A clock is such that it loses 4 minutes each day. The clock is set right on February 25, 2008 at 2 p.m. How many minutes should be added to get the right time when the clock shows 9 am on 3rd March, 2008?

As clock loses 4 minutes each day . 
Feb 25th, 2008 at 2pm it was set right . 2008 is a leap year so it will have 29 days in Feb .
So no of days between 25th Feb and 3rd march = 7 days .
In 7 days clock loses 7x4 = 28 minutes .

So now clock shows 1:32 pm on 3rd March 2008 at 2 pm.

As clock loses 4 minutes each day , it implies clock loses 4/24 minutes per each hour i.e it loses 1/360 minute for every one minute . This can imply that for every one hour clock moves only {60-$$\dfrac{\ 4}{24}$$} = $$\dfrac{\ 359}{6}$$ minutes for every hour.

lets assume 'X'  hours before 2 pm the clock shows 9 am , this implies :
1:32 pm - $$\dfrac{\ 359\left(X\right)}{6}$$ minutes = 9:00 am
$$\dfrac{\ 359\left(X\right)}{6}$$ minutes = 4 hrs 32 mins 
$$\dfrac{\ 359\left(X\right)}{6}$$ minutes = 272 mins
Therefore, X = $$\dfrac{\ 6\left(272\right)}{359}$$ hours = $$\dfrac{\ 1632}{359}\ $$ hours.

So, the time at which clock shows 9 am is actually {2 pm - $$\dfrac{\ 1632}{359}\ $$ hours } .
Now, lets the value which we need to add to get the right time at 9 am be "Y" minutes , then it implies :
    {2 pm - $$\dfrac{\ 1632}{359}\ $$ hours } = 9 am + Y minutes

   300 = $$\dfrac{\ 1632}{359}\ $$ hours  + Y minutes 

Hence, Y = 300 - $$\dfrac{\ 1632\left(60\right)}{359}\ \min$$
           Y =  $$\dfrac{\ 9780}{359}$$ minutes.