Skip to content

1/09 Math B, some good questions

January 28, 2009 am31 9:34 am

Even though Math B is winding down, I didn’t expect a good exam. And New York State did not fail to disappoint. 13 had ridiculous context. 33 included math from outside of the curriculum. Some reasonable questions, and answers:

2 .The expression \frac{5}{3+\sqrt{2}} is equivalent to…

Normal solution: multiply numerator and denominator by 3 - \sqrt{2} (the conjugate)

Cheat solution: plug the expression into a calculator. Equals about 1.1. Check the four answers. Only choice (3) is close.

Correct solution: \frac{15 - 5\sqrt{2}}{7}

4. What is the solution of the inequality x^2 - x - 6 < 0 ?

Cheat. Check a couple of numbers.  0 works, and eliminates 2 choices. 2 works and eliminates another choice.

Normal solution: Factor the quadratic: (x-3)(x+2) < 0. Sign changes at 3 and -2.

Correct solution: -2 < x < 3

20. The accompanying diagram [not here, just imagine] shows part of the architectural plans for a structural support of a building. PLAN is a rectangle and \overline{AS} \perp \overline{LN}

My comment: The context, as usual, is contrived, silly, and distracting. LN is a diagonal. AS is an altitude to LN at S.

Which equation can be used to find the length of \overline{AS} ?

(1) \frac{LS}{AS} = \frac{AS}{SN} (3) \frac{AS}{SN} = \frac{AS}{LS}

(2) \frac{AN}{LN} = \frac{AS}{LS} (4) \frac{AS}{LS} = \frac{LS}{SN}

I’ve got no cheat here. \triangle ASL ~ \triangle NSA. Sides are in proportion, so choice (1).

21. Solve for x: \sqrt{x+18} - 2 = 2

Transpose: \sqrt{x+18} = 4

Square both sides: x + 18 = 16, and x = -2. Checks, too.

25. What is the solution of the inequality |2x – 5| ≤ 11/2 ?

Abnormal solution: Factor out a 2: |x – 5/2| ≤ 11/2 This is the interior of a circle with center 5/2 and radius 11/2, so -3 ≤ x ≤ 8

26. The volume of Earth can be calculated by using the formula V = \frac{4}{3}\pi r^3 . Solve for r in terms of V.

Multiply both sides by 3/4pi, then take the cube root: r = \root 3\of{\frac{3V}{4\pi}}

28. Perform the indicated operation and express in simplest form:

\frac{3x^2+12x-15}{x^2+2x-15} \div \frac{3x^2-3x}{3x-x^2}

A former student of mine explained this one well: You factor, you cancel. She used to abbreviate it on the board, and her kids never forgot (UF, UC). I choose not to write that in class. Here goes:

\frac{3(x+5)(x-1)}{(x+5)(x-3)} \div \frac{3x(x-1)}{x(3-x)}

\frac{3(x+5)(x-1)}{(x+5)(x-3)} \times \frac{x(3-x)}{3x(x-1)}

Everything cancels. That x-3 over 3-x gives -1. The answer.

31. Find the roots of the equation x^2 + 7 = 2x

Cheat solution. Complete the square. x^2 - 2x + 1 = -6

(x - 1)^2 = -6

x - 1 = \pm \sqrt{-6}

x = 1 \pm 6i

34. Given PROE is a rhombus, \overline{SEO}, \overline{PEV}, \angle SPR \cong \angle VOR

Prove \overline{SE} \cong \overline{EV}

Ok, I am missing the diagram, but the description is sufficient.

Take those congruent angles, subtract the congruent opposite angles from the rhombus, that gives us a pair of congruent angles (SPE and VOE)

PE = OE (sides of a rhombus)

\angle SEP \cong \angle VEO (opp angles)

Now the two little triangles are congruent by ASA, and what we want is congruent by corresponding parts.

Advertisements
5 Comments leave one →
  1. daettil permalink
    January 28, 2009 am31 10:37 am 10:37 am

    I’m curious. Why is completing the square a “cheat solution”?

  2. January 28, 2009 pm31 5:12 pm 5:12 pm

    I called it a cheat, but you are right, it’s not. Just, most kids will go straight to the formula. I get the sense that many think it is unfair that completing the square works… I used “cheat” on that one in jest.

  3. Anonymous permalink
    January 28, 2009 pm31 7:07 pm 7:07 pm

    Can you put up more questions?

  4. Anonymous permalink
    January 29, 2009 am31 8:29 am 8:29 am

    If you don’t mind me asking, where are you getting these questions, and can you please post more. And about the log regression one, do you think they should excuse that one?

Trackbacks

  1. My favorite bad exam target: Math B (and a puzzle) « JD2718

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: