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Nice question...

This topic has 4 expert replies and 7 member replies
rishianand7 Senior | Next Rank: 100 Posts Default Avatar
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Nice question...

Post Tue Aug 13, 2013 1:21 pm
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?

1/4
4/5
1/5
1/6
1/7

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Post Tue Aug 13, 2013 1:25 pm
rishianand7 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?

A. 1/4
B. 4/5
C. 1/5
D. 1/6
E. 1/7
Here's one approach.

Let's plug in a nice value for the total distance traveled.
If Derek's average speed is 2.8 mph, then let's say that he traveled a total of 28 miles.
At an average rate of 2.8 mph, a 28 mile trip will take 10 hours.

Since Derek's average speed is between 2 and 3 mph, we can conclude that Derek walked 2 mph when it was sunny and he walked 3 mph when it was cloudy.

Let's t = number of hours walking while sunny
So, 10 - t = number of hours walking while cloudy

We'll begin with a word equation: (distance traveled while sunny) + (distance traveled while cloudy) = 28
Since distance = (speed)(time), we can now write:
(2)(t) + (3)(10 - t) = 28
Expand: 2t + 30 - 3t = 28
Solve: t = 2
In other words, Derek walked for 2 hours while sunny.

At a walking speed of 2 mph, Derek walked for 4 miles while sunny.
So, Derek walked 4/28 of the total distance while the sun was shining on him.
4/28 = 1/7 = E

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topspin20 Junior | Next Rank: 30 Posts Default Avatar
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Post Tue Aug 13, 2013 5:50 pm
Very nice, Brent.

At the risk of coming across as sadistic, does anybody have a purely algebraic solution to this?

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vipulgoyal Master | Next Rank: 500 Posts Default Avatar
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Post Tue Aug 13, 2013 8:31 pm
Alt approach

We know that the average speed is 2.8 mph. Thus,

as s<2.8 and S is int, so only possible value for S could be 2

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ganeshrkamath Master | Next Rank: 500 Posts
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Post Tue Aug 13, 2013 10:30 pm
rishianand7 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?

1/4
4/5
1/5
1/6
1/7
average speed = 2.8 miles/hr
Clearly, since s is an integer, s = 2 miles/hr and (s+1) = 3 miles/hr
So 2x + 3(1-x) = 2.8 __________ x is the fraction of time it was sunny.
2x + 3 - 3x = 2.8
x = 0.2

So required ratio = 2x/2.8 = 0.4/2.8 = 1/7
Choose E

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Post Wed Aug 14, 2013 4:16 am
Quote:
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?

A. 1/4
B. 4/5
C. 1/5
D. 1/6
E. 1/7
The average speed -- 2.8 miles per hour -- must be BETWEEN the two individual rates (s and s+1).
Thus, s = 2 miles per hour and s+1 = 3 miles per hour.

This is a MIXTURE problem.
A rate of 2 miles per hour is being combined with a rate of 3 miles per hour to yield an average speed of 2.8 miles per hour.
To determine how much WEIGHT must be given to each rate, we can use ALLIGATION:

Step 1: Plot the 3 rates on a number line, with the two individual rates (2 miles per hour and 3 miles per hour) on the ends and the average speed for the whole trip (2.8) in the middle.
2------------------2.8------------3

Step 2: Calculate the distances between the rates.
2--------.8--------2.8------.2-----3

Step 3: Determine the ratio of the rates.
The required ratio is the RECIPROCAL of the distances in red.
(2 miles per hour) : (3 miles per hour) = .2 : .8 = 1:4.

Here, the weight given to each rate is the amount of TIME spent at each rate.
The ratio above implies the foliowing:
For every 1 hour spent traveling at 2 miles per hour, 4 hours must be spent traveling at 3 miles per hour.

Distance traveled in 1 hour at rate of 2 miles per hour = r*t = 2*1 = 2 miles.
Distance traveled in 4 hours at a rate of 3 miles per hour = r*t = 3*4 = 12 miles.
Of the total distance, the fraction traveled at 2 miles per hour = 2/(2+12) = 2/14 = 1/7.

The correct answer is E.

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Post Wed Aug 14, 2013 9:05 pm
topspin20 wrote:
Very nice, Brent.

At the risk of coming across as sadistic, does anybody have a purely algebraic solution to this?
Only since you asked for it ...

Sunny portion of the trip:

Rate = s
Time = t

Cloudy portion of the trip

Rate = (s+1)
Time = z

Total trip:

Rate = 2.8
Time = t + z

Since D = RT, 2.8(t+z) = st + (s+1)z, or

2.8t + 2.8z = st + sz + z, or
2.8t - st = sz + z - 2.8z, or
(2.8t - st)/(s - 1.8) = z

We want the ratio of sunny distance to total distance, so

(st)/(2.8(t+z)), or

(st)/(2.8(t + (2.8t - st)/(s - 1.8))), or

(st)/(2.8t/(s-1.8)), or

(st * (s-1.8)) / 2.8t

At this point, however, we're in trouble - there's no way to tell our algebra that s must be an integer. (Even if we knew that this equation was equal to 1/7, we'd still get two solutions for s, one of which is negative.)

One thing we do notice, though, is that we have (s - 1.8) in our numerator. Since that has to be a positive number, we can conclude that s is at least 2. Since the average rate was 2.8 and is somewhere between s and (s+1), s can't be greater than 3, so s = 2.

... but as many other experts have already explained, we can deduce that without all this algebra.

Funny note on this question: I encountered it when beta-testing our Veritas exams, and TRIED TO SOLVE ALGEBRAICALLY. Whoops! Very Happy

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ela07mjt Junior | Next Rank: 30 Posts
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Post Sat Jul 04, 2015 3:50 am
Hi,

I am confused. Why does the min and max speed has to be 2 and 3? I know it has to be an integer.

Your average speed would be closer to the speed you traveled at for the larger duration of time.

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Post Sat Jul 04, 2015 10:05 am
Hi ela07mjt,

The prompt provides a few clues about why the speeds have to be 2 miles/hour and 3 miles/hour:

1) The two speeds are S miles/hour and (S+1) miles/hour, so the two numbers must differ by EXACTLY 1.
2) We're told that S is an INTEGER, so (S+1) will ALSO be an integer.
3) The average speed is 2.8 miles/hour. Since this is an AVERAGE speed, the value for S must be BELOW 2.8 and the value of (S+1) must be ABOVE 2.8

There is only one pair of numbers that fit all of these factoids: 2 and 3

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Post Thu Jul 09, 2015 2:44 am
vipulgoyal wrote:
Alt approach

We know that the average speed is 2.8 mph. Thus,

as s<2.8 and S is int, so only possible value for S could be 2
Good point as average speed of 2.8 is between s & s+1

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nikhilgmat31 Legendary Member Default Avatar
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Post Thu Jul 09, 2015 2:50 am
speed in sun S =2
speed in clouds s+1 = 3

total distance = d
distance in sun = x
distance in clouds = d-x

x/2 + (d-x)/3 = d/2.8

3x+2d -2x = 6d/2.8
28(2d+x) = 60d
56d + 28x = 60d

28x=4d
x=4/28d


x=1/7d

so in sun he travelled 1/7 of distance.

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nikhilgmat31 Legendary Member Default Avatar
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Post Thu Jul 09, 2015 2:52 am
ela07mjt wrote:
Hi,

I am confused. Why does the min and max speed has to be 2 and 3? I know it has to be an integer.

Your average speed would be closer to the speed you traveled at for the larger duration of time.
It is because speed in sun is s & speed in cloud is s+1 & we know average is 2.8

so to get average of 2.8 -- we need to pick 2 & 3.

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