August 2018 Algebra I Regents, Part III & IV

The following are some of the multiple questions from the August 2018 New York State Common Core Algebra I Regents exam.
The answers to Part I can be found here
The answers to Part II can be found here

August 2018 Algebra I, Part III

Each correct answer is worth up to 4 credits. Partial credit can be given. Work must be shown or explained.

33. Sarah wants to buy a snowboard that has a total cost of $850, including tax. She has already save $135 for it. At the end of each week, she is paid $96 for babysitting and is going to save three-quarters of that for the snowboard.
Write an inequality that can be used to determine the minimum number of weeks Sarah needs to babysit to have enough money to purchase the snowboard.
Determine and state the minimum number of full weeks Sarah needs to babysit to have enough money to purchase this snowboard.

Answer:
Very simply: The first want you to just write the inequality, which you then use to answer the second part of the question.

She has 135, she is saving 3/4 of 96 per week for n weeks and needs more than 580.
135 + (3/4)(96)n > 580.

135 + 72n > 580
72n > 445
n > (445/72)
n > (445/72)
n > 6.18
Sarah needs to babysit for 7 full weeks to pay for the snowboard.
Remember: You cannot round *down* because then she will not have enough money. She will be a few dollars short.

34. A car was purchased for $25,000. Research shows that the car has an average yearly depreciation rate of 18.5%.
Create a function that will determine the value, V(t), of the car t years after purchase.
Determine, to the nearest cent, how much the car will depreciate from year 3 to year 4.

Answer:
1.00 - 0.185 = 0.815
V(t) = 25000(0.815)^t

Use this formula to find the value of V(3) and V(4) and subtract them to find the difference. (You can also graph this in your calculator and check the Table of Values)
V(3) - V(4) = 25000(0.815)⁴ - 25000(0.815)³
= 13533.584 - 11029.871 = 2503.713 = $2,503.71 depreciation between years 3 and 4.

35. Graph the following systems of inequalities on the set of axes below:

2y > 3x - 16
y + 2x > -5

Based upon your graph, explain why (6, 1) is a solution to this system and why (-6, 7) is not a solution to this system.

Answer:
If you plan on using the graphing calculator, you will want to rewrite these inequalities in slope-intercept form. (At the very least, you will want to isolate the y on the left side.)

2y > 3x - 16
y > 3/2 x - 8
y + 2x > -5
y > -2x - 5
Update: Image added

Notice that in the first case, there will be a solid line, and in the second, it will be broken. In both cases, you will be shading about the line, and the section shaded twice will be the solution. Remember to label both lines.

(6, 1) is in the solution because it is on the solid line, and points on the solid line are part of the solution.
(-6, 7) is not a solution because it is on the broken line, which is a boundary and not part of the solution set.

Note: if you draw you graph incorrectly, your explanation would need to match your graph. If somehow (-6, 7) was a part of your solution, you could try to explain that, but the better move would be to realize that you made a mistake and fix it!

36. Paul plans to have a rectangular garden adjacent to his garage. He will use 36 feet of fence to enclose three sides of the garden. The area of the garden, in square feet, can be modeled by f(w) = w(36 - 2w), where w is the width in feet.
On the set of axes below, sketch the graph of f(w).
Explain the meaning of the vertex in the context of the problem.

Answer:
Update: image added

In the context of the problem, the vertex is the width of the garden that will produce the greatest area. When the width is 9 feet, the area will be 162 square feet.

August 2018 Algebra I, Part IV

A correct answer is worth up to 6 credits. Partial credit can be given. Work must be shown or explained.

37. At the present time, Mrs. Bee’s age is six years more than four times her son’s age. Three years ago, she was seven times as old as her son was then.
If b represents Mrs. Bee’s age now and s represents her son’s age now, write a system of equations that could be used to model this scenario.
Use this system of equations to determine, algebraically, the ages of both Mrs. Bee and her son now.
Determine how many years from now Mrs. Bee will be three times as old as her son will be then.

Answer:
Mrs. Bee (b) is (=) six years more than (+ 6 to the next item) four times her son's age (4s).
b = 4s + 6
Three years ago (b - 3), she was (=) seven times (7 *) as old as her son was then( (s-3) ).
b - 3 = 7(s - 3) -- Note the parentheses around (s-3)! Those are important!

b = 4s + 6
b - 3 = 7(s - 3) means b = 7(s - 3) + 3
So 4s + 6 = 7(s - 3) + 3)
4s + 6 = 7s - 21 + 3
4s + 6 = 7s - 18
24 = 3s
8 = s
b = 4(8) + 6 = 32 + 6 = 38.
Mrs. Bee is 38 now and her son is 8.
Check: 38 - 3 = 35. 7(8 - 3) = 7(5) = 35. Check!

In n years, when Mrs. Bee is three times older than her son, then her age then divided by the son's age then will equal three.
(38 + n) / (8 + n) = 3, or
(38 + n) = 3(8 + n)
38 + n = 24 + 3n
14 = 2n
7 = n. In seven years, Mrs. Bee will be three times her son's age.
Check: 38 + 7 = 45. 3(8 + 7) = 3(15) = 45. Check!

End of Part Exam

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