GreenlightTestPrep wrote:
On a partly cloudy day, Derek decides to walk back from work. When it is sunny, he walks at a speed of s miles/hr (s is an integer) and when it gets cloudy, he increases his speed to (s + 1) miles/hr. If his average speed for the entire distance is 2.8 miles/hr, what fraction of the total distance did he cover while the sun was shining on him?
NOTE: Enter your response as a
fraction.
Answer:
Here's another approach that doesn't assign a "nice value" for the distance traveled...
Let's first start with Derek's regular speed and increased speed (in mph)
We know that these two speeds differ by 1 mph, and we know that Derek's average speed is 2.8 mph.
This tells us that Derek walked 2 mph when it was sunny, and he walked 3 mph when it was cloudy.
Okay now let's deal with the rest of the problem.....
Let's say
d = the distance Derek travel
Time = distance/speed So, Derek's travel time =
d/2.8Since Derek's TOTAL travel time =
d/2.8, let's say that:
t = number of hours walking while sunny
So
d/2.8 - t = number of hours walking while cloudy
We'll begin with a word equation: (distance traveled while sunny) + (distance traveled while cloudy) = TOTAL distance traveled
Since
distance = (speed)(time), we can now write:
(2)(t) + (3)(
d/2.8 - t) =
dExpand: 2t + 3d/2.8 - 3t =
dSimplify: 3d/2.8 - t =
dMultiply both sides by 2.8 to get: 3d - 2.8t = 2.8d
Rearrange to get: 0.2d = 2.8t
Solve: t = 0.2d/2.8 = 2d/28 =
d/14 In other words, Derek walked for
d/14 hours while sunny.
Distance = (speed)(time)At a walking speed of 2 mph, the distance Derek walked while sunny = (2)(
d/14) = 2d/14 = d/7
So, Derek walked d/7 miles of the total distance while the sun was shining on him.
In other words, Derek walked 1/7 of the total distance while the sun was shining on him.
Answer: 1/7
ASIDE: As you can see the problem is much harder to solve when we don't have signed a nice value for the distance in this case
Cheers,
Brent