Quantitative Aptitude

Clocks and Calendars

428 Questions
All Angle between hands 150 Clock gaining or losing time 132 Clock hands coincidence 69 Right angle times 30 Find time from angle 21 Decoding mixed date formats 7 Find day using reference date 6 Repeating same calendar 6 Date without reference day 3

Clocks and Calendars Questions

Multiple choice

How many times a day does the hands of a clock meet each other at 180 degrees?

  1. 11 times

  2. 22 times

  3. 44 times

  4. 12 times
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The hour and minute hands are 180 degrees apart 22 times in 12 hours. In 24 hours (a full day), this occurs 44 times. However, the hour and minute hands actually form 180° 11 times every 12 hours (not 22), giving 22 times per day. The hands coincide 11 times per 12 hours and are 180° apart 11 times per 12 hours.

Multiple choice

If 2 o'clock, the hands of a clock make an acute angle and a reflex angle. How much is the reflex angle (in degrees)?
  1. $225^{\circ}$
  2. $275^{\circ}$
  3. $300^{\circ}$
  4. $180^{\circ}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

At 2 o'clock, the hour hand is at 60° (12×5 = 60 marks from 12) and the minute hand is at 0° (pointing to 12). The angle between them is |60° - 0°| = 60°, which is the acute angle. The reflex angle is 360° - 60° = 300°. Option A (225°) and B (275°) are incorrect calculations. Option D (180°) would be for 6 o'clock.

Multiple choice

At what time between 9 and 10 o'clock will the hands of a watch be together?

  1. 45 min. past 9

  2. 50 min. past 9

  3. 49 1/11 min. past 9

  4. 48 2/11 min. past 9
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The hands overlap every 65 5/11 minutes. At 9:00, the minute hand is at 12 and hour hand at 9. Using the formula: time past 9 = (5 × 60) / (11/2) = 300 × 2/11 = 600/11 = 54 6/11 minutes. But since the hour hand moves, we use the standard formula: (5/11) × 60 = 300/11 = 27 3/11 minutes past 9 for first overlap. For 9-10 o'clock overlap: 60 × 9/11 = 540/11 = 49 1/11 minutes past 9. Option C is correct.

Multiple choice

The questions given below contain two statements giving certain data. You have to decide whether the data given in the statements are sufficient for answering the question? Is the time in the clock 3 O’clock now? I. Exactly 15 minutes back, the hour and the minute hand of the clock coincide with each other. II. After half an hour, the minute and the hour hands of the clock will make an angle of exactly 90° with each other Mark answer-

  1. If statement I alone is sufficient but statement II alone is not sufficient.

  2. If statement II alone is sufficient but statement I alone is not sufficient.

  3. If each statement alone (either I or II) is sufficient.

  4. If statement I and II both are not sufficient.

  5. If both statements I and II together are sufficient, but neither statement alone is sufficient.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Statement I: The hour and minute hands coincide at exactly 2:38:11 (approximately). If this was 15 minutes ago, the current time would be 2:53:11, not 3:00. But more critically, the hands coincide EVERY 65+ minutes, so there are multiple possible times when this could be true going backward from any given moment - this alone is insufficient. Statement II: At 3:30, the hands are exactly 90 degrees apart (hour hand at 3.5, minute hand at 6). If 30 minutes from now they'll be 90 degrees apart, then now must be 3:00. Each statement alone is sufficient to answer the question.

Multiple choice

A clock pendulum having coefficient of linear expansion $\alpha =9\times 10^{-7}/^o$C has a period of $0.5$s at $20^o$C. If the clock is used in a climate where the temperature is $30^o$C, how much times does the clock lose in each oscillation? (g$=$constant)
  1. $2.5\times 10^{-7}$s
  2. $5\times 10^{-7}$s
  3. $1.125\times 10^{-6}$s
  4. $2.25\times 10^{-6}$s
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The change in period is given by delta T = (1/2) * alpha * delta theta * T. delta T = 0.5 * 9 * 10^-7 * 10 * 0.5 = 2.25 * 10^-6 seconds.

Multiple choice

Observe that, at any instant, the minute and hour hands of a clock make two angles between them whose sum is $360^o$. At $6:15$ what is the difference between these two angles is$?$
  1. $165^o$
  2. $170^o$
  3. $175^o$
  4. $180^o$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

At 6:15, the hour hand is at 6 + 15/60 = 6.25 hours. Position in degrees = 6.25 * 30 = 187.5 degrees. The minute hand is at 15 minutes, which is 15 * 6 = 90 degrees. The difference is 187.5 - 90 = 97.5 degrees. However, the question asks for the difference between the two angles (the reflex and the acute), which is 360 - 2*97.5 = 165 degrees.

Multiple choice

Find the time period of the meeting of minute hand and second hand of a clock.
  1. $\Rightarrow t = \dfrac{20}{59} min$
  2. $\Rightarrow t = \dfrac{60}{59} min$
  3. $\Rightarrow t = \dfrac{60}{60} min$
  4. $\Rightarrow t = \dfrac{60}{61} min$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The minute hand moves at 1/60 rev/min. The second hand moves at 1 rev/min. Relative speed = 1 - 1/60 = 59/60 rev/min. Time for one full relative revolution = 1 / (59/60) = 60/59 min.

Multiple choice

A pendulum clock loses 12 s a day if the temperature is ${40^ \circ }C$ and gains 4 s a day if the temperature is ${20^ \circ }C$. The temperature at which the clock will show correct time, and the co-efficient of linear expansion ()of the metal of the pendulum shaft are respectively.
  1. ${60^ \circ }C;\alpha = 1.85 \times {10^{ - 4}}/{}^oC$
  2. ${30^ \circ }C;\alpha = 1.85 \times {10^{ - 3}}/{}^oC$
  3. ${55^ \circ }C;\alpha = 1.85 \times {10^{ - 2}}/{}^oC$
  4. ${25^ \circ }C;\alpha = 1.85 \times {10^{ - 5}}/{}^oC$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice

A clock, with a brass pendulum, keeps correct time at $20^\circ C$, but losses $8.212 s$ per day, when the temperature rises to $30^\circ C$. The coefficient of linear expansion of brass is:-
  1. $25 \times 10^{-6}/^\circ C$
  2. $19 \times 10^{-6}/^\circ C$
  3. $20 \times 20^{-6}/^\circ C$
  4. $11 \times 10^{-6}/^\circ C$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The time lost is delta_t = (1/2) * alpha * delta_T * T, where T is the total time in seconds per day (86400). 8.212 = 0.5 * alpha * (30-20) * 86400. Solving for alpha: alpha = 8.212 / (0.5 * 10 * 86400) = 8.212 / 432000 = 1.9 * 10^-5 = 19 * 10^-6.

Multiple choice

A seconds pendulum clock has a steel wire. The clock shows correct time at $25^0C$. How much time does the clock loose or gain, in one week, when the temperature is increased to $35^0C$? ($\alpha{steel} = 1.2 \times 10^{-5} / ^0 C$)

  1. 321.5 s

  2. 3.828 s

  3. 82.35 s

  4. 36.28 s
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Time lost/gained = (1/2) * alpha * deltaT * T_total. DeltaT = 10 degrees. T_total = 7 days * 24 * 3600 seconds = 604800 seconds. Change = 0.5 * 1.2 * 10^-5 * 10 * 604800 = 0.5 * 1.2 * 60.48 = 36.288 seconds.

Multiple choice

A second pendulum clock having steel wire is calibrated at ${ 20 }^{ 0 }C$. When temperature is increased to ${ 30 }^{ 0 }C$, then how much time does the clock lose or gain in one week ? $[{ \alpha }_{ steel }=1.2\times { 10 }^{ -5 }{ (^{ 0 }C) }^{ -1 }]$ :
  1. $0.3628 s$
  2. $3.626 s$
  3. $362.8 s$
  4. $36.28 S$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A pendulum's period increases by approximately half the fractional increase in wire length. The fractional period increase is 0.5 x 1.2 x 10^-5 x 10 = 6 x 10^-5, so the clock loses 604800 x 6 x 10^-5 = 36.28 seconds in one week.