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 mins does the watch gain per day.
a).2
b).5
c).6
d).4
e).75

• 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.
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