hours at a constant speed. What is the constant of proportionality (unit rate)? (a) 30 miles per hour (b) 45 miles per hour (c) 50 miles per hour (d) 60 miles per hour
Solution:
Ans: (c) Explanation: The constant of proportionality or unit rate is found by dividing the total distance by the total time: \(\frac{150 \text{ miles}}{3 \text{ hours}} = 50\) miles per hour. This is the speed of the car.
Ans: (b) Explanation: A proportional relationship has the form \(y = kx\), where \(k\) is a constant and there is no added or subtracted term. Option (b) \(y = \frac{x}{5}\) can be written as \(y = \frac{1}{5}x\), which fits the form. Options (a) and (d) have added or subtracted constants, and option (c) is quadratic, not linear.
Sub-section A2: Fill in the Blank
Q6: In a proportional relationship, the ratio \(\frac{y}{x}\) is always __________ and is called the constant of proportionality.
Solution:
Ans: constant Explanation: The defining feature of a proportional relationship is that the ratio between the two quantities remains the same for all values, which is why it's called the constant of proportionality.
Q7: The equation of a proportional relationship can be written in the form \(y =\) __________, where \(k\) represents the constant of proportionality.
Solution:
Ans: \(kx\) Explanation: A proportional relationship between \(x\) and \(y\) is always represented by the equation \(y = kx\), where \(k\) is a non-zero constant.
Q8: If 5 pounds of apples cost $8, then the unit price (price per pound) is __________ dollars per pound.
Solution:
Ans: 1.60 Explanation: The unit price is found by dividing the total cost by the number of pounds: \(\frac{8}{5} = 1.60\) dollars per pound. This represents the constant of proportionality in the relationship between cost and weight.
Q9: A proportional relationship graphed on a coordinate plane will always be a straight line that passes through the point __________.
Solution:
Ans: \((0, 0)\) Explanation: All proportional relationships pass through the origin because when one quantity is zero, the other must also be zero. This is a key feature that distinguishes proportional from non-proportional linear relationships.
Q10: If \(y\) is proportional to \(x\) and \(y = 24\) when \(x = 6\), then the constant of proportionality is __________.
Solution:
Ans: 4 Explanation: The constant of proportionality \(k\) is found by dividing \(y\) by \(x\): \(k = \frac{24}{6} = 4\). This means \(y = 4x\) represents the proportional relationship.
Section B: Apply Your Learning
Q11: Maria is training for a marathon. She runs at a constant speed and covers 7.5 miles in 1.5 hours. At this rate, how far will she run in 4 hours? Show your work and explain your reasoning.
Solution:
Ans: Step 1: Find the constant of proportionality (speed) by dividing distance by time. \(k = \frac{7.5 \text{ miles}}{1.5 \text{ hours}} = 5\) miles per hour
Step 2: Use the constant of proportionality to find the distance for 4 hours. \(d = 5 \times 4 = 20\) miles
Alternatively, we can set up a proportion: \(\frac{7.5}{1.5} = \frac{d}{4}\) Cross-multiply: \(7.5 \times 4 = 1.5 \times d\) \(30 = 1.5d\) \(d = \frac{30}{1.5} = 20\) miles Final Answer: Maria will run 20 miles in 4 hours.
Q12: A grocery store sells oranges in bags. A 3-pound bag costs $4.50, and a 5-pound bag costs $7.50. Determine if the relationship between weight and cost is proportional for both bags. If so, what is the unit price?
Solution:
Ans: Step 1: Find the unit price for the 3-pound bag. Unit price = \(\frac{\$4.50}{3 \text{ pounds}} = \$1.50\) per pound
Step 2: Find the unit price for the 5-pound bag. Unit price = \(\frac{\$7.50}{5 \text{ pounds}} = \$1.50\) per pound
Step 3: Compare the unit prices. Since both bags have the same unit price of $1.50 per pound, the relationship is proportional with a constant of proportionality of 1.50. Final Answer: Yes, the relationship is proportional. The unit price is $1.50 per pound.
Q13: A scale model of a building uses a scale where 2 inches on the model represents 15 feet on the actual building. If the model is 8 inches tall, how tall is the actual building in feet?
Solution:
Ans: Step 1: Identify the constant of proportionality. The scale shows \(\frac{15 \text{ feet}}{2 \text{ inches}} = 7.5\) feet per inch
Step 2: Use the constant to find the actual height. Actual height = \(8 \text{ inches} \times 7.5 \frac{\text{feet}}{\text{inch}} = 60\) feet
Alternatively, set up a proportion: \(\frac{2 \text{ inches}}{15 \text{ feet}} = \frac{8 \text{ inches}}{h \text{ feet}}\) Cross-multiply: \(2h = 15 \times 8\) \(2h = 120\) \(h = 60\) feet Final Answer: The actual building is 60 feet tall.
Q14: The table below shows the relationship between hours worked and money earned. Determine if this is a proportional relationship. If so, write the equation and find how much money would be earned for 12 hours of work.
Hours Worked (\(x\))
Money Earned (\(y\))
2
$26
4
$52
6
$78
Solution:
Ans: Step 1: Calculate the ratio \(\frac{y}{x}\) for each pair. For \(x = 2\): \(\frac{26}{2} = 13\) For \(x = 4\): \(\frac{52}{4} = 13\) For \(x = 6\): \(\frac{78}{6} = 13\)
Step 2: Since the ratio is constant, this is a proportional relationship. The constant of proportionality is \(k = 13\), so the equation is \(y = 13x\).
Step 3: Find the earnings for 12 hours. \(y = 13 \times 12 = 156\) dollars Final Answer: Yes, it is proportional with equation \(y = 13x\). For 12 hours, the money earned is $156.
Q15: A water tank is being filled at a constant rate. After 8 minutes, the tank contains 120 gallons of water. After 14 minutes, it contains 210 gallons. Is the relationship between time and water volume proportional? Explain your answer using calculations.
Solution:
Ans: Step 1: Calculate the rate (gallons per minute) for both time points. At 8 minutes: \(\frac{120 \text{ gallons}}{8 \text{ minutes}} = 15\) gallons per minute At 14 minutes: \(\frac{210 \text{ gallons}}{14 \text{ minutes}} = 15\) gallons per minute
Step 2: Compare the rates. Since both rates equal 15 gallons per minute, the constant of proportionality is the same.
Step 3: Verify that the relationship passes through the origin. If the tank started empty \((0, 0)\) and fills at a constant rate, the relationship is proportional. The equation is \(y = 15x\), where \(y\) is gallons and \(x\) is minutes. Final Answer: Yes, the relationship is proportional with a constant rate of 15 gallons per minute. The equation is \(y = 15x\).
Q16: A printer can print 18 pages in 3 minutes. At this rate, how many minutes will it take to print 84 pages? Set up a proportion and solve.
Solution:
Ans: Step 1: Set up a proportion comparing pages to minutes. \(\frac{18 \text{ pages}}{3 \text{ minutes}} = \frac{84 \text{ pages}}{t \text{ minutes}}\)
Step 2: Cross-multiply to solve for \(t\). \(18t = 3 \times 84\) \(18t = 252\)
Step 3: Divide both sides by 18. \(t = \frac{252}{18} = 14\) minutes
Alternatively, find the unit rate: \(\frac{18}{3} = 6\) pages per minute Then: \(\frac{84 \text{ pages}}{6 \text{ pages per minute}} = 14\) minutes Final Answer: It will take 14 minutes to print 84 pages.
Section C: Evidence-Based Reasoning (CER)
Q17: A student claims that the equation \(y = 3x + 5\) represents a proportional relationship between \(x\) and \(y\). Do you agree or disagree with this claim? Support your answer with evidence and reasoning that explains the characteristics of proportional relationships.
Solution:
Ans: Claim: I disagree with the student's claim that \(y = 3x + 5\) represents a proportional relationship. Evidence: A proportional relationship must have the form \(y = kx\), where \(k\) is a constant and there is no added or subtracted term. The given equation \(y = 3x + 5\) includes a constant term of +5. Additionally, when graphed, a proportional relationship must pass through the origin \((0, 0)\). If we substitute \(x = 0\) into the equation, we get \(y = 3(0) + 5 = 5\), which means the graph passes through \((0, 5)\), not the origin. Finally, the ratio \(\frac{y}{x}\) is not constant. For example, when \(x = 1\), \(y = 8\), so \(\frac{y}{x} = 8\). When \(x = 2\), \(y = 11\), so \(\frac{y}{x} = 5.5\). These ratios are different. Reasoning: These three pieces of evidence demonstrate that the equation does not meet the fundamental requirements of a proportional relationship. The presence of the constant term +5 shifts the entire graph upward, preventing it from passing through the origin. This means there is not a constant ratio between \(y\) and \(x\) for all values. While this equation represents a linear relationship, it is specifically a non-proportional linear relationship because of the non-zero y-intercept. Therefore, the student's claim is incorrect.
Q18: Two students are comparing the prices of trail mix at different stores. Store A sells 2 pounds for $7, and Store B sells 3 pounds for $10.50. One student claims that Store B has a better deal because you get more trail mix. Using the concept of proportional relationships and unit rates, evaluate this claim and determine which store actually offers the better price per pound.
Solution:
Ans: Claim: The student's reasoning is flawed. Store A actually offers the better deal based on the unit price (price per pound). Evidence: To compare prices fairly, we must calculate the unit rate (price per pound) for each store, which represents the constant of proportionality in the relationship between cost and weight. For Store A: \(\frac{\$7}{2 \text{ pounds}} = \$3.50\) per pound. For Store B: \(\frac{\$10.50}{3 \text{ pounds}} = \$3.50\) per pound. Both stores actually charge exactly $3.50 per pound. The total amount purchased does not determine which is a better deal; rather, the unit price determines value. Since the unit rates are identical, neither store offers a better price-they are equally priced on a per-pound basis. Reasoning: This problem demonstrates the importance of using unit rates to compare proportional relationships. The student's error was focusing on the total quantity rather than the rate of change (price per unit). In proportional relationships, the constant of proportionality allows us to make fair comparisons between different quantities. When comparing prices, we must always calculate the unit price to determine which option provides better value. In this case, since both stores have the same constant of proportionality ($3.50 per pound), they offer equivalent deals, and the student's claim that Store B is better simply because it offers more quantity is mathematically incorrect.
Section D: Extended Thinking
Q19: A bicycle shop rents bicycles using a pricing system. The shop charges a flat fee plus an hourly rate. The total cost for a 3-hour rental is $22.50, and the total cost for a 5-hour rental is $32.50. Determine whether this pricing system represents a proportional relationship. If not, find the flat fee and the hourly rate. Then, write an equation that represents the total cost \(C\) as a function of hours \(h\), and use it to find the cost of an 8-hour rental.
Solution:
Ans: Step 1: Test if the relationship is proportional by checking if the ratio \(\frac{C}{h}\) is constant. For 3 hours: \(\frac{22.50}{3} = 7.50\) dollars per hour For 5 hours: \(\frac{32.50}{5} = 6.50\) dollars per hour Since the ratios are different, this is not a proportional relationship. There must be a flat fee in addition to an hourly rate.
Step 2: Set up a system of equations. Let \(f\) = flat fee and \(r\) = hourly rate. For 3 hours: \(f + 3r = 22.50\) ... equation (1) For 5 hours: \(f + 5r = 32.50\) ... equation (2)
Step 3: Solve the system by subtracting equation (1) from equation (2). \((f + 5r) - (f + 3r) = 32.50 - 22.50\) \(2r = 10\) \(r = 5\) dollars per hour
Step 5: Write the equation for total cost. \(C = 7.50 + 5h\) This is a linear equation but not proportional because of the flat fee (y-intercept of 7.50).
Step 6: Find the cost for 8 hours. \(C = 7.50 + 5(8) = 7.50 + 40 = 47.50\) dollars Final Answer: This is not a proportional relationship. The flat fee is $7.50, the hourly rate is $5.00 per hour, the equation is \(C = 7.50 + 5h\), and an 8-hour rental costs $47.50.
Q20: Consider three different proportional relationships: Relationship A: \(y = 2x\) Relationship B: The graph passes through \((3, 12)\) Relationship C: A table shows that when \(x = 5\), \(y = 15\)
Compare these three relationships by finding the constant of proportionality for each. Then, determine which relationship has the steepest graph and explain what this means in terms of rate of change. Finally, if all three relationships represent the speed of different vehicles (in miles per hour), which vehicle is traveling fastest?
Solution:
Ans: Step 1: Find the constant of proportionality for Relationship A. The equation \(y = 2x\) is already in the form \(y = kx\), so \(k_A = 2\).
Step 2: Find the constant of proportionality for Relationship B. Since the graph passes through \((3, 12)\) and proportional relationships pass through the origin, we use the point to find \(k\). \(k_B = \frac{y}{x} = \frac{12}{3} = 4\) So the equation is \(y = 4x\).
Step 3: Find the constant of proportionality for Relationship C. From the table, when \(x = 5\), \(y = 15\): \(k_C = \frac{y}{x} = \frac{15}{5} = 3\) So the equation is \(y = 3x\).
Step 4: Compare the constants of proportionality. \(k_A = 2\), \(k_B = 4\), \(k_C = 3\) Relationship B has the largest constant of proportionality: \(k_B = 4\).
Step 5: Interpret the meaning of the steepest graph. In a proportional relationship graphed on a coordinate plane, the constant of proportionality represents the slope of the line. A larger constant means a steeper graph, which represents a greater rate of change. This means that for every unit increase in \(x\), \(y\) increases more in Relationship B than in the other relationships.
Step 6: Apply to the context of vehicle speed. If each relationship represents speed (miles per hour), then Relationship B with \(k = 4\) represents a vehicle traveling at 4 miles per hour, compared to 2 mph for A and 3 mph for C. However, this interpretation assumes \(x\) represents hours and \(y\) represents miles. The vehicle represented by Relationship B is traveling the fastest at 4 miles per hour. Final Answer: The constants of proportionality are: A = 2, B = 4, and C = 3. Relationship B has the steepest graph with \(k = 4\), meaning it has the greatest rate of change. If these represent vehicle speeds, Relationship B represents the fastest vehicle at 4 miles per hour.
The document Mixed Questions Set: Proportional Relationships is a part of the Grade 7 Course Math Grade 7.
FAQs on Mixed Questions Set: Proportional Relationships
1. What is the main purpose of Section A: Quick Check in the exam?
Ans. The main purpose of Section A: Quick Check is to assess students' understanding of the key concepts covered in the curriculum. It typically includes straightforward questions that gauge comprehension and retention of the material.
2. How can students effectively use Section B: Apply Your Learning?
Ans. Students can effectively use Section B: Apply Your Learning by engaging with practical scenarios or problems that require them to apply the concepts they have learned. This section often encourages critical thinking and real-world application of knowledge.
3. What does Section C: Evidence-Based Reasoning (CER) entail?
Ans. Section C: Evidence-Based Reasoning (CER) entails students constructing a clear argument based on evidence. It requires them to make a claim, provide reasoning, and support their claim with appropriate evidence from the texts or experiments studied.
4. What type of questions can be found in Section D: Extended Thinking?
Ans. Section D: Extended Thinking includes more complex, open-ended questions that challenge students to think critically and deeply about a subject. This section often requires synthesis of information, evaluation of different viewpoints, and the application of knowledge to new situations.
5. How should students prepare for the different sections of the exam?
Ans. Students should prepare for the different sections of the exam by reviewing their notes, engaging in group discussions, practising past papers, and focusing on understanding concepts rather than rote memorisation. They should also ensure they can apply knowledge in practical contexts, particularly for Sections B and D.
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