Clock Reasoning

Clock reasoning questions are a common feature in various aptitude tests. They assess your ability to understand the relationship between the hour and minute hands of a clock, calculate angles, determine time based on given conditions, and solve problems involving the speed of the hands. Clock reasoning is an essential topic in logical reasoning and aptitude tests, involving problems related to the angles between clock hands, overlapping hands, mirror/water images, and faulty clocks.

I. Basic Concepts:

1.     The Clock Face: A circular dial divided into 12 equal parts, representing the hours from 1 to 12.

2.     Hour Hand: The shorter hand that indicates the hour. It completes one full circle (360°) in 12 hours.

3.     Minute Hand: The longer hand that indicates the minutes. It completes one full circle (360°) in 60 minutes.

4.     Second Hand: The thinnest and fastest hand, completing one full circle (360°) in 60 seconds. While less common in reasoning problems, it's good to be aware of.

II. Angular Movement of the Hands:

Understanding the speed at which the hour and minute hands move is crucial for solving many clock problems.

·         Minute Hand:

o    In 60 minutes, it moves 360°.

o    In 1 minute, it moves 360°/12​=6°.

·         Hour Hand:

o    In 12 hours (720 minutes), it moves 360°.

o    In 1 hour (60 minutes), it moves 360°​/12=30°.

o    In 1 minute, it moves 30°​/12=1​/2°=0.5°.

·         Second Hand:

o    In 60 seconds, it moves 360°.

o    In 1 second, it moves 360°/60​=6°.

III. Relative Speed of the Hands:

The relative speed is the difference in the speeds of the two hands. This is important when considering when the hands coincide, are opposite, or are at a certain angle.

·         Relative speed of the minute hand with respect to the hour hand:

o    In one minute, the minute hand gains 6°−0.5°=5.5° on the hour hand.

Trick Questions in Clock Reasoning

1.   When will the clock look the same in AM and PM? Answer: At symmetrical times like 10:10 AM and 10:10 PM.

2.   When do the hands move at the same speed? Answer: The hands never move at the same speed, since the minute hand is always faster.

3.   How many times a day do the hands coincide? Answer: 22 times (not 24, because at 12:00 both hands overlap once).

Types of Clock Reasoning Problems:

Here are the common types of clock reasoning problems you might encounter:

1. Finding the Angle between the Hands:

To find the angle between the hour and minute hands at a given time (H hours and M minutes):

·  Position of the minute hand: The minute hand moves 6° per minute. So, at M minutes past the hour, its position from the 12 mark is 6M degrees.

·  Position of the hour hand: The hour hand moves 30° per hour and 0.5° per minute. At H hours and M minutes, the hour hand has moved 30H degrees from the 12 mark (for the complete hours) and an additional 0.5M degrees (for the minutes past the hour). So, its position from the 12 mark is (30H+0.5M) degrees.

Angle between the hands: The angle is the absolute difference between the positions of the two hands. The formula for the angle between the hour (H) and minute (M) hands is: ∣Angle∣=∣30H−5.5M∣=∣30H−211​M∣

Where:

·  H is the hour (using 12-hour format).

·  M is the minutes past the hour.

Note

·  The formula gives the smaller angle between the two hands. The reflex angle (the larger angle) can be found by subtracting the smaller angle from 360°.

·  When applying the formula, take the absolute value to ensure a positive angle.

Example:

Question 1: What is the angle between the hour and minute hands at 7:20?

Solution:

Here, H = 7 and M = 20.

Angle =∣30×7−5.5×20∣

Angle =∣210−110∣

Angle =∣100∣=100°

Answer: The angle between the hands at 7:20 is 100°.

Question 2: What is the reflex angle between the hour and minute hands at 2:30?

Solution:

Here, H = 2 and M = 30.

Smaller angle =∣30×2−5.5×30∣

Smaller angle =∣60−165∣

Smaller angle =∣−105∣=105°

Reflex angle =360°−105°=255°

Answer: The reflex angle between the hands at 2:30 is 255°.

Question 3: What is the angle between the hour and minute hands at 3:40?

Here, H = 3 and M = 40.

Angle =∣30×3−5.5×40∣=∣90−220∣=∣−130∣=130°.

The smaller angle is 130°. The larger angle is 360°−130°=230°.

2. Finding the Time When the Hands Coincide (Overlap):

The hands coincide when the angle between them is 0°. For a given hour H (from 1 to 11), you can find the exact minute when the hands coincide. Using the formula derived earlier: 30H−5.5M=0⟹M=1160H​

Let the time be H hours and M minutes.

30H−5.5M=0

30H=5.5M

M=5.530H​=1160H​                                        

To find the time between two given hours (e.g., between 3 and 4 o'clock), substitute the smaller hour value for H in the formula.

Example:

Question 1: At what time between 5 and 6 o'clock will the hands of a clock coincide?

Solution:

Here, H = 5.

M=1160×5​=11300​=27113​ minutes.

Answer: The hands will coincide at 27113​ minutes past 5, or approximately 5:27:16.

Question 2: How many times do the hands of a clock coincide in a day?

Solution:

The hands coincide approximately every 65 minutes and 27 seconds. In a 12-hour period, they coincide 11 times (not at 1 o'clock). Therefore, in a 24-hour day, they coincide 22 times.

Answer: The hands of a clock coincide 22 times in a day.

Question 3: At what time between 4 and 5 o'clock will the hands coincide?

Here, H = 4.

M=1160×4​=11240​=21119​ minutes.

So, the hands coincide at approximately 4:21 and 119​ minutes.

3. Finding the Time When the Hands are Opposite (180° apart):

The hands are opposite when the angle between them is 180°.

∣30H−5.5M∣=180

This gives two possibilities:

·  30H−5.5M=180⟹5.5M=30H−180⟹M=1160H−360​

·  30H−5.5M=−180⟹5.5M=30H+180⟹M=1160H+360​

We need to choose the solution that gives a valid minute value (0 to 59).

Example:

Question 1: At what time between 7 and 8 o'clock will the hands be opposite?

Here, H = 7.

M=1160×7−360​=11420−360​=1160​=5115​ minutes.

So, the hands are opposite at approximately 7:05 and 115​ minutes.

Question 2: At what time between 8 and 9 o'clock will the hands of a clock be in a straight line but opposite to each other?

Solution:

Here, H = 8.

Using equation 1: M=1160×8−360​=11480−360​=11120​=101110​ minutes.

Using equation 2: M=1160×8+360​=11480+360​=11840​=76114​ minutes (Invalid, as minutes exceed 59).

Answer: The hands will be opposite at 101110​ minutes past 8, or approximately 8:10:55.

Question 3: How many times in a day are the hands of a clock in a straight line?

Solution:

The hands are in a straight line when they are either coinciding (0°) or opposite (180°). They coincide 22 times a day and are opposite 22 times a day. Therefore, they are in a straight line 22+22=44 times a day.

Answer: The hands of a clock are in a straight line 44 times in a day.

4. Finding the Time When the Hands are at a Specific Angle:

Let the required angle be θ. For a required angle θ, we use:

∣30H−5.5M∣=θ

This gives two possibilities:

·  30H−5.5M=θ⟹M=1160H−2θ​

·  30H−5.5M=−θ⟹M=1160H+2θ​

Again, ensure the minute value is within the valid range. Ensure the value of M is within the valid range (0 to 59).

Example:

Question 1: At what time between 2 and 3 o'clock will the hands be at a 90° angle?

Here, H = 2 and θ=90°.

·  M=1160×2−2×90​=11120−180​=11−60​ (Invalid, as minutes cannot be negative)

·  M=1160×2+2×90​=11120+180​=11300​=27113​ minutes. So, the hands are at a 90° angle at approximately 2:27 and 113​ minutes.

Question 2: At what time between 3 and 4 o'clock will the hands of a clock be at a 90° angle?

Solution:

Here, H = 3 and θ=90°.

Using equation 1: M=1160×3−2×90​=11180−180​=0 minutes. (3:00)

Using equation 2: M=1160×3+2×90​=11180+180​=11360​=32118​ minutes. (Approximately 3:32:44)

Answer: The hands will be at a 90° angle at 3:00 and at 32118​ minutes past 3.

Question 3: How many times in 12 hours are the hands of a clock at a right angle?

Solution:

The hands are at a right angle twice every hour, except between 2 and 4 o'clock and between 8 and 10 o'clock, where it happens only three times in each 2-hour interval. So, in 12 hours, it happens 2×10+2×1=22 times.

Answer: The hands of a clock are at a right angle 22 times in 12 hours.

5. Problems Involving Faulty Clocks:

These problems involve clocks that gain or lose time. Problems involving faulty clocks require calculating the effective time shown by the clock based on its gain or loss.

·       To solve these problems:

1.     Calculate the total time the clock has been running.

2.     Determine the amount of time the clock has gained or lost during this period.

3.     Adjust the time shown by the faulty clock to find the correct time.

A faulty clock gains or loses time. If a clock gains or loses XX minutes every YY hours, the correct time is found by:

Correct Time =  Actual Time + Time elapsed  (If Gains)

                        Actual Time - Time elapsed  (If Loses)

Example:

Question 1: A clock gains 15 minutes every hour. If it was set right at 12 noon, what time will it show at 4 pm?

·         Time elapsed = 4 hours.

·         Gain per hour = 15 minutes.

·         Total gain in 4 hours = 15×4=60 minutes = 1 hour.

·         So, at 4 pm, the faulty clock will show 4 pm + 1 hour = 5 pm.

Question 2: A clock loses 10 minutes every 24 hours. If it was set right at 8 am yesterday, what time will it show at 8 am today?

·         Time elapsed = 24 hours.

·         Total loss in 24 hours = 10 minutes.

·         So, at 8 am today, the faulty clock will show 8 am - 10 minutes = 7:50 am.

Question 3: A clock gains 10 minutes every 24 hours. If it was set right at 8 am on Monday, what time will it show at 8 pm on Wednesday?

Solution:

·  Time elapsed from 8 am Monday to 8 pm Wednesday = 2 days and 12 hours = 2×24+12=60 hours.

·  Gain per 24 hours = 10 minutes.

·  Gain per hour = 2410​ minutes.

·  Total gain in 60 hours = 2410​×60=24600​=25 minutes.

·  So, the time shown by the faulty clock will be 8 pm + 25 minutes = 8:25 pm.

Answer: The faulty clock will show 8:25 pm.

Question 4: A watch loses 5 seconds in 3 minutes. If it was set right at 7 am, what time will it show at 7 pm on the same day?

Solution:

·         Time elapsed from 7 am to 7 pm = 12 hours = 12×60=720 minutes.

·         Loss in 3 minutes = 5 seconds.

·         Loss in 1 minute = 35​ seconds.

·         Total loss in 720 minutes = 35​×720=5×240=1200 seconds.

·         Convert seconds to minutes: 601200​=20 minutes.

·         So, the time shown by the faulty watch will be 7 pm - 20 minutes = 6:40 pm.

Answer: The faulty watch will show 6:40 pm.

Question 1: A clock gains 10 minutes every 5 hours. If it shows 8:00 PM now, what is the real time after 10 hours?

Calculate the total gain in 10 hours:

• The clock gains 10 minutes every 5 hours.
• In 10 hours, the clock will gain twice that amount: 10 minutes×2=20 minutes.

Determine the real time:

• The clock currently shows 8:00 PM.
• Since the clock gains time, the real time must be earlier than what the clock shows.
• Subtract the total gain from the time shown by the clock: 8:00 PM - 20 minutes = 7:40 PM.

Therefore, the real time after 10 hours will be 7:40 AM.

6. Mirror Images of Clocks

Mirror image problems involve determining the time shown in a mirror when a clock displays a certain time. The key concept is that the image is a reflection along the vertical axis.

·  For times within a 12-hour clock, subtract the given time from 11:60.

·  For times within a 24-hour clock, subtract the given time from 23:60.

·  To find the mirror image of a given time:

·  Mirror Image Time=11:60−Given Time

Example 5: Mirror Image of 4:20

=11:60−4:20= 11:60 - 4:20 =7:40= 7:40

So, the mirror image of 4:20 is 7:40.

Question 11: What is the mirror image of the time 3:15?

Solution:

Subtract 3:15 from 11:60:

11 hours 60 minutes

3 hours 15 minutes

8 hours 45 minutes

Answer: The mirror image of 3:15 is 8:45.

Question 12: What is the mirror image of the time 10:35?

Solution:

Subtract 10:35 from 11:60:

11 hours 60 minutes

10 hours 35 minutes

1 hour 25 minutes

Answer: The mirror image of 10:35 is 1:25.

7. Water Images of Clocks

Water image problems involve the reflection of the clock across a horizontal axis. This is slightly more complex as the hour and minute hands' behaviour in a water reflection is not as straightforward as in a mirror reflection. A general approximation for the water image can be found by subtracting the time from 18:30 or 17:90. However, these are approximations and might not be perfectly accurate for all positions.

·  The water image is an inverted reflection in water, calculated by:

·  Water Image Time=6:00−Given Time

Example

Question 1: Water Image of 2:30

=6:00−2:30= 6:00 - 2:30 =3:30= 3:30

So, the water image of 2:30 is 3:30.

Question 2: What is the water image of the time 6:20?

Solution (Approximation):

Subtract 6:20 from 18:30:

18 hours 30 minutes

6 hours 20 minutes

12 hours 10 minutes

Answer (Approximation): The water image of 6:20 is approximately 12:10.

Question 3: What is the water image of the time 9:40?

Solution (Approximation):

Subtract 9:40 from 17:90:

17 hours 90 minutes

9 hours 40 minutes

8 hours 50 minutes

Answer (Approximation): The water image of 9:40 is approximately 8:50.

This detailed guide with questions and answers should further enhance your understanding of clock reasoning problems. Practice these types of questions regularly to improve your speed and accuracy.