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Re: On a partly cloudy day, Derek decides to walk back from work [#permalink]
emmak 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

Since the average speed is 2.8, s has to be 2. The speed ratio is 2:3 and the time ratio is 1:4; therefore, the required ratio is 2/(2+12) = 1/7.
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emmak 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's assign a nice value to 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 mph 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

Answer: E

Cheers,
Brent
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Re: On a partly cloudy day, Derek decides to walk back from work [#permalink]
Hi GMATPrepNow & VeritasPrepKarishma

I couldn't recognize that I can get S equal 2 from the beginning, so I used the weighted average as follows:

Let ds is the distance walked when it's sunny while dt is the total distance.

2.8 (Average speed) = S x ds/dt + (S+1)(dt-ds)/dt

Let ds/dt = K

2.8 = SK + (S+1)(1-X)
S = X + 1.8

The only answer choice that keeps S as an integer (2) is X = 1/5 => answer C

May I know what's wrong in my approach?
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hisho wrote:
Hi GMATPrepNow & VeritasPrepKarishma

I couldn't recognize that I can get S equal 2 from the beginning, so I used the weighted average as follows:

Let ds is the distance walked when it's sunny while dt is the total distance.

2.8 (Average speed) = S x ds/dt + (S+1)(dt-ds)/dt

Let ds/dt = K

2.8 = SK + (S+1)(1-X)
S = X + 1.8

The only answer choice that keeps S as an integer (2) is X = 1/5 => answer C

May I know what's wrong in my approach?


Your solution assumes that average speed is based on the proportions of the distances traveled at each speed.
So, for example, if we traveled half the distance at 40 mph and traveled the other half the distance at 60 mph, then the average speed would be 50 mph.
However, this is not how average speeds work.

For more on this, watch the following video:


Cheers,
Brent
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hisho wrote:
Hi GMATPrepNow & VeritasPrepKarishma

I couldn't recognize that I can get S equal 2 from the beginning, so I used the weighted average as follows:

Let ds is the distance walked when it's sunny while dt is the total distance.

2.8 (Average speed) = S x ds/dt + (S+1)(dt-ds)/dt

Let ds/dt = K

2.8 = SK + (S+1)(1-X)
S = X + 1.8

The only answer choice that keeps S as an integer (2) is X = 1/5 => answer C

May I know what's wrong in my approach?



Average Speed = Total Distance / Total Time

Average Speed = (Distance1 + Distance2) / (Time1 + Time2)

Average Speed = (Speed1*Time1 + Speed2*Time2) / (Time1 + Time2)

What you have done is

Average Speed = (Speed1*Distance1 + Speed2*Distance2) / (Distance1 + Distance2)
which is illogical.
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Re: On a partly cloudy day, Derek decides to walk back from work [#permalink]
VeritasPrepKarishma wrote:
swarman wrote:
What i m not getting is that why are we considering average speed as arithmetic mean of the speeds?
average speed is calculated by total distance divided by total time taken, isnt it??

kindly help :)


Average Speed lies in between the two speeds. It may not be in the center since the time taken at the two speeds might be different but it does lie somewhere in between them. You cannot drive at two speeds: 50 mph and 60 mph and still expect to average 70 mph. Your average will lie somewhere between 50 and 60.

Similarly, if the average speed is 2.8 and the two speeds are consecutive integers, the speeds must be 2 and 3. You cannot have the speeds as (1 and 2) or (3 and 4) since they cannot average out to be 2.8.


Thanks a lot Karishma :). I like the way you explain things :thumbup:
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emmak 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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Don't fall for the trap! This problem is A LOT easier than many of the mathematical solutions in this forum seem to show. You just need to visualize what is happening, and use the leverage the problem gives you to strategically attack the question. It isn't about the math. Remember: the GMAT is a critical-thinking test. For those of you studying for the GMAT, you will want to internalize strategies that actually minimize the amount of math that needs to be done, making it easier to manage your time (giving you more time for harder questions.) The tactics I will show you here will be useful for numerous questions, not just this one. My solution is going to walk through not just what the answer is, but how to strategically think about it. Ready? Let's talk strategy here. Here is the full "GMAT Jujitsu" for this question:

Let's start with the embedded leverage. The phrase "\(s\) is an integer" is what makes this entire problem tick. If the only two rates that Derek can travel are \(s\) and \(s+1\), then the average rate must be between \(s\) and \(s+1\). Since the question clearly states that his "average speed for the entire distance is 2.8 miles/hr," then \(s=2\) and \(s+1 = 3\).

Using a strategy I call in my classes "Weight Balancing", it is easy to determine the ratio between \(s\) and \(s+1\). The attached image below shows what it looks like. Since \(2.8\) is closer to \(3\), this means that the average is weighted towards the \(3\). The distances from the average give us the ratio.
Attachment:
WeightBalancing.png
WeightBalancing.png [ 11.6 KiB | Viewed 2239 times ]

Thus, Derek walks back at a ratio of \(0.8:0.2\) or \(4:1\). It is cloudy 4 times as much as it is sunny. (Sounds like Derek lives in on the Oregon Coast!) :)

Since we have the ratio of time, it is very easy to determine the distance, since Distance = Rate * Time.

The distance traveled while it is sunny is: \(D = RT = 2*1x = 2x\)
The distance traveled while it is cloudy is: \(D = RT = 3*4x = 12x\)
(Note: I am including the scaling factor, "\(x\)" in these calculations to show we are dealing with a ratio of unknown amounts, but as you will see, the scaling factor will disappear...)

Since the problem asks us for the fraction of the total distance that Derek covered while the sun was shining on him, this would be:
\(\frac{2x}{2x+12x}=\frac{2x}{14x} = \frac{1}{7}\)

The answer is "E".

Now, for those of you that are preparing to take the GMAT, let’s look back at this problem through the lens of strategy. Your job as you study isn't to memorize the solutions to specific questions; it is to internalize strategic patterns that allow you to solve large numbers of questions. This problem can teach us patterns seen throughout the GMAT. First, whenever the problem gives us leverage such as the word "integer", PAY ATTENTION. This is often a hint that you will be using logic as much as math. This solution also uses a strategy I call in my classes "Weight Balancing." The idea is simple: whenever you are given an "average" value between two groups, see if it might be useful to know the ratio between the groups. In the case of this problem, that ratio is what allows you to solve the problem quickly and efficiently without any unnecessary math. And that is how you think like the GMAT.
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emmak 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


Given, average speed = 2.8
So we know s = 2, s+1 = 3 since the average speed must be between them
s * t(s) = d(s)
(s+1) * t(c) = d(c)

Avg speed = Total Distance / Total time
2.8 = Td / t(s) + t(c)
2.8t(s) + 2.8t(c) = Td
Td = 2t(s) + 3t(c)
2.8t(s) + 2.8t(c) = 2t(s) + 3t(c)
.8t(s) = .2t(c)
t(s) / t(c) = 1/4 (trap answer A, total 1/5 is trap C because this is the ratio of the times not the distances covered)

d(c) = 3*4, d(s) = 2*1 -- > d(total) = 14

d(s) / d(total) = 2/14 = 1/7
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On a partly cloudy day, Derek decides to walk back from work [#permalink]
I originally answered the question wrong (of course, as others did, I choose the Ratio of Time spent walking at Speed S)



I used Smart Numbers to answer. Average Speed is a Weighted Average that uses TIME as the Relative Weighting. The Actual Data Points of Speeds are less important than are the RELATIVE TIMES spent at Each Speed.


However, if the Average Speed over the Distance = 2.8 m.p.h.:

then this means that the 2 Speeds --- S and (S +1)--- must be such that the Time-Weighted Average Speed lies BETWEEN the 2 Integers.



Numbers Chosen:


Let the Time he traveled = 10 hours

Total Distance therefore = (2.8 m.p.h. Avg. Speed) * (Total Time of 10 hours) = 28 miles


AND, since the Average Speed/Weighted Average must ALWAYS fall between the 2 Data Points, it must be true that:

S = 2 m.p.h.

S + 1 = 3 m.p.h.



Since the Average Speed is a "Time-Weighted" Average, I utilized the Weighted Average formula to find the Relative Time spent walking at each Speed (such that the Weighted Average will become 2.8)



(Weighting 1) / (Weighting 2) = (Data Point 2 - Wghtd. Avg.) / (Wghtd. Avg. - Data Point 1)


In this circumstance:

(Relative Time spent @ S- Speed) / (Relative Time spent @ S+1 - Speed) = (S+1 - 2.8) / (2.8 - S)


Using the Smart Numbers Chosen:

T1 / T2 = (3 - 2.8) / (2.8 - 2) = (.2) / (.8) = 2/8 = 1/4


Thus, the Ratio of:

(Time Spent at 2 mph) : (Time spent at 3 mph) : (Total Time)
____________________________________________________
1x : 4x : 5x




Since we assumed the Actual TOTAL Time = 10 hours ------ 5x = 10 ----- x = 2

The Time Spent at S-Speed = 2 mph is = 1x = 1(2) ------> 2 hours


Distance Traveled when it was Sunny = (S) * (Time spent @ S) = (2 mph) * (2 hours) = 4 miles

Assumed TOTAL Distance = 28 miles


Fraction of Total Distance spent walking when it was Sunny = (4 miles) / (28 miles) = 1/7

-E-



Looking forward to seeing the Algebriac Solutions.....
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Re: On a partly cloudy day, Derek decides to walk back from work [#permalink]
We're told the average speed for the entire distance is 2.8 miles/hr.

Given the average speed of 2.8 miles/hr, s and s+1 must equal 2 and 3 respectively.

We can create the following equation:

average speed = total distance / total time

total time = total distance / average speed

total time = \(\frac{d }{ 2.8}\)

\(\frac{d }{ 2.8} = \frac{d1}{2} + \frac{d2}{3}\)

Let k = time spent when the sun was shining
let 1 - k = time spent when cloudy

\(\frac{d }{ 2.8} = \frac{d1}{2} + \frac{d2}{3} = \frac{kd}{2} + \frac{(1-k)d}{3}\)

\(7(k+2) = 15\)
\(7k = 1\)
\(k = \frac{1}{7}\)

Answer is E.
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On a partly cloudy day, Derek decides to walk back from work [#permalink]
Why we cannot use mixture tools here?
Attachment:
photo5125353228333066530.jpg
photo5125353228333066530.jpg [ 62.08 KiB | Viewed 1616 times ]


Edit: it was already answered above! =)
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Re: On a partly cloudy day, Derek decides to walk back from work [#permalink]
What if we instead were asked for the time walked while the sun was shining on him. How would we do then?

d = vt -->

We have v = 1/5 * 2 = 2/5

If we want the distance we take d = 2/5*t

But if we want the time we take t = d/(2/5) and in this case we get 5d/2 (sunny) + 5d/12 (cloudy) = 30d/12 + 5d/12 --> 30/35 = 6/7

So the time has the opposite fraction of the distance?

Correct me if I'm wrong.
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Re: On a partly cloudy day, Derek decides to walk back from work [#permalink]
Hello,

Is the algebra good here?

I do not get how do we go from the 1st line to the second?

2.8 = (x+y) / (x/2 + y/3)
2.8x/2 + 2.8y/3 = (x+y)

In other terms, if we multiply : (x+y)/(x/2 + y/3) by (x/2 + y/3) do we obtain (x+y)?
Or is (x/2 + y/3)/ (x/2 + y/3)= 1 ?

Thanks

WholeLottaLove 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?

Looking for the fraction of his total distance that was covered when the sun was shining on him. We're going to have to assign variables to sun distance and cloud distance so we can ultimately plug into the formula for average speed. We will also need to get the time taken for each leg of the journey which means we also need to know the speed traveled for each leg.
x=sun
y=clouds
The fraction of time spent walking when it is sunny = x/(x+y) (x divided by the total distance)

We have established that the speed for the first part = 2 and the speed for the second part =3

We are given average speed so it is fair to assume that we will apply it to a formula: Average speed = total distance/total time taken
Time taken:
sunny part of the journey: x/s (distance/speed)
cloudy part of the journey: x/s+1

2.8 = (x+y) / (x/2 + y/3)
2.8x/2 + 2.8y/3 = (x+y)
8.4x/6 + 5.6y/6 = x+y
8.4x+5.6y/6 = x+y
8.4x+5.6y = 6x+6y
2.4x = .4y
6x=y
x=1/6y
x/y=1/6

Answer: E. 1/7
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Re: On a partly cloudy day, Derek decides to walk back from work [#permalink]
Official Explanation: Since the speed in the sun, s, is an integer, s + 1 must also be an integer. The average speed should lie somewhere in between these two integers. Since the average speed is 2.8 miles/hr, this implies that s and (s + 1) must be 2 and 3 miles/hr respectively.

Now, Average Speed = [(2 × Sunny Time) + (3 × Cloudy Time)]/ (Sunny Time + Cloudy Time) 2.8 = [(2 × Sunny Time) + (3 × Cloudy Time)]/ (Sunny Time + Cloudy Time)

Sunny Time/ Cloudy Time = 1/4.

Hence ratio of time spent under the sun to time spent while it was cloudy is t : 4t

Be careful here – the ratio of Time Spent in the sun to time spent in the clouds is 1:4, but the question asks for the fraction of the total distance. Rate = Distance/Time, so:

Rate Sunny = 2 = D/(1/5) time

Rate Cloudy = 3 = D / 4/5 time

2/5 = Sunny Distance

12/5 = Cloudy Distance

The distances have a ratio of 2:12, which means that the distance he covered in the sun is 2/14 of the total distance, and the answer reduces to 1/7.
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Naptiste wrote:
Hello,

Is the algebra good here?

I do not get how do we go from the 1st line to the second?

2.8 = (x+y) / (x/2 + y/3)
2.8x/2 + 2.8y/3 = (x+y)

In other terms, if we multiply : (x+y)/(x/2 + y/3) by (x/2 + y/3) do we obtain (x+y)?
Or is (x/2 + y/3)/ (x/2 + y/3)= 1 ?

Thanks

WholeLottaLove 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?

Looking for the fraction of his total distance that was covered when the sun was shining on him. We're going to have to assign variables to sun distance and cloud distance so we can ultimately plug into the formula for average speed. We will also need to get the time taken for each leg of the journey which means we also need to know the speed traveled for each leg.
x=sun
y=clouds
The fraction of time spent walking when it is sunny = x/(x+y) (x divided by the total distance)

We have established that the speed for the first part = 2 and the speed for the second part =3

We are given average speed so it is fair to assume that we will apply it to a formula: Average speed = total distance/total time taken
Time taken:
sunny part of the journey: x/s (distance/speed)
cloudy part of the journey: x/s+1

2.8 = (x+y) / (x/2 + y/3)
2.8x/2 + 2.8y/3 = (x+y)
8.4x/6 + 5.6y/6 = x+y
8.4x+5.6y/6 = x+y
8.4x+5.6y = 6x+6y
2.4x = .4y
6x=y
x=1/6y
x/y=1/6

Answer: E. 1/7


Yes, Naptiste

\(\frac{a}{b} * b = \frac{ab}{b} = a\)

The b in the denominator get's cancelled with the b in the numerator.

So

\(\frac{(x+y)}{(x/2 + y/3)} * (x/2 + y/3) = (x + y)\)

Though the intent of this question is to use reason and ratios and avoid algebra.

With an average speed of 2.8 and two consecutive integer speeds, we know that the two speeds are 2 and 3.

If the average is 2.8, ratio of time taken at speed 2 and speed 3 = (3 - 2.8)/(2.8 - 2) = 1/4

(Check this post if you are not sure why: https://anaprep.com/arithmetic-weighted-averages/
and this video: https://www.youtube.com/watch?v=_GOAU7moZ2Q )

Say time taken was t and 4t.

Then ratio of distance covered at speed 2 and speed 3 = 2*t/3*4t = 1/6

So he covered 1/7th of the distance at speed 2 i.e. when the sun was shining on him.

Answer (E)

Originally posted by KarishmaB on 30 Jan 2022, 22:26.
Last edited by KarishmaB on 08 Aug 2023, 04:30, edited 1 time in total.
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