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e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3367

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Updated on: 13 Aug 2018, 05:07
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Difficulty:

85% (hard)

Question Stats:

56% (03:23) correct 44% (03:57) wrong based on 73 sessions

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3 common mistakes you must avoid in Distance questions – Practice question 2

A bus made a roundtrip journey from Madison towards Chicago, which is 240 km away. In the onward journey the bus divided the journey time in 3 equal blocks. Started at 50 kph, it took stoppages after every block and increased its speed by 10 kph after every stoppage. At Chicago, it took a halt of 7 hrs 44 minutes. The return journey was divided into two equidistant blocks, covered at 45 kph and 75 kph respectively. Find by what amount the average speed of the whole journey of the bus is less than the sum of the average speeds of onward and return journey;

A. 75 kph
B. 86.25 kph
C. 97.5 kph
D. 105 kph
E. 120 kph

To solve question 3: Question 3

To read the article: 3 common mistakes you must avoid in Distance questions

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Originally posted by EgmatQuantExpert on 16 May 2018, 06:37.
Last edited by EgmatQuantExpert on 13 Aug 2018, 05:07, edited 6 times in total.
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16 May 2018, 06:40
1
1

Solution

Given:
• Distance between Madison and Chicago = 240 km
• In the onward journey, the total journey time was divided into 3 equal blocks
• The bus increased its speed by 10 kph after every stoppage
o Speed in the 1st time block = 50 kph
o Speed in the 2nd time block = 60 kph
o Speed in the 3rd (final) time block = 70 kph
• Halt time in Chicago = 7 hrs 44 minutes = $$\frac{116}{15}$$ hrs
• In the return journey, the total journey was divided into 2 equal distance blocks
o Speed in the 1st distance block = 45 kph
o Speed in the 2nd distance block = 75 kph

To find:
• The value of the average speed of the bus in the whole journey is what amount less than the sum of the average speeds of the onward and return journeys

Approach and Working:
• In the onward journey, let us assume each time block takes t hrs
o Therefore, we can say 50t + 60t + 70t = 240
Or, 180t = 240
Or, t = $$\frac{240}{180}$$ hrs = $$\frac{4}{3}$$ hrs
• Hence, the total time taken in the onward journey = 3t = $$3 * \frac{4}{3}$$ hrs = 4 hrs
• Therefore, the average speed in the onward journey = $$\frac{240}{4}$$ kph = 60 kph
[as the time taken in each block is same, we could have also calculated the average speed by taking the arithmetic mean of the individual speeds, i.e. $$\frac{(50 + 60 + 70)}{3}$$ = 60 mph]

• In the return journey, because the total distance is equally divided into 2 parts, we can say the average speed is = $$\frac{(2 * 45 * 75)}{(45 + 75)}$$ = 56.25 kph
• Also, the time taken in the return journey = ($$\frac{120}{45} + \frac{120}{75}$$) hrs = ($$\frac{8}{3} + \frac{8}{5}$$) hrs = $$\frac{64}{15}$$ hrs

• Now, if we consider the whole journey,
The total time taken = ($$4 + \frac{116}{15} + \frac{64}{15}$$) hrs = 16 hrs
o Total distance travelled = (240 + 240) km = 480 km
• Therefore, the average speed of the whole journey = $$\frac{480}{16}$$ kph = 30 kph

• So, the amount by which overall average speed is less than the sum of the individual average speeds = (60 + 56.25) – 30 kph = 116.25 – 30 kph = 86.25 kph

Hence, the correct answer is option B.

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Joined: 30 Jun 2019
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04 Mar 2020, 23:24
EgmatQuantExpert wrote:

Solution

Given:
• Distance between Madison and Chicago = 240 km
• In the onward journey, the total journey time was divided into 3 equal blocks
• The bus increased its speed by 10 kph after every stoppage
o Speed in the 1st time block = 50 kph
o Speed in the 2nd time block = 60 kph
o Speed in the 3rd (final) time block = 70 kph
• Halt time in Chicago = 7 hrs 44 minutes = $$\frac{116}{15}$$ hrs
• In the return journey, the total journey was divided into 2 equal distance blocks
o Speed in the 1st distance block = 45 kph
o Speed in the 2nd distance block = 75 kph

To find:
• The value of the average speed of the bus in the whole journey is what amount less than the sum of the average speeds of the onward and return journeys

Approach and Working:
• In the onward journey, let us assume each time block takes t hrs
o Therefore, we can say 50t + 60t + 70t = 240
Or, 180t = 240
Or, t = $$\frac{240}{180}$$ hrs = $$\frac{4}{3}$$ hrs
• Hence, the total time taken in the onward journey = 3t = $$3 * \frac{4}{3}$$ hrs = 4 hrs
• Therefore, the average speed in the onward journey = $$\frac{240}{4}$$ kph = 60 kph
[as the time taken in each block is same, we could have also calculated the average speed by taking the arithmetic mean of the individual speeds, i.e. $$\frac{(50 + 60 + 70)}{3}$$ = 60 mph]

• In the return journey, because the total distance is equally divided into 2 parts, we can say the average speed is = $$\frac{(2 * 45 * 75)}{(45 + 75)}$$ = 56.25 kph
• Also, the time taken in the return journey = ($$\frac{120}{45} + \frac{120}{75}$$) hrs = ($$\frac{8}{3} + \frac{8}{5}$$) hrs = $$\frac{64}{15}$$ hrs

• Now, if we consider the whole journey,
The total time taken = ($$4 + \frac{116}{15} + \frac{64}{15}$$) hrs = 16 hrs
o Total distance travelled = (240 + 240) km = 480 km
• Therefore, the average speed of the whole journey = $$\frac{480}{16}$$ kph = 30 kph

• So, the amount by which overall average speed is less than the sum of the individual average speeds = (60 + 56.25) – 30 kph = 116.25 – 30 kph = 86.25 kph

Hence, the correct answer is option B.

why do you assume the time take for the bus to travel each block is equal? It doesn't say that in the problem, just that the distances were equal.
first trip = 240/(80/50+80/60+80/70) = 58.87
2nd trip = 240/(120/45+120/75) = 56.25
Average of the whole trip 480/(80/50+80/60+80/70+120/45+120/75) = 57.53

58.87+56.25 - 57.53 =~ 59+56-58 = 57
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Joined: 13 Feb 2020
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09 Mar 2020, 12:14
The bus took stoppages after every block, but there is no mention of those stoppages in the solution.
Manager
Joined: 30 Jun 2019
Posts: 208

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09 Mar 2020, 12:53
1
Bunuel can you check out this problem? I disagree with the provided solution (i explained my thought in the post above).
The solution seems flawed based on
1. the assumption that the first part of the journey has an even distribution of time across equal distance segments, but different rates of travel
2. The average speed of the whole trip accounts for time when the bus isn't moving.
Intern
Joined: 28 May 2017
Posts: 9

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28 Mar 2020, 22:16
fireagablast wrote:
Bunuel can you check out this problem? I disagree with the provided solution (i explained my thought in the post above).
The solution seems flawed based on
1. the assumption that the first part of the journey has an even distribution of time across equal distance segments, but different rates of travel
2. The average speed of the whole trip accounts for time when the bus isn't moving.

fireagablast ,
1. If we carefully decode the word problem, we can see that "In the onward journey the bus divided the journey time in 3 equal blocks."

2. When the bus started, the "odometer" was ON & would be stopped ONLY at the return/ destination point. So the whole time (including the halt time) will be considered here.

P.S - As there is no mention for the stoppage time, for the sake of the question, we can move on without considering the stoppage time.

Bunuel , EgmatQuantExpert and VeritasKarishma am I correct ?
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Joined: 30 Jun 2019
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29 Mar 2020, 20:29
bobnil wrote:
fireagablast wrote:
Bunuel can you check out this problem? I disagree with the provided solution (i explained my thought in the post above).
The solution seems flawed based on
1. the assumption that the first part of the journey has an even distribution of time across equal distance segments, but different rates of travel
2. The average speed of the whole trip accounts for time when the bus isn't moving.

fireagablast ,
1. If we carefully decode the word problem, we can see that "In the onward journey the bus divided the journey time in 3 equal blocks."

2. When the bus started, the "odometer" was ON & would be stopped ONLY at the return/ destination point. So the whole time (including the halt time) will be considered here.

P.S - As there is no mention for the stoppage time, for the sake of the question, we can move on without considering the stoppage time.

Bunuel , EgmatQuantExpert and VeritasKarishma am I correct ?

I see the issue with the first highlighted part, thanks.

As for the second issue, i'm still unconvinced that the time the car is parked should be considered as part of the average speed.
If i drive from point A to point B, but stop at a hotel to sleep along the way, why would i count that time as a part of my average speed?
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Posts: 10435
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29 Mar 2020, 21:54
2
fireagablast wrote:
bobnil wrote:
fireagablast wrote:
Bunuel can you check out this problem? I disagree with the provided solution (i explained my thought in the post above).
The solution seems flawed based on
1. the assumption that the first part of the journey has an even distribution of time across equal distance segments, but different rates of travel
2. The average speed of the whole trip accounts for time when the bus isn't moving.

fireagablast ,
1. If we carefully decode the word problem, we can see that "In the onward journey the bus divided the journey time in 3 equal blocks."

2. When the bus started, the "odometer" was ON & would be stopped ONLY at the return/ destination point. So the whole time (including the halt time) will be considered here.

P.S - As there is no mention for the stoppage time, for the sake of the question, we can move on without considering the stoppage time.

Bunuel , EgmatQuantExpert and VeritasKarishma am I correct ?

I see the issue with the first highlighted part, thanks.

As for the second issue, i'm still unconvinced that the time the car is parked should be considered as part of the average speed.
If i drive from point A to point B, but stop at a hotel to sleep along the way, why would i count that time as a part of my average speed?

The question gives you:
"A bus made a roundtrip journey ..."
"Find by what amount the average speed of the whole journey of the bus is less ..."

The "journey" is the roundtrip journey.
Total "journey distance" is the total distance travelled.
Total "journey time" is the total time taken to return.
It will include the stoppage time until and unless mentioned otherwise.

If you drive from point A to B, your average speed of day 1 or day 2 will not take your overnight stay at hotel time. But your average speed for the entire journey will include the time taken for the entire journey irrespective of what you were doing in that time.
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Karishma
Veritas Prep GMAT Instructor