Clock — Part 1 MCQs with Explanations

$$\eqalign{ & {\text{Angle}}\,{\text{traced}}\,{\text{be}}\,{\text{the}}\,{\text{hour}}\,{\text{hand}}\,{\text{in}}\,{\text{6}}\,{\text{hours}} \cr & = {\left( {\frac{{360}}{{12}} \times 6} \right)^ \circ } = {180^ \circ } \cr} $$

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 $$\eqalign{ & = {\left( {\frac{{360}}{{60}} \times 25} \right)^ \circ } \cr & = {150^ \circ } \cr} $$ $$\eqalign{ & \therefore {\text{Reflex angle}} \cr & = {360^ \circ } - {\left( {312\frac{1}{2} - 150} \right)^ \circ } \cr & = {360^ \circ } - 162{\frac{1}{2}^ \circ } \cr & = 197{\frac{1}{2}^ \circ } \cr} $$

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

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.

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

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

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

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

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

In 12 hours, they are at right angles 22 times. ∴ In 24 hours, they are at right angles 44 times.

$$\eqalign{ & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{hour}}\,{\text{hand}}\,{\text{in}}\,\frac{{17}}{2}\,{\text{hrs}} \cr & {\text{ = }}\,{\left( {\frac{{360}}{{12}} \times \frac{{17}}{2}} \right)^ \circ } \cr & = 255^ \circ \cr & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{min}}{\text{.}}\,{\text{hand}}\,{\text{in}}\,{\text{30}}\,{\text{min}} \cr & = {\left( {\frac{{360}}{{60}} \times 30} \right)^ \circ } \cr & = 180^ \circ \cr & \therefore {\text{Required}}\,{\text{angle}} \cr & = {\left( {255 - 180} \right)^ \circ } = {75^ \circ } \cr} $$

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

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

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.

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.

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

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

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.

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 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.

Number of rotation $$ = \frac{{72}}{{12}} = 6$$

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 $$\eqalign{ & {\text{ = }}{\left( {6 \times 30} \right)^ \circ } \cr & = {180^ \circ } \cr & {\text{Required angle}} \cr & = {255^0} - {180^0} \cr & = {75^0} \cr} $$

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 $$\eqalign{ & \Leftrightarrow \frac{{649}}{{30}}{\text{hours}} \cr & = {\left( {\frac{{360}}{{12}} \times \frac{{649}}{{30}}} \right)^ \circ } \cr & = {649^ \circ } \cr} $$

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 $${\text{ = }}\left[ {77 \times \left( {2 \times \frac{{22}}{7} \times 10} \right) - 6\frac{5}{{12}} \times \left( {2 \times \frac{{22}}{7} \times 7} \right)} \right]{\text{cm}}$$ $$\eqalign{ & = \left( {4840 - 282.33} \right){\text{ cm}} \cr & = 4557.67{\text{ cm}} \cr} $$

One clock show 10 pm, on 21 st January 2010 One clock gains = 2 minutes Other clock loses = 5 minutes Time period between 10 pm and 4 pm = 18 hours ∴ Required difference = (2 × 18 + 5 × 18 ) minutes = 126 minutes

Time gained in 1 hour = 6 minutes Time gained in 6 hours = (6 × 6) minutes = 36 minutes After 6 hours, the correct time is 11 : 00 am and the watch will show 11 : 36 am.

Time from 1 pm on Wednesday to 1 pm on Thursday = 48 hours So, the watch gains (1 + 2) minute or 3 minutes in 48 hours. Now, 3 minutes are gained in 48 hours So, 1 minute is gained in $$\left( {\frac{{48}}{3}} \right)$$ = 16 hours. Thus, the watch showed the correct time 16 hours after 1 pm on Tuesday, i.e., 5 am on Wednesday

To be together between 8 am and 9 am, the minute hand has to gain 40 minutes spaces. 55 minutes spaces are gained in 60 minutes. 40 minutes space are gained in $$\left( {\frac{{60}}{{55}} \times 40} \right)$$ minutes = $${\text{43}}\frac{7}{{11}}$$ minutes So, Henry started his trip at $${\text{43}}\frac{7}{{11}}$$ minutes past 8 am. Now, to be 180° apart, the hands must be 30 minutes spaces apart. At 2 pm, they are 10 minutes spaces apart. ∴ The minute hand will have to gain (10 + 30) = 40 minutes spaces. As calculate above, 40 minutes spaces are gained in $${\text{43}}\frac{7}{{11}}$$ minutes. So, Henry's trip ended at $${\text{43}}\frac{7}{{11}}$$ minutes past 2 pm ∴ Duration of travel = Duration from $${\text{43}}\frac{7}{{11}}$$ minutes past 8 am to $${\text{43}}\frac{7}{{11}}$$ minutes past 2 pm = 6 hours

Since the time read by the lady was 57 minutes earlier than the correct time, so the minute hand is (60 - 57) = 3 minutes spaces behind the hour hand. Now, at 5 o'clock, the minute hand is 25 minutes spaces behind the hour hand. To be 3 minutes spaces behind, it must gain (25 - 3) = 22 minutes spaces. 55 minutes spaces are gained in 60 minutes. 22 minutes spaces are gained in $$\left( {\frac{{60}}{{55}} \times 22} \right)$$ = 24 minutes Hence, the correct time was 24 minutes past 5.

Explanation in Detail: To determine how many times the hour hand and minute hand of a clock form a right angle between 1:00 pm and 10:00 pm, we analyze their angular positions over this period. 1. Calculating Angular Movements: - Minute Hand: The minute hand moves 360 degrees in 60 minutes, so in 9 hours (from 1:00 pm to 10:00 pm), it covers: \[ 360 \text{ degrees/hour} \times 9 \text{ hours} = 3240 \text{ degrees} \] - Hour Hand: The hour hand moves 30 degrees in 60 minutes (or 0.5 degrees per minute), covering: \[ 30 \text{ degrees/hour} \times 9 \text{ hours} = 270 \text{ degrees} \] 2. Relative Angular Distance: - The difference in their angular positions over 9 hours is: \[ 3240 \text{ degrees (minute hand)} - 270 \text{ degrees (hour hand)} = 2970 \text{ degrees} \] 3. Calculating Right Angles: - A right angle is formed every 180 degrees. - Therefore, the number of times they form a right angle is: \[ \frac{2970 \text{ degrees}}{180 \text{ degrees}} = 16.5 \] Rounding down, they form a right angle 16 times. 4. Identifying Times of Right Angles: - The first right angle occurs at 1:21.8181... pm. - Subsequent right angles occur approximately every 32.7878... minutes until the last right angle at 9:32.7272... pm. 5. List of Times: - The hour and minute hands form a right angle at the following times: - 1:21.8181 pm - 1:54.5454 pm - 2:27.2727 pm - 3:00 pm - 3:32.7272 pm - 4:05.4545 pm - 4:38.1818 pm - 5:10.9090 pm - 5:43.6363 pm - 6:16.3636 pm - 6:49.0909 pm - 7:21.8181 pm - 7:54.5454 pm - 8:27.2727 pm - 9:00 pm - 9:32.7272 pm Therefore, the hour hand and minute hand of the clock form a right angle 16 times between 1:00 pm and 10:00 pm, as calculated based on their angular movements and the criteria for forming right angles.

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

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

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 $$\eqalign{ & = \left( {\frac{9}{8} \times 10} \right){\text{ seconds}} \cr & = 11.25{\text{ seconds}} \cr} $$

$$\eqalign{ & {\text{Time lost in }}6\frac{1}{2}{\text{ hours}} \cr & = {\text{ }}\left( {6\frac{1}{2} \times 4} \right)\sec \cr & = 26\sec \cr} $$ 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