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

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

Answer: Option C. -> 18
The duration from 1 : 00 pm to 10 : 00 pm is 9 hours and during each of these 9 hours the hands of the clock are at right angles twice.
So, required number = 9 × 2 = 18

Answer: Option B. -> 14 hours 9 minutes 40 seconds
$$\eqalign{
& {\text{Time lost in }}6\frac{1}{2}{\text{ hours}} \cr
& = {\text{ }}\left( {6\frac{1}{2} \times 4} \right)\sec \cr
& = 26\sec \cr} $$
Correct time after $${\text{6}}\frac{1}{2}$$ hours
= 7 hours 40 minutes 6 seconds + 6 hours 30 minutes
= 14 hours 10 minutes 6 seconds
Time show by the clock
= 14 hours 10 minutes 6 seconds - 26 sec
= 14 hours 9 minutes 40 seconds

Answer: Option B. -> 12 noon, after 135 days
After 12 days, i.e., after 12 × 24 hours clock A will gain 48 minutes and will show 12 : 48 noon.
After 12 days, i.e., after 12 × 24 hours clock B will loose 16 minutes and will show 11 : 44 am
The two clocks will show the same time after time after 135 days.
The time difference has to be 12 hours between then = 720 minutes.
A will gain 540 minutes in 135 days.
B will loose 180 minutes in 135 days, total 720 minutes.
Further if we consider only time then the problem becomes simpler Total difference of minutes between the times shown by the clocks after 36 hours
⇒ $$\frac{{16}}{3}$$   minutes difference in 1 day
⇒ 12 × 60 minutes difference in $$\frac{3}{{16}}$$ × 12 × 60 = 135 days
∴ 12 noon, after 135 days

Answer: Option D. -> 11.25 seconds
There are 8 intervals in 9 tinging 10 intervals in 11 tinging.
Time duration of 8 intervals = 9 seconds
∴ Required time = Duration of 10 intervals
$$\eqalign{
& = \left( {\frac{9}{8} \times 10} \right){\text{ seconds}} \cr
& = 11.25{\text{ seconds}} \cr} $$

Answer: Option C. -> 9 pm
Time interval from 9 am on Monday to 8 : 30 pm on Wednesday.
$$\eqalign{
& {\text{ = }}\left( {24 \times 2.5} \right) - {\text{0:30 hours }} \cr
& {\text{ = 60}} - {\text{0}}{\text{:30 hours}} \cr
& {\text{ = 59 hours 30 minutes}} \cr
& = 59\frac{{30}}{{60}} \cr
& = 59\frac{1}{2} \cr
& = \frac{{119}}{2}{\text{ hours}} \cr
& {\text{Also 24 hours}} - {\text{12 minutes}} \cr
& = {\text{23 hours 48 minutes}} \cr
& = 23 + \frac{{48}}{{60}} \cr
& = 23\frac{4}{5} \cr
& = \frac{{119}}{5}{\text{ hours}} \cr
& \therefore \frac{{119}}{2}{\text{ hours of this clock}} \cr
& = \frac{{24 \times 5}}{{119}} \times \frac{{119}}{2} \cr
& = 60{\text{ hours}} \cr
& \left( {60 - \frac{{119}}{2}} \right){\text{ hours}} \cr
& {\text{ = }}\frac{{120 - 119}}{2}{\text{ hours}} \cr
& {\text{ = }}\frac{1}{2}{\text{ hours}} \cr
& = 30{\text{ minutes}} \cr} $$
Hence, the correct time is 30 minutes after 8:30 pm i.e., 9 pm

Answer: Option B. -> 2 p.m. on Wednesday
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.

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

Answer: Option D. -> $$16\frac{4}{{11}}$$ min. past 3
At 3 o'clock, the minute hand is 15 min. spaces apart from the hour hand.
To be coincident, it must gain 15 min. spaces.
55 min. are gained in 60 min.
15 min. are gained in $$\left( {\frac{{60}}{{55}} \times 15} \right)$$   min.
= $$16\frac{4}{{11}}$$ min.
∴ The hands are coincident at $$16\frac{4}{{11}}$$ min. past 3