<p>The measure of the largest angle in a certain triangle is twice the sum of the measures of the remaining angles. What is the measure of the largest angle.
(A) 45
(B) 60
(C) 90
(D) 120
(E) 150
The correct answer is (D), but I cannot figure it out. Any help would be greatly appreciated.</p>
<p>Assume that the largest angle has measure x deg. Then the sum of the other two is 180-x. We have x = 2(180-x), or x = 120.</p>
<p>If you don’t like the algebra, you can also do this by “Starting with choice (C).”</p>
<p>If we start with choice (C) and guess that the largest angle has measure 90 degrees, then the sum of the other two is 180 - 90 = 90. This is NOT twice the largest.</p>
<p>Let’s try (D) next. If the largest angle has measure 120 degrees, then the sum of the other 2 is 180 - 120 = 60. So twice the sum of the measures of the other 2 is 120. This matches so that the answer is choice (D).</p>
<p>Or just use reasoning:</p>
<p>The largest angle in a triangle is twice the sum of the remaining angles simply means that the largest angle is none other than 2/3 of the sum of all angles.</p>
<p>Hence … 2/3 of 180 is 120! </p>
<p>By the way, DrSteve, I (finally) found a problem where plugging number is one of the best alternatives.</p>
<p>[SECTION</a> 3 QUESTION 17 MATH SAT QAS MAY 2013 - SAT May 2013 Question and Answer Service - SATQuantum](<a href=“http://www.satquantum.com/satqasmay2013journal/section-3-question-17-math-sat-qas-may-2013.html]SECTION”>http://www.satquantum.com/satqasmay2013journal/section-3-question-17-math-sat-qas-may-2013.html)</p>
<p>Fwiw, there is a straightforward approach that yields the equation (x^2+9x=18) to solve. However, an interesting approach is to recognize that the square in the middle of the table is 9, and that the “complete” figure is a rectangle of 27, which area is thus (3+x)(3+2x). </p>
<p>That is where the plugging in really shines as x =1 yields 20 and x =2 yields 30.</p>
<p>Now, now Xiggi… It is not hard to find lots of problems where plugging in is the best strategy for MOST solvers. You can find them on just about every QAS. Nothing beats a lightning flash of insight. But plugging in numbers is often an almost mindless way to grind out an answer in a way that is practically relaxing – like a vacation in the middle of the test. Just saying…</p>
<p>Hey, I used words such as “where the plugging in really shines” as a true token of my appreciation for the method. My point here was pluggind in numbers in THIS case worked really well. As an example, it took me about 40 seconds to solve the algebra way, and half that time with trying 1, 2, and then 1.5. </p>
<p>And, I happen to agree that the plugs might amount to a leisurely walk in the park. All roads lead to Rome, or all paths lead to the top of the mountain. Unfortunately, taking to the slow boat to China is often exactly why the demons in Princeton hope to see the test takers do on the SAT.</p>
<p>All in all, I am far to be opposed to rely on plugging in numbers. It is, however, a method that is, except for the rare occasions, the least effective method.</p>
<p>PS I’d be interested to know how you and DS approached the linked problem. ;)</p>
<p>Actually, I did what you did! When I got (2x+3)(x+3)=27, my laziness kicked in. If I had a ti89 within arm’s reach, I would have used it. But instead, I tried the same numbers you did even in the same order.</p>
<p>For the record, though I enjoy (and recommend) SATQuantum’s videos, I’ve never actually used the quadratic formula on the SAT. Still, it’s a great site. When I do my video presentations on how to self study, I always include a link to that site. (And of course, this one!)</p>
<p>Fwiw, I wrote down the equation incorrectly. It should be 2x^2 and x^2. </p>
<p>I also forget to mention a different approach. </p>
<p>The area between the table is 3 x 3 or 9. The tables account for 18. If one could divide the 3 x 3 in four squares … you could also divide the tables in 8 squares. This means that each square should be 18/8 or 2.25. The square root of 2.25 is 3/2. And you’re done. </p>
<p>Actually, with a bit of logic, you could realize that each of the four squares inside the table area is also 1.5. </p>
<p>Don’t think there is anything faster. :)</p>
<p>@xiggi,</p>
<p>Thank you for posting the two alternate methods for this difficult problem. I agree that solving the (2x+3)(x+3)=27 with either a calculator or guessing nice numbers would do the job. The only problem is that this is a grid in problem, and the answer could be say 1.6 or 2.8, which would be a pain to guess by plugging numbers. In general, SAT uses “nice” numbers for answers, but this is not guaranteed, and I would personally not bank on that. The calculator approach or solving the quadratic by factoring would not have this disadvantage. </p>
<p>The second approach of recognizing that the area of the table is twice the area of the square is also very clever. But again I wonder if one would be guaranteed of such simple relationships between the areas. What if I drew the same table and said the area of the table is 170 square feet and instead of the 3 feet by 3 feet square we have an 8 feet by 8 feet square. In this case, the area of the inside square of 64 square feet is not an integer multiple of 170 square feet and it would be impossible to make a jump to the answer x=5 ft using the method you proposed. (Perhaps there is a way but I don’t see it currently). So again I am not sure how easy it would be to always look for such connections and rely on such approaches. I know for sure that my odds of doing this under exam conditions would be low.</p>
<p>Having said all that I appreciate you presenting the other two angles of approaching this problem, and I am sure that developing habits to look for such connections would only strengthen one’s problem solving skills.</p>
<p>
</p>
<p>Humm, typing on a iPhone when running around can be problematic. Something when amiss in the above sentence. What was meant is that the four squares that have a total area of 9 should have a side (which is x) of 1.5. The area of each little square is, of course, 2.25.</p>
<p>@Xiggi</p>
<p>Is this a grid in problem? </p>
<p>When I just looked at the problem, this is the first way I solved it:</p>
<p>I first observed that the area of the little square is 9 so that the area of the big square is 9 + 18 = 27. So we have (2x+3)(x+3)=27. I then used my graphing calculator (yes I actually stood up, walked across the room and got it - a rare thing for me). I put Y1 = (2x+3)(x+3) and Y2 = 27, changed the viewing window to go -50 to 50 in both the x and y direction, and used the intersect function to get x = 1.5.</p>
<p>Now, after I did it this way, I realized guessing isn’t a bad way to go either:</p>
<p>If we guess that x=1, then we get the total area is (2+3)(1+3) = 20. If we guess that x=2, then the total area is (4+3)(2+3) = 42. Finally, x=1.5 gives (3+3)(1.5+3) = 27, bingo!</p>
<p>Guessing is a bit risky here since there is no guarantee the number is nice.</p>
<p>I then also did it another way - by moving the pieces in the picture around I noticed that you can form a 3x by 3 rectangle and 2 x by x rectangles to get a total area of 9x+2x^2 which is equal to 18. So 2x^2 + 9x - 18 = 0. This is not too hard to factor, and the positive solution is x = 3/2.</p>
<p>I also notice you can form one long rectangle that is x by (3+3+3+x+x) or x by (9+2x), and continue as above.</p>
<p>Okay I wrote all this before looking at everyone else’s solutions, and I don’t think I contributed anything new, but this is one of those rare instances where you get to experience my thought process as it’s happening. So I’ve decided not to edit this post at all.</p>
<p>By the way, I would never use the quadratic formula for anything except getting decimal approximations. As long as the coefficients of a quadratic equation are not too messy, completing the square is much faster and cleaner than the quadratic formula. </p>
<p>But I have never taught either of these to an SAT student except one that was already in the mid 700s. </p>
<p>And by the way, I wouldn’t consider what we did in this problem “plugging in” since there are no answer choices. I would call it “guessing.”</p>
<p>Dr Steve, to be fair, after I solved it with one method, I applied a number of others. Always looking for shorter alternatives. I probably compiled all the above, in addition to one that was based on playing on 3 x 9 = 27 and introducing “multipliers” and this lead to an easy (3x2) * (9*1/2) = 27 and the answer! </p>
<p>It is indeed a grid-in and I forget if the figure drawn is to scale or not. I remember also trying to “flip” around the two legs of the table … as I figured that it would have to form a new rectangle with a total of 3 x 6 = 18. The hunch was that 2x had to equal 18. But that might not have been available with a proof. </p>
<p>Fwiw, SQ, I think that each problem has a weak spot. Your new problem with a table being 170 and the middle section 8 x 8 could be solved with some … elegance.</p>
<ol>
<li>Add 8 x 8 to 170 to get 234.</li>
<li>Factor 234 to get 1, 2, 3, 6, 9, 13, 18, 26, 39, 78, 117, and 234</li>
<li>Realize that only 9, 13, 18, 26 could be the sides of a rectangle of 234 if the middle section is 8.<br></li>
<li>13 x 18 is a very plausible answer and since it is a grid-in, I can pick one answer</li>
<li> And x = 5 it is. </li>
</ol>
<p>All in all, while it is a different relation than the one with the original problem, the theory of using the entire area (including the empty middle) is quite similar. </p>
<p>For the calculators’ gurus, it is also obvious that the equation (8+x)(8+2x)= 234 is all they’d need. And the same for the “pluggers” or “very smart guessers!” :)</p>
<p>@Xiggi</p>
<p>Here is an example of a problem where I thing “plugging in” is one of the better ways to solve it. I’d like to hear your thoughts, and possibly “steal” a really clever solution from you. :)</p>
<p>When each side of a given square is lengthened by 3 inches, the area is increased by 45 square inches. What is the length, in inches, of a side of the original square?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7</p>
<p>I’m not a huge fan of plugging in, but I’ll do it in rare cases, like possibly the one above.
An alternative to plugging in is to write out the squares:
3^2 = 9
4^2 = 16
5^2 = 25
6^2 = 36
7^2 = 49
8^2 = 64
9^2 = 81
We see that 6^2 and 9^2 differ by 45, so the answer is 6.</p>
<p>In reality, I would most likely solve it algebraically (personal preference). After all, it reduces to 6s + 9 = 45.</p>
<p>At first blush, the algebra is quite easy:</p>
<p>x^2 + 6x + 9 = x^2 + 45
6x + 9 = 45
6x = 36
x = 6</p>
<p>I am not sure there is a shortcut that would drastically be faster, with perhaps the exception of being in the SAT mode and having some standard squares memorized. In this case, 36 and 81 came to mind, but that might have been a mental bank fluke. An orderly approach might have been the following:</p>
<p>Add 3 and square up, and compare to original squares</p>
<p>(A) 3 > 36 - 9
(B) 4 > 49 - 16
(C) 5 > 64 - 25
(D) 6 > 81 - 36
(E) 7 > 100 - 49</p>
<p>But that is … plugging in! And I think that your approach would be one that starts with C. :)</p>
<p>Perhaps, there is something in the pattern (the differences increase by 6) that could lead to a shortcut, but nothing “popped” in a reasonable timeframe.</p>
<p>A similar solution: just draw a picture. Suppose we have a square of side s, and we extend the sides of s by 3. We split the region of area 45 into three rectangles of area 3s, 3s, and 3^2 = 9. So 6s + 9 = 45, s = 6.</p>
<p>However, I feel that all these solutions take about the same amount of time since the problem is relatively straightforward. So it’s more a matter of personal preference which one you should go for (or, which one you’re less likely to make a mistake on). For harder problems, finding the best solution will save more time than that on an easier problem.</p>
<p>@MITer</p>
<p>I would personally also do this algebraically. But my point is that most students can’t handle (or even formulate) the algebra necessary to solve this problem. For most students plugging in (starting with choice (C)) seems to be the best way to solve this problem.</p>
<p>I also like your last solution, but I do think it’s a bit too clever for the average student to come up with in the moment.</p>
<p>I agree that it’s a matter of personal preference which method to use here, but I think it’s important for every student to know about plugging in to guarantee that they can solve this problem.</p>
<p>The easiest and the best explanation…</p>