Latest Elitmus Aptitude Question SOLUTION: Two trains A and B seperated by a distance of 1200m approach each other at 30m/min.A bird flying at 60 m/min flies from train A to B, returns to A flies back to B and so on till th

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(#M40032881) ELITMUS Numerical Ability question Time and distance Keep an EYE

Two trains A and B seperated by a distance of 1200m approach each other at 30m/min.A bird flying at 60 m/min flies from train A to B, returns to A flies back to B and so on till the trains meet.
What is the total distance travelled by the bird?
How many trips between the train does the bird make in all?

Asked In Elitmus Ankit Boae (10 years ago)

Unsolved Read Solution (10)

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Read Solution (10) :

Dr.Dipin Singh  (10 years ago)

1200m
Time taken by trains before meeting = 1200/(30+30)= 20 mins
Distance travelled by bird during this time = 20*60=1200 mtrs

Is this solution Helpfull? Yes (24) | No (3)  

ruchi shah  (10 years ago)

A--------1200 m------------B
A's speed = B's speed =30 m/min
let A and B meet at point C and both take same time to meet at point C which is t
so,
x/30=(1200-x)/30
=>2x=1200
x=600 m at point c from A
so , t=600/30=20min
in this time bird take total distance is =60*20=1200m
and total no of turn is one

Is this solution Helpfull? Yes (10) | No (7)  

Pankaj Kumar  (9 years ago)

Meeting time of 2 train 30 + 30 = 1200/t
t = 20 min;
Bird traveling time in 20 min = 60*20 =1200 meter and only 1 trip ;

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DHAKAD NEETENDRA SINGH  (10 years ago)

in first trip
let t1 is the time taken to cover the distance of 1200m by bird and first train
30t1+60t1=1200
t1=40/3
then,
distace travelled by first train=30*40/3=400m
..........................bird = 60*40/3=800m
now,remaining=400m
distance between two trains = 400m
in second trip bird takes let t2 time
30*t2+60*t2=400
t2=40/(3^2) similarly,
t3=40/(3^3)
.
.
.
tn=40/(3^n)
.
.
to infinity

total distance travelled by the bird
=60(t1+t2+t3+........infinite)
=60[40/(3^1)+40/(3^2)+40/(3^3)+..............]
=60*40/3[1+1/3+1/9+1/27+................]
=60*40/3[1/(1-1/3)]
=1200m
trips
the no of trips is infinite ans...

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harsh  (9 years ago)

use relative concept ..
let train B at rest

BIrd
AAAA(train)...........B(rest)

speed of train a will be 60 km/hr
and speed of bird also 60 km/hr.
after 1st meet distance will be 1200m ;)
so

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ANUP KUMAR  (8 years ago)

total distance=1200
no. of trips =4

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ashish kumar  (10 years ago)

total 1200m will be traveled by bird till both the trains approaches each other..

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Birjendra Rawat  (10 years ago)

let B is in rest then bird's speed wrt B is 90 m/min and A's speed is 30 m/min wrt B . So time when bird reaches train B is 1200/90 , actual distance covered by bird is 60*1200/9=800 ,till then train A covers 30*1200/90=400m , remaining distance=1200-400=800m.
in 2nd trip let A be in rest then Bird's speed wrt A is 90 & B's speed is 30 wrt A so time taken = 800/90 ,distance covered by bird is 800*60/90=800*2/3 , train covers 30*800/9 ,distance remaining=>800-30*800/90= 800(1-1/3)=800*2/3
so this will form an GP 800 + 1600/3 + 3200/9 + 6400/27 + 12800/81 ...
S=800/(1-2/3)=2400 ans.
No. of trips made by bird are infinite

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Kranthi Kumar Kota  (9 years ago)

Time taken by two trains before meeting
= 1200/(30+30)
= 20 mins..

Distance traveled by bird during this time
= 20*60
=1200 mts..

If we consider Half time...ie., 10min
Both A & B are 600mts apart by moving 300mts each.. At that moment bird reaches Mid point..
In similar pattern, After 5more min, Bird reaches train B... In next 5min both trains and bird meets again..

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ramanuj  (8 years ago)

total distance is 1200m
time to two train meet each other= 20
time of first trip=1200/60+30=40/3
total distance reduced=400 to each train travelled
distance between two train before second tour=1200-800=400
time req for second tour =400/90...
continue it till distance b/w train is approx to 0

total distance travelled=60*20=1200m

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