Answer: Option D. -> 54 6 min. past 4 11

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

60
x 50
min. or 54
6
min.
55
11

Required time = 54
6
min. past 4.
11

Answer: Option C. -> 49 1 min. past 9 11

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

60
x 45
min or 49
1
min.
55
11

The hands are together at 49
1
min. past 9.
11

Answer: Option C. -> 155º
Angle traced by hour hand in 12 hrs = $${360^ \circ }$$
Angle traced by hour hand in 5 hrs 10 min. i.e.,
$$\eqalign{
& \frac{{31}}{6}{\text{hrs}} = {\left( {\frac{{360}}{{12}} \times \frac{{31}}{6}} \right)^ \circ } \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {155^ \circ } \cr} $$

Answer: Option D. -> $${\text{197}}{\frac{1}{2}^ \circ }$$
Angle traced by hour hand in $$\frac{{125}}{{12}}$$ hrs
$$\eqalign{
& = {\left( {\frac{{360}}{{12}} \times \frac{{125}}{{12}}} \right)^ \circ } \cr
& = 312{\frac{1}{2}^ \circ } \cr} $$
Angle traced by minute hand in 25 min
$$\eqalign{
& = {\left( {\frac{{360}}{{60}} \times 25} \right)^ \circ } \cr
& = {150^ \circ } \cr} $$
$$\eqalign{
& \therefore {\text{Reflex angle}} \cr
& = {360^ \circ } - {\left( {312\frac{1}{2} - 150} \right)^ \circ } \cr
& = {360^ \circ } - 162{\frac{1}{2}^ \circ } \cr
& = 197{\frac{1}{2}^ \circ } \cr} $$

Answer: Option D. -> 180º
$$\eqalign{
& {\text{Angle}}\,{\text{traced}}\,{\text{be}}\,{\text{the}}\,{\text{hour}}\,{\text{hand}}\,{\text{in}}\,{\text{6}}\,{\text{hours}} \cr
& = {\left( {\frac{{360}}{{12}} \times 6} \right)^ \circ } = {180^ \circ } \cr} $$

Answer: Option B. -> 4 p.m.
Time from 7 a.m. to 4:15 p.m. = 9 hrs 15 min. = $$\frac{{37}}{4}$$ hrs
3 min. 5 sec. of this c;ocl = 3 min. of the correct clock
⇒ $$\frac{{37}}{{720}}$$ hrs. of this clock = $$\frac{1}{{20}}$$ hrs of the correct clock
⇒ $$\frac{{37}}{4}$$ hrs. of this clock = $$\left( {\frac{1}{{20}} \times \frac{{720}}{{37}} \times \frac{{37}}{4}} \right)$$     hrs. of the correct clock
= 9 hrs. of the correct clock
∴ The correct time is 9 hrs. after 7 a.m. i.e., 4 p.m.

Answer: Option A. -> $$32\frac{8}{{11}}$$ min.
$$\eqalign{
& 55\,\min .\,{\text{spaces}}\,{\text{are}}\,{\text{covered}}\,{\text{in}}\,60\,\min \cr
& 60\,\min .\,{\text{spaces}}\,{\text{are}}\,{\text{covered}}\,{\text{in}} \cr
& = \left( {\frac{{60}}{{55}} \times 60} \right)\,\min . \cr
& = 65\frac{5}{{11}}\,\min . \cr
& {\text{Loss}}\,{\text{in}}\,64\,\min . \cr
& = {65\frac{5}{{11}} - 64} = \frac{{16}}{{11}}\,\min . \cr
& {\text{Loss}}\,{\text{in}}\,24\,hrs. \cr
& = \left( {\frac{{16}}{{11}} \times \frac{1}{{64}} \times 24 \times 60} \right)\,\min. \cr
& = 32\frac{8}{{11}}\,\min. \cr} $$

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

Answer: Option B. -> $$43\frac{7}{{11}}$$ min. past 5
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
$$\eqalign{
& = \left( {\frac{{60}}{{55}} \times 40} \right){\kern 1pt} {\kern 1pt} \min . \cr
& = 43\frac{7}{{11}}{\kern 1pt} {\kern 1pt} \min . \cr
& \therefore {\text{Required time}} = 43\frac{7}{{11}}{\kern 1pt} {\kern 1pt} \min .\,{\text{past}}\,5 \cr} $$

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