Asked by jaylen
Joshua is going kayaking with a group during one of his vacation days. In his vacation planning, he budgeted $50 for a kayak rental.
A. How can Joshua determine the number of hours he can rent a kayak for himself? Describe two different options.
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B. Joshua found out that there is a $25 nonrefundable equipment fee in addition to the hourly rates. How does this requirement change the mathematics of the situation?
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C. Choose Efficient Methods How do the processes you used for parts A and B differ? How are they the same?
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All Answers 1
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A.
Let r be the hourly rental rate (dollars per hour) and h the number of hours.
Option 1 — algebraic: set up the cost equation r·h = 50 and solve for h:
h = 50 / r.
(If Joshua must rent an integer number of hours, take h = floor(50 / r).)
Option 2 — concrete/counting or graphical: make a table or repeatedly subtract the hourly rate from $50 until the remaining money is less than r. The number of subtractions is the number of whole hours he can afford. Or plot cost C(h) = r·h and find where C(h) = 50 (intersection with the horizontal line C = 50).
(Example: if r = $10/hr then h = 50/10 = 5 hours; repeated subtraction 50 → 40 → 30 → 20 → 10 → 0 gives 5 subtractions/hours.)
B.
With a $25 nonrefundable equipment fee the total cost is 25 + r·h. The budget constraint becomes
25 + r·h ≤ 50.
Solve for h:
r·h ≤ 25 → h ≤ 25 / r.
So Joshua can rent at most 25/r hours (or floor(25 / r) whole hours). If the $25 fee alone exceeds $50, he cannot afford any rental.
C.
Similarities: Both parts model cost as a linear function of hours and use the same kinds of methods (algebraic division, repeated subtraction, tables, or graphs) to find h. In both cases you solve a linear inequality or equation of the form (constant) + r·h ≤ budget.
Differences: Part B adds a fixed cost term, so you must subtract the equipment fee from the budget before dividing by the hourly rate (h = (budget − fixed fee)/r) instead of dividing the entire budget by r. Algebraic division is the most efficient method in both parts; repeated subtraction or tables are less efficient but give the same answer and can be helpful for whole-hour counting.
Let r be the hourly rental rate (dollars per hour) and h the number of hours.
Option 1 — algebraic: set up the cost equation r·h = 50 and solve for h:
h = 50 / r.
(If Joshua must rent an integer number of hours, take h = floor(50 / r).)
Option 2 — concrete/counting or graphical: make a table or repeatedly subtract the hourly rate from $50 until the remaining money is less than r. The number of subtractions is the number of whole hours he can afford. Or plot cost C(h) = r·h and find where C(h) = 50 (intersection with the horizontal line C = 50).
(Example: if r = $10/hr then h = 50/10 = 5 hours; repeated subtraction 50 → 40 → 30 → 20 → 10 → 0 gives 5 subtractions/hours.)
B.
With a $25 nonrefundable equipment fee the total cost is 25 + r·h. The budget constraint becomes
25 + r·h ≤ 50.
Solve for h:
r·h ≤ 25 → h ≤ 25 / r.
So Joshua can rent at most 25/r hours (or floor(25 / r) whole hours). If the $25 fee alone exceeds $50, he cannot afford any rental.
C.
Similarities: Both parts model cost as a linear function of hours and use the same kinds of methods (algebraic division, repeated subtraction, tables, or graphs) to find h. In both cases you solve a linear inequality or equation of the form (constant) + r·h ≤ budget.
Differences: Part B adds a fixed cost term, so you must subtract the equipment fee from the budget before dividing by the hourly rate (h = (budget − fixed fee)/r) instead of dividing the entire budget by r. Algebraic division is the most efficient method in both parts; repeated subtraction or tables are less efficient but give the same answer and can be helpful for whole-hour counting.
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