38 Replies to “WitCH 103: You’re Blocked”

  1. Okay. What’s the issue? The questions are answerable, but not for the non-mathematically minded student.

    1. It is incredibly confusing, and most of it’s irrelevant. Took me a few minutes to understand what was actually going on.

      This is one of those questions that tests reading comprehension more than mathematics. And I’m bloody sure on purpose, too.

      The actual mathematics is pretty straightforward and trivial once you parse through the mess of a question to get what little’s left.

      1. Thanks, James. I gave up: my girlfriend figured out what the set-up was. Do you really think it was on purpose? I just took it to be incompetence.

        1. I’d like to think that there are limits to incompetence…

          I have a hard time believing that someone could write this question and honestly think that it’s clear and readable. It’s like they tried to make a trick question, but instead of adding a “trick”, they just made it unreadable.

            1. They have to rank the students somehow…

              For what it’s worth, I’ve seen (or rather, been subjected) to worse stuff in SACs… unfortunately we weren’t allowed to keep them.

              1. Of course. And the whole SAC system is a recipe for mediocrity. But, given the idiotic system, a stupid question on a SAC is understandable. A question(s) this stupid on the GAT is a disgrace.

              2. Thats a shame. I think schools should hand back all assessments so that students can learn from them and reflect on them and refer back to them. I have heard of lots of schools that keep sacs just so they can keep using them year after year. I think its lazy and Ive heard about more than one school getting a nasty bite on the bum.

                You said “they have to rank students somehow”. Thats what exams are for!

  2. The algebraically correct answer to 46 is B, 4L/5, but this implies you could get wall lengths with of some metres plus 0.2, or 0.4, or 0.6, or 0.8 metres. Impossible if using blocks 0.5 metres long.
    And just annoying and dumb.

    1. That’s pretty funny. Thanks, Ferdy, I hadn’t noticed that. I guess it’s slightly dumber than you suggest, since L need not be an integer.

  3. Just had this problem in the Specialist Maths 3 and 4 textbook. Myself and my teacher think its unsolvable- it has too many unknowns. Wonder if this is the influence of VCAA’s erroneous exams?

        1. Thanks. I’m out and about but will look. In general best to just email me first. Then I’ll look to see if it’s worth hammering.

          1. I’ll hammer it because dynamics and statics is no longer in the Study Design. I dont support this being deleted, but it was. Are you working from the new edition textbook or an old edition? It shouldn’t be getting taught anymore?
            PS Maybe the fifth constraint is (T1)^2 = (T2)^2 + (T3)^2 since its a righteous triangle.

        2. Just checking, is this stuff still examinable? And, can you tell me the answer(s) given in the textbook? (I don’t think the problem, or topic, is in the current edition.)

          Too tired to look today, but I’ll look tomorrow.

            1. Thanks, fred. I’ll look tomorrow (if A doesn’t look earlier). And again, why are you bothering to look at this, given those morons at VCAA have killed the topic?

              1. According to those answers, Tac (the tension in the string upon which the 6 kg mass is suspended) is zero. This does not seem reasonable.
                (T1)^2 = (T2)^2 + (T3)^2 cannot be used as a fifth equation because of the horizontal forces so I dont know where a fifth equation comes from (but Im confident if there is one it wont lead to Tac = 0).

                  1. OK, so equilibrium of the system requires 0 leq Tac < 6g so Tac = 0 is possible (that is the string from A to C is not needed). But I dont see how Tac = 0 is justified here and hence the books answers.

                1. The problem is the diagram. The string goes from A to the mass at B and then to the mass at C. I can only assume that the line AC was drawn as a guide, and has ended up just muddying the waters. There is no string directly connecting A to C.

                  1. Yes thats the likely explanation. Thankyou Banjaxed. I wonder how much time that question has wasted (and is still wasting) over the years. How hard for the guideline to be dashed?

                2. Thanks, Banjaxed. Yes, that sounds like a, and maybe the, mistake. I guess the string still could be there, with either zero tension (unique answer) or positive tension (many answers).

                  1. Yes, I think that’s their point. If there is no string then there is just one solution. If there is a string (as the diagram incorrectly suggests) then there is a range of solutions, including a solution with zero tension in the AC string.

                    1. So, given there is no string from A to C, what is the answer, and how is it worked out? Its been driving us nuts.

                    2. Still not sure why you and your teacher are working on dead material. But, trusting my infallible commenters, I assume the answer is as given in the text. Did you+teacher assume no AC string and still come up with multiple solutions?

                    3. 6 kg mass:
                      Vertical equilibrium: Tbc/2 -6g = 0
                      Horizontal equilibrium: F2 – sqrt{3}Tbc/2 = 0

                      4 kg mass:
                      Vertical equilibrium: sqrt{3}Tab/2 – Tbc/2 – 4g = 0
                      Horizontal equilibrium: Tab/2 + sqrt{3}Tbc/2 – F1 = 0

                      Books answers are correct.

  4. Can “estimate” mean the same as calculating the number exactly?

    It is a crime that they laid out the general formula and a particular example with no separation.

    1. Thanks, Tungsten. That “estimate” is not the worst aspect of the question, but it’s probably the most irritating, since it’s so gratuitously stupid. And yes, the layout of the “equation” (singular) is atrocious.

  5. Of the many things I could raise (for example, I pity the ESL student), I will simply say that there is no mention anywhere that the “simple wall formula” only applies for a wall that is two blocks ‘deep’ and that it is totally confusing to have walls two metres long but talk about the number of blocks per metre. In fact, I think its unfair to not mention the ‘depth’ of the walls at all. And I pity the student trying to visualise the triangular wall from the front on view. Why not include an attempt at a ‘three-dimensional ‘ picture. It would have been much better to define a triangular number instead of giving this blockheaded context.

    1. Thanks, A. I agree with all. In particular, the two-meter diagram to illustrate the per-meter computation is hilarious. As a consequence, they have 16 blocks in the diagram, but they’re not the same 16 blocks that they’re counting in the computation. That is, in its way, masterful.

  6. 45. How many more concrete blocks are needed to increase the height of this wall by 2 concrete blocks?

    Um… 2?

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