Algebra Problems of the Day (Integrated Algebra Regents, January 2012)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

Integrated Algebra Regents, January 2012

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


37. The sum of three consecutive odd integers is 18 less than five times the middle number. Find the three integers. [Only an algebraic solution can receive full credit.]

Answer:

Consecutive integers are one after another, like n, n+1, n+2... Consecutive odd integers are each two more than the previous one, like n, n+2, n+4, where n is an odd integer.

Change the sentence in the question into an equation:

n + n + 2 + n + 4 = 5(n + 2) - 18

3n + 6 = 5n + 10 - 18

3n + 6 = 5n - 8

14 = 2n

7 = n

The numbers are 7, 9 and 11.

You could also have used n - 2, n and n + 2 and save yourself some work:

n - 2 + n + 2 = 5n - 18

3n = 5n - 18

-2n = - 18

n = 9

The numbers are still, 7, 9, and 11 becuase you used n-2, n and n+2.

End of Exam.

More to come. Comments and questions welcome.

More Regents problems.

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