Time from 12 noon to 9 : 38 A.M.
= 12 hours + 9 hours 38 minutes
= 21 hours 38 minutes
$$\eqalign{ & {\text{ = 21}}\frac{{38}}{{60}}{\text{ hours}} \cr & {\text{ = 21}}\frac{{19}}{{30}}{\text{ hours}} \cr & {\text{ = }}\frac{{649}}{{30}}{\text{hours}} \cr} $$
Angle traced by the hour hand in 12 hours = 360°
Angle traced by the minute hand in
$$\eqalign{ & \Leftrightarrow \frac{{649}}{{30}}{\text{hours}} \cr & = {\left( {\frac{{360}}{{12}} \times \frac{{649}}{{30}}} \right)^ \circ } \cr & = {649^ \circ } \cr} $$

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.

Time from 12 pm on Monday to 2 pm on the following Monday = 7 days 2 hours
= 170 hours
∴ The watch gains \[\left( {2 + 4\frac{4}{5}} \right)\]   minutes or \[\frac{{34}}{5}\]   minutes in 170 hours.
Now, $$\frac{{34}}{5}$$ minutes are gained in 170 hours
∴ 2 minutes are gained in \[\left( {170 \times \frac{5}{{34}} \times 2} \right)\]   hours = 50 hours
∴ Watch is correct 2 days 2 hours after 12 pm on Monday i.e., it will be correct at 2 pm on Wednesday.

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

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°
$$\eqalign{ & = \frac{{\pi {r^2}\theta }}{{360}} \cr & = \left( {\frac{{22}}{7} \times 10 \times 10 \times \frac{{210}}{{360}}} \right){\text{c}}{{\text{m}}^2} \cr & = \frac{{550}}{3}{\text{c}}{{\text{m}}^2} \cr & = 183\frac{1}{3}{\text{c}}{{\text{m}}^2} \cr} $$

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

The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours. (Because between 5 and 7 they point in opposite directions at 6 o'clcok only).
So, in a day, the hands point in the opposite directions 22 times.

At 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
$$\eqalign{ & \left( {\frac{{60}}{{55}} \times 50} \right){\text{min}}{\text{.}}\,{\text{or}}\,54\frac{6}{{11}}\,{\text{min}}. \cr & \therefore {\text{Required}}\,{\text{time}} \cr & = 54\frac{6}{{11}}{\text{min}}{\text{.}}\,{\text{past}}\,4 \cr} $$

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
$$\left( {\frac{{60}}{{55}} \times 45} \right){\text{min}}{\text{.}}{\kern 1pt} \,{\text{or}}{\kern 1pt} \,49\frac{1}{{11}}{\kern 1pt} {\text{min}}.$$
∴ The hands are together at $$49\frac{1}{{11}}$$ min. past 9

Number of rounds completed by the minute hand in 3 days 5 hours
$$\eqalign{ & = \left( {3 \times 24 + 5} \right) \cr & = 77 \cr} $$
Number of rounds completed by the hour hand in 3 days 5 hours
$$\eqalign{ & = \left( {3 \times 2 + \frac{5}{{12}}} \right) \cr & = 6\frac{5}{{12}} \cr} $$
∴ Difference between the distance traversed
$${\text{ = }}\left[ {77 \times \left( {2 \times \frac{{22}}{7} \times 10} \right) - 6\frac{5}{{12}} \times \left( {2 \times \frac{{22}}{7} \times 7} \right)} \right]{\text{cm}}$$
$$\eqalign{ & = \left( {4840 - 282.33} \right){\text{ cm}} \cr & = 4557.67{\text{ cm}} \cr} $$

Number of minutes in a day
= (24 × 60) = 1440
∴ Required percentage
$$\eqalign{ & = \left( {\frac{5}{{1440}} \times 100} \right)\% \cr & = \frac{{50}}{{144}}\% \cr} $$

When the hands of the clock are in the same straight line but not together, they are 30 minute spaces apart.
At 7 o'clock, they are 25 min. spaces apart.
∴ Minute hand will have to gain only 5 min. spaces.
55 min. spaces are gained in 60 min.
5 min. spaces are gained in
$$\eqalign{ & = \left( {\frac{{60}}{{55}} \times 5} \right){\kern 1pt} \min . \cr & = 5\frac{5}{{11}}{\kern 1pt} \min . \cr & \therefore {\text{Required time}} = 5\frac{5}{{11}}{\kern 1pt} \min .{\kern 1pt} \,{\text{past}}{\kern 1pt} 7 \cr} $$

At 5 o'clock, the hands are 25 min. spaces apart.
To be at right angles and that too between 5.30 and 6, the minute hand has to gain (25 + 15) = 40 min. spaces.
55 min. spaces are gained in 60 min.
40 min. spaces are gained in
$$\eqalign{ & = \left( {\frac{{60}}{{55}} \times 40} \right){\kern 1pt} {\kern 1pt} \min . \cr & = 43\frac{7}{{11}}{\kern 1pt} {\kern 1pt} \min . \cr & \therefore {\text{Required time}} = 43\frac{7}{{11}}{\kern 1pt} {\kern 1pt} \min .\,{\text{past}}\,5 \cr} $$

$$\eqalign{ & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{hour}}\,{\text{hand}}\,{\text{in}}\,\frac{{13}}{3}\,{\text{hrs}} \cr & = {\left( {\frac{{360}}{{12}} \times \frac{{13}}{3}} \right)^ \circ } = {130^ \circ } \cr & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{min}}{\text{.}}\,{\text{hand}}\,{\text{in}}\,{\text{20}}\,{\text{min}} \cr & = {\left( {\frac{{360}}{{60}} \times 20} \right)^ \circ } = {120^ \circ } \cr & \therefore {\text{Required}}\,{\text{angle}} \cr & = {\left( {130 - 120} \right)^ \circ } \cr & = {10^ \circ } \cr} $$

$$\eqalign{ & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{hour}}\,{\text{hand}}\,{\text{in}}\,\frac{{21}}{4}\,{\text{hrs}} \cr & = {\left( {\frac{{360}}{{12}} \times \frac{{21}}{4}} \right)^ \circ } = 157{\frac{1}{2}^ \circ } \cr & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{min}}{\text{.}}\,{\text{hand}}\,{\text{in}}\,15\,{\text{min}} \cr & = {\left( {\frac{{360}}{{60}} \times 15} \right)^ \circ } = {90^ \circ } \cr & \therefore {\text{Required}}\,{\text{angle}} \cr & = {\left( {157\frac{1}{2}} \right)^ \circ } - {90^ \circ } \cr & = 67{\frac{1}{2}^ \circ } \cr} $$

$$\eqalign{ & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{hour}}\,{\text{hand}}\,{\text{in}}\,12\,{\text{hrs}}\,{\text{ = }}\,{\text{36}}{{\text{0}}^ \circ } \cr & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{it}}\,{\text{in}}\,\frac{{11}}{3}\,{\text{hrs}} \cr & = {\left( {\frac{{360}}{{12}} \times \frac{{11}}{3}} \right)^ \circ } = {110^ \circ } \cr & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{min}}{\text{.}}\,{\text{hand}}\,{\text{in}}\,60\,{\text{min}}\,{\text{ = }}\,{\text{36}}{{\text{0}}^ \circ } \cr & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{it}}\,{\text{in}}\,{\text{40}}\,{\text{min}}. \cr & = {\left( {\frac{{360}}{{60}} \times 40} \right)^ \circ } = {240^ \circ } \cr & \therefore {\text{Required}}\,{\text{angle}} \cr & = {\left( {240 - 110} \right)^ \circ } = {130^ \circ } \cr} $$