<p>Hey guys!</p>
<p>Sorry to interrupt! I never was good with rate problems! Would anyone help me with this LEVEL 5 problem ?</p>
<p>Not sure but is the answer 16 miles? Do you have the answer or just the question?</p>
<p>The answer is 80/3 sir</p>
<p>Calculate the length of time, and divide by the distance. 10 miles at 20 mph would be half an hour. 10 miles at 40 mph would then be half of that, 15 minutes, giving you 45 minutes, or .75 hour of total travel time. 20 miles/.75 hour = 80/3, as noted above, or 26 and 2/3 miles per hour.</p>
<p>Avg speed = (total distance) / (total time)</p>
<p>2.20.40 / (20+40) = 80/3</p>
<p>Do not reduce fraction. Takes 10 seconds.</p>
<p>Easier way…</p>
<p>20 mph = 10 miles = 30 minutes of travel
40 mph = 10 miles = 15 minutes of travel</p>
<p>45 minutes of travel (TOTAL TIME FOR THE 20 MILES)</p>
<p>Now, you know that it takes you 45 minutes to travel 20 miles. You have to find out how long you can travel in 1 hour, or 60 minutes. So, CROSS MULTIPLY.</p>
<p>45 / 60 = 20 / x </p>
<p>(make them vertical, they’re fractions, hard to do on forums tho.)</p>
<p>You will get this.
45x = 1200</p>
<p>Then just divide
x = 26 2/3 miles OR 80/3 miles</p>
<p>^that is not an easier way. U complicated it. Why apply proportions and bring a variable x when u already calculated the total time. The total distance is given as 20. Distance/time. 20/(3/4)=80/3</p>
<p>^ The proportions is basically the same thing you did, except it’s safer. Especially for a question near the end of a math section. Your way is Distance / time, which is complicated because the time is not 3/4. The “time”, you meant, is the proportion between the total time over an hour. Which is what I did. If you try to teach a guy who doesn’t know how to do the problem, skipping a step won’t be the best way to teach him. Btw your post above Avg speed = (total distance) / (total time) is misleading. Total distance, 20 divided by total time, 45, gives you 0.4. So, the avg. speed is 0.4? per second? per year? Either way it’s the wrong explanation.</p>
<p>/smile at the argument above. </p>
<p>Try your “easier” ways with speeds of 39 mph and 11 mph. Or 11 and 5 mph.</p>
<p>There is a fast and idiot-proof approach to such problems on the SAT, and I posted it.</p>
<p>@johntee, i think if we’re dividing by hours, we will get the answer in terms of hours</p>
<p>@Xiggi, I see what you’re trying to say. My “easier” way is quite a misleading assertion. What I was trying to say is that it works well for me so I won’t get caught up in converting miles to hours, hours to miles, etc. Anyways, your way is definitely less time consuming and a better way if you already understand the concept behind the question . As for your 39 and 11 conversion, yes, it would not work well with my way but the SAT wouldn’t be that cruel to give you those data :)</p>
<p>@Muhammad The hour thing was just an addition to my response. The point is, your formula total distance / total time would get you the wrong answer.</p>
<p>Sorry for the math war, but in my pot, i clearly got the right answer.
@xiggi, that is the faster and convenient way, but not everyone knows how to utilize it.</p>
<p>^ Although you got the right answer, your formula makes no sense because it is badly explained. Let me show you the formula that you told WorkOverTalent.</p>
<p>muhammad9211:
Avg speed = (total distance) / (total time)</p>
<p>This is what WorkOverTalent will think…</p>
<p>Ok, speed = distance / time, seems easy enough.
Let me just calculate the total distance, oh 20 miles, not too bad.</p>
<p>Now, let me calculate the total time. Ok if you are going 10 miles driving 20mph… easy, 30 minutes, ok now 10 miles driving 40 mph… 15 minutes, even easier. 30 + 15 = 45 minutes. </p>
<p>Ok now that I found the total minutes and total distance, let me plug them in.
Avg speed = total distance / total time
Avg speed = 20 miles / 45 minutes
Ok, Avg speed = 0.4 mph</p>
<p>Right, now that I got the answer let me just… oh wait it’s not on here. Dang, should’ve used JohnTee’s formula.</p>
<p>Now do you understand what I’m trying to say?</p>
<p>U assumed he would convert time to minutes. Why assume he wont convert back. Time = 10 / 20 = 1/2. U converted that number to 30 min. U didnt convert back to hours. I think that numbet shouldnt even be converted in the first place. Total time is 3/4. Now apply formula s=d/t
S=20/(3/4)=80/3
Bottom line is both methods are correct. It depends on which ppl prefer. Whether they prefer using proportions which u should know many ppl do struggle with. Or applying a simple formula .</p>
<p>“Avg speed = 20 miles / 45 minutes
Ok, avg speed = 0.4 mph”
Yes, this is what u wrote. U divided miles by minutes, yet wrote ur answer in miles per hour <em>facepalm</em>. Speed is 0.4 miles per minute. Thats why i said dont convert from hr to min.</p>
<p>Okay I think I understand it pretty well! Thank you Muhammed and JohnTee! I tried out both of your methods and they worked pretty well. So when I tried this today again. What I did was use RT=D and then I got 1/2 for the time it took for 20 miles and 1/4 for the time it took for 40 miles. Together thats 1/2 and 1/4. Since that is the total time I used fractions and added it to get 3/4;thus, giving you the time it took for both 20 miles. Since 20 miles is the total distance, TOTAL DISTANCE/TIME= AVG SPEED so 20/(3/4)= 80/3 or 26.667. Totally got it now. I guess it was the avg speed formula that threw me off guard. Xiggi, Im not fully understanding your method. If you could clarify on it, that would be great. Thank you guys so much! I appreciate it</p>
<p><em>sits back…eats popcorn</em></p>
<p>WOT, check the stickied thread or search this forum for Xiggi method. I have explained it many times. It is a very simple application of the harmonic mean formula, which itself is derived from the d=rt formula. </p>
<p>Simply stated, the TCB is HOPING student waste time on unneeded steps or mess up with a straight average. Fwiw, the formula is Answer is 2<em>S1</em>S2 / S1+S2.</p>
<p>@xiggi, ur method is valid if distance travelled is the same? What if he travelled 15 miles at 20mph and 10 miles at 15mph. Would it work?
I only use that formula for rate of work. Good to know it works for these as well.</p>