Algebra Problems of the Day (Integrated Algebra Regents, August 2011)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones. The Integrated Algebra Regents covered most of the same material as the current Algebra Regents, with a few differences.

More Regents problems.

Integrated Algebra Regents, August 2011

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


37. Vince buys a box of candy that consists of six chocolate pieces, four fruit-flavored pieces, and two mint pieces. He selects three pieces of candy at random, without replacement.

Calculate the probability that the first piece selected will be fruit flavored and the other two will be mint.

Calculate the probability that all three pieces selected will be the same type of candy.

Answer:

There is a total of 12 pieces of candy. You are selecting without replacement. That means that the number of items left will decrease.

Part 1: P(fruit then mint then mint), order matters:

= 4/12 * 2/11 * 1/10 = 8/1320

Part 2: P(all the same) = P(chocolate, chocolate, chocolate) + P(fruit, fruit, fruit). Note that the probability of 3 mint is zero because there are only two mint.

= 6/12 * 5/11 * 4/10 + 4/12 * 3/11 * 2/10

= 120 / 1320 + 24 / 130

= 144 / 1320

These can be reduced to 1/165 and 6/55, respectively, but isn't necessary. In probability, it is useful to have the same denominator for easy comparisons.

38. 8 On the set of axes below, solve the following system of equations graphically and state the coordinates of all points in the solution set.

y = −x² + 6x − 3
x + y = 7

Answer:

If you are using a graphing calculator, rewrite x + y = 7 as y = -x + 7. If you are not using one, then you know that it's in Standard Form and the x-intercept and the y-intercept are both 7.

If you are plugging in numbers, you know that the vertex will be when x = -b/(2a) = -(6)/(2(-1)) = 3.

The vertex, therefore, is y = -(3)² + 6(3) - 3 = 6, (3, 6).

Don't forget to lavel the lines and the points of intersection (which are the solutions).

39. Solve for m: m / 5 + 3(m - 1) / 2 = 2(m - 3)

Answer:

If you multiply the equation by 10 (or by (5) and (2)), you can remove the denominators from the equation.

(10) ( m / 5 + 3(m - 1) / 2 ) = 2(m - 3) (10)

2m + 15m - 15 = 20m - 60

17m - 15 = 20m - 60

45 = 3m

m = 15

Check: 15 / 5 + 3(15 - 1) / 2 ?= 2(15 - 3)
3 + 3(14)/2 ?= 2(12)
3 + 21 = 24
CHECK!

End of Part Exam.

More to come. Comments and questions welcome.

More Regents problems.

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