The hour hand and the minute hand of the clock make an angle of 180° with each other 11 times during 12 hours.

Options clearly mentioned we need to find the time where hour hand and minute hand make an angle of 180° after 7 o'clock.

Angle covered by minute hand in 1 minute\(= ({360 \over 60})^\circ = 6^\circ\)

Angle covered by hour hand in 1 hour  \(= ({360 \over 12})^\circ = 30^\circ\)

Angle covered by hour hand in 1-minute  \(= ({30 \over 60})^\circ = ({1 \over 2})^\circ\)

From the above equation, we can say that the minute hand goes ahead by  \(5 {1\over 2}^\circ\)  in comparison to the hour hand in 1 minute.

At, 7 o'clock angle made between the hour hand and minute hand =  7 × 30° = 210°. (or it is 360° - 210° = 150°)

For making an angle of 180° between the hour hand and minute hand, the Minute hand needs to cover 30° with respect to the hour hand.

So, the time required to travel 30° by minute hand with respect to hour hand past 7 o'clock  \(= {30^\circ \over ({11\over 2})^\circ} minutes = {60\over 11}minutes = 5 {5 \over 11} minutes\)

Therefore, the hour hand and the minute hand of the clock make an angle of 180° with each other Between 7:05 hours and 7:10 hours.

• (5n - 30) ×\(12\over11\)  minutes past n, when n > 6  
• (5n + 30) ×  \(12\over11\)  minutes past n, when n < 6  

we simply put n = 7, we get

((5 × 7) - 30) ×  \(12\over11\)  minutes past 7 o'clock

= (35 - 30) ×  \(12\over11\)  minutes past 7 o'clock

=  \(5 ×12\over11\)  minutes past 7 o'clock 

=  \(60 \over 11\)  minutes past 7 o'clock  

=  \(5{5 \over 11} minutes\)  past 7 o'clock .

Therefore, the hour hand and the minute hand of the clock make an angle of 180° with each other Between 7:05 hours and 7:10 hours.