<p>These are the problems I get consistently wrong the most simply because I don't know how to deal with them. I know that I can use the formula (for avg speed):</p>
<h2>2(R1*R2)</h2>
<p>(R1+R2)</p>
<p>If the time frame is one hour, but what if it isn't? Here is a question that I have found that I'd have no idea how to do:</p>
<p>Calvin traveled from his home to another city at an avg speed of 32 mph and then returned home along the same route at an avg speed of 40 mph. If his total traveling time for the trop was 3 hours, how many MINUTES did it take Calvin to drive from his home to the other city?</p>
<p>A 100
B 90
C 80
D 72
E 60</p>
<p>I got this from RocketReview (no it's not from a practice test made by them, I know only to do blue book ones). It states that you can simply see that if the first half of the trip is traveled slower than the second, the time going to the other city must be greater than 90 minutes, leaving only choice A.</p>
<p>That seems like good advice, but what would've happened if there were two choices greater than 90, or worse, if it was a grid in question? It also shows a long solution and states that there actually is a quicker solution (I'm guessing involving the above formula), but doesn't even state what it is. I'm at a loss here with these types of problems, so can anyone tell me, or direct me to a site, that can tell me how to solve these problems regardless of whether there's a one hour time frame or ten hour, or any other changes. It would be nice this is based off of the formula rather than solving with many variables and equations open to error.</p>
<p>Also, what is the formula for rate of work problems. Example:</p>
<p>Steven can do a certain job in 3 hours and Gary can do the same job in 2 hours. How many hours will it take Steven and Gary to do the job working together?</p>
<p>Um ya, RR is really a great book and thinking for a second to rationalize is important but anyways let me see if I can make this simple for you...</p>
<p>Now, since the questions ask how many minutes did it take Calvin to drive from his house to the other city, you must divide x by his speed to the other city. So... (--- is dividing)
53 1/3
-------- = 1 2/3
32 mph</p>
<p>Finally, since it ask in Minutes, 1 2/3 hours = 100 Minutes</p>
<p>I tried working this out from the above formula for one hour (2 times the product of speeds over the sum) and it works if 3 is used as a constant instead of 2. Does this work so that for every hour added, .5 is added to 2? That's probably hard to understand...</p>
<p>That last part does help though since it's basically like a simple physics problem of velocity, displacement, and time.</p>
<p>So basically, to put it simply if you can't find a shortcut, you take sum of the x over each velocity and set them equal to time in terms of hours in order to get displacement. Then to get the time you plug your known variables into Velocity = Displacement / Time and convert to what is asked.</p>
<p>IF there is any quicker way to do it that would be great; thanks for the help.</p>
<p>Any info on the formula for work rate problems? </p>
<p>I hate these rate problems; they're my one true problem on the math SAT.</p>
<p>Calvin traveled from his home to another city at an avg speed of 32 mph and then returned home along the same route at an avg speed of 40 mph. If his total traveling time for the trop was 3 hours, how many MINUTES did it take Calvin to drive from his home to the other city?</p>
<p>Well, we know that V=D/T, so T=D/V</p>
<p>So, D/V + D/V = 3 hours. Now plug in 32 and 40 into the velocities:</p>
<p>D/32 + D/40 = 180 minutes => D= 3200</p>
<p>So, from his home to the other city, the velocity was 32 mph.</p>
<p>While RR has a few good tips, in this case, Adam Robinson simply wrote one more of his usual horrible math examples. It fails to follow the VERY obvious patterns of the way ETS writes this type of questions and ... proposed answers. </p>
<p>All problems of average speed on the SAT use the same logic. The subject travels the same distance at different speed. The faster he goes on one leg, the less amount of time it will take. Since the distance of each leg of a round trip is the same (duh) the speed must be the proportional reverse of the time. </p>
<p>The answer is EXTREMELY easy to identify in a few seconds by merely looking at the numbers. Here's the ONLY thing that needs to be done:</p>
<ol>
<li><p>Total time is 3 hours or 180 minutes. Remember the number 180 </p></li>
<li><p>Look at the two speeds 40 and 32. Now write them in the form of a ratio ... 40/32 which is obviously the same as 10/8. 10 and 8 ... how convenient since it adds up to 18. Could it be that it takes 100 minutes one way and 80 minutes the other way for a total of 180. Of course! Now, we just have to make sure to follow the logic of the problem to pick which one is 80 and 100. </p></li>
</ol>
<p>My final recommendation: drop this poor example of a problem and stick to the VERIFIED shortcuts discussed many times on CC:</p>
<ol>
<li>Use the formula for the grid-ins</li>
<li>Pick the correct answer by correctly identifying the 3 trick answers and apply the very simple test to the remaining answers. It takes about ten seconds to do it.<br></li>
</ol>
<p>As long as ETS continues to trip students with the simple form of this problem, there won't be a reason to make it more difficult. So, in the meantime, the simple answer will continue to work and every time.</p>
<p>For the work rate problems, when it comes to the SAT, try to keep it simple by drawing a diagram or using real numbers. </p>
<p>Using your example of "Steven can do a certain job in 3 hours and Gary can do the same job in 2 hours. How many hours will it take Steven and Gary to do the job working together?" </p>
<p>Let's say Steven (L1) and Gary (L2) are two lumberjacks who are cutting trees down. L1 would chop down 30 trees in three hours and L2 would only need two hours. So, here's what you can write down:</p>
<p>L1 30/3 = 10 trees per hour
L2 30/2 = 15 trees per hour</p>
<p>L1 and L2 working together .... 25 trees per hour. </p>
<p>Now, how long would it take them to cut down 60 trees (the total output if they worked separately) if working together they can cut down 25 trees per hour? Now, how long would it be for 30 trees? </p>
<p>Not so hard when you rephrase the problem like that! </p>
<p>PS There are other approaches that use variations. For instance, you could say that L1 would cut 40% of the trees and L2 60% (2-3 ratio) This will tell you that L1 cut 12 out of the 30 trees. Plugging L1 speed (rate) would give you the answer.</p>
<p>hahaha Xiggi, I have been on CC a while and although I have heard that your methods are great this one just made meeh laugh, haha I love the way you solved the problem while making fun of ETS</p>
<p>xiggi could not help but take another swing at the slacker/worker question.</p>
<p>As a rule, in questions like that both guys perform some task in a whole number of hours (T1=3 and T2=2 respectively in our case).
Let Steven (D1) and Gary (D2) be very grave diggers now (no more trees left!).
If you multiply their times (T1 times T2, or 3x2=6 hours) and make them work for that number of hours, each of them will dig a whole number of graves: 6/3=2 and 6/2=3 respectively.<br>
That means in 6 hours, working together theyll dig 2+3=5 graves.
They will spend (6 hours) / (5 graves) = 1.2 hours per grave as long as they dont dig their own grave.
In other words, itll take Steven and Gary 1.2 hours to do the job.</p>
<p>Sounds long, but you can try to keep it simple by drawing a diagram:
3 x 2 = 6.
D1: U U (2 graves in 6 hours)
D2: U U U (3 graves in 6 hours)
+++++++
U U U U U (5 graves in 6 hours) - D1 and D2 working together.
6 / 5 = 1.2 hours.</p>
<p>Now, how about two pipes filling a tank: T1=7 hours, T2=3 hours.
7 x 3 = 21
P1: O O O (3 tanks in 21 hours)
P2: O O O O O O O (7 tanks in 21 hours)
++++++++++++++
O O O O O O O O O O (10 tanks in 21 hours) - P1 and P2.
21 / 10 = 2.1 hours.</p>
<p>Did you notice something neat?
(T1xT2) / [(T1xT2) / T1 + (T1xT2) / T2] =
(T1xT2)/ (T1+T2).
Looks familiar, does not it?
The beauty of this formula is that it'll work even if given times are not integers.</p>