<p>Can you guys provide questions regarding rate,distance,and time? i want to practice
these type of questions and please type in the answer and explanation below.</p>
<p>Random question I wrote up:</p>
<p>Alfred starts at the origin and walks north at a rate of 3 mph. At the same time, Bailey starts at the origin and walks east at 4 mph. After how many hours will they be 15 miles apart?</p>
<p>A) 15/7
B) 3
C) 5
D) 7
E) 15</p>
<p>is the answer B?</p>
<p>^ Yes it is.</p>
<p>Yep, B is correct.</p>
<p>Here’s another: </p>
<p>A bicyclist races around a track, averaging 10 mph for the first lap, 20 mph for the next two laps and 30 mph for the last 3 laps. What was the cyclist’s average speed for the 6-lap race?</p>
<p>thats not really a rate,distance,and time question i want questions that include all of that</p>
<p>pckeller, that’s a great question. I certainly used distance = rate * time to solve it. </p>
<p>Did anyone find this intuitively easy? It only is to me after I worked it out. </p>
<p>Everyone knows the answer is not (10 + 2<em>20 + 3</em>30)/6, right?</p>
<p>@jeffgordon: yes, it was pretty easy. You can “assume” that a lap is 60 miles long (or 60 units, whatever). Then it takes the bicyclist 6+3+3+2+2+2 = 18 hours (using D = r*t) to complete the course. He rode 360 miles, so the average speed is (360 mi)/(18 hr) = 20 mph.</p>
<p>Good question, but those kind of “trick” questions seem to appear too frequently on the SAT…</p>
<p>How about this:</p>
<p>Every day I commute from my house to work. If I travel at an average speed 50 mph, I arrive x minutes early, but if I travel at an average speed of 40 mph, I arrive 2x minutes late. Assuming I always leave my house at the same time, what average speed should I travel so that I arrive exactly on time?</p>
<p>@jeffgordon: did you discover the intuitive way after getting 20 as the answer? True confession – when I wrote the problem, I didn’t intend for there to be an easy way. The answer caught me by surprise! But after I thought about it, it was clear:</p>
<p>In general, you can NOT average average speeds to get an average speed. For example, if you go to work at 20 mph and return at 40 mph, your average speed is not 30 mph. As has been discussed here MANY times, you need the harmonic mean – Xiggi’s formula. But as for WHY you cannot just take the averages, it’s because you spend more time at 20 mph than you do at 40 mph when you are traveling the same distance. You can only average average rates when you do them for the same amount of TIME, not the same amount of distance.</p>
<p>So when I made up this problem, I happened to choose an initial rate, then double it and then triple it. But I also made the distances double and triple as well. So you do in fact spend the same amount of time at each rate and you can in fact just average the three rates. </p>
<p>Having said all that, I should add that as a result of that little coincidence, my problem would NOT make it through the screening processs to get on an SAT. There would be too many people who got it right by just averaging 10, 20 and 30 without knowing why. That’s one of the things the experimental section is for. They don’t just want to see what percent of kids solve a problem – they also want to see that harder problems are answered correctly by higher scorers and incorrectly by lower scorers.</p>
<p>@rspence: also a nice problem. I can do it w/ algebra but I don’t see an intuitve way. I needed a system of 3 eqns…no real SAT problem will do that! In fact, I have never seen a real SAT question that required solving two simultaneous eqns. There’s always a wise-guy way to work around it…</p>
<p>Did you make this one up? And were x and 2x chosen arbitrarily? Also the switching from minutes to hours throws in another wrinkle…</p>
<p>@pckeller Yes I made it up arbitrarily. I figured, x and x would make it fairly easy. But you don’t need a system of three equations to solve it.</p>
<p>Without loss of generality we can “assume” the commute is 200 miles long (pick an easy number). Therefore at 50 mph you can arrive in 4 hr, but at 40 mph you can arrive in 5 hr. Note that a 4 hr commute results in being x min early and a 5 hr commute results in being 2x min late. You kind of have to intuit this one and say that x = 20 (since x + 2x should equal 1 hr or 60 min).</p>
<p>Therefore the desired length of commute is 4 hr 20 min = 4 1/3 hr. The desired speed is (200 mi)/(4 1/3 hr) = 600/13, approx. 46.2 mph.</p>
<p>If you want to generalize, you can assume that the commute is 200k miles long, where k is some real number. You can work it through using the same method, and the k’s should cancel out.</p>
<p>I don’t know if mine is messier:</p>
<p>Let d = the distance, v = the “right” speed, t = the “right” time.</p>
<p>d=vt</p>
<p>But if you travel too fast, you get there early, too slow and you get there late:</p>
<p>vt/50 = t - x/60</p>
<p>vt/40 = t + 2x/60</p>
<p>Then, double the first eq, add them…the x’s drop out and you can divide out the t’s as well…</p>
<p>Yeah, you should get 13vt/200 = 3t, 13v = 600, v = 600/13 (mph). Little messy setting it up, but it’s a good solution overall.</p>
<p>rspence, pckeller, you guys are good! </p>
<p>Re: post #9, “assuming” a convenient constant to work with is classic. And 60 is nice with all the factors. You know that’s why base 60 has survived the test of time 
</p>
<p>Re: post #11, it’s not good when one can reason incorrectly and arrive at a correct answer. I’ve seen a couple interesting occasions before, this one is cool. Also hadn’t seen the harmonic mean attributed to anyone before. Never heard of Xiggi.</p>
<p>I like your work on the commuting problem. </p>
<p>Here’s one that appeared in Liber Abacci (by Leonardo of Pisa) in 1202: A lion can devour a goat in 4 hours, a leopard in 5 hours, and a bear in 6 hours. If they share a goat and eat at the above rates, how long will it take to eat the goat? NOTE: I know this is a common type of problem nowadays, and we are taught a way to do this in algebra. But the method Fibonacci advocates, regula falsi, which you two appear to be familiar with, is much easier.</p>