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P.W= Amount – (T.D)

Time = 100×TDPW×R

Sum ⇒ Amount 

A = (SI x TD)/(SI - TD)

Given a x a = 12

asked to find a x a x a x a =  (a x a) x (a x a)

a⁴ = a² x a² 

= 12 x 12

= 144.

More :: Mathematical operations questions and answers.

At 2 o'clock, the hour hand is at 2 and the minute hand is at 12, i.e. they are 10 min spaces apart. 

To be together, the minute hand must gain 10 minutes over the hour hand. 

Now, 55 minutes are gained by it in 60 min.

  10 minutes will be gained in 6055×10 min. = 101011 min. 

The hands will coincide at 101011 min. past 2.

At 3 O'clock, Minute hand is at 12 while the Hour hand is at 3. Again the minute hand has to sweep through ( 30 x 5 ) ie 150° for reaching the figure 5 to show 25 mins. 

Simultaneously the Hour hand will also rotate for 25 mins. Thus starting from the mark, 3 the hour hand will cover an angle = (25 x 30) / 60 = 12.5° 

Hence, Angle between Hour and the Minute hand = ( 60 - 12.5 ) = 47.5°

The hands of a clock coincide 11 times in every 12 hours (Since between 11 and 1, they
coincide only once, i.e., at 12 o'clock).

The hands overlap about every 65 minutes, not every 60 minutes.
The hands coincide 22 times in a day.

To be together between 9 and 10 o'clock, the minute hand has to gain 45 min. spaces.
55 min. spaces gained in 60 min.

45 min. spaces are gained in 6055×45 min  or 49111min

4 o'clock, the hands of the watch are 20 min. spaces apart.
To be in opposite directions, they must be 30 min. spaces apart.
Minute hand will have to gain 50 min. spaces.
55 min. spaces are gained in 60 min 

50 min. spaces are gained in 6055×50 min. or 54611 

Required time = 54611 min. past 4.