<p>A car crash causes a family 6000 worth of damage to their car, which gets an average of 20 miles per gallon. They can buy a new car for 7000 that will give 30 miles per gallon. Since gas price is fixed a 1.50 per gallon, how many miles will they have to drive to make the buying the car more profitable that fixing the old one? (Round to the nearest mile)</p>
<p>The answer choices are :
1. 13,333
2. 16.667
3. 40,000
4. 240,000
5. 280,000</p>
<p>Start with choice (C) as your first “guess”</p>
<p>So we’re guessing that the answer is 40,000 miles. </p>
<p>The old car will need 40,000/20 = 2000 gallons of gas. So the cost for the old car is 6000 + 2000*1.50 = $9000.</p>
<p>The new car will need 40,000/30 = 1333.3333… gallons of gas. So the cost of the new car is 7000 + 1333.3333*1.50 = $9000.</p>
<p>Since these 2 answers agree, the answer is choice (C).</p>
<p>Remark: Make sure you do the above computations in your calculator. Don’t round the decimal during your computations.</p>
<p>Here is an algebraic solution (note:I do NOT recommend you do it this way on the SAT - my method above is much better during the test. It is useful to UNDERSTAND this method however.)</p>
<p>6000 + 1.50 (x/30) = 7000 + 1.50 (x/20)</p>
<p>Multiply both sides of the equation by the lcd (which is 60)</p>
<p>360,0000 +3x = 420,000 + 4.5x</p>
<p>1.5x = 60,000</p>
<p>x= 60,000/1.5 = 40,000</p>
<p>Remark:</p>
<p>x represents the unknown, that is x is the number of miles they have to drive so that buying the car is more profitable than fixing the old car (technically this is incorrect - we actually found the value at which the two costs are equal - this is a flaw in the statement of the problem, not in the method of solving the problem - I’m guessing this is NOT an actual SAT problem, but comes from some test prep company that is not the College Board).</p>
<p>Very interesting, as I was just about to post an answer that started with this. </p>
<p>As always, there are plenty of paths to the right answer. Some are safer; some are faster.</p>
<p>You could plug in a figure, but I do not recommend to this because this is typical example of a … timesink. It is easy to solve. From start to finish, it takes about 15 seconds. Anyone who starts plugging in would still be calculating his or her FIRST plug-in by the time I start the next question. ETS also EXPECTS and HOPES for people to grab that TI-89 and start playing. That is why calculators should be the weapon of LAST resort. Your brain and pencil work a lot faster. By the way, I talk about ETS although I doubt that this question originated at ETS. </p>
<p>It is also important to note that there is NO need to play with the cost of the cars. This is an attempt by ETS to make your equations more complicared (as shown above). The 7,000 and 6,000 should NOT be part of your (easy) equations.</p>
<p>Please remember that LONG problems (almost) always have a short and easy answer. This is all you need on your paper. Mine only has the numbers in bold. </p>
<p>Cost to offset is $1,000
Cost per mile at 20 mpg = $1.50 / 20 = .075</p>
<h2>Cost per mile at 30 mpg = $1.50 / 30 = .050</h2>
<p>Savings per mile = .025 </p>
<p>Answer is 1000/.025 = 40,000.</p>
<p>===========</p>
<p>PS Fwiw, had I decided to plug in a number, I would have used a different number. I know I could use 40,000 as it is C, but I always try to do the SIMPLEST plug-ins, and if I can, do this in my head. Writing a lot is a waste of time. </p>
<p>In this case, 30,000 because of the easy calculations with 20 and 30. If I can, I do not want to have fractions and decimals. My plug would have shown this</p>
<p>30,000 at 30mpg 1,<a href=“mailto:000@1.50”>000@1.50</a> = 1,500
30,000 at 20mpg 1,<a href=“mailto:500@1.50”>500@1.50</a> = 2,250 ===> Savings is 750 and it should be 1000</p>
<p>Thus 30,000 x 4/3 (same proportion a 1000/750) and the answer would be 40,000</p>
<p>But, again, this is **NOT the best way **to solve this easy problem.</p>
<p>Yea that’s correct. This is not from college board. you seem very good at these stuff. Can you give me some pointers on how to tackle ETS problems? Thank you very much for the help.</p>
<p>Xiggi –</p>
<p>I have to agree with Dr Steve on this one. I understand your solution and I agree it is insightful and fast. A kid who is heading for really high scores might do it that way. But of the kids I teach, only a small fraction will think of your way. And if they don’t have that kind of “math mind”, then this particular approach will not generalize. I’ve had students who try to memorize the problem type: “OK, so if they give me different mileages and a gas price, I can divide to find the cost per mile with each mileage and then see how far it takes for the differential to offset the cost difference. Got it.” But in fact, they don’t have it enough to apply it in even a slightly different situation.</p>
<p>On the other hand, the try-the-answers approach takes maybe a minute and requires almost no insight at all. It’s like a vacation in the middle of the test. Especially if you look at the numbers – a and b were just to small to bother with. C was the first one I tried and I was pretty sure it was going to work. And this method works on a great variety of problems. I absolutely include it in my course and book. (I think it’s in just about every SAT book out there!) Now it is true that sometimes when I am showing this method, a sharp student will look at the problem and say “couldn’t you just…” – to which I almost always reply: Yes, if you think of it. My method works for the times when you didn’t have the flash of insight.</p>
<p>Pckeller, as I said there are several paths to the right answer. Just as you, I believe that there are times to use the (often) efficient method of plugging in numbers. I agree that books almost always include the method in their panoply of answers. This said, it is qually obvious that an excessive reliance on this method produces a LOT of bad coaching. Take a look at the current offerings of Showme/Easel for the iPad for an illustration.</p>
<p>There are times, however, when setting up the correct equations is straightforward and simple. This is one of those times. I am sure how someone could fault an answer that requires a few seconds and correct reasoning. What am I missing here? </p>
<p>As far as looking at the numbers, I could have written that I ALSO would have guessed 40,000 because the other four were poorly selected (again showing how poor non-ETS questions are.) But it would not help anyone to read that one could pick the correct answer because the question was so poorly written in the first place.</p>
<p>Fwiw, how would approach the same problem with the following choices</p>
<p>The answer choices are :
<p>Still picking the middle value at the start?</p>
<p>This is an interesting discussion, and everyone has good points. Let me first say that I like Xiggi’s solution, BUT I agree with pckeller that I would generally only teach Xiggi’s solution to more advanced students (if a student comes to this solution themselves, then of course that’s great). </p>
<p>Due to the fact that the original poster (OP) posted this question (which I would consider about level 2 or 3 out of 5), my intuition tells me that my method of “starting with choice (C)” is the best for him. This should answer your question in your last post…if those were the answer choices I would still start with the middle one - $42,000 in this case. For students up to about a 550 (PSAT or SAT score) I always recommend starting with the middle answer choice (choice (C)) UNLESS the word “least” or “greatest” appears in the problem (sometimes it helps eliminate 2 other choices, sometimes it doesn’t).</p>
<p>As far as the issue of “wasting time,” I actually find that when I’m using the right strategies with the right level of student, the appropriate amount of time is spent. There’s nothing wrong with a 400 student spending a little extra time on each question since I don’t want them to get more than two thirds through the section anyway. </p>
<p>The one issue I see with Xiggi’s solution is that the strategy is “problem specific.” By this I mean it only works in a very specific kind of problem which is fairly rare. This means if I teach this strategy it is unlikely the student will be able to implement it on the test unless I go over it at least a few times. The strategy of starting with choice (C) works on so many problems that it’s worth knowing by everyone - a standard method for lower scoring students, and a “fallback” method for very strong students. If I were going over this problem with a 600 or higher student, then I would use Xiggi’s method.</p>
<p>And finally to demonstrandum, keep posting your questions, and I (and others) will keep answering them. If different people post different solutions, choose the solution that works best for you. But always make sure to keep redoing the problem at least once a week until you can get the answer on your own.</p>
<p>I think we agree about most of this. And yes, the number choices with this problem were mostly unreasonable, marking it as non-ETS.</p>
<p>Just one other note: I didn’t start with c because it was in the middle. I started with c because this time it looked best. In general, when you are plugging in answers, I recommend that you start with whatever looks easiest. My feeling is that if you are using this method, it’s because you don’t really know what’s going on yet. While you are in the process of figuring it out, why not give yourself easier arithmetic? (Xiggi – you took that idea even farther – you plugged in a number that wasn’t even an answer choice! Nice…)</p>
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<p>Many SAT problems or faux-SAT problems like this one can be solved most efficiently with algebra, without a calculator, and without resorting to using answers, etc. In my experience, however, many SAT students are weak enough in algebra that the efficient use of it is not practical for a problem like this. The answers strategy is certainly slower, but these students only need to be working on about half the problems that strong students need to. Xiggi, your solution is excellent and could warm a bitter math teacher’s soul, but may not be the best way to go for everyone, as straightforward and simple as the solution seems to you. I think that’s what drsteve and pckeller are trying to say.</p>
<p>I got a question; since it asks when it will be MORE profitable isn’t the answer technically D then? Because at 40,000 miles both cost exactly the same: 9000.</p>
<p>Needshelp - As I stated in my post above this question is flawed. What the answer should actually be is not really clear. The words “have to” imply that there should only be one answer. But any number greater than 40,000 would work. Thus there isn’t even a least answer (for example 40,000.0000001 is a possible answer). In any case since they say to round to the nearest mile, and an answer just slightly bigger than 40,000 would work, it follows that 40,000 should be the answer. But I agree - it is a poorly written question. Clearly it is not an actual SAT question and was not written by the College Board. Unfortunately many SAT Prep books have such flawed questions in them.</p>
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<p>I do understand what other say here. For the record, I was indeed answering this ONE question and suggested that it was (IMHO) the best way to attack THIS problem. I was not trying to design a teaching method or implying that this was a universal answer. </p>
<p>I really believe that almost all SAT questions can be solved through various methods. I really enjoy to find a solution that allows someone to solve a problem without calculating a single thing, and I particularly enjoy reading such solutions when presented by others. I particularly like algebra problems that can be solved graphically or through a mere visual evaluation. </p>
<p>Here’s an example of a GRE QC problem, (sorry for the lack of relevance)</p>
<p>Col A 100,210 x 90,021
Col B 100,021 x 90,210</p>
<p>Is A < B, or A > B, or A = B, or is it undefined?</p>
<p>Without a calculator, this is hard to solve in 60 seconds or less. However, with a bit of logic AND the knowledge of mathematical properties of … geometry, this can solved in a few seconds. </p>
<p>Fwiw, I also know that many solutions that appear simple to one remain nebulous to someone else. Hence the value of different approaches.</p>
<p>^ It might help to think of those products as the areas of rectangles, and then notice that the two rectangles have the same perimeters…</p>