Question 1161456: I'm studying for the SAT's and this question popped up.
Corrine drives d miles to her office at an average speed of 50 miles per hour. Returning home, she travels by the same route and averages 60 miles per hour. If her trip home is 10 minutes shorter than her trip to her office, what is the value of d?
The practice book shows the equations would then be  and  . It then shows it would substitute  into  .
I just don't understand where they got the 1/6 from in the word problem. Could you please explain where the 1/6 came from so I can better understand how the equation was set up?
Found 5 solutions by ikleyn, Theo, Edwin McCravy, solver91311, MathTherapy:
Answer by ikleyn(52921)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! the t + 1/6 comes from the following.
you're dealing in miles per our.
10 minutes / 60 = 1/6 of an hour.
rate * time = distance.
when traveling at 60 miles per hour, the formula becomes 60 * T = D
when traveling at 50 mles per hour, the formula becomes 50 * (T + 10/60) = D
the reason is because you have to make sure your times are consistent.
since the rate of travel is in hours, then the time nees to be in hours as well.
you have two equations that need to be solved simultaneously.
they are:
60 * T = D
50 * (T + 1/6) = D
the 1/6 means 1/6 of an hour.
you should be able to solve the problem from this.
if you, however, still have problems, let me know and i'll help you find the solution.
my own calculations tell me that the solution is D = 50 miles.
let me know if you agree once you worked it out.
Answer by Edwin McCravy(20065)  (Show Source):
You can put this solution on YOUR website!
10 minutes is 10/60th or 1/6th of an hour.
The way the problem is stated it should be worked this way:
distance rate time
going to office d 50 t
returning home d 60 t-1/6
This will give the correct answer for d also.
The way the practice book did it, requires interpreting it as if were
worded this way:
Corrine returns d miles from her office at an average speed of 60 miles per
hour. When she left from home, she traveled by the same route and averaged
50 miles per hour. If her trip to the office took her 10 minutes longer
than her trip home her office, what is the value of d?
The practice book reinterpreted the problem. The difference is that the
practice book considered t to be the time returning so the 10 minutes
1/6 of an hour had to be added to make the time going longer. But the
way the problem is stated, we should let t be the time going, so we would
subtract the 1/6th of an hour to make the time retuning be shorter by 1/6th
of an hour.
Both ways are correct, but the best way is to solve it as the problem is
stated literally. Since the problem states that the time is shorter
returning, then you should pick t as the time going, for that's the time
that we are to make shorter by subtracting. IOW, why change the problem
from one where you make the time shorter to a problem to one where you make
the time longer, when the one that was given was the one where you make the
time shorter?
The moral of the story is: Don't assume the practice book always gives the
most obvious, and best, way to work a problem.
Edwin
Answer by solver91311(24713)  (Show Source):
Answer by MathTherapy(10557)  (Show Source):
You can put this solution on YOUR website!
I'm studying for the SAT's and this question popped up.
Corrine drives d miles to her office at an average speed of 50 miles per hour. Returning home, she travels by the same route and averages 60 miles per hour. If her trip home is 10 minutes shorter than her trip to her office, what is the value of d?
The practice book shows the equations would then be and . It then shows it would substitute into .
I just don't understand where they got the 1/6 from in the word problem. Could you please explain where the 1/6 came from so I can better understand how the equation was set up?
The 10 minutes, when converted to hours gives you: 
If that's how the book showed the setup-equations, I totally disagree. To me, it's unnecessary to find "t" or time, and then use that to find d, the distance.
Unless, of course, you're asked to find the time, but if it's just "d" or the distance you need, I'd do it differently, and go STRAIGHT for the answer.
I'm an avid proponent of getting straight to the answer, and so, my preference would be the following TIME equation: 
I'd then solve that for d.
The time on each section of the SAT is short - it's not like the Regents where you have like a "ton" of time - and the less time you spend on each problem, the better off you are!
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