CAT Exam Model Paper 4 with solutions for free online practice

The minute-hand of a normal clock covers $\frac{360}{60}=6°$ per minute. The hour-hand of a normal clock covers $\frac{30}{60}=\frac{1°}{2}$ per minute. So once they are together, in every minute the minute hand gains $6-\frac{1}{2}=\frac{11°}{2}$ over the hour hand.
So, time between two meetings $\frac{360}{11}\times 2=\frac{720}{11}$ minutes
So, in any clock the hour-hand and the minute-hand meet after every $\frac{720}{11}$ minutes
If 60 minutes have passed in a normal clock then time passed in the faulty clock is 70 minutes.
If 24 hrs (or 24 x 60 minutes) have passed in a normal clock then time passed in the faulty clock must be 24 x 70 = 1680 minutes.
Number of times the hands meet $=\frac{1680}{72}\times 11=25.67=25$