| Content On This Page | ||
|---|---|---|
| Clock: Concepts of Angles and Relative Speed of Hands | Problems involving Coincidence, Opposite, and Right Angles of Hands | Problems on Faulty Clocks |
| Solving Clock Problems | ||
Clocks
Clock: Concepts of Angles and Relative Speed of Hands
Clock problems are a common type of quantitative reasoning question. They require understanding the circular motion of the hour hand and minute hand on a standard analog clock, their speeds, and their relative movement to determine the angles between them or the times at which they are in specific relative positions.
Clock Face and Angles:
A standard clock face is circular, representing a full angle of 360 degrees. The dial is marked with 12 hour divisions and 60 minute divisions.
- The angle between two consecutive hour marks (e.g., from 12 to 1, 1 to 2, ..., 11 to 12) is obtained by dividing the full circle's angle by the 12 hour divisions.
Angle per hour division $= \frac{360^\circ}{12} = 30^\circ$
- The angle between two consecutive minute marks is obtained by dividing the full circle's angle by the 60 minute divisions.
Angle per minute division $= \frac{360^\circ}{60} = 6^\circ$
Each minute mark corresponds to 6 degrees. Since there are 5 minute divisions between two consecutive hour marks, the angle between two consecutive hour marks is $5 \text{ minute divisions} \times 6^\circ\text{/minute division} = 30^\circ$, which is consistent.
Speed of the Hands:
The hour hand and the minute hand move at constant but different speeds.
Speed of the Minute Hand:
The minute hand completes one full rotation (360 degrees) in 60 minutes.
Speed of Minute Hand $= \frac{360^\circ}{60 \text{ minutes}} = 6^\circ \text{ per minute}$
So, the minute hand moves $6^\circ$ for every minute that passes.
Speed of the Hour Hand:
The hour hand completes one full rotation (360 degrees) in 12 hours. To find its speed per minute, convert 12 hours to minutes: $12 \text{ hours} = 12 \times 60 = 720$ minutes.
Speed of Hour Hand $= \frac{360^\circ}{12 \text{ hours}} = \frac{360^\circ}{720 \text{ minutes}}$
Simplify the fraction:
$\frac{360}{720} = \frac{36}{72} = \frac{1}{2} = 0.5$
Speed of Hour Hand $= 0.5^\circ \text{ per minute}$
So, the hour hand moves $0.5^\circ$ for every minute that passes.
Relative Speed of the Hands:
The minute hand moves faster than the hour hand. The relative speed of the minute hand with respect to the hour hand (the rate at which the minute hand gains on the hour hand) is the difference between their speeds.
Relative Speed $= \text{Speed of Minute Hand} - \text{Speed of Hour Hand}$
Relative Speed $= 6^\circ\text{/min} - 0.5^\circ\text{/min} = 5.5^\circ \text{ per minute}$
$\boldsymbol{\text{Relative Speed} = 5.5^\circ \text{ per minute}}$
... (i)
This means the minute hand gains $5.5^\circ$ on the hour hand every minute. This relative speed is crucial for solving problems about when the hands will be together or at a specific angle.
Angle between the Hands at a Given Time:
To find the angle between the hour and minute hands at a specific time (let's say H hours and M minutes, where H is between 1 and 12, and M is between 0 and 59), we can calculate the angular position of each hand relative to the 12 o'clock mark (which we consider as 0 degrees or 360 degrees) and then find the difference.
- Position of the Minute Hand: The minute hand moves $6^\circ$ per minute. At M minutes past the hour, the minute hand's position from the 12 mark is $6M$ degrees.
- Position of the Hour Hand: The hour hand moves based on both the hours and the minutes. It moves $30^\circ$ per hour. In H hours, it moves $30H$ degrees from 12. Additionally, in M minutes, it moves $0.5^\circ$ per minute. So, in M minutes, it moves an additional $0.5M$ degrees from the hour mark it was at.
Angular position of Hour Hand $= (30H + 0.5M)^\circ$ from 12.
Angular position of Minute Hand $= (6M)^\circ$ from 12.
The angle between the hands is the absolute difference between their angular positions. Since a clock face is circular, there are two angles between the hands (the interior angle and the reflex angle). We usually calculate the smaller angle (between 0 and 180 degrees).
$\boldsymbol{\text{Angle between hands} = |(30H + 0.5M) - 6M|^\circ}$
Simplify the expression inside the absolute value:
$\boldsymbol{\text{Angle between hands} = |30H - 5.5M|^\circ}$
... (ii)
Where H is the hour (use 12 for 12 o'clock) and M is the minutes. If the calculated angle from formula (ii) is greater than 180 degrees, subtract it from 360 degrees to get the smaller angle.
Example 1. Find the angle between the hour hand and the minute hand at 4:40.
Answer:
Given time: H $= 4$, M $= 40$.
Using the formula (ii): Angle $= |30H - 5.5M|^\circ$
Angle $= |30 \times 4 - 5.5 \times 40|^\circ$
Calculate the values:
$30 \times 4 = 120$
$5.5 \times 40 = 5.5 \times 4 \times 10 = 22 \times 10 = 220$
Angle $= |120 - 220|^\circ = |-100|^\circ = 100^\circ$.
The angle between the hands at 4:40 is $\boldsymbol{100^\circ}$.
Example 2. Find the angle between the hour hand and the minute hand at 10:10.
Answer:
Given time: H $= 10$, M $= 10$.
Using the formula (ii): Angle $= |30H - 5.5M|^\circ$
Angle $= |30 \times 10 - 5.5 \times 10|^\circ$
Calculate the values:
$30 \times 10 = 300$
$5.5 \times 10 = 55$
Angle $= |300 - 55|^\circ = |245|^\circ = 245^\circ$.
Since the calculated angle ($245^\circ$) is greater than 180 degrees, the smaller angle between the hands is $360^\circ - 245^\circ = 115^\circ$.
Smaller Angle $= 360^\circ - 245^\circ = 115^\circ$.
The smaller angle between the hands at 10:10 is $\boldsymbol{115^\circ}$.
Competitive Exam Notes:
Clock problems often involve calculating angles or times when hands are in specific relative positions (together, opposite, right angle, etc.).
- Speeds: Minute hand: $6^\circ$/min. Hour hand: $0.5^\circ$/min.
- Relative Speed: Minute hand gains $5.5^\circ$ (or $11/2^\circ$) per minute on the hour hand.
- Angle Formula: Angle $= |30H - 5.5M|^\circ$. Remember to find the smaller angle (between 0 and 180).
- Angular Positions: Minute hand position from 12 $= 6M$. Hour hand position from 12 $= 30H + 0.5M$.
- Clock as Relative Speed Problem: Problems about when hands meet or are at a certain angle can be solved using relative speed. For example, to find when hands meet, they need to cover the initial angle between them at the relative speed. To be $X^\circ$ apart, the faster hand needs to be $X^\circ$ ahead or behind.
Problems involving Coincidence, Opposite, and Right Angles of Hands
These problems involve finding the exact time instances within an hour interval when the hour hand and minute hand of a clock are in specific standard relative positions. These positions correspond to particular angles between the hands: 0 degrees (coincidence), 180 degrees (opposite), and 90 degrees (right angle).
Finding the Time for a Specific Angle ($\theta$):
The core principle is based on the relative movement of the hour and minute hands. The minute hand moves at $6^\circ$ per minute, and the hour hand moves at $0.5^\circ$ per minute. The minute hand gains $6^\circ - 0.5^\circ = 5.5^\circ$ on the hour hand every minute.
To find the time at which the hands form a specific angle $\theta$ (in degrees) between H and H+1 o'clock (e.g., between 4 and 5 o'clock, H=4), we can consider their positions relative to the 12 o'clock mark (0 degrees).
At the start of the hour (exactly H o'clock), the hour hand is at $30H$ degrees from 12, and the minute hand is at $0$ degrees from 12. The initial angle between them is $30H$ degrees.
Let M be the number of minutes past H o'clock when the hands form an angle of $\theta$.
In M minutes, the minute hand moves $6M$ degrees from 12.
In M minutes, the hour hand moves an additional $0.5M$ degrees from its position at H o'clock. So its position from 12 is $(30H + 0.5M)$ degrees.
The angle between the hands is the absolute difference between their positions:
Angle $= |(30H + 0.5M) - 6M|^\circ = \theta$
Simplify the expression inside the absolute value:
$\boldsymbol{|30H - 5.5M|^\circ = \theta}$
To solve for M, we consider two cases for the absolute value: $30H - 5.5M = \theta$ or $30H - 5.5M = -\theta$.
Case 1: $30H - 5.5M = \theta \implies 30H - \theta = 5.5M \implies M = \frac{30H - \theta}{5.5}$.
Case 2: $30H - 5.5M = -\theta \implies 30H + \theta = 5.5M \implies M = \frac{30H + \theta}{5.5}$.
Combining these, we get the formula for M minutes past H o'clock:
$\boldsymbol{\text{M} = \frac{30H \pm \theta}{5.5} = \frac{2}{11} (30H \pm \theta) \text{ minutes}}$
... (iii)
Where H is the starting hour of the interval (1 to 12), and $\theta$ is the desired angle (between 0 and 180 degrees, usually). We use $\pm \theta$ because the minute hand can form the angle $\theta$ with the hour hand in two ways (minute hand ahead or behind). For $\theta = 0^\circ$ or $180^\circ$, one of the $\pm$ cases might yield an invalid time (outside 0-60 minutes) or be a duplicate of the other, indicating only one such instance per hour interval (with some exceptions around 12 and 6).
Specific Angle Cases:
Let's apply the formula (iii) to find the times for the most common specific angles:
1. Coincidence (0 degrees)
The hands are together, forming an angle of $\theta = 0^\circ$.
Using formula (iii): $M = \frac{2}{11} (30H \pm 0) = \frac{2}{11} (30H) = \frac{60H}{11}$ minutes past H.
- The hands coincide once in every hour interval, except for the interval between 11 and 1 (over the 12 o'clock period), where they coincide only exactly at 12:00.
- In a 12-hour period, the hands coincide 11 times.
- In a 24-hour period, they coincide 22 times.
2. Opposite (180 degrees)
The hands are in a straight line, pointing in opposite directions, forming an angle of $\theta = 180^\circ$.
Using formula (iii): $M = \frac{2}{11} (30H \pm 180)$ minutes past H.
- The hands are opposite once in every hour interval, except for the interval between 5 and 7 (over the 6 o'clock period), where they are opposite only exactly at 6:00.
- In a 12-hour period, the hands are opposite 11 times.
- In a 24-hour period, they are opposite 22 times.
3. Right Angle (90 degrees)
The hands are perpendicular to each other, forming an angle of $\theta = 90^\circ$.
Using formula (iii): $M = \frac{2}{11} (30H \pm 90)$ minutes past H.
- The hands form a right angle twice in every hour interval. One time when the minute hand is approximately 90 degrees ahead of the hour hand, and another time when it is approximately 90 degrees behind.
- Exceptions: Around 3 o'clock and 9 o'clock, the instances of 90 degrees overlap. Between 2 & 3 and 3 & 4, there are 3 instances of 90 degrees total (one at 3:00 sharp). Similarly, between 8 & 9 and 9 & 10, there are 3 instances total (one at 9:00 sharp).
- In a 12-hour period, the hands form a right angle 22 times.
- In a 24-hour period, they form a right angle 44 times.
Example 1. At what time between 5 and 6 o'clock will the hands of a clock coincide?
Answer:
We need to find the time between 5:00 and 6:00 when the hands coincide, so the angle $\theta = 0^\circ$.
The starting hour for the interval is 5, so H $= 5$.
Using the formula (iii): $\text{M} = \frac{2}{11} (30H \pm \theta)$ minutes past H.
$\text{M} = \frac{2}{11} (30 \times 5 \pm 0)$ minutes past 5.
$\text{M} = \frac{2}{11} (150)$ minutes past 5.
$\text{M} = \frac{300}{11}$ minutes past 5.
Convert the improper fraction to a mixed number:
$\frac{300}{11} = 27\frac{3}{11}$ minutes.
$\boldsymbol{\text{M} = 27\frac{3}{11}}$ minutes.
The hands will coincide at $\boldsymbol{27\frac{3}{11}}$ minutes past 5 o'clock.
Example 2. At what time between 8 and 9 o'clock will the hands of a clock be in a straight line but not together?
Answer:
We need to find the time between 8:00 and 9:00 when the hands are in a straight line but not together. This means the angle between them is $\theta = 180^\circ$.
The starting hour for the interval is 8, so H $= 8$.
Using formula (iii): $\text{M} = \frac{2}{11} (30H \pm \theta)$ minutes past H.
$\text{M} = \frac{2}{11} (30 \times 8 \pm 180)$ minutes past 8.
$\text{M} = \frac{2}{11} (240 \pm 180)$ minutes past 8.
We consider the two possibilities:
Possibility 1 (using +180): Time $= \frac{2}{11} (240 + 180) = \frac{2}{11} (420) = \frac{840}{11}$ minutes.
Convert to mixed number: $\frac{840}{11} = 76\frac{4}{11}$ minutes. This is greater than 60 minutes, so it falls into the next hour (past 9 o'clock). This is not the solution between 8 and 9.
Possibility 2 (using -180): Time $= \frac{2}{11} (240 - 180) = \frac{2}{11} (60) = \frac{120}{11}$ minutes.
Convert to mixed number: $\frac{120}{11} = 10\frac{10}{11}$ minutes. This is between 0 and 60 minutes, so it is valid for the 8-9 interval.
$\boldsymbol{\text{M} = 10\frac{10}{11}}$ minutes.
The hands will be exactly opposite to each other at $\boldsymbol{10\frac{10}{11}}$ minutes past 8 o'clock.
Example 3. At what time between 6 and 7 o'clock will the hands of a clock be at right angles?
Answer:
We need to find the times between 6:00 and 7:00 when the hands are at right angles, so the angle $\theta = 90^\circ$.
The starting hour for the interval is 6, so H $= 6$.
Using formula (iii): $\text{M} = \frac{2}{11} (30H \pm \theta)$ minutes past H.
$\text{M} = \frac{2}{11} (30 \times 6 \pm 90)$ minutes past 6.
$\text{M} = \frac{2}{11} (180 \pm 90)$ minutes past 6.
We get two possibilities:
Possibility 1 (using +90): Time $= \frac{2}{11} (180 + 90) = \frac{2}{11} (270) = \frac{540}{11}$ minutes.
Convert to mixed number: $\frac{540}{11} = 49\frac{1}{11}$.
This is within the 6-7 hour range.
$\boldsymbol{\text{Time}_1 = 49\frac{1}{11} \text{ minutes past 6}}$
Possibility 2 (using -90): Time $= \frac{2}{11} (180 - 90) = \frac{2}{11} (90) = \frac{180}{11}$ minutes.
Convert to mixed number: $\frac{180}{11} = 16\frac{4}{11}$.
This is also within the 6-7 hour range.
$\boldsymbol{\text{Time}_2 = 16\frac{4}{11} \text{ minutes}}$.
The hands will be at right angles at $\boldsymbol{16\frac{4}{11}}$ minutes past 6 o'clock and $\boldsymbol{49\frac{1}{11}}$ minutes past 6 o'clock.
Competitive Exam Notes:
These problems are solvable by understanding the relative motion of the hands and applying the formulas derived from relative speed or angular positions.
- Key Formula: $\text{M} = \frac{2}{11} (30H \pm \theta)$ minutes past H. H is the hour, $\theta$ is the desired angle.
- Coincidence ($\theta=0^\circ$): M = $\frac{60H}{11}$. Occurs once per hour (except around 12).
- Opposite ($\theta=180^\circ$): M = $\frac{2}{11} (30H \pm 180)$. Usually solved using $M = \frac{60}{11} (H \pm 6)$. Use $H+6$ or $H-6$ such that the result is between 0 and 60. Occurs once per hour (except around 6).
- Right Angle ($\theta=90^\circ$): M = $\frac{2}{11} (30H \pm 90)$. Usually solved using $M = \frac{60}{11} (H \pm 3)$ or $M = \frac{60}{11} (H \pm 9)$. Occurs twice per hour (except around 3 and 9).
- Total Occurrences: Coincidence/Opposite: 11 times in 12 hours, 22 times in 24 hours. Right Angle: 22 times in 12 hours, 44 times in 24 hours.
- The $2/11$ Factor: Comes from $\frac{1}{5.5} = \frac{1}{11/2} = \frac{2}{11}$. It represents the time it takes for the minute hand to gain 1 degree on the hour hand.
Problems on Faulty Clocks
Faulty clock problems involve clocks that do not keep accurate time. These clocks either gain time (run faster than a correct clock) or lose time (run slower than a correct clock) at a constant rate. Problems require calculating the total error accumulated over a period or determining the correct time when the faulty clock shows a certain time, or vice versa.
Calculating the Total Error:
If a clock gains or loses time at a constant rate, we can calculate the total time gained or lost over any given period.
Suppose a faulty clock gains $x$ minutes in $T$ hours of correct time.
The rate of gain is $\frac{x}{T}$ minutes per hour of correct time.
The total time gained after $T_{\text{total}}$ hours of correct time is given by:
$\boldsymbol{\text{Total Gain} = \left(\frac{x \text{ minutes gain}}{T \text{ hours}}\right) \times T_{\text{total}} \text{ hours}}$
... (iv)
If the clock loses $y$ minutes in $T$ hours of correct time, the rate of loss is $\frac{y}{T}$ minutes per hour. The total time lost after $T_{\text{total}}$ hours of correct time is:
$\boldsymbol{\text{Total Loss} = \left(\frac{y \text{ minutes loss}}{T \text{ hours}}\right) \times T_{\text{total}} \text{ hours}}$
... (v)
Alternatively, we can think in terms of ratios of time elapsed:
- If a clock gains $x$ minutes in $T$ hours of correct time, then when $T$ hours have passed on a correct clock, the faulty clock shows $T$ hours and $x$ minutes. In hours, this is $T + \frac{x}{60}$ hours. The ratio of time on Faulty Clock : Time on Correct Clock $= \left(T + \frac{x}{60}\right) : T$. This ratio is constant.
- If a clock loses $y$ minutes in $T$ hours of correct time, then when $T$ hours have passed on a correct clock, the faulty clock shows $T$ hours minus $y$ minutes. In hours, this is $T - \frac{y}{60}$ hours. The ratio of Time on Faulty Clock : Time on Correct Clock $= \left(T - \frac{y}{60}\right) : T$. This ratio is constant.
Finding the Correct Time or Faulty Clock Time:
If a faulty clock was set to the correct time at a specific moment, the time it shows at a later point can be calculated by adding the total gain or subtracting the total loss from the correct time that should have elapsed.
- If the clock gains time:
$\text{Time shown by Faulty Clock} = \text{Correct Time elapsed} + \text{Total Gain}$
$\text{Correct Time elapsed} = \text{Time shown by Faulty Clock} - \text{Total Gain}$
- If the clock loses time:
$\text{Time shown by Faulty Clock} = \text{Correct Time elapsed} - \text{Total Loss}$
$\text{Correct Time elapsed} = \text{Time shown by Faulty Clock} + \text{Total Loss}$
Alternatively, use the constant ratio of time elapsed. Let $T_C$ be the duration elapsed on a correct clock and $T_F$ be the duration elapsed on the faulty clock since they were set right.
- If clock gains $x$ mins in $T$ hours (correct time): $\frac{T_F}{T_C} = \frac{T + x/60}{T}$. So, $T_C = T_F \times \frac{T}{T + x/60}$.
- If clock loses $y$ mins in $T$ hours (correct time): $\frac{T_F}{T_C} = \frac{T - y/60}{T}$. So, $T_C = T_F \times \frac{T}{T - y/60}$.
These ratios relate the durations elapsed. The final time shown is the starting time plus the elapsed duration.
Example 1. A clock gains 3 minutes in every hour. If the clock was set right at 9 AM, what time will it show at 2 PM on the same day?
Answer:
Given: Clock gains 3 minutes in every hour (of correct time).
The clock was set right at 9 AM.
We need to find the time shown by the faulty clock at 4 PM correct time.
Time elapsed from 9 AM to 2 PM on the same day $= 5$ hours of correct time.
Error rate $= 3$ minutes gain per hour.
Total time gained in 5 hours $= \text{Error rate} \times \text{Time elapsed} = 3 \text{ minutes/hour} \times 5 \text{ hours} = 15$ minutes.
The faulty clock gains 15 minutes in 5 hours.
Correct time at the end of the period is 2 PM.
Since the clock gains time, the time shown by the faulty clock will be ahead of the correct time.
Time shown by Faulty Clock = Correct Time (2 PM) + Total Gain (15 minutes)
Time shown $= 2:00 \text{ PM} + 15 \text{ minutes} = \boldsymbol{2:15 \text{ PM}}$.
The faulty clock will show $\boldsymbol{2:15 \text{ PM}}$.
Example 2. A clock loses 5 minutes in every hour. If the clock was set right at 10 AM on a Tuesday, what will be the correct time when the faulty clock shows 4 PM on the same day?
Answer:
Given: Clock loses 5 minutes in every hour (of correct time).
The clock was set right at 10 AM on Tuesday.
The faulty clock shows 4 PM on the same day.
Time elapsed on the faulty clock from 10 AM to 4 PM $= 6$ hours.
Let $T_C$ be the correct time elapsed in hours since 10 AM Tuesday. Let $T_F$ be the time elapsed on the faulty clock in hours since 10 AM Tuesday. We know $T_F = 6$ hours.
In 1 hour (60 minutes) of correct time, the faulty clock loses 5 minutes, so it shows 55 minutes (60 - 5) of elapsed time.
The ratio of time on Faulty Clock : Time on Correct Clock is constant: $\frac{T_F}{T_C} = \frac{55 \text{ minutes}}{60 \text{ minutes}} = \frac{55/60 \text{ hours}}{1 \text{ hour}} = \frac{55}{60} = \frac{11}{12}$.
So, $\frac{T_F}{T_C} = \frac{11}{12}$.
We are given $T_F = 6$ hours. We need to find $T_C$.
$\frac{6 \text{ hours}}{T_C} = \frac{11}{12}$
... (vi)
Solve for $T_C$ from equation (vi):
$\boldsymbol{T_C = 6 \times \frac{12}{11} = \frac{72}{11}}$ hours.
Convert the improper fraction to a mixed number:
$\frac{72}{11} = 6\frac{6}{11}$
$\boldsymbol{T_C = 6\frac{6}{11}}$ hours.
This is the total correct time elapsed since 10 AM Tuesday. The correct time is 10 AM Tuesday + $6\frac{6}{11}$ hours.
10 AM + 6 hours = 4 PM.
The extra time is $\frac{6}{11}$ hours. Convert to minutes: $\frac{6}{11} \times 60 = \frac{360}{11}$ minutes.
$\frac{360}{11} = 32\frac{8}{11}$ minutes.
Correct time $= 4:32\frac{8}{11} \text{ PM}$.
When the faulty clock shows 4 PM, the correct time is $\boldsymbol{4:32\frac{8}{11} \text{ PM}}$.
Competitive Exam Notes:
Faulty clock problems are about calculating and applying the error accumulated based on time elapsed.
- Error Rate: Determine how much time is gained/lost per unit of correct time (e.g., minutes per hour).
- Total Error: Total Error = Rate $\times$ Total Correct Time elapsed.
- Gain/Loss affects Time Shown: A clock that gains is ahead (Faulty Time = Correct Time + Gain). A clock that loses is behind (Faulty Time = Correct Time - Loss).
- Ratio Method: Use the ratio of elapsed time on the faulty clock to elapsed time on the correct clock. This ratio is constant. If clock gains $x$ mins in $T$ hrs (correct), ratio $= (T + x/60) / T$. If loses $y$ mins in $T$ hrs (correct), ratio $= (T - y/60) / T$. This is often useful when the time shown by the faulty clock is given.
- "Set Right": This is the starting point for calculating accumulated error.
- When Hands Meet on Faulty Clock: This is a variation where you first find the time interval between successive coincidences on the faulty clock itself (which will be slightly more or less than $65\frac{5}{11}$ minutes), then use this interval to find specific times.
Solving Clock Problems
Solving clock problems requires the application of the fundamental concepts of the angular speeds of the hour and minute hands, their relative speed, and the calculation of angles between them at specific times or finding times for specific angles. Consistency in units (degrees and minutes or hours) is essential.
Example 1. Find the angle between the hour hand and the minute hand at 5:10.
Answer:
Given time: H $= 5$, M $= 10$.
Using the formula for the angle between the hands: Angle $= |30H - 5.5M|^\circ$.
Angle $= |30 \times 5 - 5.5 \times 10|^\circ$
Calculate the values:
$30 \times 5 = 150$
$5.5 \times 10 = 55$
Angle $= |150 - 55|^\circ = |95|^\circ = 95^\circ$.
The angle between the hands at 5:10 is $\boldsymbol{95^\circ}$.
Example 2. At what time between 7 and 8 o'clock will the hands of a clock be at right angles?
Answer:
We need to find the times between 7:00 and 8:00 when the angle is $\theta = 90^\circ$.
The starting hour for the interval is 7, so H $= 7$.
Using the formula for M minutes past H o'clock for angle $\theta$: M $= \frac{2}{11} (30H \pm \theta)$ minutes.
M $= \frac{2}{11} (30 \times 7 \pm 90)$ minutes past 7.
M $= \frac{2}{11} (210 \pm 90)$ minutes past 7.
We get two possible times:
Possibility 1 (using +90):
M $= \frac{2}{11} (210 + 90) = \frac{2}{11} (300) = \frac{600}{11}$ minutes.
Convert to mixed number: $\frac{600}{11} = 54\frac{6}{11}$ minutes.
This time ($54\frac{6}{11}$ minutes past 7) is between 7:00 and 8:00.
$\boldsymbol{\text{Time}_1 = 54\frac{6}{11} \text{ minutes past 7}}$.
Possibility 2 (using -90):
M $= \frac{2}{11} (210 - 90) = \frac{2}{11} (120) = \frac{240}{11}$ minutes.
Convert to mixed number: $\frac{240}{11} = 21\frac{9}{11}$ minutes.
This time ($21\frac{9}{11}$ minutes past 7) is also between 7:00 and 8:00.
$\boldsymbol{\text{Time}_2 = 21\frac{9}{11} \text{ minutes past 7}}$.
The hands will be at right angles at $\boldsymbol{21\frac{9}{11}}$ minutes past 7 o'clock and $\boldsymbol{54\frac{6}{11}}$ minutes past 7 o'clock.
Example 3. A watch which gains 5 seconds in 3 minutes of correct time was set right at 7 AM. What time will it show at 4 PM on the same day?
Answer:
Given: The clock gains 5 seconds in 3 minutes of correct time.
The clock was set right at 7 AM.
We need to find the time shown by the faulty clock at 4 PM correct time.
Time elapsed from 7 AM to 4 PM on the same day $= 9$ hours of correct time.
Convert the gain rate to a consistent unit, e.g., seconds per minute or minutes per hour or hours per hour.
Gain = 5 seconds in 3 minutes.
Gain per minute $= \frac{5}{3}$ seconds/minute.
Gain per hour $= \frac{5}{3} \text{ seconds/minute} \times 60 \text{ minutes/hour} = 100$ seconds/hour.
100 seconds $= \frac{100}{60} = \frac{5}{3}$ minutes.
So, the clock gains $\frac{5}{3}$ minutes in every hour (of correct time).
Total time elapsed in hours $= 9$ hours.
Total time gained in 9 hours $= \text{Rate of gain} \times \text{Time elapsed}$
Total Gain $= \frac{5}{3} \text{ minutes/hour} \times 9 \text{ hours} = \frac{5}{3} \times 9 = 5 \times 3 = 15$ minutes.
The faulty clock gains 15 minutes in 9 hours.
Correct time at the end of the period is 4 PM.
Since the clock gains time, the time shown by the faulty clock will be ahead of the correct time.
Time shown by Faulty Clock = Correct Time (4 PM) + Total Gain (15 minutes)
Time shown $= 4:00 \text{ PM} + 15 \text{ minutes} = \boldsymbol{4:15 \text{ PM}}$.
The faulty clock will show $\boldsymbol{4:15 \text{ PM}}$.
Competitive Exam Notes:
Solving clock problems involves applying the standard formulas and being careful with calculations and unit conversions. Practice different types of problems to become comfortable.
- Angular Speed: Minute hand $6^\circ$/min, Hour hand $0.5^\circ$/min.
- Relative Speed: $5.5^\circ$/min (minute hand gains on hour hand).
- Angle Calculation: Angle $= |30H - 5.5M|^\circ$. Always give the smaller angle (0-180).
- Time for Specific Angle: M $= \frac{2}{11} (30H \pm \theta)$. Understand when to use + and - (gives two times per hour for $\theta \neq 0, 180$).
- Faulty Clocks:
- Calculate the constant rate of gain or loss per unit of correct time.
- Calculate total gain/loss over the given period of *correct* time.
- Add gain or subtract loss from the correct time to find the faulty time.
- If the faulty time is given, use the ratio of faulty time elapsed to correct time elapsed to find the correct time elapsed.
- Unit Consistency: Be consistent with degrees for angles and minutes for time, or convert everything to a single system.