Latest Capgemini Aptitude Question SOLUTION: Two trains for Palwal leave Kanpur at 10 a.m and 10:30 am and travel at the speeds of 60 Kmph and 75 Kmph respectively. After how many kilometers from Kanpur will the two trains b

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Two trains for Palwal leave Kanpur at 10 a.m and 10:30 am and travel at the speeds of 60 Kmph and 75 Kmph respectively. After how many kilometers from Kanpur will the two trains be together?

Asked In Capgemini Sasidhar Reddy (8 years ago)

Solved Taheer Read Solution (12)

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Read Solution (12) :

Nazmul Hasan  (8 years ago)

Let two train meet at X km distance.
Time taken by first train to cover X km=x/75 h
Time taken by 2nd train to cover X km=x/60 h
Now, x/60 - x/75=1/2
thus, x=150.

Is this solution Helpfull? Yes (80) | No (6)  

Taheer  (8 years ago) Best Solution

If two trains leave say 'P' for 'Q' at time 't1' 't2' and with speeds 'a' and 'b' ,
Then the distance D from p where they meet, is given by
D = diff in time (t2-t1) X a*b / b-a
There u go ...

Is this solution Helpfull? Yes (33) | No (18)  

vikas patta  (8 years ago)

relative speed =75-60=15kmph
in half an hour 1 train travelled 30km.
to reach I st train 2 nd train need 30/15=2 hours.
in two hours 2 nd train travelled 75*2=150km

Is this solution Helpfull? Yes (23) | No (3)  

sudha singh  (8 years ago)

K----------------------------------------------------------P
T11 10:00 60 km/h
TT12 10:30 75 km/h
Suppose they meet after t time from 10.00
75(t- 1/2)=t *60
t=2.5
Distance travelled till meeting=75*(2.5-.5)=150 km

Is this solution Helpfull? Yes (13) | No (8)  

hemant  (6 years ago)

As we know that Train A and Train B leaves from the same station ,
simple trick for solving that is find the factors of both .
5 |60 , 75
------------------
3 | 12 , 15
------------------
4 | 4 , 5
-----------------
5 | 1 , 1

(5*3*4*5 ) / 2= 150

Is this solution Helpfull? Yes (9) | No (3)  

akhil  (5 years ago)

travelled distance is same so
60*t=75*(t-1/2)
t=5/2
s=60*5/2=150

Is this solution Helpfull? Yes (5) | No   

sumitgoyal  (7 years ago)

x+1x=3

⟹(x+1x)2=9

⟹x2+1x2+2=9

⟹x2+1x2=7

Again,

(x+1x)3=27

⟹x3+1x3+3(x+1x)=27

⟹x3+1x3=27−3×3=18

Now,

(x2+1x2)(x3+1x3)=(x5+1x5)+(x+1x)

⟹18×7=x5+1x5+3

⟹x5+1x5=18×7−3=123

Is this solution Helpfull? Yes | No (16)  

madhav vishwakarma  (6 years ago)

2.5 dm = .0.25m
19.25 m × 2 m = 38.5 m^2
12.50 m × 2 m = 25 m^2
Since they will overlap therefore we have to subtract 2×2 m^2
Therefore answer will be
(25 + 38.5 - 4 ) × 1.32
I.e 78.65

Is this solution Helpfull? Yes | No (2)  

madhav vishwakarma  (6 years ago)

Solving logically .
When train A = 30
Train B = 0

Train A = 31
Train B = 1.25

Train A = 32
Train B = 2.5

Train A = 34
Train B = 5

Similarly @ a at 38 and B at 10

Now if we subtract 5 from 34 that will be 29.
The 10 from 38 ie 28...and this will continue. When it will be 0? After 30 steps.
Therefore 5 × 30 will come 150. Or 30 (train A)+ (30×4(common d)) will also give 150.

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radha  (5 years ago)

First train start in 10 am and 2nd train start train 10:30
So the 1st train cover 30 km distance in 10:30.
Now, both train time are equal
Let the train distance = x
time 1= time 2
x-30/60=x/75
So x=150 km

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sanjana k  (4 years ago)

A-10am
B-10:30 am
Speed(A)=60kmph So when B starts A would have covered 30km and when they meet they will have same time. let x be the distance covered by b when they meet.
(30+x)/60=x/75
x=150 km.

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Manojkumar Baskaran  (4 years ago)

S1*S2*t / S1-S2
(60*75*1/2 )/60-75
(60*75*1/2)/15
60*5*1/2
60*2.5
150.0km

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