Latest Elitmus Aptitude Question SOLUTION: Two swimmers start swimming in a swimming pool from opposite ends. they met first at a distance of 50m from east and return back, they met again for second time at 20m from west th

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(#M40036061) ELITMUS Numerical Ability question Numbers Keep an EYE

Two swimmers start swimming in a swimming pool from opposite ends. they met first at a distance of 50m from east and return back, they met again for second time at 20m from west then find the length of pool. Assume their speeds are constant.

Asked In Elitmus (10 years ago)

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VINOTH KUMAR  (10 years ago)

ans:70 let a swimmer start at the east(e) and another at west(w).now,during the opposite direction of swimming,e will cover 50m(as e and w meet 50m from east)and w will cover x meters.now while returning back,e will cover 50m and w will cover the same x meters.now,during the second time,w will cover 20m(as speeds of w and e are constant).as long as e and w don't increase their speed,e and w will meet 50m from the east and 20m from west.therefore, total length of the pool=50+20=70.click yes if u understood r else click no.

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Vedika Tripathi  (10 years ago)

Let thr be 2 swimmer E n W swimming from east and west direction respectively and thr speed be Se and Sw.
CASE 1 :
Time taken by E = 50/Se
Time taken by W = d-50/Sw
Clearly, 50/Se=d-50/Sw => Sw/Se=d-50/50 .... eq(1)

CASE 2:
Dist. covered by E = 50+50+(d-50-20)
= d+30 Since, he return back n thn meet at 20m frm west
Dist. covered by W = d-50+20
= d-30
Time taken by E = d+30/Se
Time taken by W = d-30/Sw
Therefore, Sw/Se= d-30/d+30
From eq(1),
d-30/d+30=d-50/50
=> d=70 (ans)

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Debabrata Nayak  (10 years ago)

options are:
a-70
b-100
c-130
d-not remember.
answer ll be (d)130 ...try to solve using concept v directl proportional to distance...it is easy...

Is this solution Helpfull? Yes (14) | No (9)  

Nitish Kumar  (10 years ago)

in 1st case from east, distance =50 and rest is d-50(d is total length), now as speed is constant so t1*s1=50----eqn1 and t2*s2=d-50-----eqn2 as time taken is equal so we get s1/s2=50/d-50 now in second case from west distance is 20 and rest is d-20 so now s2*t2=20 and s1*t1=d-20 time taken same so we get s1/s2=d-20/20 after solving we get d=0 and d=70 , so d=0 not possible so we get 70 as answer

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dhamu  (9 years ago)

ans=130
ratio of speeds of 2 persons is same for the 2 meets.
50/(d-50)=(d-30)/d+30)
solving this we get d= 130

Is this solution Helpfull? Yes (5) | No (2)  

RAJAT PACHOURI  (10 years ago)

As the speed is constant and they return back , so they will cover same distance both of the times.
first time, the distance was given from the east side ie 50m.
second time, the distance was given from the west side ie 20m.
so total distance will be 70m.
i think this is the simplest explanation.

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AMEER ALI KHAN  (8 years ago)

answer will be 130.
ms vedika got confused with east side and west side so she made the eq opposite
the eq will come as (d-30)/(d+30)=50/(d-50)

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Auromika Parida  (10 years ago)

as it is given speed is constant
1st time
so in time t swimmer 'a' covers 'x=50' from east and
swimmer 'b' covers 'y' distance so the total distance = 50+y
similarly
2nd time
swimmer 'a' covers 'x' distance from east
swimmer 'b' covers 'y=2o' from west
as there speed are constant and nothing is given about time
let us assume that time is also constant
so in constant speed and time both of them will cover same distance both the time
thus total distance = X+Y= 50+20=70

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Manisha Choudhary  (7 years ago)

W−−−−−−−−> ∙ ∙ ====> swimmer.

In the first leg she swam 50m50m, so in the 2nd leg she must have swum 100m100m
In the 2nd leg, she swam L−50+20,L−50+20, so equating

L−50+20=100L−50+20=100, which yields L=130m

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Adil khan  (6 years ago)

Ans will be 130 u have perhaps u have mistaken to solving the equation. and m not solving just explaining many of them solve already whose ans is 130.
i wll tell u why 70 wont be ans.
east denote with E and west denotes with W. accord to ques 1st meeting is 50 from east so if u are saying 70 is length then W travell 20 meter so its means E is faster than W cs when W cover 20 E cover 50.
2nd meeting 20 meter from west so its means E travell 40 meter and W travel 100 meter its not possible. E is faster than W.
now assume 130 when E cover 50 w cover 80 in 1st meeting
2nd meeting E cover 80+20=100
W cover 50+110=160
because W is faster i hope u ll understand.

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shivendra pratap singh  (9 years ago)

answer is 70 since they both have taken same time
let total distance between them is x
so d1/s1=d2/s2
(x-50)/u=50/u .........i)

20/u=(x-20)/u .........ii)
solve above two equations
we get 70 as a ans

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Saroj kumar  (8 years ago)

Mmmmmmmmnmmm

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khalada tabassum  (7 years ago)

let the two swimmers be at east as e and at west be w..
initially they started swimming in opposite direction to each other..
the total length of pool be x..
they meet at 50m from e and x-50 m from w..
after meeting they return in opposite direction and again meet at 20m from west..
distance traveled by e is 50+50+(x-50-20)..
distance traveled by w is x-50-20..
for first meet
50/Ve=(x-50)/Vw............1
for second meet
(50+50+(x-50-20))/Ve=(x-50-20)/Vw.........2
solving 1 and 2 we get x=70

Is this solution Helpfull? Yes | No (1)  

Ankit Yadav  (5 years ago)

W−−−−−−−−>∙∙ swimmer.

In the first leg she swam 50m, so in the 2nd leg she must have swum 100m
In the 2nd leg, she swam L−50+20, so equating

L−50+20=100, which yields L=130m

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Kartik Choudhary  (5 years ago)

W−−−−−−−−>∙∙ swimmer.

In the first leg she swam 50m, so in the 2nd leg she must have swum 100m
In the 2nd leg, she swam L−50+20, so equating

L−50+20=100, which yields L=130m

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