<p>Ok So I got this first one but it took some Guess and Check, but I just completely didnt do the other 2. I just wanted to know the method for doing these by maybe using some formula and not Guess and Check which takes way to much time.</p>
<ol>
<li><p>The sum of 5 consecutive integers is 1,000. What is the value of the greatest of these integers. Its 202</p></li>
<li><p>In the xy-coordinate plane, the graph of x=y²-4 intersects line l at (0,p) and (5,t). What is the greatest possible value of the slope of l?</p></li>
<li><p>Esther drove to work in the morning at an average speed of 45 MPH. She returned home in the evening along the same route and averaged 30 MPH. If Esther spent a total of one hour commuting to and from work, how many miles did Esther drive to work in the morning?</p></li>
</ol>
<p>I would just like to know if their is any quicker way to do these, or any formula, and if you could working out using these methods. Thanks mucho!</p>
<ol>
<li>The sum of 5 consecutive integers is 1,000. What is the value of the greatest of these integers.
x + (x+1) + (x+2) + (x+3) + (x+4) = 1000
5x=990
x=198
x+4=202</li>
</ol>
<ol>
<li><p>either of the two above methods works fine (for the record, I used the second one)</p></li>
<li><p>First fine p and t. 0=p^2-4, p^2=4, so p=+ or - 2. 5=t^2-4, 9=t^2, t=+ or -3. Now, the equation for slope is (y1-y2)/(x1-x2)=(t-p)/(5-0). You want to find the slope that will give you the greatest slope. You could plug in all the possibilities for p and t and find four slopes. Or, you could realize that t-p alone has to be the greatest, hence (+3) - (-2)=5 would give you the greatest slope, which corresponds to 5/5=1. </p></li>
<li><p>There is a formula. It is d=(2<em>t</em>mph1<em>mph2)/(mph1+mph2), where d is total distance both ways, t is time (both ways), and mph1 and mph2 are the speeds. Here is how that formula is derived: x/mph 1 + x/mph 2= time total (t). You divide by miles per hour (or multiply by hours per mile) because you want the mile units to cancel. Also, x is the distance for the first leg, so dividing mph1 from x gets you the time for the first leg and dividing mph2 from x gets you the time for the second leg. Getting a least common denominator, you get (x</em>mph2)/(mph1<em>mph2)+(x</em>mph1)/(mph1<em>mph2)=t. Adding fractions: (x</em>mph2+x<em>mph1)/(mph1</em>mph2)=t. x<em>(mph1+mph2)/(mph1</em>mph2)=t. x=t<em>(mph1</em>mph2)/(mph1+mph2). Since the distance (d) is 2x, d=(2<em>t</em>mph1<em>mph2)/(mph1+mph2). Plugging in your numbers, one gets 2</em>1<em>45</em>30/(45+30)=2700/75=36 miles. Generally, questions ask for total distance, but this asks for the distance of one leg. So just divide by 2 to get 18 miles. </p></li>
</ol>
<p>^If you don’t want to remember the formula, just do unit conversions. Convert miles into hours using what is given (miles per hour). If you don’t know what a unit conversion is, just ask and I’ll give you a quick explanation. It is a very helpful tool for many types of problems (not just SAT).</p>
<h1>3 took me forever, because I had to remember all those god damn equations.</h1>
<p>This is using the more obvious method. I have no hell of an idea what the person above me is doing.
rt = d, rate * time = distance.</p>
<p>So in the morning, rate = 45 mi\h. The distance traveled is, 45<em>X = D, where x is the time it took in the morning. Later, the rate = 30 mi\h, so 30</em>Y = D, where y is the time it took in the afternoon. The times added equal 1 hour, so X+Y = 1. </p>
<p>Make 45<em>X and 30</em>Y equal, so 45X = 30Y. Now, use substitution with the last equation.</p>
<p>Y = 1-X, replace Y with that, so 45X = 30(1-X) = </p>
<p>45X = 30-30X 75X = 30 Solve. X = .4, so that means it took .4 of an hour or 2\5s of an hour. So plug it in, 45* (2\5) = 18 miles.</p>
<p>If you use this formula, x will equal the length of one leg of her trip. Since it asks for the length of her commute to work, the answer is 18. But if it asked for her total commute, you would have to multiply x by 2 and in that case the answer would be 36. But here, it is 18.</p>
<p>EDIT:cjgone beat me to it. But it goes to show there is almost always more than one way to do a math problem on the SAT. Just remember the method you find easiest.</p>
<p>^Ah, still more difficult then most of any of the other SAT questions. I completely forgot the equation and mixed up things like t*d=r, or some other non-working variations taking me much longer then it should have. O_o</p>
<p>I never knew that the SAT required equation memorization because I don’t remember any other questions that required something like this.</p>
<p>@your method is the fastest it seems. I wish I saw that one :D.</p>
<p>Oh, all what GammaGrozza tried to do was not only give the formula but also show how it was derived. This could have been helpful for someone who does not “buy” the quick and dirty formula for this type of problems. </p>
<p>There are indeed several methods to attack such problems and TCB hopes that the student will follow one of the “standard” timesinks. After all, the reason it remains one of the most difficult problems is that it DOES trip many people or forces them to abandon because of excessive time wasted. </p>
<p>It’s a shame because knowing an incredibly simple formula plus paying a modicum of attention to the problem statement transforms a perennial trap of the College Board into one of its easiest questions. And, to boot, when appearing in a standard MC, a question that can be solved in 4-5 seconds without even using a pencil. But that is another story.</p>
<p>Eh, just use proportions. (I hope this isn’t a repeat solution)</p>
<ol>
<li>Driving time is inversely related to driving speed (obviously, if you drive faster, it takes proportionatle less time). Thus, her drive times to and from work are in the ratio 30:45, that is, she spent 30/(30+45) of her total time driving to work. Notice that simplifies to 2/5. Thus, she spends (2/5)60 = 24 minutes driving to work. If she drove at 60 mph, then she drove would drive 1 mile per minute. Since 45 is 3/4 of that, she drove (3/4) mile per minute, or (3/4)24 = 18 miles.</li>
</ol>
<p>and xiggi, thanks for defending me :). I didn’t read cjone’s post. A lot of people just post the formula. But I think it’s helpful to see where it comes from. </p>
<p>So, there are a lot of ways to do this problem:
-unit conversions (1253729 and one of my listed methods).
-THE formula (the one I and yankee posted).
-System of Equations (cjone’s method)
-Proportions (generallyrong’s)
-If it is an MC, the one xiggi breifly alluded to. If we are thinking about the same thing, the method where you know that average speed both ways will be slightly less than the average between 30 & 45. If you have answer choices of 37.5, 19.25, 18, 15, and 22, you know the answer is 18 because 18*2 is slightly less than the average of 30 & 45 (since time is an hour, distance travelled=average speed).</p>
