If U can.....answer this
- Thread starter vssoma
- Start date
hi guys...
after long time,,,,, i have got some time.....so here goes next puzzle....
Ram and Shyam take part in a two leg (after reaching the other end, one should come back to start) swimming race.
Both start from opposite ends of the pool.
On the first leg, the boys pass each other at 18m from the deep end of the pool.
During the second leg they pass at 10m from the shallow end of the pool.
Both go at constant speed but one of them is faster.
Each boy rests for 4 seconds at the end of the first leg.
What is the length of the pool?
after long time,,,,, i have got some time.....so here goes next puzzle....
Ram and Shyam take part in a two leg (after reaching the other end, one should come back to start) swimming race.
Both start from opposite ends of the pool.
On the first leg, the boys pass each other at 18m from the deep end of the pool.
During the second leg they pass at 10m from the shallow end of the pool.
Both go at constant speed but one of them is faster.
Each boy rests for 4 seconds at the end of the first leg.
What is the length of the pool?
Tried to solve it by taking Length of the Pool as D, and Speed as X and Y , x fast and Y slow, and Time t, when they meet.... afer lotsof try got the answer as : 0 (Zero)
Whoever posts the answer, please post the calculations also!!
Whoever posts the answer, please post the calculations also!!
Speed of one starting from shallow end be S1
from Deep end be S2
Both meet at 18 metres from deep end.
If L is the length of the pool,
Time taken by deep end starter is....
T1 = (L-18)/S1 ........(a)
& starter from shallow end is...
T2 = 18/S2 ................(b)
Since they meet at 18m therefore time taken is equal....
(L-18)/S1 = 18/S2
or
S1/S2 = (L-18)/18 ..............(c)
Deep end one travels 18m with speed S1 to reach the deep end, rests for 4 sec, travels (L-10)m with speed S1, where they meet each other again...
Time taken to travel the distance is....
T3 = (18/S1) + 4 + ((L-10)/S1) .......(d)
Shallow end swimmer travels (L-18) with speed S2 to reach the shallow end & resting for 4 seconds, travels 10m with speed S2, where they meet each other again.....
Time taken to travel this distance would be....
T4 = (L-18/S2) + 4 + (10/S1) ........(e)
Since they cross each other at T3 & T4 being equal....
therefore....
(18/S1) + 4 + ((L-10)/S1) = (L-18/S2) + 4 + (10/S1)
Simplifying it further...
S1/S2 = (L+8)/(L-8)
Substituting the values....S1/S2 from above....
(L-18)/18 = (L+8)/(L-8)
L can be either 0 or 44
Since L cannot be 0 therefore length must be 44m
https://www.google.com.sa/search?hl=en&site=&source=hp&q=On+the+first+leg%2C+the+boys+pass+each+other+at+18m+from+the+deep+end+of+the+pool.&oq=On+the+first+leg%2C+the+boys+pass+each+other+at+18m+from+the+deep+end+of+the+pool.&aq=f&aqi=&aql=&gs_l=hp.12...1438.1438.0.2625.1.1.0.0.0.0.187.187.0j1.1.0...0.0.xA7LXbz4oEY
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