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# A train after traveling at 50kmph meets with an accident and then

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Intern
Joined: 01 Jan 2016
Posts: 27
Re: A train after travelling 50 km meets with an accident  [#permalink]

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29 Jun 2017, 18:36
1
theperfectgentleman wrote:
A train after travelling 50 km meets with an accident and as a result, it proceeds at 3/4 of its former speed
and arrives at its destination 35 minutes late. Had the accident occurred 24 km further the train would have reached its
destination 25 minutes late. The original speed of the train is:
(1) 48 km/h
(2) 36 km/h
(3) 54 km/h
(4) 58 km/h
(5) 50 km/h

The algebraic solution for this question is pretty long and inefficient. Tried the following logic and my answer is wrong.

Case 1: Accident @ 50 Kms
-----------A1-------------
Case 2: Accident 24 Kms
----A2--------------------

Difference in time = 10 mins
Difference between A1 and A2 = 26 Kms

Ratio of speeds between case 1 and case 2 is 1:3/4 or, 4:3
Ratio of time (Inverse) 3:4

Time saved is 4-3 = 1 part; 1 part leads to difference of 10 mins

Time taken in case 1,
3 parts = 3*10= 30 mins

Original speed of train: 26/(30/60) = 42Kmph

Can someone highlight what is wrong in my approach?
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Re: A train after traveling at 50kmph meets with an accident and then  [#permalink]

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29 Jun 2017, 21:16
Let the distance be $$D$$, and the original speed of the train before the accident be $$S$$ km/h.
$$\frac{50}{S} + \frac{4(D-50)}{3S} = \frac{35}{60}$$ hours late

Had the accident occurred 24 km farther - had the train traveled 74 km while the accident took place.
$$\frac{74}{S} + \frac{4(D-74)}{3S} = \frac{25}{60}$$ hours late

(i) - (ii):

$$(\frac{50}{S} + \frac{4D-200}{3S}) - (\frac{74}{S} + \frac{4D-296}{3S}) = \frac{1}{6}$$

$$(\frac{150 + 4D - 200}{3S}) - (\frac{222 + 4D - 296}{3S}) = \frac{1}{6}$$

$$\frac{150 + 4D - 200 - 222 - 4D + 296}{3S} = \frac{1}{6}$$

$$\frac{24}{3S} = \frac{1}{6}$$

$$3S = 24*6$$

$$S = 48 km/h$$.
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A train after traveling at 50kmph meets with an accident and then  [#permalink]

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29 Jun 2017, 21:43
1
1
Blackbox wrote:
A train after traveling for 50km meets with an accident and then proceeds at 3/4 of its former speed and arrives at its destination 35 minutes late . Had the accident occurred 24 km farther, it would have reached the destination only 25 minutes late. What is the speed of the train.

a) 45
b) 33
c) 48
d) 55
e) 61

1. Let y be the speed of the train
2. The train can save 10 minutes by traveling the 24 km at its original speed instead of the reduced speed
3. 24/(3y/4)-24/y=10/60, So y=48km/hr
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Re: A train after traveling at 50kmph meets with an accident and then  [#permalink]

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15 Jul 2017, 09:14
desaichinmay22 wrote:
Let y be the balance distance to be covered and x be the former speed.
A train after traveling for 50km meets with an accident and then proceeds at 3/4 of its former speed and arrives at its destination 35 minutes late
so, y/(3x/4) - y/x = 35/60
4y/3x - y/x = 7/12
y/x(4/3-1)=7/12
y/x*1/3=7/12
y/x=7/4
4y-7x=0 ........ 1

Had the accident occurred 24 km farther, it would have reached the destination only 25 minutes late
so, (y-24)/(3x/4) - (y-24)/x = 25/60
4(y-24)/3x - (y-24)/x = 5/12
(y-24)/x (4/3-1) = 5/12
(y-24)/x *1/3 = 5/12
(y-24)*12 = 3x*5
(y-24)*4 = 5x
4y-5x = 96 ....... 2

eq2 - eq1
2x=96
x=48

Ans = C

i must say wonderful ...
but i highly doubt whether it is workable under examination condition in less than 2 minutes ...
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Re: A train after traveling at 50kmph meets with an accident and then  [#permalink]

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15 Jul 2017, 10:26
Blackbox wrote:
Chinmay - Sorry for being thick, but I struggle to understand why you would subtract the times if there were no accident. Is it because they say the train arrived late? What if there was a case when the train arrives early? Would you add times?

hi

if you subtract the usual time caused by usual speed from the extra time caused by low speed, you certainly will get 35 minutes that is caused by accident ...

hope this helps...
cheers..
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Re: A train after traveling at 50kmph meets with an accident and then  [#permalink]

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22 Jul 2018, 17:35
Blackbox wrote:
A train after traveling for 50km meets with an accident and then proceeds at 3/4 of its former speed and arrives at its destination 35 minutes late . Had the accident occurred 24 km farther, it would have reached the destination only 25 minutes late. What is the speed of the train.

a) 45
b) 33
c) 48
d) 55
e) 61

We can let the speed of the train be r kmph and the distance to its destination be d km. We can create the following equations for the time:

50/r + (d - 50)/(3r/4) = d/r + 35/60

and

74/r + (d - 74)/(3r/4) = d/r + 25/60

Multiplying both equation by 60r, we have:

3000 + 80(d - 50) = 60d + 35r

and

4440 + 80(d - 74) = 60d + 25r

Subtracting these two equations, we have:

-1440 + 1920 = 10r

480 = 10r

48 = r

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Re: A train after traveling at 50kmph meets with an accident and then  [#permalink]

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06 Jan 2019, 19:43
Let speed be 4v, so 3/4th is 3v
The difference 10 mins(35 mins - 25mins), is the difference in time travelling 24 km at 4v and 3v speed.

(24/3v) - (24/4v) = 10/60 (in hours)
solving we get v = 12, so speed of train 4v = 48 km/hr
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Re: A train after traveling at 50kmph meets with an accident and then  [#permalink]

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17 Jan 2020, 06:54
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Re: A train after traveling at 50kmph meets with an accident and then   [#permalink] 17 Jan 2020, 06:54

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