<p>This is a math question from one of the practice tests in the BB:</p>
<p>The figure is an arrangement of 10 squares - 4 in the bottom, then three over those 4, then two and one on top. They are all lined up to the right. </p>
<p>Q: "The figure above shows an arrangement of 10 squares, each with side of length k inches. The perimeter of the figure is p inches. The area of the figure is a square inches. If p=a, what is the value of k?"</p>
<p>I keep getting k=0 so I really can't solve this one. Help is very much appreciated.</p>
<p>The length of one side of a small square is k. Therefore the perimeter of the figure is 16k. The area of one small square is k^2. Since there are 10 small squares in the figure, the total area is 10k^2. We are given that the perimeter is equal to the area. So we solve the following equation for k.</p>
<p>10k^2 = 16k
10k^2 – 16k = 0
2k(5k – 8) = 0
2k = 0 or 5k – 8 = 0
k = 0 or 5k = 8
k = 0 or k = 8/5<br>
We reject k = 0 (since the length of the side of a square must be positive). The answer is therefore k = 8/5 or 1.6.</p>
<p>oh i thought that since they said “the area of the figure is a square inches” they meant that the area was a^2 (stupid - slaps hand against forehead) Thanks!</p>
<p>As a physics teacher, I have always disliked this problem. It seems like they are saying that a perimeter is equal to an area, something that is dimensionally impossible. I know that they actually worded it quite carefully to avoid saying that, but still…</p>
<p>CHD’s method is just a bit dangerous. Dividing by k is only allowed if k is not 0. And in fact, k = 0 is a solution to the equation. It’s true that we need to reject that solution in this problem (otherwise the figure would cease to be), but in general it is possible to get a Level 5 question wrong if you forget about this solution.</p>
<p>@pckeller I actually have no issues with this problem. They are saying that p = a, not that the perimeter is equal to the area. I certainly see your point, but the problem is perfectly sound mathematically, and as a mathematician, that’s all that matters to me. </p>
<p>But then doesn’t the rule that forces you to reject the Zero response for the equation … also allows you to divide by k? In both cases, we know that k > 0 from the problem statement. </p>
<p>@Xiggi Yes of course. For this question it’s perfectly fine to divide by k (and that’s how I would do it).</p>
<p>My point is that most students divide by k blindly without realizing that there even could be a problem (in fact, this fallacy is used in many “proofs” to show that 0 = 1). </p>
<p>For example, on a college precalculus test, suppose I were to ask to solve the equation k^2 = k. Many students would answer k = 1, and would get the question wrong. </p>
<p>In other words, a large number of students are still making this mistake in college. They are simply not aware that when dividing by an expression with a variable in it there may be a case where they are dividing by 0. This special case needs to be considered separately, or the equation needs to be solved without dividing. </p>
<p>Simply telling students to divide by k without more explanation is doing them a disservice - most students will then think it’s ALWAYS okay to do this, and may wind up getting a level 5 question wrong on their actual SAT.</p>
<p>@Dr Steve, for the record, I liked the fact that you did show the extra steps of creating an equation leading to the two possible answers. I liked because it could help someone in the future. </p>
<p>My main point here, if there is one, is that the correct reading of the problem statement helps eliminating … impossible answers. In this case, because of the geometry context, we should KNOW before testing any possibility that zero is not a possible answer. This is a bit along the lines of the plugging of numbers when people do try ALL five choices when 2 or 3 are immediately excludable. </p>
<p>Fwiw, there is the issue of having to pick an answer that is correct. In fact, were the equation to yield two possible answers, it would not change anythig as long as one works for a grid-in. </p>
<p>I understand that the division by k if k=0 is not an universal approach, but as you said, in this case it works elegantly. When it comes to the SAT, and in a departure from the typical HS math, the key is to find an answer that works for that particular problem with the fewest steps possible. I happen to think that many do waste a a lot of precious time in verifying that other possible answers might exist and try them. Of course, we should also remember that in the case of MC there is ONLY ONE answer that works. If one finds two, chances are that he or she made a careless errors. Possibly by dividing by zero :)</p>