2 math questions

<p>1) A fellow travels from city A to city B. The first half of the way, he drove at the constant speed of 20 miles per hour. Then he increased his speed and traveled the remaining distance at 30 miles per hour. Find the average speed of the motion.</p>

<p>(A) 24 --> Answer
(B) 25
(C) 26
(D) 28
(E) 30</p>

<p>2) Suppose we're driving a car from Amherst to Boston at a constant speed of 60 miles per hour. On the way back from Boston to Amherst, we drive a constant speed of 30 miles per hour. What is the average speed (mph) for the round trip?</p>

<p>(A) 40 --> Answer
(B) 43
(C) 44
(D) 45
(E) 50</p>

<p>For #1:
Average speed = total distance/total time
Total time = (d/rate1) + (d/rate2)</p>

<p>In this case, let’s have half the distance = d, which means the distance of the whole trip = 2d. The time equals (d/30 + d/20). Therefore, the average speed = 2d/(d/30 + d/20), which simplifies down to 24</p>

<p>For #2:
Use the same formulas but just change the times: 2d/(d/30 + d/60), which simplifies down to 40</p>

<p>I hope that made sense. If not, let me know and I can clarify.</p>

<p>(1) To make the problem a bit easier I would choose a distance. Let’s say the total distance is 120 miles. Then for the first 60 miles, he drove at 20 miles per hour, so it took 3 hours. For the last 60 miles he drove at 30 miles per hour, so it took 2 hours. So the total distance is 120 miles and the total time is 3 + 2 = 5 hours. Using d = rt, we have 120=r(5), or r = 120/5 = 24 mph.</p>

<p>Note: I chose 120 because it made the reasoning very easy since 60 is divisible by both 20 and 30 (to pick 120 I took the lcm of 20 and 30 and then doubled this - try to think of why I did this). Any number will work, but the computations might be messier and common sense might fail.</p>

<p>Now see if you can use a similar procedure to do the second one.</p>

<p>different method</p>

<p>1.
x = total distance</p>

<p>x/2 = 20 distance/time = rate
x/2 = 30</p>

<p>then</p>

<p>x/40 + x/60 = 1 (1 equals the entire trip) distance/time*rate = 1</p>

<p>solve</p>

<p>3x/120 + 2x/120 = 1</p>

<p>5x/120 = 1</p>

<p>5x = 120 </p>

<p>x = 24</p>

<p>Pretty much the same for the second one too. Just use the d=rt formula cuase its easy and it flows well once you know it.</p>

<p>DrSteve- Thank you so much, I had never even thought of making up variables in these types of questions.</p>

<p>And by the way OP- there’s a formula for the type of question in the second problem: 2 times speed1 times speed 2 divided by speed 1 + speed 2</p>

<p>So it would be 2x60x30 which is 3600, and then 3600 divided by 60+30 equals 40.</p>

<p>Thank you guys so much =D</p>

<p>This is the SAT. While one can start the tedious process of reconstructing the d=rt formulas, such approach is EXACTLY what ETS hopes one would do, and that is WASTING time.</p>

<p>All those problems can be solved in fewer than 10 seconds. All the time and without any change to make a silly mistake in setting up the equations. </p>

<p>And it does not get simpler than making a problem that remains hard for anyone who follow the “high school” approach and trivial to anyone who knows how to recognize such problem. </p>

<p>1) A fellow travels from city A to city B. The first half of the way, he drove at the constant speed of 20 miles per hour. Then he increased his speed and traveled the remaining distance at 30 miles per hour. Find the average speed of the motion.</p>

<p>2.20.30 … 120
-------- = …------ = 24
20 + 30 … 5</p>

<p>2) Suppose we’re driving a car from Amherst to Boston at a constant speed of 60 miles per hour. On the way back from Boston to Amherst, we drive a constant speed of 30 miles per hour. What is the average speed (mph) for the round trip?</p>

<p>2.60.30 … 120
-------- = …------ = 40
60 + 30 … 3</p>

<p>PS The astute reader might notice how the solution that starts by taking 120 as basis ends with the same number as the solution above. </p>

<p>PPS Picking numbers is not a bad idea, but how does one go about with a slightly different problem such as:</p>

<p>Suppose we’re driving a car from Amherst to Boston at a constant speed of 61 miles per hour. On the way back from Boston to Amherst, we drive a constant speed of 23 miles per hour. What is the average speed (mph) for the round trip?</p>

<p>Good luck with the other “shortcuts!”</p>

<p>I really like Dr Steves method of approach. It is much easier and less messy then others. Just make up a number for total mile between two points, and you can use that as a distance for the equation d=rt.</p>

<p>I never thought of this. I will use this forever now.</p>

<p>^ Xiggi, can you elaborate on your process? I do not understand anything you did… With the “-----…----” you cut out much of the process, so I do not understand what you did to get the answer.</p>

<p>All he’s saying is 2xspeed1xspeed2 divided by speed1+speed2= average speed</p>

<p>^ Oh, then I am sorry. </p>

<p>I agree with Xiggi then, it is a much faster process using that simple equation. </p>

<p>I tried to use d=rt on xiggi’s problem at the bottom, it’s going to take years. Alright, I guess I will use that equation.</p>

<p>Wait. </p>

<p>120/50=2.4</p>

<p>Not 24. Im more confused…</p>

<p>You added in the numerator; you’re supposed to multiply everything in the numerator. So you would get 2x20x30 which equals 1200.</p>

<p>2.20.30 … 120
-------- = …------ = 24
20 + 30 … 5</p>

<p>I presented it that way because that is what your paper should show. I did not show the reductions steps. 2X20X30 / 50 is 1200/50 or 120/5 or 24. </p>

<p>2X20X30 … 120
-------- = …------ = 24
20 + 30 … 5</p>

<p>

</p>

<p>And I have been saying it for years. Thanks for the quotation!</p>

<p>Xiggi, forever a genius.
Thanks!</p>

<p>(2AB)/(A+B) is also known as the “harmonic mean” of A and B. But it is only a matter of time before it shows up in textbooks as “Xiggi’s Formula”. Afterall, if silverturtle made it into Direct Hits, can this be far behind?</p>

<p>“Xiggi’s Formula” is nice and gets you the answer to this and similar questions very quickly. I am certainly not opposed to his solution (and I would almost say I’m for it), but I prefer the solution I have given because my method applies to a wide range of problems, whereas Xiggi’s is specific to a certain type of problem which most likely will not show up on your SAT. I tend to only suggest memorizing a formula if it’s highly likely to be useful on your test. Students on this forum can probably benefit from memorizing this formula (since most of you are very good students), but if I gave this formula to the average student that I tutor I feel like it might be a waste of time for everyone. This is just my opinion - feel free to disagree.</p>

<p>And one more remark: In the real world Xiggi’s objections to my method are valid, but in SAT world they are not. On all the SATs I have seen the numbers have been “nice.” There is simply no evidence to suggest that messy numbers will ever appear. Of course they COULD appear, but I prefer to use the past as the best indication of the future (at least when it comes to SAT prep). This attitude has not failed me or my students so far.</p>