At a certain moment a watch shows 2 min lag although it is running fast.If it showed a 3 min lag at that moment, but also gains by 1/2 min more a day than its current speed it would show the true time one day sooner than it usually does.How many minutes does the watch gain per day.

Interview Candidate
Jan 30th, 2006
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C. Sivasankaran

May 18th, 2006

Let fast per day be x, no of days required to show correct time
for condition 1 is     2 / x.
For condition 2 is     3 / (x + 1/2)
Condition 2 is one day sooner hence 2 / x = 3 / (x + 1/2)   + 1
Solving this we will get x^2 +1.5 x - 1 = 0
x = -2 or +1/2. It is running fast hence x = 1/2
Hence the answer is 1/2 min per day.
To verify this for condition 1 it require 4 days ( 2 % 1/2 )
Condition 2 it require 3 days [ 3 % (1/2 + 1/2 )] i.e 1 day less than condition 1.