June 2014 NY Integrated Algebra Regents, Thread 2

Parts 2, 3 and 4 of the Algebra test were open-ended and you had to show your work to get full credit. I don't know the raw score required for passing, but every point here, even partial credit helps. In general, over the last few years, a score of 30-32 is a passing grade. That means you got at least half of the multiple-choice questions correct. However, if you were able to get that many of those questions correct, you have no excuse for not showing some work on these problems.

To answer those who asked on yesterday's Part 1 thread and on Twitter, for the past year I worked at Tottenville High School in Staten Island, and before that I was at William E. Grady High School in Coney Island, Brooklyn, with some time at Boys High School in Bed-Stuy.

Okay, so here we go. Illustrations may get added later, but students seem to be in a hurry for answers:

June 2014 Integrated Algebra Regents, Parts 2, 3 and 4

31. A patio consisting of two semicircles and a square is shown in the diagram below. [ A SQUARE with a semicircle on top and right sides.] The lenght of each side of the square region is represented by 2x. Write an expression for the area of the entire patio, in terms of x and pi.

In terms of x and pi means that you aren't solving for x and you aren't replacing pi with 3.14 or anything else.

Two semicircles make one whole circle. The area of a circle is pi*r². The radius is half the diameter. The diameter is the side of a square, which is 2x. So the radius is 1x. The area of a square is s².
The area of the entire patio is (2x)² + (pi)(x)², which is 4x² + (pi)(x)². That should be enough for full credit.
You could also represent it as (4 + pi)(x)², but that shouldn't be necessary.

32. Clayton is performing some probability experiments consisting of flipping three coins. What is the probability that when Clayton flips the three coins, he gets two tails and one heads?

I did a very similar problem in class recently. Hopefully, you made a tree diagram or a sample space because there aren't that many possible outcomes: 2 X 2 X 2 = 8 outcomes.

On your diagram you'll see the only possibilities are that the first is H and the others tails, the second is H and the others tails or the third is H and the others tails. Therefore there are 3 positive outcomes, so the probability is 3/8.

This was much easier than a Common Core two-point question!

33. Ross is installing edging around his pool, which consists of a rectangle and a semicircle, as shown in the diagram below. [RECTANGLE WITH A SEMICIRCLE ON TOP, WITH A DIAMETER THE WIDTH OF THE SMALLER END OF THE RECTANGLE.] Determine the length of edging, to the nearest tenth of a foot, that Ross will need to go completely around the pool.

Two irregular shapes in one test? That's new. And both contain semicircles. However, this is a perimeter/circumference problem, not an area problem.

The three sides of the rectangle that need edging are 30 + 15 + 30, which is 75 feet. The semicircle is (1/2)(pi)(d), which is (1/2)(3.141592...)(15), which is 23.56..., which rounds to 23.6.

The total is 75 + 23.6 = 98.6 feet.

Part 3

34. Solve the following system of equations algebraically for all values of x and y. y = x² + 2x - 8; y = 2x + 1

Algebraically means that you can't use your graphing calculator to find the answer and then use "guess and check" on the paper.

Using substitution, you get x² + 2x - 8 = 2x + 1
Subtract 2x and subtract 1 from both sides and you have x² - 9 = 0
This is a Difference of Squares and can be factored as: (x + 3)(x - 3) = 0
So either x + 3 = 0 or x - 3 = 0
Your solution set for x is {-3, 3}

DO NOT STOP HERE. You need the values of y as well.

If x = -3, y = 2(-3) + 1 = -6 + 1 = -5. (-3, -5)
If x = 3, y = 2(3) + 1 = 6 + 1 = 7, (3, 7)

If you plug these into the other equations, they will check.

35. A storage container in the form of a rectangular prism is measured to be 12 inches by 8 inches by 4 inches. Its actual measurements are 11.75 inches by 7.75 inches by 4 inches. Find the relative error in calculating the volume of the container, to the nearest thousandth.

Volume = length X width X height.
Relative error = (difference in the values) / (actual value)

So this problem is ( (12)(8)(4) - (11.75)(7.75)(4) ) / ( (11.75)(7.75)(4) ). If you are using a graphing or scientific calculator, and doing this in one step, those extra parentheses are needed.

This becomes ( 384 - 364.25 ) / 364.25 = 19.75 / 364.25 = 0.054221..., which is 0.054 to the nearest thousandth.
DO NOT make it into a percentage.

36. Perform the indicated operations and express the answer in simplest radical form.

This is new, using the Distributive Property with radicals. Good luck.
Keep in mind, this is how I would do it. As long as you did similar steps, possibly in the same order, and achieved the same solution, you're fine. This would've taken longer if I hadn't seen a number pattern and multiplied the sevens before finding the largest perfect squares.

Again, you didn't have to do it my way, as long as you multiplied, factored, added and came up with 189 times the square root of 2.

If I made a mistake here, feel free to point it out and I'll correct it.

Part 4

37. During its first week of business, a market sold a total of 108 apples and oranges. the second week, five times the number of apples and three times the number of oranges were sold. A total of 452 apples and oranges were sold during the second week. Determine how many apples and how many oranges were sold the first week. [Only an algebraic solution can receive full credit.]

We'll let x = the number of apples sold the first week and y = the number of oranges sold the first week. I won't use A and O, because O looks like 0 and that can be confusing.

We know that x + y = 108 and 5x + 3y = 452. Multiply (x + y = 108) by (-3) and you get -3x + -3y = -324.
Combine 5x + 3y = 452 and -3x + -3y = -324 and you get 2x = 128. The y (number of oranges) is eliminated.
If 2x = 128, then x = 64.

64 + y = 108, so y = 44.

Check: 5(64) + 3(44) = 452?
320 + 132 = 452?
452 = 452 (check!)

38. On the set of axes below, solve the following systems of inequalities graphically. Label the solution set S.

2x + 3y < -3; y - 4x > 2

If you want to use your calculator as a guide, you need to rewrite these equations in slope-intercept form.

y < (-2/3)x - 1; y > 4x + 2

Note that the first line will be a broken or dashed line. The second inequality has a solid line.

A picture of the graph will have to wait until later but, you will shade below the first line and above the second line. The solution set will be the section on the left side of the graph between the two lines. That's where the S goes.

39. During the last 15 years of his baseball career, Andrew hit the following number of home runs each season: 35, 24, 32, 36, 40, 32, 40, 38, 36, 33, 11, 20, 19, 22, 8. State and label the values of the minimum, 1st quartile, median, 3rd quartile and maximum. Using the line below, construct a box-and-whisker plot for this set of data.

First, put the numbers in order to make life easier. Your graphing calculator can do this. (It can also make the box-and-whisker plot, too!)

The numbers, in order, are: 8, 11, 19, 20, 22, 24, 32, 32, 33, 35, 36, 36, 38, 40, 40.

There are 15 numbers, so the 8th number is the median (32). The middle of the first 7 numbers, the 4th number, is Q1 (20). The middle of the back 7 numbers, the 12th number, is Q3 (36). The minimum is 8 and the maximum is 40.

I would use a scale of 2 on the number line, which should allow the numbers from 0 to 50 if you wanted. Plot the five numbers we found above. Draw the box around the middle three and the extend the whiskers to the first and last.