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Cloudy and Sunny

by srcc25anu » Tue Apr 30, 2013 4:25 am

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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

[spoiler]OA:E[/spoiler]
Last edited by srcc25anu on Tue Apr 30, 2013 7:17 am, edited 1 time in total.
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by Brent@GMATPrepNow » Tue Apr 30, 2013 7:15 am
srcc25anu 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

[spoiler]OA:E[/spoiler]
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 = [spoiler]1/7 = E[/spoiler]

Cheers,
Brent
Last edited by Brent@GMATPrepNow on Mon Aug 08, 2016 1:45 pm, edited 1 time in total.
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by GMATGuruNY » Tue Apr 30, 2013 8:21 am
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--------0.8--------2.8------0.2-----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) = 0.2 : 0.8 = 1:4.

Here, the weight given to each rate is the amount of TIME spent at each rate.
The ratio above implies the following:
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.
Last edited by GMATGuruNY on Tue Dec 02, 2014 7:02 am, edited 1 time in total.
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by BlackDog » Fri Aug 16, 2013 10:00 am
This may seem like a bit of a stupid question, but why isn't this a "typical" average speed problem?

I tried solving this out like a more typical average speed problem. My incorrect methodology is as follows:



There are two portions to Derek's walk: the sunny portion (S) and the cloudy portion (C).

speed (S): s/1hour
speed (C): (s+1)/1hour

what fraction of the total distance did he cover while the sun was shining on him?

We need to find total distance, then solve for the distance he covered while the sun was shining.

Average speed = total distance/total time
2.8 = d / (t)
The problem is, we don't know how long it took Derek to walk the distance.

(s is an integer)

If s=2 then on the first leg of the trip he walked at 2 miles/hour and on the return he walked at (2+1) = 3 miles/hour. This would allow for an average speed of 2.8 miles/hour.

speed (S): 2/1hour
speed (C): (3)/1hour

Average speed = total distance/total time taken
time = distance/speed

2.8 = d/[ (d/2) + (d/3) ]
2.8 = d/[ (3d/6) + (2d/6) ]
2.8 = d/[5d/6]
2.8 = d/1 * [6/5d]
2.8 = 6d/5d
2.8 = d

He walked a total of 2.8 miles. If his average speed was 2.8 miles then the entire walk took 1 hour.

Thanks!!
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by davidetav » Mon Dec 08, 2014 10:39 am
GMATGuruNY wrote:
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--------0.8--------2.8------0.2-----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) = 0.2 : 0.8 = 1:4.

Here, the weight given to each rate is the amount of TIME spent at each rate.
Hi, awesome approach.
But why does that technique give the amount of time spent? I have some difficulties to understand the mathematical logic in that technique. How do we know that it gives the time rate and not the distance rate?
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by Naznin » Sat Nov 26, 2016 5:02 am
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by Mo2men » Sun Jun 11, 2017 9:33 am
GMATGuruNY 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
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--------0.8--------2.8------0.2-----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) = 0.2 : 0.8 = 1:4.

Here, the weight given to each rate is the amount of TIME spent at each rate.
The ratio above implies the following:
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.
Dear GMATGuru,

When dealing with average speed problems as mixture problem, you used the speeds which to place up the alligation.
In general, Can I use times or distances to place the alligation? or use only times or speeds?


Thanks
Last edited by Mo2men on Sat Jun 24, 2017 3:42 pm, edited 2 times in total.
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by [email protected] » Sun Jun 11, 2017 12:16 pm
Hi All,

This question can be solved by TESTing THE ANSWERS. To start, 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. Knowing that, we can now use the answer choices 'against' the question and figure out how an average of 2.8 mph could occur. Since that average speed is relatively close to 3 mph, we're clearly spending far MORE time 'in cloudy' than we do 'in the sun.' We're asked for the fraction of the total distance he was traveling 'in the sun', so we're likely going to be dealing with a relatively small fraction.

Let's TEST Answer D: 1/6 of the distance

IF... we spent 2 miles traveling in the sun, then there was a total of 12 miles (and we spent 10 miles when it was cloudy).
Sunny = 2 miles @ 2mph = 1 hour traveled
Cloudy = 10 miles @ 3mph = 10/3 hours traveled

Average Speed = 12 miles/(1 + 10/3 hours) = 12/(13/3) = 36/13 = 2 10/13 mph. This is NOT 2.8 mph (it's a little less than that) - thus the correct answer is NOT Answer D. To increase the average speed to 2.8 mph, we need to spend MORE time at 3 mph. Based on the remaining answer choices, there's only one way that can occur....

Final Answer: E

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by Matt@VeritasPrep » Thu Jun 22, 2017 5:51 pm
Mo2men wrote:
GMATGuruNY 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
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--------0.8--------2.8------0.2-----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) = 0.2 : 0.8 = 1:4.

Here, the weight given to each rate is the amount of TIME spent at each rate.
The ratio above implies the following:
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.
Dear GMATGuru,

When dealing with average speed problems as mixture problem, you used the speeds which to place up the alligation.
In general, Can I use times or distances to place the alligation? or use only times or speeds?

Thanks
R can be expressed in terms of T and D, so depending on what facts you're given in the stem there might be some way to use T and D in a similar fashion.
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by Jeff@TargetTestPrep » Mon Jul 02, 2018 8:45 am
srcc25anu 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
Let the total distance = d and x = the distance he covered when it was sunny. Thus, the distance he covered when it was cloudy is d - x. We need to determine the value of x/d.

Furthermore, we see that s must be 2. That is because if s = 1, then s + 1 = 2. We can see that if both speeds are 2 mph or less, then the average speed can never be 2.8 mph. A similar situation exists if s = 3 (or more), and s + 1 = 4 (or more). We can see that if both speeds are 3 mph or greater, then the average speed again can never be 2.8 mph. Therefore, s must be 2 since it given that it's an integer. Therefore, his "sunny" speed is 2 mph, and his "cloudy" speed is 3 mph.

Since (total distance)/(total time) = average speed, we have:

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

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

Multiplying the equation by 84, we have:

30d = 42x + 28(d - x)

30d = 42x + 28d - 28x

2d = 14x

2/14 = x/d

1/7 = x/d

Answer: E

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