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A bus completed first 50 miles of a 120-mile trip at an average speed

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A bus completed first 50 miles of a 120-mile trip at an average speed  [#permalink]

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New post Updated on: 11 Aug 2018, 12:41
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E

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  35% (medium)

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75% (01:43) correct 25% (02:28) wrong based on 65 sessions

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A bus completed first 50 miles of a 120-mile trip at an average speed of 20 mph. Then it took a halt of 30 minutes and completed the half of the remaining journey at an average speed of 35 mph. At what average speed it should complete the remaining journey so that the overall average speed of the whole journey becomes 20 mph?

    A. 40 mph
    B. 35 mph
    C. 30 mph
    D. 22.5 mph
    E. 17.5 mph


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Originally posted by EgmatQuantExpert on 30 Jul 2018, 21:55.
Last edited by EgmatQuantExpert on 11 Aug 2018, 12:41, edited 1 time in total.
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Re: A bus completed first 50 miles of a 120-mile trip at an average speed  [#permalink]

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New post 30 Jul 2018, 22:16
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A bus completed first 50 miles of a 120-mile trip at an average speed of 20 mph. Then it took a halt of 30 minutes and completed the half of the remaining journey at an average speed of 35 mph. At what average speed it should complete the remaining journey so that the overall average speed of the whole journey becomes 20 mph?

Wanted overall average speed of journey = 20 mph
Total distance = 120 miles
Time required to cover for total distance = 120/20 = 6 Hours

First 50 mile 20 mph => Time taken for trip = 50/20 = 2.5 Hours
Halt of 30 minutes or 0.5 Hour.

Remaining distance = 120 - 50 = 70 miles.
Half of remaining journey = 70/2 = 35 miles at 35 mph => Time taken for trip = 35/35 = 1 Hour

Time Elapsed so far = 2.5 + 0.5 + 1 = 4 Hours
Time that should be spent to achieve 20 mph = Time required to cover for total distance - Time Elapsed so far => 6 - 4 = 2

Remaining distance 70 - 35 = 35 miles
Average Speed = 35/2 = 17.5 mph

Answer E
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Re: A bus completed first 50 miles of a 120-mile trip at an average speed  [#permalink]

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New post 07 Aug 2018, 18:28
Well if you have 20mph, then 35 mph, then x mph.
You want the average to be 20 again, then x has to be under 20. Hence E.
I think this is a poor quality question, you should update the answers to be more misleading.
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Re: A bus completed first 50 miles of a 120-mile trip at an average speed  [#permalink]

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New post 09 Aug 2018, 07:27
total time according to 20mph and 120 mi distance= 120/20=6hrs
let speed in final leg be x
now
time of leg1(50 mi at the speed of 20mi/hr)+wait time +time of leg 2 (35 mi at speed of 35mi/hr)+ time of remaining leg=6 hrs

(50/20)+0.5+(35/35)+((120-85)/x)=6 hrs
4+(35/x)=6
x=17.5mi/hr
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A bus completed first 50 miles of a 120-mile trip at an average speed  [#permalink]

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New post 10 Aug 2018, 06:26

Solution



Given:
    • A bus completed first 50 miles of a 120-mile trip at an average speed of 20 mph.
    • Then it took a halt of 30 minutes and completed the half of the remaining journey at an average speed of 35 mph.
    • The overall average speed of the whole journey is 20 mph

To find:
    • Average speed of the bus in the remaining part of the journey

Approach and Working:
We know, average speed of the journey is the ratio of total distance covered and the total time taken to cover that distance.

Total distance = 120 miles
Average speed of the whole journey = 20 mph
    • Total journey time = \(\frac{120}{20}\) = 6 hours

Now, as per the individual journey times,
    • Time taken in first part of journey = \(\frac{50}{20}\) = 2.5 hours
    • Time taken in second part of journey = \(\frac{0.5(120-50)}{35}\) = 1 hour
    • Halt time = 0.5 hours
    • Hence, time taken in the last part of journey = 6 – (2.5 + 1 + 0.5) = 2 hours
    • Distance covered in the last part of journey = 120 – [50 + 0.5 (120 – 50)] = 35 miles

Therefore, the average speed in the last part of journey = \(\frac{35}{2}\) = 17.5 mph

Hence, the correct answer is option E.

Answer: E

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A bus completed first 50 miles of a 120-mile trip at an average speed &nbs [#permalink] 10 Aug 2018, 06:26
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