<p>Ellen takes a trip that is y miles long in total, where y>20. She travels the first 15 miles at an average speed of 30 miles per hour and the rest of the trip at an average speed of 40 miles per hour. Which of the following represents the total time of the trip, in hours?</p>
<p>(A) 1/2 + y-15/40</p>
<p>Every car at a certain dealership is either a convertible, a sedan, or both. If one-fifth of the convertibles are also sedans and one-third of the sedans are also convertibles, which of the following could be the total number of cars at the dealership?</p>
<ol>
<li><p>Divide the 15 miles by 30 miles per hour to see that the first part of the trip takes 1/2 hour. Now for the next part, you will be dividing the distance by 40 mph, but that distance is no longer 15 miles – it’s y - 15 (because of the original y miles, you already completed 15 of them in the first stage).</p></li>
<li><p>Let x be the number that are both sedans and convertibles. Since that represents 1/5th of the convertibles, then the convertibles that are NOT sedans = 4x (so that the total is 5x and the x will be 1/5th of that total). Similarly, the sedans that are NOT convertibles must be 2x (so that the total is 3x and the x represents 1/3 of that total)</p></li>
</ol>
<p>So all together, you have x + 4x + 2x = 7x. I’m betting that none of the other answer choces were multiples of 7…</p>
<p>Pc gave you some algebraic solutions. I might do the first one by plugging in a value for y: let’s say y=55. For the first 15 miles at 30 mph, the total time is 15/30 = .5 hr. For the last 55-15=40 miles at 40 mph, the total time is 40/40=1. Thus the total time is 1.5. Now plug y=55 into all 5 answer choices. Since you only gave us choice (A), I’ll just plug into that one (but on the test you MUST do all 5). </p>
<p>As long as none of the other choices came out correct, this shows choice (A) is the answer.</p>
<p>Note that sharper students can certainly do this problem Pc’s way. Students having a little trouble might want to plug in a value for y as I did.</p>
<p>Yeah, this time I did algebra. I don’t really know why. Anyone who reads my posts or The New Math SAT Game Plan (my book) knows that I almost always favor making up numbers. But this time, it was such a short distance from the variable to the final expression…</p>
<p>But if you do make up #s, be sure you understand why DrSteve chose 55 – that is not an obvious choice but it really made the rest of your calculation easy. You wanted a number that was divisible by 40 after you subtract out the inital 15 miles. So 55, 95, 135 – any of those would be a good choice.</p>
<p>Let’s also be fair: a student who realizes that 55 is a good choice has already figured out the heart of this problem and is likely to be able to choose the correct answer algebraically.</p>
<p>Fwiw, this us yet another example of a bad question, and why one should avoid tests written by McGrawHill. It is extremely doubtful that such question would appear in this format on the SAT. </p>
<p>I will (again) offer a different solution. </p>
<p>This solution does not require using plug numbers nor using algebra. It only requires reasoning and reading the problem. </p>
<p>Ellen takes a trip that is y miles long in total, where y>20. She travels the first 15 miles at an average speed of 30 miles per hour >> **Mark 1/2 on your paper **
and the rest of the trip at an average speed of 40 miles per hour. >> Mark y-15/40 </p>
<p>Which of the following represents the total time of the trip, in hours? Haha, you just GAVE the answer. Total means to add up! </p>
<p>I don’t think I see a difference between your solution and mine. And I agree, I would not call this “doing algebra” – more like “writing algebra” in the sense that as soon as you write the expression, you are done! That’s what I meant when I said it was a short distance from the expression to the answer.</p>
<p>I actually think that the algebraic solution is pretty straightforward in this case, and I would probably go with that for students at about a 500 or higher (as per PC and/or Xiggi). I think it’s a little easier to pick numbers here, but definitely more time consuming. And Pc is right when he says that I chose a particularly nice value for y. I do however think it’s slightly easier to pick a “correct” value for y than to do the algebra. </p>
<p>And finally I want to point out that it’s not really necessary to choose such a nice value for y. Any value will work (of course y must be greater than 20). You would just need to use your calculator and the basic formula d=rt.</p>
<p>And Xiggi, I was torn between 35 and 55, but decided to go with the number that made the division easier.</p>