Latest CTS Aptitude Question SOLUTION: 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 w

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(#M40007286) CTS Numerical Ability question numerical problem Keep an EYE

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

Asked In CTS (12 years ago)

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Read Solution (1) :

preyksha rajan  (8 years ago)

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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