<p>Corinne travels home to work at an average speed of 50 miles per house, and returns home by the same route at 60 miles per hour. It takes her 10 more minutes to get to work than it takes her to get home. How many miles is it from Corinne's home to work? </p>
<p>Some of you might get the answer just by reading the question but if you could show how you solved it step by step, that'd be great. </p>
<p>Thanks</p>
<p>Answer:3000</p>
<p>x=time it takes to reach work at 60 mph
y=distance betweeen work and home</p>
<p>60x=y and 50(x+10)=y</p>
<p>Solve the system.
y=3000 (x=50)</p>
<p>as technologic said. just keep in mind distance = rate * time. </p>
<p>both distances going from work and coming back will be the same, so rate<em>going*time</em>going = rate<em>coming*time</em>coming, solve for what you can.</p>
<p>Answer is 3000 miles? What?
The answer in the book is nowhere near that.</p>
<p>EDIT: My bad.</p>
<p>60x=y
and
50(x+10/60)=y</p>
<p>answer is 5/6</p>
<p>The answer in the book is 50.</p>
<p>Here is how it is done:</p>
<p>D=r*t</p>
<p>for coming:
D=60x<--time unknown</p>
<p>for going:
D=50(x+10)<--time is unknown but 10 min more than coming</p>
<p>so:</p>
<p>60x=50(x+10)<--because distance is same going and coming</p>
<p>solve the above and x = 50 miles.</p>
<p>BTW you said 50 miles per house! :-)</p>
<p>The formula you need is:
2* (speed 1) * (speed 2) / (speed 1 + speed 2 )</p>
<p>Speed 1 is 50
Speed 2 is 70 since it takes 10 more minutes to get from work so you would add 10 more miles because her rate was 1 mile a minute.</p>
<p>ps. use the speeds given infront of the divsion.</p>
<p>2<em>50</em>60/(50+70)</p>
<p>^^ that formula gives you the harmonic mean of the speeds. The questions is asking for distance...</p>
<p>To Din123, in case gk23's explanation wasn't clear enough (written at the time he/she posted, it seems):</p>
<p>Fifty miles per house, eh? She must be going awfully fast, considering the number of houses there are in the world. :D</p>
<p>Answer: 50 (Duh, seeing as how you gave it)</p>
<p>Distance = Rate * Time; d = rt
=> 50(t+10) = 60t
=> 50t + 500 = 60t
=>500 = 10t
=>t = 50 minutes
=>t = 5/6 of a hour</p>
<p>Now, here's where technologic seems to have run afoul. Second time around, he didn't finish the question...</p>
<p>Plug and Chug (sorry, math teacher-ism :D)
rt = 60(5/6) = 50 miles.</p>
<p>Cool?</p>
<p>I always hated these. My teacher gave us the following mnemonic - direction questions are "DRT"y. As in, d = rt.</p>
<p>i know but it works both ways with adjustment</p>
<p>Yeah, but way to overdo it, Just2Fitz. If someone is asking for help on a question like this without mentioning harmonic speed, I think it's fair to assume that they are asking after the distance formula.</p>
<p>This is the SAT, not Alg2. We're looking for the best and quickest way to solve a problem.</p>
<p>You're right, this is the SAT, which only requires math knowledge up to Alg II. Not Calculus. Sorry, but that's above the average SAT taker's ability. Maybe you learned it in Alg I for whatever reason, but I'm learning it in Calculus now.</p>
<p>Well, perhaps you learned the simple, two number form...</p>
<p>In any case, it's not wise to expose someone to something they don't know in this informal manner. Memorizing the formula is meaningless if they don't understand why it works. Also, I'm pretty sure CB expects people to use d = rt, which is standard Alg II stuff, whereas harmonic means are way above that, bar the two number form, which isn't uniformly taught.</p>
<p>If only ETS cared how it worked. </p>
<p>So what, you're in calculus... So am I, but I don't run around flaunting and ****.</p>
<p>You sit for an SAT and figure out why/how each problem works and see where it takes you.</p>
<p>just to add:</p>
<p>take your two speeds and divide them and then multiply by 60.</p>
<p>Guess and check also works.</p>
<p>You don't give people who don't have calculus background formulas arbitrarily that they won't understand. The SAT anticipates d = rt, so it must work into the time. Not like it takes much time to use, anyway.</p>
<p>The figuring out process should've been done beforehand; that's what the SAT tests for, and that's what is taught in classes. If calculus wasn't one of those classes, then they shouldn't need to use that formula.</p>
<p>And I'm not flaunting; my calculus is pretty paltry. How is what I said flaunting? Just to mention that as proof that it isn't below Alg II material? I merely wanted to point out that it was a calculus concept - far above the standards. I am sorry if you find some offense in that.</p>
<p>So kind of you to presumably use an expletive. Is there some way we can diffuse this? I don't want to enter some kind of back-and-forth malicious banter with you.</p>
<p>No banter, but how is calculus paltry?</p>
<p>
[quote]
[...]my calculus is pretty paltry.
[/quote]
</p>
<p>That's how it is. :p</p>
<p>The answer is 50. I was using the 'solver' on my calculator the second time around and didn't keep track of the variables well enough.</p>