An urn contains 1 liter of water while a second urn nearby is empty. after 1/2 of the water in the first urn is emptied into the second urn, 1/3 of the water in the second urn is emptied into the first urn. then, after 1/4 of the contents of the first urn is poured into the second urn, 1/5 of the contents of the second urn is poured into the first urn. at each successive pouring from alternate urns, the denominator of the fractional part poured increases by 1. how many liters of water remain in both urns after 1997th pourings

We are given that there will be (1/2) a litre after the first pouring, so considering two successive pourings (n and (n+1)) with 1/2 litre in each before the nth pouring occurs:

1/2 × (1/n) = 1/(2n)

1/2 - 1/(2n) = (n-1)/2n

1/2 + 1/(2n) = (n+1)/2n

(n-1)/2n and (n+1)/2n in each urn after the nth pouring

Then now consider the (n+1)th pouring

(n+1)/2n × 1/(n+1) = 1/(2n)

(n+1)/(2n) - 1/(2n) = n/(2n) = 1/2

Therefore this means that after an odd number of pouring, there will be 1/2 a litre in each urn

Since 1997 is an odd number, then there will be 1/2 a litre of water in each urn.

1/2

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