<p>Hey .. I just need help with this question that was on a grid-in on a pratice SAT (its easy, I just hate these problems and I hate the explanation the book gives even more)!</p>
<p>A woman drove part of a 185mi trip at 50mi/h and the rest at 55mi/h. Find the distance she travled at 50mi/h if her total driving time was 3 h and 30 min. </p>
<p>I need the correct answer - but also a clear explanation. Thanks guys!</p>
<p>Use a system of equations:</p>
<p>x and y are minutes.</p>
<p>(5/6 miles per minute)x + (11/12)y = 185 miles
x + y = 210 minutes (3hr 30min)</p>
<p>Solve for x first.</p>
<p>So then using the 2nd equation, y=210-x
plug that into the first equation.</p>
<p>So (-1/12)x = -7.5 .... x = 90</p>
<p>So the time she went 50 mph is 90 min.</p>
<p>90 min = 1.5 hours</p>
<p>so 50*1.5 = 75 miles is the answer.</p>
<p>i can explain... but first.. is the answer 75 miles?</p>
<p>oh nvm. then.</p>
<p>yes 75 is the answer ..</p>
<p>do you understand his reasoning, vinny? I did it a bit differently, if you want I could show you my method.</p>
<p>Did my explanation help? Or should I re-explain, because now that I look at my post, things are kind of cluttered up.</p>
<p>wait....I dont understand how you got the 1st equation ... where did you get 5/6 or 11/12 <--- good explanation.. but can u explain again ... and you can explain ur method if you want son of liberty... the more the better!</p>
<p>okay so you start with the equation d = rt... that equation will be the core of your solution.</p>
<p>Now. There are 2 separate distances. 1 distance was traveled to at 50 mph... the other was traveled to at 55 mph. so... you set up 2 separate equations</p>
<p>d1 = r1 x t1</p>
<p>d2 = r2 x t2</p>
<p>Now... you know that the total distance of the trip was 185 miles so... you set up another equation</p>
<p>d1 + d2 = 185
This is because the total of the 2 distances mentioned above equals the total of the entire trip.</p>
<p>You can also set up this other equation...</p>
<p>t1 + t2 = 3.5 hrs</p>
<p>All right then. Refer back to the original equation... there are some variables we can plug in, so let's update.</p>
<p>d1 = 50t1
d2 = 55t2</p>
<p>This, of course, means that d1 + d2 = 50t1 + 55t2... which can be further reduced to...</p>
<p>50t1 + 55t2 = 185.</p>
<p>Go back to t1 + t2 = 3.5... and solve for either t1 or t2... Now... plug this number in to the above equation and solve.</p>
<p>I don't know if that was clear or not... ask questions if you're confused.</p>
<p>Ok, I'll clear up some confusions in my original explanation.</p>
<p>I usually like to convert all the time into minutes, which is what I originally did. But actually you don't have to. It's actually wasier to use hours. So I'll do it in all hours.</p>
<p>First, 3 hours 30 min. = 3.5 hours.</p>
<p>now knowing that, I will set up a system of equations for 2 variables.</p>
<p>x and y are in hours.</p>
<p>mph*hours=miles. So...</p>
<ol>
<li><p>(50 mph)x + (55 mph)y = 185 miles</p></li>
<li><p>x + y = 3.5 hours</p></li>
</ol>
<p>using the second equation: y = 3.5 - x</p>
<p>plug into first equation: 50x + 55(3.5-x) = 185</p>
<p>solve for x: -5x = -7.5
So, x = 1.5</p>
<p>To find total distance you take 50 mph * 1.5 hours.</p>
<p>Answer = 75 miles.</p>
<p>Sorry about the first one. This way is a lot easier. I just have a habit of using all minutes.</p>
<p>Thanks!!! Both of your explanations were clear and not too hard (although Eugenes was a little bit easier). Anyway, thanks again and I will prob. be posting soon with more math question!! Why cant the SAT be all verbal?</p>
<p>Where did you get that question?</p>
<p>Princeton Review - 2005 (The NEW SAT) ........ why?</p>
<p>Here is all that I did...</p>
<p>Using the equation r = d/t, let's rearrange to represent that we know what the total time is...</p>
<p>t = d/r</p>
<p>The manipulation...</p>
<p>t = [d/r1] + [(185 - d)/r2]
t = (dr2 + 185r1 - dr1)/(r1r2)
tr1r2 = dr2 + 185r1 - dr1
tr1r2 - 185r1 = dr2 - dr1
(tr1r2 - 185r1)/(r2 - r1) = d</p>
<p>So, we start with something like this</p>
<p>3.5 = [d/50] + [(185 - d)/55]</p>
<p>Plugging in,</p>
<p>(3.5<em>50</em>55 - 185*50)/(55 - 50) = d
75 = d</p>
<p>What if you reverse the d and (185 - d) terms with their respective denominators? Nothing...you just have to remember to plug back in appropriately at the very end...</p>
<p>3.5 = [(185 - d)/50] + [d/55] <===NOTE where the 50 is!
3.5 = (10175 - 55d + 50d)/2750
9625 = 10175 - 5d
5d = 550
d = 110</p>
<p>Since 185 - d was the numerator in that set-up where 50 was in the denominator, just plug back in to that numerator's expression: 185 - 110 = 75.</p>
<p>Well, matter of preference again.
Most of the time (not always!) it makes sense to use a variable that directly answers the question.
x miles - she traveld with 50 mph
it took her x/50 h
(185-x) miles - the rest - with 55 mph
it took her (185-x)/55 h.
altogether 3.5 h
x/50 + (185-x)/55 = 3.5
(11x + 1850 - 10x)/550 = 3.5
x+1850 = 1925
x=75</p>
<p>My solution was basically the same as setzwxman's, just broken into steps.
++++++++++++++++
What's interesting, slight modification to the original question allows you use a very cool formula (I know that xiggi brought it up; could be somebody else before).
++++++++++++
Length of the trip not given.
First half of the distance - 50 mph, second - 55 mph.</p>
<h2>Total driving time 3.5 h.</h2>
<h1>How long was the trip?</h1>
<p>Average speed (2<em>r1</em>r2)/(r1 + r2) = (2<em>50</em>55)/(50+55) = 52.381 mph.
In 3.5 h she drove 3.5 * 52.381 = 183.334 mi. That's the answer.
Let's check:
Half of the trip 183.334 / 2 = 91.67.
91.67 / 50 + 91.67 / 55 = 3.5. Bingo.
++++++++++++++
One more modifiaction.
A woman drove 185 miles.
The first and the third quarter of the time of the trip her speed was 50 mph,
the rest 55 mph.</p>
<h1>How long was she driving?</h1>
<p>Now that driving times are equal
average speed is (50 + 55)/2 = 52.5 mph
Travel time 185 / 52.5 = 3.52 h = 1h 31min (approximately).
Your turn to check.</p>
<p>Yet another expansion:
+++++++++
A student drove the first part of a 210mi trip home at 50mi/h.
Find the speed at which she/he drove the remaining part, if the average speed on this trip was 52.5mi/h, and it took the student 1 hour to finish the second part.</p>
<p>hm... that you'd be likely to say 55 since 52.5 is the average of 50 and 52.5?</p>
<p>but it's NOT the right answer because the lengths of time were different.</p>
<p>if that's what you're talking about, it was also in the intro section of the PR SAT book a few years ago. i'm not sure if it still is, though. something about a lady driving 20 mph one way and 30 mph back. joe bloggs would say the average speed is 25 mph, but it's really 24 5/7 or something like that.</p>
<p>Right, this is the classic trap.</p>
<p>What's the answer, anyway?</p>