Algebra Problems of the Day (Integrated Algebra Regents, June 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, June 2011

Part III: Each correct answer will receive 3 credits. Partial credit is available.


34. Given the following list of students’ scores on a quiz: 5, 12, 7, 15, 20, 14, 7

Determine the median of these scores.

Determine the mode of these scores.

The teacher decides to adjust these scores by adding three points to each score. Explain the effect, if any, that this will have on the median and mode of these scores

Answer:

The median is the middle number when in order. The mode is the number that appears the most frequently.

Putting the numbers in order, you have 5, 7, 7, 12, 14, 15, 20.

The median is 12. The mode is 7.

If three points were added to each score, then the median and the mode would both increase by three. The median would be 15, and the mode would be 10.

35. Chelsea has $45 to spend at the fair. She spends $20 on admission and $15 on snacks. She wants to play a game that costs $0.65 per game. Write an inequality to find the maximum number of times, x, Chelsea can play the game.

Using this inequality, determine the maximum number of times she can play the game.

Answer:

She must spend less than or equal to $45. She'll spend 20 + 15 = 35 one time, which is the initial value. The only variable amount is how many times she spends 0.65.

The inequality would be, therefore, 0.65x + 35 < 45

Solving for x:

0.65x + 35 < 45

0.65x < 10

x < 15.38...

She can play the game at most 15 times.

36. A plastic storage box in the shape of a rectangular prism has a length of x + 3, a width of x − 4, and a height of 5.

Represent the surface area of the box as a trinomial in terms of x.

Answer:

The Surface Area of a rectangular prism is SA = 2lw + 2hw + 2lh

SA = 2(x + 3)(x - 4) + 2(5)(x - 4) + 2(x + 3)(5)

SA = 2(x² - x - 12) + 10x - 40 + 10x + 30

SA = 2x² - 2x - 24 + 20x - 10

SA = 2x² + 18x - 34

End of Part III.

More to come. Comments and questions welcome.

More Regents problems.

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