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Bunuel
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A man leaves his home and walks at a speed of 12 km per hour, reaching 02 Apr 2020, 09:36
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C
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74% (02:31) correct 26% (02:34) wrong based on 242 sessions

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A man leaves his home and walks at a speed of 12 km per hour, reaching the railway station 10 minutes after the train had departed. If instead he had walked at a speed of 15 km per hour, he would have reached the station 10 minutes before the train's departure. What is the distance (in km) from his home to the railway station ?

A. 16
B. 18
C. 20
D. 25
E. 30
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C
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varbe
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A man leaves his home and walks at a speed of 12 km per hour, reaching 02 Apr 2020, 10:17
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We are given speeds and travelling time for two cases:
First, when the man travels at 12 km/hour he reaches 10 minutes late.
Second, when the man travels at 15 km/hour he reaches 10 minutes early.
So, the time difference between travelling at 12 km/hour and 15km/hour is 20 minutes.

As, time1-time2=20 minutes and time=distance/speed,
The difference in time can be expressed as:
D/12 -D/15=20/60 (20/60 because the terms are in hours)
D/60=20/60
D=20
Hence, the distance is (C) 20 Kms.
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TheMongoose
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A man leaves his home and walks at a speed of 12 km per hour, reaching 02 Apr 2020, 09:45
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Bunuel
A man leaves his home and walks at a speed of 12 km per hour, reaching the railway station 10 minutes after the train had departed. If instead he had walked at a speed of 15 km per hour, he would have reached the station 10 minutes before the train's departure. What is the distance (in km) from his home to the railway station ?

A. 16
B. 18
C. 20
D. 25
E. 30

We Know,

D = S*T

So

D = 12*(T+ (10/60)) - (1)
D = 15*(T-(10/60)) - (2)

Solve (1) and (2) and plug value of T back in either (1) or (2)

OA - C
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A man leaves his home and walks at a speed of 12 km per hour, reaching 03 Apr 2020, 05:19
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Bunuel
A man leaves his home and walks at a speed of 12 km per hour, reaching the railway station 10 minutes after the train had departed. If instead he had walked at a speed of 15 km per hour, he would have reached the station 10 minutes before the train's departure. What is the distance (in km) from his home to the railway station ?

A. 16
B. 18
C. 20
D. 25
E. 30

from given info we know
12*(t+1/6)=15 * ( t-1/6)
solve for t = 3/2
we get distance
12 * (3/2+1/6)
; 2*10 ; 20
IMO C
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A man leaves his home and walks at a speed of 12 km per hour, reaching 03 Apr 2020, 10:33
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Bunuel
A man leaves his home and walks at a speed of 12 km per hour, reaching the railway station 10 minutes after the train had departed. If instead he had walked at a speed of 15 km per hour, he would have reached the station 10 minutes before the train's departure. What is the distance (in km) from his home to the railway station ?

A. 16
B. 18
C. 20
D. 25
E. 30

Ans C. Expl: Let assume t1= time wasted in 12 km per hour and t2= time wasted in 15 km per hour. In both cases distance is equal; we can write
12*t1 = 15*t2; then t1/t2 = 5/4. In 15 km per hour speed, the total time wasted is 20 minutes less than that of 12 km per hour as per the statement. In our calculated ratio, t1 is 1 unit higher than t2. For 1 unit difference, the man needs 20 minutes; for 5 unit time, he needs 20*5=100 minutes (1.67 hour), which is the total time wasted actually. So distance = speed (12 km per hour) * time (1.67); = 20 km
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A man leaves his home and walks at a speed of 12 km per hour, reaching 28 Jun 2021, 10:18
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When distance is same and different time and speeds are given

Distance =s1*s2*change in time/(s1-s2)

=12*15*20/60*3
=20 km

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A man leaves his home and walks at a speed of 12 km per hour, reaching 28 Jun 2021, 12:11
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Distance =speed * time

Lets assume T as correct time to reach station

case 1) D=12* (t+10)/ 60 ( 10 mins late so t+10 and converting 10 mins to hours)

case 2) D= 15 * (t-10)/60 (10 mins early so t-10)

Distance from home to station is same

hence 12* (t+10)/60 =15*(t-10)/60
we get t= 90 mins

we need Distance so substitute t in case 1
D = 12kmhr * (90+10)/60
D = 20 km
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A man leaves his home and walks at a speed of 12 km per hour, reaching 29 Sep 2024, 23:53
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