Algebra Problems of the Day (Integrated Algebra Regents, June 2013)

While I'm waiting for new Regents exams to come along, I'm revisiting some of the older NY Regents exams.

More Regents problems.

Administered June 2013

Part IV: Each correct answer will receive 4 credits. Partial credit is possible.


37. Solve algebraically:

2 / 3x + 4 / x = 7 / (x + 1)

[Only an algebraic solution can receive full credit.]

Answer:

You need to eliminate the denominators however you can. You could multiply each side by (3x) and (x + 1), for instance. After that, you will have a simpler equation to solve.

You can also combine the two fractions on the left, leaving you with a proportion, and then you can cross-multiply, which amounts to the same thing as above.

2 / 3x + 4 / x = 7 / (x + 1)
2 / 3x + 12 / 3x = 7 / (x + 1)
14 / 3x = 7 / (x + 1)
(cross-multiply)
21x = 14 (x + 1)
21x = 14x + 14
7x = 14
x = 2

Checking your answer: (not required, but a good idea)
2 / (3(2)) + 4 / 2 ?= 7 / (2 + 1)
2/6 + 2 = 7/3
1/3 + 2 = 7/3
7/3 = 7/3 (check)

35.A jar contains five red marbles and three green marbles. A marble is drawn at random and not replaced. A second marble is then drawn from the jar.

Find the probability that the first marble is red and the second marble is green.

Find the probability that both marbles are red.

Find the probability that both marbles are the same color.

Answer:

The total number of marbles for the first draw is 8 and for the second draw is 7. These are dependent events.

P(R, G) = 5 / 8 * 3 / 7 = 15/56

P(R, R) = 5/ 8 * 4 / 7 = 20/56

P(both the same) = P(R, R) + P(G, G) = 20/56 + 3/8 * 2/7 = 20/56 + 6/56 = 26/56.

The second and third answers could be simplified, but didn't need to be. Also, tree diagrams would have been acceptable for showing work.

39.In the diagram below of rectangle AFEB and a semicircle with diameter CD, AB = 5 inches, AB = BC = DE = FE, and CD = 6 inches. Find the area of the shaded region, to the nearest hundredth of a square inch.

Answer:

First, since it is a rectangle, then side AF = BC + CD + DE = 5 + 6 + 5 = 16. Second, the radius of the semicircle is half of CD, which would be 3.

The area of the semicircle is 1/2 π r² = 1/2 (3.141592)(3)² = 14.137164.

The area of the rectangle is (5)(16) = 80

The shaded region is the area of the rectangle minus the area of the semicircle: 80 - 14.137164 = 65.862836, or 65.86 to the nearest hundredth.

Note that if you used 3.14 for pi, you would not have had enough accuracy. Use the pi key on your calculator.

End of Part IV



More to come. Comments and questions welcome.

More Regents problems.