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# On a partly cloudy day, Derek decides to walk back from work

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21 Aug 2013, 19:56
Simplest way to the answer that I found.....

s just has to equal 2, and s+1 just has to equal 3. If average speed = 2.8, he has to be at s(2) for 1/5 of time, and s+1(3) for 4/5 of time.

Time is not defined, only the average, so lets say the total time was 5 hours, so in one hour he walked 2 miles, then in each of the other four hours he walked 3 miles, 2+3+3+3+3=14.

What fraction of the total distance did he walk at s(2), is equal to 2 miles / 14 miles = 1/7
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06 Mar 2014, 22:55
Is there an algebraic way to solve for s (or even solve directly for the distance ratio) if it wasn't immediately apparent to us that 2.8 is between s and s+1 and s is an integer so it has to be 2?

I attempted to do this and it not only turned into a mess, but also didn't seem possible? Knowing that s=2 is definitely the key, but I was curious if there is a feasible workaround, thanks.
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Re: On a partly cloudy day, Derek decides to walk back from work  [#permalink]

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07 Mar 2014, 08:04
2
m3equals333 wrote:
Is there an algebraic way to solve for s (or even solve directly for the distance ratio) if it wasn't immediately apparent to us that 2.8 is between s and s+1 and s is an integer so it has to be 2?

I attempted to do this and it not only turned into a mess, but also didn't seem possible? Knowing that s=2 is definitely the key, but I was curious if there is a feasible workaround, thanks.

I wrote this question and I will tell you what my intent was in writing it - I wanted to make people use logic. I wanted them to use deduction on the given data. Use of logic greatly simplifies many high level GMAT questions. So I tried to force them to deduce this vital piece of information - the two speeds must be 2 and 3. You can't put everything given to you in terms of an equation and that is what I tried to exploit.

Even if you don't start with it, at some point you will require some deduction.

For example, say you use weighted averages. 2.8 is the average speed. Weights in case of speeds are the time taken in the two cases.

Ratio of time spent while its sunny and time spent while its cloudy = $$\frac{Ts}{Tc} = \frac{(s+1) - 2.8}{2.8 - s}$$

$$\frac{Ts}{Tc}= \frac{s - 1.8}{2.8 - s}$$

Now what can you say about s?

s must be greater than 1.8 since the fraction cannot be negative. It must also be less than 2.8 since the fraction cannot be negative.

Since s must be an integer, it must be 2. So (s+1) must be 3.
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13 Aug 2014, 15:14
sunny speed (s) < 2.8< (s+1) cloudy speed
As S is integer. The only possible value is S=2.

As avg speed is 2.8 .
i.e total distance/ total time = 2.8
Let the total distance is 28 and time taken 10 hrs.

Out of 10 hours X hours in Sunny and 10-X hours in cloudy

So total distance S*X + (S+1) * (10-x) = 28
2x + 30 - 3x = 28 ( substituted s = 2)
x = 2

In sunny at speed 2 miles/hr , the person walked for 2 hrs. Hence distance covered = 2*2 = 4 miles
Ration = 4miles ( distance in sunny) / 28miles( total distance) = 1/7

Ans E.
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10 Sep 2014, 05:52
A slightly slightly different way...

Since s must be 2 we can set up an equation relating the two speeds:

2.8 = 2x + 3y

x and y being the fraction of time walked for each speed therefore:

x + y = 1

Solving, x= .2 and y = .8 .

Then 2 becomes .4 and 3 becomes 2.4 .

.4/2.8 = 1/7 or E
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06 Nov 2014, 00:59
total distance=k

sunny..
speed=s
distance=a

cloudy..
speed=s+1
d2=k-a

a/k=?

k/ [(a/s)+{(k-a)/(s+1)}]=2.8

how can we solve this further to arrive at the right ans..
kindly help any1.
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08 Dec 2014, 10:03
Can someone please explain why that approach is not correct?

Ds = distance sunny time
Dc = distance cloudy time

Average speed = 2.8 = 14/5

and we know, Average speed = total distance / total time
then, why can't we simply express that with Ds and Dc?

I mean : $$14/5 = \frac{(Ds+Dc)}{(Ds/2)+(Dc/3)}$$

and then try to obtain the ratio asked in the question
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08 Dec 2014, 20:19
1
oss198 wrote:
Can someone please explain why that approach is not correct?

Ds = distance sunny time
Dc = distance cloudy time

Average speed = 2.8 = 14/5

and we know, Average speed = total distance / total time
then, why can't we simply express that with Ds and Dc?

I mean : $$14/5 = \frac{(Ds+Dc)}{(Ds/2)+(Dc/3)}$$

and then try to obtain the ratio asked in the question

This approach is not wrong. You get Dc/Ds = 6/1 which means Ds = (1/7)th of the entire distance.
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27 Apr 2015, 23:53
sambam wrote:
If s is an integer and we know that the average speed is 2.8, s must be = 2. That means "s+1" = 3. This implies that the ratio of time for s=2 is 1/4 of the total time. The formula for distance/rate is D=Rt...so the distance travelled when s=2 is 2t. The distance travelled for s+1=3 is 3*4t or 12t. Therefore, total distance covered while the sun was shining over him is 2/14 = 1/7. Answer: E

This implies that the ratio of time for s=2 is 1/4 of the total time
I didn't understand the Bold part...
i am missing something here.. how come we know this??
Please explain. may be i am missing some concept here..
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On a partly cloudy day, Derek decides to walk back from work  [#permalink]

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Updated on: 06 May 2015, 22:53
dkumar2012 wrote:
sambam wrote:
If s is an integer and we know that the average speed is 2.8, s must be = 2. That means "s+1" = 3. This implies that the ratio of time for s=2 is 1/4 of the total time. The formula for distance/rate is D=Rt...so the distance travelled when s=2 is 2t. The distance travelled for s+1=3 is 3*4t or 12t. Therefore, total distance covered while the sun was shining over him is 2/14 = 1/7. Answer: E

This implies that the ratio of time for s=2 is 1/4 of the total time
I didn't understand the Bold part...
i am missing something here.. how come we know this??
Please explain. may be i am missing some concept here..

Hi dkumar2012,

Average speed is 2.8miles/hours
=> s < 2.8 < s + 1
Given s is an integer, s = 2 and s+1 = 3

Let time traveled with speed (s = 2) be t1
Let time traveled with speed (s + 1 = 3) be t2
As the average speed is 2.8,
2t1 + 3t2 = 2.8(t1 + t2)
=> t2(3-2.8) = t1(2.8 - 2)
=> t1/t2 = 0.2/0.8 = 1/4
=> t2/t1 = 4

Distance traveled with speed (s = 2) is 2t1
Distance traveled with speed (s + 1 = 3) is 3t2
Fraction of the distance traveled with speed (s = 2) is 2t1/(2t1 + 3t2)
2t1/(2t1 + 3t2) = 2/(2 + 3t2/t1) = 2/(2+3*4) = 2/14 = 1/7

Originally posted by PrepTap on 28 Apr 2015, 03:48.
Last edited by PrepTap on 06 May 2015, 22:53, edited 1 time in total.
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28 Apr 2015, 10:45
dkumar2012 wrote:
PrepTap wrote:
. ..2t1 + 3t2 = 2.8(t1 + t2) ...

this is what I missed here..
Thank you got it now...

Good to know it helped.
In general if you know the weighted average N of two number N1 and N2
The ratio of their weights $$\frac{w1}{w2}= \frac{(N - N2)}{(N1 - N)}$$
(This is derived from the equation w1N1 + w2N2 = N(w1 + w2)
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Updated on: 07 Aug 2015, 05:37
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

consider total distance as 28 miles and total time taken as 10 hours, so that the average speed is 2.8
let x is the first distance and y is the second distance
x/s +y/(s+1) = 10 and x+y = 28
s being an integer, this is possible only when s=2; implies x = 4 and y = 24
correct option is (E)

Originally posted by mathivanan on 13 Jun 2015, 02:51.
Last edited by mathivanan on 07 Aug 2015, 05:37, edited 1 time in total.
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13 Jun 2015, 02:59
matvan wrote:
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

I should admit that I could not solve this mathematically. I assumed total distance as 2.8 miles. Then I tried from the options to get 's' as an integer. Option E gives an integer value for s. Therefore, that should be the answer.

.4/s + 2.4/(s+1) = 1
solving this you get s=2 and s+1 = 3
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Updated on: 05 Jul 2017, 18:47
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 x=hours walked while sunny
y=hours walked while cloudy
we know that 2 mph and 3 mph are the respective rates for sunny and cloudy walking
2x+3y=total distance walked
x+y=total time walked
(2x+3y)/(x+y)=2.8 average mph
y=4x
ratio of sunny distance to total distance=2x/(2x+3y)
substituting 4x for y,
2x/(2x+12x)=1/7
1/7=fraction of total distance walked while sunny
E

Originally posted by gracie on 30 Jun 2016, 21:17.
Last edited by gracie on 05 Jul 2017, 18:47, edited 1 time in total.
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Re: On a partly cloudy day, Derek decides to walk back from work  [#permalink]

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29 Nov 2016, 08:16
VeritasPrepRon wrote:
sambam wrote:
If s is an integer and we know that the average speed is 2.8, s must be = 2. That means "s+1" = 3. This implies that the ratio of time for s=2 is 1/4 of the total time. The formula for distance/rate is D=Rt...so the distance travelled when s=2 is 2t. The distance travelled for s+1=3 is 3*4t or 12t. Therefore, total distance covered while the sun was shining over him is 2/14 = 1/7. Answer: E

The key to this question is indeed that s must be 2, and therefore s+1 must be 3. It is impossible to average out a speed of 2.8 with any other two consecutive integers. The algebraic solution outlined above using D=RT gives the correct answer very quickly, but this can also be solved by using the concept and backsolving.

First of all, the classic trap of 1/4 of the time of the total time above is misleading. In fact, it implies that Derek spent 1/5 of his time at the slower speed and 4/5 at the higher speed. This is because 0.8 is analogous to 4/5, and can be demonstrated by (1/5 * 2) + (4/5 * 3) = 2/5 + 12/5 = 14/5 or 2.8. Once we know that Derek spent 1/5 of his time walking at his sunny-weather rate, and we know that the other 4/5 of the time he was walking faster, we can deduce that he covered less than 1/5 of the distance at the sunny-weather rate. If we understand this concept, we are down to two answer choices, D or E.

We can now backsolve by converting time into distance. Using D (1/6 of distance), we can assume 6 miles of distance, 1 of which at 2 mph and the other 5 at 3 mph. This leaves us with 0.5 hours of timeat speed s and 1.67 hours at speed s+1. These need to be in the ratio of 1:4 (or 1/5 to 4/5), and therefore don't work. Once this doesn't work, we know the answer is E. (Note: Backsolving for E gives 7 miles, 1 at 2 mph and 6 at 3 mph, yielding totals of 0.5 hours and 2 hours, exactly what we're looking for.

Undoubtedly the algebraic solution is faster, however, the concept alone leaves this at a 50/50 choice between D and E. Afterward, solving using one answer choice will confirm which one of the two must be correct.

Thanks !! I understood everything but I could not understand " we can deduce that he covered less than 1/5 of the distance at the sunny-weather rate." When we found out that the ratio of the time is 1:4, what logic did you use to say that sunny weather rate will cover less than 1/5 distance.
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Re: On a partly cloudy day, Derek decides to walk back from work  [#permalink]

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06 Dec 2016, 01:14
spaceman29spiff wrote:
Bunuel 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

Where, in which post, is the average speed calculated as the average of the two speeds?

Hi Bunuel,
I took the following classical equations route to solve

Sunny:
Speed = s
Distance = x
hence, time = x/s

Cloudy:
Speed = s+1
dist = y
hence, time = y/(s+1)

Average:
Speed = 2.8
Total dist = x+y
total time = (x/s) + y/(s+1)

solving,
x+y = 2.8 * [(x/s) + (y/s+1)]

i got the final equation as;

s*(s-1.8) / 2.8 = x/(x+y)

How to solve it from here?

thanks,
Rohan

HI, I started with the classical way too and ended up with a messy equation. However you can still manage to do it classical way and get to the right answer.

Avg speed = X+Y/T1+T2

(And As S=2 , S+1 =3) [ As S is Integer and <2.8 and >1.8 :. S=2]

Here T1 = X/2 and T2 = Y/3

:. 2.8 = $$\frac{X+Y}{x/2+y/3}$$

Put X+Y = Dt (total distance) and Y= Dt - X

Upon Solving you will find a relation between Dt and X as $$\frac{1}{7}$$

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

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08 Dec 2016, 23:17
Solved this question by considering 'proper' numbers: Since Avg. speed is 2.8 mph, took distance as 28 miles and therefore total time to cover the distance is 10 hours.

Assumed distance walked when sunny as 'x' and distance walked when cloudy becomes '28-x'. Therefore [x/s+28-x/s+1 = 10] , solving for 'x': x=2s[5s-9]. Consider 's' as 2, (as s=1 gives distance in negative value) , x=4, hence 4/2+24/3 = 10. Hence answer is 4/28 = 1/7 [E]
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08 Dec 2016, 23:31
I guess that the trick is AS is 2.8 so value of elements must be around it or 2 and 3
and required fraction is of the total distance not the distance walked on cloudy time.
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18 Jan 2017, 02:29
VeritasPrepRon wrote:
sambam wrote:
If s is an integer and we know that the average speed is 2.8, s must be = 2. That means "s+1" = 3. This implies that the ratio of time for s=2 is 1/4 of the total time. The formula for distance/rate is D=Rt...so the distance travelled when s=2 is 2t. The distance travelled for s+1=3 is 3*4t or 12t. Therefore, total distance covered while the sun was shining over him is 2/14 = 1/7. Answer: E

The key to this question is indeed that s must be 2, and therefore s+1 must be 3. It is impossible to average out a speed of 2.8 with any other two consecutive integers. The algebraic solution outlined above using D=RT gives the correct answer very quickly, but this can also be solved by using the concept and backsolving.

First of all, the classic trap of 1/4 of the time of the total time above is misleading. In fact, it implies that Derek spent 1/5 of his time at the slower speed and 4/5 at the higher speed. This is because 0.8 is analogous to 4/5, and can be demonstrated by (1/5 * 2) + (4/5 * 3) = 2/5 + 12/5 = 14/5 or 2.8. Once we know that Derek spent 1/5 of his time walking at his sunny-weather rate, and we know that the other 4/5 of the time he was walking faster, we can deduce that he covered less than 1/5 of the distance at the sunny-weather rate. If we understand this concept, we are down to two answer choices, D or E.

Could you explain the deduction in the bolded part?
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Re: On a partly cloudy day, Derek decides to walk back from work  [#permalink]

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18 Jan 2017, 04:11
1
sharkr wrote:
VeritasPrepRon wrote:
sambam wrote:
If s is an integer and we know that the average speed is 2.8, s must be = 2. That means "s+1" = 3. This implies that the ratio of time for s=2 is 1/4 of the total time. The formula for distance/rate is D=Rt...so the distance travelled when s=2 is 2t. The distance travelled for s+1=3 is 3*4t or 12t. Therefore, total distance covered while the sun was shining over him is 2/14 = 1/7. Answer: E

The key to this question is indeed that s must be 2, and therefore s+1 must be 3. It is impossible to average out a speed of 2.8 with any other two consecutive integers. The algebraic solution outlined above using D=RT gives the correct answer very quickly, but this can also be solved by using the concept and backsolving.

First of all, the classic trap of 1/4 of the time of the total time above is misleading. In fact, it implies that Derek spent 1/5 of his time at the slower speed and 4/5 at the higher speed. This is because 0.8 is analogous to 4/5, and can be demonstrated by (1/5 * 2) + (4/5 * 3) = 2/5 + 12/5 = 14/5 or 2.8. Once we know that Derek spent 1/5 of his time walking at his sunny-weather rate, and we know that the other 4/5 of the time he was walking faster, we can deduce that he covered less than 1/5 of the distance at the sunny-weather rate. If we understand this concept, we are down to two answer choices, D or E.

Could you explain the deduction in the bolded part?

It comes from the fact that when you spend equal amounts of time on two different speeds, you cover less distance at the lower distance. It is logical if you think about it.
Say you spend 1 hr at 60 kmph and 1 hr at 80 kmph. So obviously, you cover a distance of only 60 km at lower speed and you cover a distance of 80 km at higher speed.

This will be true here also. If you spend 1/5 of your time (say 1 hr) at a lower speed (say at 2 mph) and 4/5 of your time (say 4 hrs) at a faster speed (say at 3 mph), the distance you will cover at lower speed will be less than 1/5th (2/14).

Does this make sense?
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