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siddhantvarma
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A train travels between two cities at an average speed of 80 k m / h . 13 Jun 2025, 04:34
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

75% (hard)

Question Stats:

63% (02:52) correct 37% (03:13) wrong based on 135 sessions

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A train travels between two cities at an average speed of 80 k m / h . Due to track maintenance, it must reduce its speed by 20 km / h for a portion of the return journey, causing the return trip to take 1 hour longer than the outbound trip. If the train were to reduce its speed by 30 km / h for the same portion of the track on a different day, how many additional minutes would the return journey take compared to the outbound journey?


(A) 60
(B) 90
(C) 108
(D) 120
(E) 144
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C
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Krunaal
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A train travels between two cities at an average speed of 80 k m / h . 13 Jun 2025, 14:26
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siddhantvarma
A train travels between two cities at an average speed of 80 k m / h . Due to track maintenance, it must reduce its speed by 20 km / h for a portion of the return journey, causing the return trip to take 1 hour longer than the outbound trip. If the train were to reduce its speed by 30 km / h for the same portion of the track on a different day, how many additional minutes would the return journey take compared to the outbound journey?


(A) 60
(B) 90
(C) 108
(D) 120
(E) 144
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Speed * Time = Distance

80*t = 60*(t+1)

t = 3 hours or 180 minutes, so the distance = 240 km

50*t = 240
t = 4.8 hours => 288 minutes

288 - 180 = 108 additional minutes.

Answer C.
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Yosemite98
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A train travels between two cities at an average speed of 80 k m / h . 14 Jun 2025, 02:20
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Hi Krunaal,

I do not understand how can you assume that the entire return trip was done at the reduced speed by using this formula: 80*t = 60*(t+1)

I do not know what am I missing, could you please elaborate?
thanks!
Krunaal
siddhantvarma
A train travels between two cities at an average speed of 80 k m / h . Due to track maintenance, it must reduce its speed by 20 km / h for a portion of the return journey, causing the return trip to take 1 hour longer than the outbound trip. If the train were to reduce its speed by 30 km / h for the same portion of the track on a different day, how many additional minutes would the return journey take compared to the outbound journey?


(A) 60
(B) 90
(C) 108
(D) 120
(E) 144
Speed * Time = Distance

80*t = 60*(t+1)

t = 3 hours or 180 minutes, so the distance = 240 km

50*t = 240
t = 4.8 hours => 288 minutes

288 - 180 = 108 additional minutes.

Answer C.
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Krunaal
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A train travels between two cities at an average speed of 80 k m / h . 14 Jun 2025, 02:50
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Yosemite98
Hi Krunaal,

I do not understand how can you assume that the entire return trip was done at the reduced speed by using this formula: 80*t = 60*(t+1)

I do not know what am I missing, could you please elaborate?
thanks!
Krunaal
siddhantvarma
A train travels between two cities at an average speed of 80 k m / h . Due to track maintenance, it must reduce its speed by 20 km / h for a portion of the return journey, causing the return trip to take 1 hour longer than the outbound trip. If the train were to reduce its speed by 30 km / h for the same portion of the track on a different day, how many additional minutes would the return journey take compared to the outbound journey?


(A) 60
(B) 90
(C) 108
(D) 120
(E) 144
Speed * Time = Distance

80*t = 60*(t+1)

t = 3 hours or 180 minutes, so the distance = 240 km

50*t = 240
t = 4.8 hours => 288 minutes

288 - 180 = 108 additional minutes.

Answer C.
Show more
Yes, it isn't fair to assume, however I tried it first (since it's quick to check) to avoid long calculation and save time - and luckily the distance for which train was slowed down turned out to be the total distance.

Ideally, outbound trip will be \(\frac{D}{80}\) hours, return will be \(\frac{x}{60}\) + \(\frac{(D-x)}{80}\) hours. Then, \(\frac{x}{60}\) + \(\frac{(D-x)}{80}\) = \(\frac{D}{80}\) + 1 solving we'll get x as 240. Then, we'll subtract \(\frac{D}{80}\) from \(\frac{240}{50}\) + \(\frac{(D-240)}{80}\) and get the time difference. A quick check initially implying the speed reduced for complete distance worked out in this case.
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Yosemite98
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A train travels between two cities at an average speed of 80 k m / h . 14 Jun 2025, 09:57
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makes total sense! thanks for the reply, I was going crazy not understanding why :)
Krunaal
Yosemite98
Hi Krunaal,

I do not understand how can you assume that the entire return trip was done at the reduced speed by using this formula: 80*t = 60*(t+1)

I do not know what am I missing, could you please elaborate?
thanks!
Yes, it isn't fair to assume, however I tried it first (since it's quick to check) to avoid long calculation and save time - and luckily the distance for which train was slowed down turned out to be the total distance.

Ideally, outbound trip will be \(\frac{D}{80}\) hours, return will be \(\frac{x}{60}\) + \(\frac{(D-x)}{80}\) hours. Then, \(\frac{x}{60}\) + \(\frac{(D-x)}{80}\) = \(\frac{D}{80}\) + 1 solving we'll get x as 240. Then, we'll subtract \(\frac{D}{80}\) from \(\frac{240}{50}\) + \(\frac{(D-240)}{80}\) and get the time difference. A quick check initially implying the speed reduced for complete distance worked out in this case.
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A train travels between two cities at an average speed of 80 k m / h . 17 Jun 2026, 12:52
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