Answer: Option C. -> 44
In 12 hours, the hands coincide or are in opposite direction 22 times.
∴ In 24 hours, the hands coincide or are in opposite direction 44 times a day.

Answer: Option C. -> 22
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).
AM
12:00
1:05
2:11
3:16
4:22
5:27
6:33
7:38
8:44
9:49
10:55
PM
12:00
1:05
2:11
3:16
4:22
5:27
6:33
7:38
8:44
9:49
10:55
The hands overlap about every 65 minutes, not every 60 minutes.
∴ The hands coincide 22 times in a day.

Answer: Option A. -> $${30^ \circ }$$
Angle traced by the minute hand in 5 minutes.
$$\eqalign{
& = {\left( {\frac{{360}}{{60}} \times 5} \right)^ \circ } \cr
& = {30^ \circ }{\text{ }} \cr} $$

Answer: Option C. -> $${\text{183}}\frac{1}{3}{\text{c}}{{\text{m}}^2}$$
Angle swept by the minute hand in 35 minutes.
$$\eqalign{
& = {\left( {\frac{{360}}{{60}} \times 35} \right)^ \circ } \cr
& = {210^ \circ } \cr} $$
∴ Required area = Area of a sector of a circle with radius 10 cm and central angle 210°
$$\eqalign{
& = \frac{{\pi {r^2}\theta }}{{360}} \cr
& = \left( {\frac{{22}}{7} \times 10 \times 10 \times \frac{{210}}{{360}}} \right){\text{c}}{{\text{m}}^2} \cr
& = \frac{{550}}{3}{\text{c}}{{\text{m}}^2} \cr
& = 183\frac{1}{3}{\text{c}}{{\text{m}}^2} \cr} $$

Answer: Option B. -> $${\text{17}}{\frac{1}{2}^ \circ }$$
Let angle between the hands of clock be x
When the time is 4 : 25 am
Where [M = minutes and H = hours]
Required angle
$$\eqalign{
& = {\text{30}}\left( {\frac{{\text{M}}}{5} - {\text{H}}} \right) - \frac{{\text{M}}}{2} \cr
& = 30\left( {\frac{{25}}{5} - 4} \right) - \frac{{25}}{2} \cr
& = 30\left( {\frac{{25 - 20}}{5}} \right) - \frac{{25}}{2} \cr
& = 30\left( {\frac{5}{5}} \right) - \frac{{25}}{2} \cr
& = 30 - \frac{{25}}{2} \cr
& = \frac{{60 - 25}}{2} \cr
& = \frac{{35}}{2} \cr
& = 17{\frac{1}{2}^{° }} \cr} $$

Answer: Option D. -> 3 : 36 pm
At 3 o'clock, the minute hand is 15 minute spaces behind the hour hand.
Thus, the minute hand has to gain (15 + 18) = 33 minute spaces
55 minutes are gained in 60 minutes.
33 minutes are gained in $$\left( {\frac{{60}}{{55}} \times 33} \right)$$   = 36 minutes
∴ The hands will be 18 minutes spaces apart at 3 : 36 pm.

Answer: Option D. -> 840°
Angle traced by the minute hand in 2 hours 20 minutes,i.e.,
$$\eqalign{
& = 140{\text{ minutes}} \cr
& = {\left( {\frac{{360}}{{60}} \times 140} \right)^ \circ } \cr
& = {840^ \circ } \cr} $$

Answer: Option C. -> 1 am
The watch loses $$\frac{1}{2}$$ hour each hour. So, it must have take 8 hours to show 4 pm from 12 noon.
Thus, it stopped at 8 pm.
So, the correct time is 5 hours ahead of 8 pm, i,e., 1 am.

Answer: Option B. -> 6
Number of rotation
$$ = \frac{{72}}{{12}} = 6$$

Answer: Option B. -> $${\text{7}}{{\text{5}}^ \circ }$$
In 1 hour, the hour hand make the angle of 30°
The hour hand make the angle of x in $${\text{8}}\frac{{30}}{{60}}$$ hours
$$\eqalign{
& \Rightarrow x = {\left( {30 \times 8\frac{1}{2}} \right)^ \circ } \cr
& \Rightarrow x = {\left( {30 \times \frac{{17}}{2}} \right)^ \circ } \cr
& \Rightarrow x = {255^ \circ } \cr} $$
The minute hand make the angle in 1 minute = 6°
Minute hand makes the angle in 30 minutes
$$\eqalign{
& {\text{ = }}{\left( {6 \times 30} \right)^ \circ } \cr
& = {180^ \circ } \cr
& {\text{Required angle}} \cr
& = {255^0} - {180^0} \cr
& = {75^0} \cr} $$