<p>Jennifer ran from her house to school at an average speed of 6 miles per hour and returned along the same route at an average speed of 4 miles per hour. If the total time it took her to run to the school and back was one hour, "how many minutes" did it take her to run rom her house to school?</p>
<p>i know how to solve the how many miles one.
the book explanation is plugging in.
is there a shortcut/formula for this problem?</p>
<p>how did you solve it??</p>
<p>there are many formulas that other have posted, but the only one I find worth remembering is D=RT (self-explanatory)</p>
<p>distances are constant in this problem, so house > school RT = school > house RT. 6x = 4(1-x). x= 2/5. so 2/5 of an hour or 24 minutes. </p>
<p>Once again, others might have different formulas but D=RT can SOLVE ANY DISTANCE PROBLEM</p>
<p>He probably knew that formula. I'll be surprised if he doesnt know rate=distance/time....Maybe he's talking about some other formula.</p>
<p>I would find the distance one way 1st. I learned this from Xiggi. To find the distance one way for average rates when the hour is one (the key is that time has to be in units of one) is:</p>
<p>(rate 1* rate 2)/(rate1 + rate 2)</p>
<p>I got 2.4 miles from her house to her school. </p>
<p>Then just use the common time= distance/rate formula. Since the problem tells you the rate from her house to her school, it is:</p>
<p>2.4 miles/6 mph= .4 hours or 24 minutes. </p>
<p>That's kind of the more technie way to do it, but snipez's way is actually simpler. He solved for time directly whereas I just took a lenthier route.</p>
<p>"Jennifer ran from her house to school at an average speed of 6 miles per hour and returned along the same route at an average speed of 4 miles per hour. If the total time it took her to run to the school and back was one hour, "how many minutes" did it take her to run rom her house to school?"</p>
<p>Rate x time = distance</p>
<p>Since the distance out is the same as the distance back, we know that:</p>
<p>rate<em>out x time</em>out = rate<em>in x time</em>back</p>
<p>We are given the ratio of rate<em>out:rate</em>back as 6:4 or 3:2.</p>
<p>Since the rate x time products must be equal,</p>
<p>time<em>out:time</em>back = 2:3. implying</p>
<p>2/5 time spent going out (and 3/5 coming back)</p>
<p>2/5 of 60 is 24 min. (Working the solution is faster than explaining it.)</p>
<p>tofulover, does this formula always work?</p>
<p>I cannot for the life of me wrap my dumb mind around this problem. Can anybody care to break this down a bit more? Specificall, I don't understand in my$0.02 where the whole "time<em>out:time</em>back = 2:3" part comes in, or in snipez90's explanation where there whole equation of 4(1-x) comes in. Anybody care to help this person who is hopeless at math?</p>
<p>Okay the best formula to use is this one. It finds the average speed; from which you can easily find the time taken/ total distance traveled. This is a formula is used in higher level physics:</p>
<p>Avg. v=2(v1*v2)/(v1+v2)<br>
v=speed </p>
<p>you can prove this formula using the force formulas from physics (can't be bothered right now lol). This works in all cases!!!</p>
<p>yeah thats the one i put on my calculator!!! but sometimes i forget exactly what it solves for.</p>
<p>People , why don't you bother doing all those problems from the Blue Booklook that are listed in </a><br>
Updated Consolidated List of Blue Book Math Solutions[/size]?
You'd find there a very involved discussion of the Blue Book question 18/page 412. That question is almost identical to the one you are wrestling with.</p>
<p>ok...lets look through every single problem till we find one identical....</p>
<p>^ ... and pay attention...
Did not I point to that one?</p>
<p>i know, but you had that "know-it-all" thing going on.</p>
<p>OK, you're right. Sorry.
Besides, my grammar sucked. :o
Thanks Tyler for generously not having me nailed for that!</p>
<p>I mainly expressed my :mad:ing frustration that same questions pop up time and time again when their solutions are right there for everybody's enjoyment.
Plus, I've invested a lot of time and efforts contributing to that stickie.</p>
<p>
</p>
<p>All of a sudden I got what you meant, Tyler.</p>
<p>I would not be original if I said that going through all the tests in the Blue Book (more than once) is the major part of prepping for the SAT (read xiggi, for example).
Using the "Updated Consolidated List of Blue Book Math Solutions" would prove extremely helpful.<br>
You work on the question, then you look up what other people had to say about it. Maybe you'll discover some interesting shortcuts or a generic approach to the whole class of questions.
Then, if you encounter a question similar to one that can be found in the "...Consolidated...", you will not have to search for it "by looking through every single problem".
I hope it makes sense.</p>