Two swimmers start from same end, A, of a 200 meter long swimming pool

Question

Two swimmers start from same end, A, of a 200 meter long swimming pool. The faster one reaches the other end, B, turns back and meets the slower one at a distance of 175 meters from the starting end, A. Where will the slower one be when faster one reaches the starting point A?

A. 89 meters from A
B. 91 meters from A
C. 93 meters from A
D. 99 meters from A
E. 111 meters from A

Are You Up For the Challenge: 700 Level Questions

A. 89 meters from A
B. 91 meters from A
C. 93 meters from A
D. 99 meters from A
E. 111 meters from A

Are You Up For the Challenge: 700 Level Questions

gurmukh · Answer

When faster covers 225 slower covers 175
When faster covers 400 slower covers 400×175÷225= 311.11
So 89 m away
Option A is the answer

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yashikaaggarwal · Answer

IMO A

The faster one reaches the other end, B, turns back and meets the slower one at a distance of 175 meters from the starting end, A. 
Let faster one be X and slower one be Y
X travelled a total of 200+25 = 225 meter
While Y did 175 meter

Distance/SPEED= TIME
225/Sx = T ------(1)
175/Sy = T ------(2)
Equating both
225/Sx= 175/Sy
Sx/Sy = 9/7 (9:7)

Let the slower one (Y) will be at Point P when faster one (X) reaches the starting point A
X completed 2 rounds(back and forth) = 400
400/9=P/7
P=311.11111
Since one round is of 200 meter. Y already started 2nd round and completed (311.11-200=111.11 meter)
P= 111.11 meter
P distance from A is 200-111.11
89 meter.

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itsabhimanyu12 · Answer

1. Second swimmer meets first swimmer at 200m - 25m = 175m from A
2. By the time first swimmer swims 175m, second swimmer swims 225m
3. T1 = T2 -> (175/S1) = (225/S2) -> S1/S2 = 7/9
4. When S2 reaches A, his distance is 400m. Let X be the distance of S1
5. X/400 = S1/S2 -> X/400 = 7/9
6. X = 311.11m -> 400 - 311.11 = 89m from A

HoneyLemon · Answer

IMO A .

----A---------------------------------------B
F---------------------- Vf

S------------Vs

Let faster one be F and slower one be S
F traveled a total of 200+25 = 225 meter
While S did 175 meter

D/S= T .. 225/Vf = T ------(1)
175/Vs = T ------(2)

225/Vf = 175/Vs == Vf = (9/7) Vs

now lets consider in t time F will cover rest 175 meter .

so t = 175 / (Vf)

so in t time , S will cover Vs * t = Vs * 175 * 7/ (9 *Vs) = 137 metrer (app. ) 
 i.e 225 - 137 = 89 meter .

So A .

lacktutor · Answer

Faster one : 
if it takes him t hour to swim for 225 meters, he will spend:

225 ----t
400 ----x
---> x =  \frac{400}{225} t=  \frac{16}{9} t hour for 400 meters.

Slower one:  
if it takes him t hour to swim for 175 meters, where will he be after  \frac{16}{9}  t hour.
175 --- t 
x -----( \frac{16}{9} )t

--->  x = \frac{(\frac{16}{9})*175t}{t} = \frac{16*175}{9}= \frac{2800}{9} = 311.(1) 
 400 -311 ≈89  meters away from A point

Answer (A).

NitishJain · Answer

IMO  A

When both swimmers meet : Faster one has covered 225 m & slower one has covered 175 m

Distance to left to reach point A = 175m

to find how much distance slower one will cover (say x) when faster one will reach point A.

225/175 = 175/x
x = 136.11

Slower one need to cover 25 m to reach point B, so distance cover by slower swimmer from point B towrds point A = 136.11 -25 = 111.11

Distance of slower swimmer from starting point A= 200 - 111.11 ~ 89 m

TarunKumar1234 · Answer

\frac{225}{S_A} =\frac{175}{S_B} or, \frac{S_A}{S_B} = \frac{9}{7} Distance travelled by B = Speed of B * time taken = S_B * \frac{400}{S_A} = \frac{2800}{9} = 311 So, distance from starting point = 400 - 311 = 89 So, It is A.

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Two swimmers start from same end, A, of a 200 meter long swimming pool

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Two swimmers start from same end, A, of a 200 meter long swimming pool

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