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# Cathy traveled half of the journey by train at 80 kmph, half of the re

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Posts: 65012
Cathy traveled half of the journey by train at 80 kmph, half of the re  [#permalink]

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17 Apr 2020, 03:37
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Question Stats:

79% (02:02) correct 21% (02:01) wrong based on 57 sessions

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Cathy traveled half of the journey by train at 80 kmph, half of the remaining journey by bus at 40 kmph and the remaining part of the journey by bicycle at 20 kmph. What was Cathy's average speed for the entire journey ?

A. 30 kmph
B. 40 kmph
C. 45 kmph
D. 50 kmph
E. 60 kmph

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Re: Cathy traveled half of the journey by train at 80 kmph, half of the re  [#permalink]

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17 Apr 2020, 04:20
Let the total distance be X
Cathy travelled half of the distance at 80kmph by train
Then time taken throughout the train journey
(X/2)/80----(1) {distance/speed}
Cathy travelled half of the rest distance at 40kmph by bus
Then time taken throughout the bus journey
(X/4)/40---(2) {distance/speed}
Cathy travelled remaining distance by bicycle at 20kmph
Then time taken throughout the bicycle journey
(X/4)/20----(3)
{distance/speed}
Total time taken throughout journey=(1)+(2)+(3)
(X/2)/80+(X/4)/40+(X+4)/20
X/160+X/160+X/80
4X/160
Total time= X/40
Total distance=X
Average Speed= Total Distance/Total time
=X/(X/40)
Average Speed = 40

OA is B

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Re: Cathy traveled half of the journey by train at 80 kmph, half of the re  [#permalink]

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17 Apr 2020, 04:23
Bunuel wrote:
Cathy traveled half of the journey by train at 80 kmph, half of the remaining journey by bus at 40 kmph and the remaining part of the journey by bicycle at 20 kmph. What was Cathy's average speed for the entire journey ?

A. 30 kmph
B. 40 kmph
C. 45 kmph
D. 50 kmph
E. 60 kmph

Solution:

Let x km be the total distance of Cathy’s journey.
• Distance travelled by train = $$\frac{x}{2}$$ km
o Time taken to cover $$\frac{x}{2} = \frac{x}{160}$$
• Distance travelled by bus = $$\frac{x}{4}$$
o Time taken to cover $$\frac{x}{4} = \frac{x}{160}$$
• Distance travelled by bicycle = $$\frac{x}{4}$$
o Time taken to cover $$\frac{x}{4} = \frac{x}{80}$$
• Total time taken = $$\frac{x}{160} + \frac{x}{160} +\frac{x}{80} =\frac{4x}{160} = \frac{x}{40}$$
• Average speed = $$\frac{distance}{time}$$ = $$\frac{x}{\frac{x}{40}} = 40$$kmph
Hence, the correct answer is Option B.
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Re: Cathy traveled half of the journey by train at 80 kmph, half of the re  [#permalink]

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17 Apr 2020, 07:08
Bunuel wrote:
Cathy traveled half of the journey by train at 80 kmph, half of the remaining journey by bus at 40 kmph and the remaining part of the journey by bicycle at 20 kmph. What was Cathy's average speed for the entire journey ?

A. 30 kmph
B. 40 kmph
C. 45 kmph
D. 50 kmph
E. 60 kmph

CONCEPT: Average Speed = $$\frac{Total DIstance }{ Total Time}$$

DIstance = Speed* Time

Total Distance = D

Time of train Journey = (D/2)/80
Time of Bus Journey = (D/4)/40
Time of Cycle Journey = (D/4)/20

Total Time = (D/160) + (D/160) + (D/80) = (4D/160) = (D/40)

Average Speed = D/(D/40) = 40 kmph

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Re: Cathy traveled half of the journey by train at 80 kmph, half of the re  [#permalink]

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17 Apr 2020, 07:28
Bunuel wrote:
Cathy traveled half of the journey by train at 80 kmph, half of the remaining journey by bus at 40 kmph and the remaining part of the journey by bicycle at 20 kmph. What was Cathy's average speed for the entire journey ?

A. 30 kmph
B. 40 kmph
C. 45 kmph
D. 50 kmph
E. 60 kmph

Let the distance be d .

Time to cover 1/2 the distance at 80 km/hr = d/160
Time to cover 1/4 the distance at 40 km/hr = d/160
Time to cover 1/4 the distance at 20 km/hr = d/80

Total Time : d/40

Therefore Average Speed = d/(d/40) =40 (B)
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Re: Cathy traveled half of the journey by train at 80 kmph, half of the re  [#permalink]

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17 Apr 2020, 07:31
Bunuel wrote:
Cathy traveled half of the journey by train at 80 kmph, half of the remaining journey by bus at 40 kmph and the remaining part of the journey by bicycle at 20 kmph. What was Cathy's average speed for the entire journey ?

A. 30 kmph
B. 40 kmph
C. 45 kmph
D. 50 kmph
E. 60 kmph

let total distance be x
so first half of distance ; x/2
2nd half of the remaining ; x/2 * 1/2 ; x/4
and balance distance ; x-(x/2+x/4) = x/4
so avg time would be ; total distance/total speed
x/2 * 1/80 + x/4 * 1/40 + x/4 * 1/20
=> 4x/160 ; x/40
avg speed of complete journey
x/x/40 ; 40 kmph
OPTION B
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Re: Cathy traveled half of the journey by train at 80 kmph, half of the re  [#permalink]

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17 Apr 2020, 07:42
Assume that the total distance is 800km

First 400km is covered in 400/80=5hours

Next 200km is covered in 200/40=5hours

Final 200km is covered in 200/20=10hours

Total time taken to cover 800km=5+5+10=20 hours

Average speed for entire journey =800/20=40kmph

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Re: Cathy traveled half of the journey by train at 80 kmph, half of the re  [#permalink]

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17 Apr 2020, 08:47
Cathy traveled half of the journey by train at 80 kmph, half of the remaining journey by bus at 40 kmph and the remaining part of the journey by bicycle at 20 kmph. What was Cathy's average speed for the entire journey ?

A. 30 kmph
B. 40 kmph --> correct: let's say, total distance=4d, so Cathy's average speed = 4d/( (2d/80)+(d/40)+(d/20) ) = 4d/( (d/40)+(d/40)+(d/20) ) = 4d/( (2d/40)+(d/20) )= 4d/( (d/20)+(d/20) ) = 4d/(2d/20)=4d/(d/10)=40kmph
C. 45 kmph
D. 50 kmph
E. 60 kmph
Re: Cathy traveled half of the journey by train at 80 kmph, half of the re   [#permalink] 17 Apr 2020, 08:47