Chapter 2 Numbers Applications (Concepts)

Averages

In statistics and mathematics, an Average is a single value that is chosen to represent the central position of a set of numbers. It provides a summary of the entire dataset and helps in understanding the typical value within the set. Averages are also known as Measures of Central Tendency. While there are several types of averages, such as the Median, Mode, Geometric Mean, and Harmonic Mean, the most commonly used and fundamental average is the Arithmetic Mean. In everyday language, when we refer to the "average", we are typically talking about the Arithmetic Mean.

Arithmetic Mean (AM)

The Arithmetic Mean is calculated by adding up all the values in a dataset and then dividing the sum by the number of values in the dataset. It is the most straightforward type of average to compute and understand.

Arithmetic Mean for Ungrouped Data

For a set of $n$ individual observations (or data points) denoted by $x_1, x_2, x_3, ..., x_n$, the Arithmetic Mean, symbolized by $\overline{x}$ (pronounced "x-bar"), is defined by the following formula:

Expressing this formula using mathematical notation, if we have $n$ observations $x_1, x_2, ..., x_n$:

Using summation notation ($\sum$), which is a compact way to represent the sum of a series of values, the formula can be written even more concisely as:

Here, $\sum\limits_{i=1}^{n} x_i$ means the sum of all the $x_i$ values, starting from the first observation ($i=1$) up to the $n$-th observation ($i=n$).

Derivation of the Formula:

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Properties of Arithmetic Mean

The Arithmetic Mean has several important properties that are useful in understanding its behaviour and for solving related problems:

• Sum of Deviations is Zero: The sum of the differences between each observation and the arithmetic mean is always zero.

  $$\sum\limits_{i=1}^{n} (x_i - \overline{x}) = 0$$

  ... (4)

  Derivation:

  We know that $\overline{x} = \frac{\sum\limits_{i=1}^{n} x_i}{n}$.

  Multiplying both sides by $n$, we get $n\overline{x} = \sum\limits_{i=1}^{n} x_i$.

  Consider the sum of deviations:

  $$\sum\limits_{i=1}^{n} (x_i - \overline{x}) = (x_1 - \overline{x}) + (x_2 - \overline{x}) + ... + (x_n - \overline{x})$$$$

  Rearrange the terms by grouping $x_i$ together and $\overline{x}$ together:

  $$= (x_1 + x_2 + ... + x_n) - (\underbrace{\overline{x} + \overline{x} + ... + \overline{x}}_{n \text{ times}})$$$$

  $$= \sum\limits_{i=1}^{n} x_i - n\overline{x}$$$$

  Since we established that $\sum\limits_{i=1}^{n} x_i = n\overline{x}$:

  $$= n\overline{x} - n\overline{x}$$$$

  [Substituting $\sum\limits_{i=1}^{n} x_i = n\overline{x}$]

  $$= 0$$

  Thus, the sum of deviations from the mean is zero.

• Effect of Addition/Subtraction of a Constant: If a constant value $k$ is added to or subtracted from each observation in a dataset, the new arithmetic mean will be the original mean plus or minus that constant, respectively.

  If $y_i = x_i + k$, then $\overline{y} = \overline{x} + k$.

  If $y_i = x_i - k$, then $\overline{y} = \overline{x} - k$.

  Derivation for addition:

  Let the original observations be $x_1, x_2, ..., x_n$ with mean $\overline{x} = \frac{\sum x_i}{n}$.

  The new observations are $y_i = x_i + k$, i.e., $y_1 = x_1+k, y_2 = x_2+k, ..., y_n = x_n+k$.

  The new mean $\overline{y}$ is the sum of new observations divided by $n$:

  $$\overline{y} = \frac{\sum\limits_{i=1}^{n} y_i}{n} = \frac{\sum\limits_{i=1}^{n} (x_i + k)}{n}$$

  Expand the summation:

  $$= \frac{(x_1+k) + (x_2+k) + ... + (x_n+k)}{n}$$$$

  Group the $x_i$ terms and the $k$ terms:

  $$= \frac{(x_1 + x_2 + ... + x_n) + (k + k + ... + k \text{ (n times)})}{n}$$$$

  $$= \frac{\sum\limits_{i=1}^{n} x_i + nk}{n}$$$$

  Separate the terms in the numerator:

  $$= \frac{\sum\limits_{i=1}^{n} x_i}{n} + \frac{nk}{n}$$$$

  Recognize the original mean and simplify the second term:

  $$= \overline{x} + k$$

  Thus, $\overline{y} = \overline{x} + k$. The derivation for subtraction is similar.

• Effect of Multiplication/Division by a Constant: If each observation in a dataset is multiplied or divided by a non-zero constant value $k$, the new arithmetic mean will be the original mean multiplied or divided by that constant, respectively.

  If $y_i = k x_i$, then $\overline{y} = k \overline{x}$.

  If $y_i = x_i / k$ ($k \neq 0$), then $\overline{y} = \overline{x} / k$.

  Derivation for multiplication:

  Let the original observations be $x_1, x_2, ..., x_n$ with mean $\overline{x} = \frac{\sum x_i}{n}$.

  The new observations are $y_i = kx_i$, i.e., $y_1 = kx_1, y_2 = kx_2, ..., y_n = kx_n$.

  The new mean $\overline{y}$ is the sum of new observations divided by $n$:

  $$\overline{y} = \frac{\sum\limits_{i=1}^{n} y_i}{n} = \frac{\sum\limits_{i=1}^{n} (kx_i)}{n}$$

  The constant factor $k$ can be taken out of the summation:

  $$= \frac{k \sum\limits_{i=1}^{n} x_i}{n}$$

  Recognize the original mean $\overline{x} = \frac{\sum x_i}{n}$:

  $$= k \left(\frac{\sum\limits_{i=1}^{n} x_i}{n}\right)$$$$

  $$= k \overline{x}$$

  Thus, $\overline{y} = k \overline{x}$. The derivation for division is similar.

• Combined Mean: If we have two or more groups of data, the overall mean (combined mean) of the entire dataset can be calculated if we know the mean and the number of observations for each group.

  If $\overline{x}_1$ is the mean of $n_1$ observations and $\overline{x}_2$ is the mean of $n_2$ observations, then the combined mean ($\overline{x}_{12}$) of all $(n_1 + n_2)$ observations is given by:

  $$\overline{x}_{12} = \frac{n_1 \overline{x}_1 + n_2 \overline{x}_2}{n_1 + n_2}$$

  ... (5)

  This formula can be extended to combine more than two groups.

  Derivation:

  Let the sum of observations in the first group be $S_1$. Since $\overline{x}_1 = \frac{S_1}{n_1}$, the sum is $S_1 = n_1 \overline{x}_1$.

  Let the sum of observations in the second group be $S_2$. Since $\overline{x}_2 = \frac{S_2}{n_2}$, the sum is $S_2 = n_2 \overline{x}_2$.

  When the two groups are combined, the total number of observations is $N = n_1 + n_2$.

  The total sum of all observations is $S_{total} = S_1 + S_2 = n_1 \overline{x}_1 + n_2 \overline{x}_2$.

  The combined mean $\overline{x}_{12}$ is the total sum divided by the total number of observations:

  $$\overline{x}_{12} = \frac{S_{total}}{N} = \frac{n_1 \overline{x}_1 + n_2 \overline{x}_2}{n_1 + n_2}$$

  This proves the formula for the combined mean of two groups.

• Arithmetic Mean is Affected by Outliers: The mean is sensitive to extreme values (observations that are much larger or much smaller than the rest of the data). A single outlier can significantly shift the value of the mean.

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Arithmetic Mean for Grouped Data (Brief Mention)

When data is presented in a frequency distribution (i.e., grouped data), calculating the arithmetic mean requires considering the frequency of each observation or class interval.

• Discrete Frequency Distribution: If we have distinct values $x_1, x_2, ..., x_k$ occurring with corresponding frequencies $f_1, f_2, ..., f_k$, the total number of observations is the sum of frequencies, $N = \sum\limits_{i=1}^{k} f_i$. The sum of all observations is $\sum\limits_{i=1}^{k} f_i x_i$ (each value $x_i$ is multiplied by how many times it appears, $f_i$). The mean is calculated as:

  $$\overline{x} = \frac{f_1 x_1 + f_2 x_2 + ... + f_k x_k}{f_1 + f_2 + ... + f_k} = \frac{\sum\limits_{i=1}^{k} f_i x_i}{\sum\limits_{i=1}^{k} f_i}$$

  ... (6)

• Continuous Frequency Distribution: When data is presented in class intervals (e.g., 0-10, 10-20), we cannot use the individual observations. Instead, we assume that the midpoint of each class interval represents all the observations within that interval. We calculate the midpoint (class mark) of each interval, denoted by $x_i$, and use the frequency of that interval, $f_i$. The formula then becomes the same as for discrete frequency distribution:

  $$\overline{x} = \frac{\sum\limits_{i=1}^{k} f_i x_i}{\sum\limits_{i=1}^{k} f_i}$$

  ... (7)

  where $x_i$ here is the midpoint of the $i$-th class interval, and $f_i$ is its frequency.

Other Measures of Central Tendency (Brief Introduction)

While the Arithmetic Mean is widely used, it's important to be aware of other measures of central tendency, each with its own strengths and uses:

• Median: The median is the middle value in a dataset that is ordered from least to greatest. If there is an odd number of observations, the median is the single middle value. If there is an even number of observations, the median is typically the average of the two middle values. The median is less affected by extreme values than the arithmetic mean, making it a better measure of central tendency for skewed data.

• Mode: The mode is the value that appears most frequently in a dataset. A dataset can have one mode (unimodal), more than one mode (multimodal), or no mode at all. The mode is the only measure of central tendency that can be used for categorical data (data that can be sorted into categories but not ordered numerically, e.g., favourite colour).

• Geometric Mean (GM): Used for data involving growth rates, percentages, or factors. For $n$ positive observations $x_1, x_2, ..., x_n$, the Geometric Mean is the $n$-th root of their product:

  $$\text{GM} = \sqrt[n]{x_1 \times x_2 \times ... \times x_n} = (x_1 x_2 ... x_n)^{1/n}$$

  It is always less than or equal to the Arithmetic Mean for positive data.

• Harmonic Mean (HM): Used particularly for averaging rates or ratios when the numerator is constant (e.g., average speed over a fixed distance). For $n$ positive observations $x_1, x_2, ..., x_n$, the Harmonic Mean is the reciprocal of the arithmetic mean of the reciprocals of the observations:

  $$\text{HM} = \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + ... + \frac{1}{x_n}} = \frac{n}{\sum\limits_{i=1}^{n} \frac{1}{x_i}}$$$$

  For positive data, the relationship between these three means is $HM \leq GM \leq AM$.

For most basic problems encountered in this chapter, when asked for the "average", the Arithmetic Mean is implied unless otherwise specified.

Clock

Clock problems are a common type of mathematical reasoning question that involves understanding the movement of the hands of a clock and calculating angles or times based on their relative positions. A standard analog clock face is a circle, representing 12 hours, which is equivalent to 360 degrees.

Understanding the Speed of Clock Hands

In a standard 12-hour analog clock, there are typically two hands we consider: the minute hand and the hour hand. Sometimes a second hand is also present, but it moves much faster and is less commonly involved in standard angle/time problems.

Speed of the Minute Hand

The minute hand completes one full rotation (360 degrees) in 60 minutes. Therefore, its speed can be calculated as:

This means for every minute that passes, the minute hand moves $6^\circ$ along the clock face.

Speed of the Hour Hand

The hour hand completes one full rotation (360 degrees) in 12 hours. This means it moves from the 12 back to the 12 in 12 hours. First, let's find its speed per hour:

Now, let's convert this to speed per minute. Since 1 hour = 60 minutes:

So, the hour hand moves $0.5^\circ$ along the clock face every minute.

Relative Speed of the Minute Hand with Respect to the Hour Hand

Since the minute hand moves faster than the hour hand, the difference in their speeds determines how quickly the minute hand gains on the hour hand. This difference is called their relative speed.

This relative speed of $5.5^\circ$ per minute is crucial because it tells us how many degrees the minute hand gains on the hour hand every minute. To gain $360^\circ$ (i.e., to complete one full relative rotation), the minute hand takes $\frac{360^\circ}{5.5^\circ/\text{minute}} = \frac{360}{11/2} = \frac{720}{11}$ minutes. This duration, $\frac{720}{11} \text{ minutes} = 65 \frac{5}{11} \text{ minutes}$, is the average time interval between the minute hand overtaking the hour hand.

Calculating the Angle Between the Hands

To find the angle between the hour hand and the minute hand at a given time, say H hours and M minutes (where H is between 1 and 12, and M is between 0 and 59), we calculate the angle each hand has moved clockwise from the 12 o'clock position.

• Angle covered by the Minute Hand: The minute hand moves $6^\circ$ per minute. So, in M minutes, it moves:

  $$\text{Angle}_{Minute} = M \times 6^\circ$$

• Angle covered by the Hour Hand: The hour hand moves $30^\circ$ per hour. In H hours, it moves $H \times 30^\circ$. Additionally, in M minutes, it moves $\frac{M}{60}$ of an hour's movement. So, in M minutes, the hour hand moves an extra $\frac{M}{60} \times 30^\circ = \frac{M}{2}^\circ$.

  The total angle covered by the hour hand from the 12 o'clock position is the angle for H complete hours plus the angle for M minutes:

  $$\text{Angle}_{Hour} = (H \times 30^\circ) + (\frac{M}{60} \times 30^\circ)$$

  $$= 30H^\circ + \frac{M}{2}^\circ$$

The angle between the hands is the absolute difference between these two angles. Since there are two angles between the hands (the interior angle and the exterior angle), we usually refer to the smaller angle, which is $\leq 180^\circ$.

The formula for the angle $\theta$ between the hands at H hours and M minutes is $\theta = |30H - \frac{11}{2}M|$. If the calculated angle is greater than $180^\circ$, the smaller angle is $360^\circ$ minus the calculated angle.

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Relative Positions of Hands and Time

Clock problems often ask for the time when the hands are in a specific relative position (e.g., coinciding, opposite, perpendicular). These problems can be solved by understanding the relative speed of the minute hand with respect to the hour hand ($5.5^\circ$ per minute).

Consider the relative distance (angle) the minute hand needs to cover or gain on the hour hand to reach a specific position after H o'clock but before (H+1) o'clock. At H o'clock, the hour hand is at $30H^\circ$ from 12, and the minute hand is at $0^\circ$ (at 12). The initial angle between them is $30H^\circ$. For the minute hand to be at a certain angle $\theta$ relative to the hour hand at M minutes past H o'clock, it needs to gain a specific angle over the initial $30H^\circ$ gap.

Alternatively, we can use the general formula $\theta = |30H - \frac{11}{2}M|$. We know $\theta$ (the desired angle) and $H$ (the hour), and we solve for $M$ (the number of minutes). The equation becomes $\pm \theta = 30H - \frac{11}{2}M$. Solving for $M$: $\frac{11}{2}M = 30H \mp \theta$, so $M = \frac{2}{11} (30H \mp \theta) = \frac{60H \mp 2\theta}{11}$. Since the angle formula uses absolute value, we consider both $30H - \frac{11}{2}M = \theta$ and $30H - \frac{11}{2}M = -\theta$, leading to $\frac{11}{2}M = 30H - \theta$ and $\frac{11}{2}M = 30H + \theta$. Thus, $M = \frac{60H - 2\theta}{11}$ or $M = \frac{60H + 2\theta}{11}$.

We need to choose the $\pm$ sign such that M is between 0 and 60. The formula might give two valid times (e.g., for perpendicular hands) or require adjusting H (e.g., for times near 12). For intervals between H and H+1, we use the value of H.

1. Hands Coinciding (Angle $\theta = 0^\circ$)

The hands coincide when the angle between them is $0^\circ$. Using formula (5) with $\theta = 0$: $M = \frac{60H \pm 2(0)}{11} = \frac{60H}{11}$.

The hands coincide approximately every $65 \frac{5}{11}$ minutes. In a 12-hour period, they coincide 11 times (at 12:00, and once between each consecutive hour pair from 1-2 up to 10-11, and between 11-12). They coincide at 12:00. Between 11 and 1, they coincide only at 12:00.

In 12 hours, they coincide 11 times.

In 24 hours, they coincide 22 times.

The times they coincide are calculated using $M = \frac{60H}{11}$ minutes past H o'clock for $H=1, 2, ..., 11$. For $H=12$ (or 0), it is 12:00.

2. Hands Opposite (Angle $\theta = 180^\circ$)

The hands are opposite when the angle between them is $180^\circ$. Using formula (5) with $\theta = 180$: $M = \frac{60H \pm 2(180)}{11} = \frac{60H \pm 360}{11} = \frac{60(H \pm 6)}{11}$.

The hands are opposite approximately every $65 \frac{5}{11}$ minutes. In a 12-hour period, they are opposite 11 times (at 6:00, and once between each consecutive hour pair). They are opposite at 6:00. Between 5 and 7, they are opposite only at 6:00.

In 12 hours, they are opposite 11 times.

In 24 hours, they are opposite 22 times.

The times they are opposite are calculated using $M = \frac{60(H \pm 6)}{11}$ minutes past H o'clock. For $H=1, 2, 3, 4, 5$, use $M = \frac{60(H+6)}{11}$. For $H=7, 8, 9, 10, 11$, use $M = \frac{60(H-6)}{11}$. For $H=6$, it is 6:00. For $H=12$ (or 0), use $M = \frac{60(0+6)}{11} = \frac{360}{11} = 32 \frac{8}{11}$ minutes past 12, i.e., approx 12:32.7, but the primary opposite time around 12 is 6:00.

3. Hands Perpendicular (Angle $\theta = 90^\circ$)

The hands are perpendicular when the angle between them is $90^\circ$. Using formula (5) with $\theta = 90$: $M = \frac{60H \pm 2(90)}{11} = \frac{60H \pm 180}{11} = \frac{60(H \pm 3)}{11}$.

The hands are perpendicular twice every hour, approximately $32 \frac{8}{11}$ minutes apart. In a 12-hour period, they are perpendicular 22 times. This is because the perpendicular positions around 3 and 9 are shared between two hour intervals (e.g., 3:00 is the end of 2-3 interval and start of 3-4 interval, 9:00 is similar). Specifically, they are perpendicular 3 times between 2 and 4 o'clock, and 3 times between 8 and 10 o'clock, instead of 4 times in each 2-hour block.

In 12 hours, they are perpendicular 22 times.

In 24 hours, they are perpendicular 44 times.

The two times they are perpendicular between H and H+1 o'clock are calculated using $M = \frac{60(H + 3)}{11}$ and $M = \frac{60(H - 3)}{11}$ minutes past H o'clock. We might need to adjust H by 12 if $H-3$ is negative (e.g., for H=1 or H=2).

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Calendar

Calendar problems are common in competitive exams and general aptitude tests. They involve calculating the day of the week for a specific date, determining the number of leap years within a period, or understanding the cyclical nature of days and dates. The core concept used to solve these problems is that of Odd Days.

Odd Days

In the context of calendars, an "odd day" refers to the number of days that remain after forming complete weeks from a given total number of days. Since a week consists of 7 days, any number of days can be expressed as a certain number of full weeks plus a remainder. This remainder is the number of odd days.

To find the number of odd days for a given number of days, we divide the total number of days by 7 and note down the remainder.

For example, if we have 10 days, dividing by 7 gives a quotient of 1 and a remainder of 3 ($10 = 1 \times 7 + 3$). So, 10 days consist of 1 complete week and 3 extra days. The number of odd days is 3.

Similarly, 20 days $= 20 \div 7 = 2$ remainder 6. So, 20 days have 6 odd days.

49 days $= 49 \div 7 = 7$ remainder 0. So, 49 days have 0 odd days.

Odd Days in Months

The number of odd days in a specific month depends on the total number of days in that month.

Understanding the number of days in each month is essential for calculating odd days over a period of months.

Leap Year and Ordinary Year

The distinction between an ordinary year and a leap year is crucial in calendar calculations because it affects the number of days in February and, consequently, the total number of days in the year.

• Ordinary Year: An ordinary year has 365 days. It consists of 52 full weeks and some extra days.

  $$365 \text{ days} = 52 \text{ weeks} + 1 \text{ day}$$

  [Since $365 = 52 \times 7 + 1$]

  Therefore, the number of odd days in an ordinary year is 1.

• Leap Year: A leap year has 366 days. The extra day is added to February, making it 29 days long in a leap year. It also consists of 52 full weeks and some extra days.

  $$366 \text{ days} = 52 \text{ weeks} + 2 \text{ days}$$

  [Since $366 = 52 \times 7 + 2$]

  Therefore, the number of odd days in a leap year is 2.

Identifying a Leap Year

To determine if a given year is a leap year, follow these rules:

A year is a leap year if:

• It is divisible by 4, AND it is not divisible by 100.
  Examples: 2008, 2012, 2020. These are divisible by 4 but not by 100.
• OR it is divisible by 400. (This rule applies specifically to century years).
  Examples: 1600, 2000, 2400. These are divisible by 400.

Conversely, a year is an ordinary year if:

• It is not divisible by 4 (e.g., 2007, 2011, 2019).
• It is divisible by 100 but not by 400 (e.g., 1700, 1800, 1900, 2100). These are called century years, and they are ordinary years unless they are divisible by 400.

Odd Days in Centuries

We can calculate the number of odd days accumulated over a period of centuries. This calculation relies on the number of ordinary and leap years within that century period.

• Odd days in 100 years: In the first 100 years (0001 to 0100), there are 24 leap years (since 100 is divisible by 100 but not by 400, it is an ordinary year) and $100 - 24 = 76$ ordinary years.

  Total odd days in 100 years $= (76 \times 1 \text{ odd day/ordinary year}) + (24 \times 2 \text{ odd days/leap year})$$$$

  $$= 76 + 48 = 124 \text{ days}$$

  ... (a)

  To find the number of odd days in 100 years, divide the total days by 7:

  $$124 \div 7 = 17 \text{ with remainder } 5$$

  So, the number of odd days in 100 years is 5.

• Odd days in 200 years: This is the odd days in 2 sets of 100 years.

  Total odd days in 200 years $= 2 \times (\text{Odd days in 100 years})$$$$

  $$= 2 \times 5 = 10 \text{ days}$$$$

  ... (b)

  Number of odd days in 200 years $= 10 \div 7 = 1$ with remainder 3.

  So, the number of odd days in 200 years is 3.

• Odd days in 300 years: This is the odd days in 3 sets of 100 years.

  Total odd days in 300 years $= 3 \times (\text{Odd days in 100 years})$$$$

  $$= 3 \times 5 = 15 \text{ days}$$$$

  ... (c)

  Number of odd days in 300 years $= 15 \div 7 = 2$ with remainder 1.

  So, the number of odd days in 300 years is 1.

• Odd days in 400 years: This is the odd days in 4 sets of 100 years, PLUS the extra day from the leap year 400 itself (since 400 is divisible by 400, it is a leap year, unlike 100, 200, 300).

  Total odd days in 400 years $= 4 \times (\text{Odd days in 100 years}) + 1 \text{ (for leap year 400)}$$$$

  $$= (4 \times 5) + 1 = 20 + 1 = 21 \text{ days}$$$$

  ... (d)

  Number of odd days in 400 years $= 21 \div 7 = 3$ with remainder 0.

  So, the number of odd days in 400 years is 0.

This establishes a cycle of odd days for centuries: 100 years have 5 odd days, 200 years have 3, 300 years have 1, and 400 years have 0. This pattern (5, 3, 1, 0) repeats for every subsequent block of 400 years (e.g., 500 years have the same odd days as 100 years, 800 years have 0 odd days, etc.).

Day of the Week based on Total Odd Days

We use a reference point, usually considering Sunday as the 0th day. The total number of odd days calculated up to a certain date corresponds to the day of the week as follows:

To find the day for a specific date, we calculate the total odd days up to the day *before* that date, starting from a historical reference (like 01/01/0001, which is considered Day 1). The remainder when the total odd days is divided by 7 gives the position of the day of the week (0 for Sunday, 1 for Monday, etc.).

Calculating Day of the Week for a Given Date

To determine the day of the week for a particular date (Day/Month/Year), we need to calculate the total number of odd days from the beginning of the calendar era (often taken as the end of year 0000, so we count days from 01/01/0001) up to the day immediately preceding the target date. The calculation typically involves breaking down the time period into completed centuries, completed years within the current century, completed months within the current year, and the specified number of days in the current month.

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Time, Work and Distance

This section of Applied Mathematics focuses on problems that involve the interrelationships between quantities like time, the amount of work done, the number of people or machines involved, the speed of movement, and the distance covered. These problems are common in various applications and often test logical reasoning and proportional understanding. We will deal with two main types of problems: Time and Work, and Time, Speed, and Distance.

Time and Work

Time and Work problems typically involve determining how long it takes to complete a task, given the rates at which individuals or groups can perform the work. The central idea is the Work Rate, which is the amount of work done per unit of time (e.g., per day, per hour).

If a person or entity can complete a certain amount of work (let's call the total work 'W') in $T$ units of time, then their work rate ($R$) is given by:

In most basic problems, the "Total Work" is considered as one complete unit, i.e., Total Work = 1. So, if a person can complete the entire work (1 unit) in $T$ days, their daily work rate is $\frac{1}{T}$ of the work per day.

The relationship can also be expressed as:

From this, we can also derive:

Formulas and Concepts for Individual and Combined Work

• If Person A can complete a specific piece of work in $d_A$ days, it means A's daily work rate is $\frac{1}{d_A}$ of the total work per day.

• If Person B can complete the same work in $d_B$ days, B's daily work rate is $\frac{1}{d_B}$ of the total work per day.

• If A and B work together on the same task, their individual work rates add up to give their combined work rate (assuming they work efficiently together without interference).
  Combined daily work rate of A and B $= \text{A's daily rate} + \text{B's daily rate}$

  $$\text{Combined Rate} = \frac{1}{d_A} + \frac{1}{d_B}$$$$

  Combining the fractions:

  $$= \frac{d_B}{d_A d_B} + \frac{d_A}{d_A d_B} = \frac{d_A + d_B}{d_A d_B}$$$$

  ... (11)

• The time taken by A and B together to complete the work is the reciprocal of their combined work rate (since Work Done = 1):

  $$\text{Time Together} = \frac{1}{\text{Combined Rate}} = \frac{1}{\left(\frac{d_A + d_B}{d_A d_B}\right)}$$$$

  $$= \frac{d_A d_B}{d_A + d_B} \text{ days}$$

  ... (12)

• If multiple persons work together, their combined rate is the sum of their individual rates.

  If persons $P_1, P_2, ..., P_k$ can complete a work in $d_1, d_2, ..., d_k$ days respectively, their combined rate is $\frac{1}{d_1} + \frac{1}{d_2} + ... + \frac{1}{d_k}$. The time taken together is $\frac{1}{\sum \frac{1}{d_i}}$.

• If Person A works for $t_A$ days and Person B works for $t_B$ days on the same work (not necessarily together for the whole time), the total work done is the sum of the work done by each person individually:

  $$\text{Total Work Done} = (\text{A's Rate} \times t_A) + (\text{B's Rate} \times t_B)$$$$

  $$= \left(\frac{1}{d_A} \times t_A\right) + \left(\frac{1}{d_B} \times t_B\right) = \frac{t_A}{d_A} + \frac{t_B}{d_B}$$$$

  If they complete the entire work together (Total Work Done = 1), then:

  $$\frac{t_A}{d_A} + \frac{t_B}{d_B} = 1$$

  ... (13)

Concept of Efficiency

Efficiency is a measure of how much work a person or machine can do per unit of time. It is directly proportional to the work rate and inversely proportional to the time taken to complete a fixed amount of work.

If the ratio of efficiencies of two persons A and B is $E_A : E_B$, then the ratio of times taken by them to complete the same work is inversely proportional, i.e., $T_A : T_B = \frac{1}{E_A} : \frac{1}{E_B}$. If $E_A : E_B = a:b$, then $T_A : T_B = b:a$.

Man-Days Concept

The concept of Man-Days (or Person-Days, or Unit-Days in general) is based on the idea that the total amount of work required to complete a task is constant. If $M$ men work for $D$ days to complete a work, the total work done is equivalent to $M \times D$ man-days. This product represents the total effort units required for the work.

If the same work is completed by a different number of men ($M_2$) in $D_2$ days, then the total work must be the same:

This relationship assumes that all men work at the same rate and for the same number of hours per day.

If the workers work for different hours per day ($H$) or complete different amounts of work ($W$), the formula can be extended:

Here, $W$ represents the amount of work done (e.g., constructing 10 km of road, digging a 5-meter deep well). If the work is the same in both cases, then $W_1 = W_2$, and the formula simplifies to $M_1 D_1 H_1 = M_2 D_2 H_2$. If the number of hours is constant, it simplifies further to $M_1 D_1 = M_2 D_2$.

Derivation of $\frac{M_1 D_1 H_1}{W_1} = \frac{M_2 D_2 H_2}{W_2}$:

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Time, Speed and Distance

Time, Speed, and Distance problems are based on a fundamental relationship between these three quantities when an object moves at a constant speed. This relationship is one of the most important concepts in physics and is widely used in various applications.

The relationship states that Speed is defined as the distance covered per unit of time.

From this basic formula, we can derive the other two related formulas:

Derivation of Distance and Time Formulas:

Consistency of Units

It is crucial that the units used for Distance, Speed, and Time are consistent. Common sets of consistent units are:

• Speed in Kilometres per Hour (km/hr), Distance in Kilometres (km), Time in Hours (hr).
• Speed in Metres per Second (m/s), Distance in Metres (m), Time in Seconds (s).

If the units are mixed (e.g., speed in km/hr, distance in metres), you must convert one or more quantities so that the units become consistent before applying the formulas.

Unit Conversions for Speed

Converting speed between km/hr and m/s is frequently required.

• Conversion from km/hr to m/s: To convert speed from km/hr to m/s, we multiply by the factor $\frac{5}{18}$.

  $$\text{Speed in m/s} = \text{Speed in km/hr} \times \frac{5}{18}$$

  Derivation:

  We know that 1 kilometre (km) = 1000 metres (m).

  And 1 hour (hr) = 60 minutes = 60 $\times$ 60 seconds = 3600 seconds (s).

  So, 1 km/hr means 1 kilometre covered in 1 hour.

  $$1 \text{ km/hr} = \frac{1 \text{ km}}{1 \text{ hr}}$$$$

  Substitute the equivalent values in metres and seconds:

  $$1 \text{ km/hr} = \frac{1000 \text{ m}}{3600 \text{ s}}$$$$

  Simplify the fraction:

  $$= \frac{\cancel{1000}^{10}}{\cancel{3600}_{36}} \text{ m/s} = \frac{10}{36} \text{ m/s}$$$$

  Further simplification:

  $$= \frac{\cancel{10}^{5}}{\cancel{36}_{18}} \text{ m/s} = \frac{5}{18} \text{ m/s}$$$$

  Thus, to convert speed from km/hr to m/s, we multiply by $\frac{5}{18}$.

• Conversion from m/s to km/hr: To convert speed from m/s to km/hr, we multiply by the reciprocal factor $\frac{18}{5}$.

  $$\text{Speed in km/hr} = \text{Speed in m/s} \times \frac{18}{5}$$

  Derivation:

  We know from the previous derivation that $1 \text{ km/hr} = \frac{5}{18} \text{ m/s}$.

  To find out how many km/hr are in 1 m/s, we can rearrange this equation. Divide both sides by $\frac{5}{18}$:

  $$\frac{1 \text{ km/hr}}{5/18} = \frac{5/18 \text{ m/s}}{5/18}$$$$

  $$1 \text{ m/s} = \frac{1}{5/18} \text{ km/hr} = \frac{18}{5} \text{ km/hr}$$$$

  Thus, to convert speed from m/s to km/hr, we multiply by $\frac{18}{5}$.

Relative Speed

When two objects are moving, their speed relative to each other is important, especially in problems where they are moving towards or away from each other, or one is overtaking the other.

• Objects moving in the Same Direction: If two objects are moving in the same direction with speeds $S_1$ and $S_2$, and $S_1 > S_2$, the relative speed of the faster object with respect to the slower object is the difference of their speeds.

  $$\text{Relative Speed} = S_1 - S_2$$

  This is the speed at which the faster object gains on the slower object.

• Objects moving in Opposite Directions: If two objects are moving towards each other or away from each other in opposite directions with speeds $S_1$ and $S_2$, their relative speed is the sum of their speeds.

  $$\text{Relative Speed} = S_1 + S_2$$

  This is the speed at which the distance between them decreases (when moving towards each other) or increases (when moving away from each other).

Problems on Trains

Train problems are a specific type of Time, Speed, and Distance problem where the length of the train is usually significant and must be accounted for in the distance calculation. When a train of length $L_T$ passes an object, the distance covered is the length the train's front end travels until the train's rear end clears the object.

• Passing a Point Object (e.g., a pole, a single person standing, a signal post): A point object has negligible length compared to the train. For the train to pass the object, its front end must reach the object, and its rear end must move past the object. The total distance covered by the train is equal to its own length $L_T$.

  $$\text{Time Taken} = \frac{\text{Distance}}{\text{Speed of train}} = \frac{L_T}{S_T}$$

• Passing a Stationary Object with Length (e.g., a platform, a bridge, a tunnel): Let the length of the stationary object be $L_O$. For the train to pass this object, the front end of the train must enter the object, and the rear end of the train must exit the object. The total distance covered by the train is the length of the object plus the length of the train itself, i.e., $L_O + L_T$.

  $$\text{Time Taken} = \frac{\text{Total Distance}}{\text{Speed of train}} = \frac{L_T + L_O}{S_T}$$

• Passing a Moving Object (e.g., another train, a person running): When a train passes a moving object, we need to consider the relative speed and the total distance. The total distance is usually the sum of the lengths of the train and the object if the object has significant length (like another train). If the object is a person running, their length is negligible.

  • Passing another train of length $L_2$ moving in the Same Direction with speed $S_2$: Let the first train have length $L_1$ and speed $S_1$. Assume $S_1 > S_2$ for the first train to overtake the second. The relative speed of the first train with respect to the second is $S_1 - S_2$. The total distance that needs to be covered for the first train to completely pass the second is the sum of their lengths, $L_1 + L_2$.

    $$\text{Time Taken} = \frac{\text{Sum of Lengths}}{\text{Relative Speed (Same Direction)}} = \frac{L_1 + L_2}{S_1 - S_2}$$

  • Passing another train of length $L_2$ moving in the Opposite Direction with speed $S_2$: The relative speed of the two trains is the sum of their speeds, $S_1 + S_2$. The total distance is again the sum of their lengths, $L_1 + L_2$.

    $$\text{Time Taken} = \frac{\text{Sum of Lengths}}{\text{Relative Speed (Opposite Direction)}} = \frac{L_1 + L_2}{S_1 + S_2}$$

  • Passing a Person running in the Same Direction as the train (speed $S_P < S_T$): The person's length is negligible. The distance covered by the train relative to the person is the length of the train, $L_T$. The relative speed is $S_T - S_P$.

    $$\text{Time Taken} = \frac{L_T}{S_T - S_P}$$

  • Passing a Person running in the Opposite Direction to the train (speed $S_P$): The distance is $L_T$. The relative speed is $S_T + S_P$.

    $$\text{Time Taken} = \frac{L_T}{S_T + S_P}$$

Problems on Boats and Streams

These problems involve the movement of a boat or swimmer in water, where the water itself might be moving (a stream or current). The speed of the water affects the effective speed of the boat.

• Let $S_b$ be the speed of the boat (or swimmer) in still water (water that is not moving).

• Let $S_s$ be the speed of the stream (or current).

• Speed Downstream: When the boat moves in the same direction as the stream, the speed of the stream adds to the boat's speed in still water. This combined speed is the effective speed downstream.

  $$\text{Speed Downstream } (S_d) = S_b + S_s$$

  ... (20)

• Speed Upstream: When the boat moves against the direction of the stream, the speed of the stream opposes the boat's speed in still water. The effective speed upstream is the difference between the boat's speed and the stream's speed.

  $$\text{Speed Upstream } (S_u) = S_b - S_s$$

  ... (21)

  Note that for upstream movement to be possible, the speed of the boat in still water must be greater than the speed of the stream ($S_b > S_s$).

• If the speed downstream ($S_d$) and speed upstream ($S_u$) are known, we can find the speed of the boat in still water and the speed of the stream by solving equations (20) and (21) simultaneously.

  Adding (20) and (21):

  $S_d + S_u = (S_b + S_s) + (S_b - S_s)$

  [Adding (20) and (21)]

  $S_d + S_u = 2 S_b$

  [Simplifying]

  $$S_b = \frac{S_d + S_u}{2}$$

  ... (22)

  Subtracting (21) from (20):

  $S_d - S_u = (S_b + S_s) - (S_b - S_s)$

  [Subtracting (21) from (20)]

  $S_d - S_u = S_b + S_s - S_b + S_s$

  $S_d - S_u = 2 S_s$

  [Simplifying]

  $$S_s = \frac{S_d - S_u}{2}$$

  ... (23)

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Mensuration

Mensuration is a fundamental branch of mathematics that focuses on the measurement of geometric figures and their properties. It involves calculating quantities such as length, perimeter, area, surface area, and volume of various shapes in two and three dimensions. These concepts have wide-ranging applications in real life, from construction and architecture to engineering and everyday tasks like calculating the amount of paint needed for a wall or the capacity of a container.

Area of 2D Shapes (Plane Figures)

The area of a two-dimensional shape is the amount of surface it covers. It is measured in square units (e.g., square centimetres $\text{cm}^2$, square metres $\text{m}^2$). Here are the formulas for calculating the area of some common plane geometric figures:

Formulas for Area

• Square: A quadrilateral with four equal sides and four right angles. If 'a' is the length of a side:

  $$\text{Area of Square} = \text{side} \times \text{side} = a^2$$

• Rectangle: A quadrilateral with four right angles and opposite sides equal in length. If 'l' is the length and 'b' is the breadth (width):

  $$\text{Area of Rectangle} = \text{length} \times \text{breadth} = l \times b$$

• Circle: The set of all points in a plane that are at a fixed distance (radius) from a central point. If 'r' is the radius:

  $$\text{Area of Circle} = \pi \times (\text{radius})^2 = \pi r^2$$

  ($\pi$ is a mathematical constant approximately equal to 3.14159 or $\frac{22}{7}$).

• Triangle: A polygon with three sides and three vertices. If 'b' is the length of the base and 'h' is the perpendicular height from the opposite vertex to the base:

  $$\text{Area of Triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} b h$$

• Triangle (using Heron's formula): If the lengths of the three sides of a triangle are 'a', 'b', and 'c', and the semi-perimeter ($s$) is half the perimeter ($s = \frac{a+b+c}{2}$), the area can be calculated using Heron's formula:

  $$\text{Area of Triangle} = \sqrt{s(s-a)(s-b)(s-c)}$$$$

  This formula is particularly useful when the height is not known but all three side lengths are.

• Parallelogram: A quadrilateral with two pairs of parallel sides. If 'b' is the length of the base and 'h' is the perpendicular height between the base and the opposite side:

  $$\text{Area of Parallelogram} = \text{base} \times \text{height} = b \times h$$

• Trapezium (or Trapezoid): A quadrilateral with at least one pair of parallel sides. If $a_1$ and $a_2$ are the lengths of the parallel sides and 'h' is the perpendicular height between them:

  $$\text{Area of Trapezium} = \frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{height} = \frac{1}{2} (a_1 + a_2) h$$

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Surface Area and Volume of 3D Shapes (Solid Figures)

Three-dimensional shapes (solids) have volume and surface area. Volume is the amount of space occupied by a 3D object, measured in cubic units (e.g., cubic centimetres $\text{cm}^3$, cubic metres $\text{m}^3$). Surface Area is the total area of all the surfaces (faces) of a 3D object. It is measured in square units, just like the area of a 2D shape. Solid shapes can have total surface area (sum of areas of all surfaces) and sometimes lateral or curved surface area (sum of areas of sides, excluding bases).

Formulas for Surface Area and Volume

• Cube: A six-sided solid figure where all sides are squares of equal size. If 'a' is the length of a side (edge):

  $$\text{Volume} = \text{side} \times \text{side} \times \text{side} = a^3$$

  $$\text{Total Surface Area (TSA)} = 6 \times (\text{Area of one face}) = 6a^2$$

  $$\text{Lateral Surface Area (LSA)} = \text{Area of 4 side faces} = 4a^2$$

• Cuboid (Rectangular Prism): A six-sided solid figure where all faces are rectangles, and opposite faces are equal. If 'l' is length, 'b' is breadth, and 'h' is height:

  $$\text{Volume} = \text{length} \times \text{breadth} \times \text{height} = l \times b \times h$$

  $$\text{Total Surface Area (TSA)} = 2(\text{lb} + \text{bh} + \text{hl})$$

  (Sum of areas of 3 pairs of opposite faces: top/bottom (lb), front/back (lh), left/right (bh). The formula in the input uses bh and hl instead of lh, correcting that.)

  $$\text{Total Surface Area (TSA)} = 2(\text{lb} + \text{lh} + \text{bh})$$

  $$\text{Lateral Surface Area (LSA)} = \text{Area of 4 side faces} = 2(l+b)h \text{ or } 2(lh + bh)$$$$

  (Sum of areas of front/back and left/right faces. The input formula $2h(l+b)$ is also correct as $2lh + 2bh$).

• Cylinder (Right Circular Cylinder): A solid with two parallel circular bases connected by a curved surface. If 'r' is the radius of the base and 'h' is the height:

  $$\text{Volume} = (\text{Area of base}) \times \text{height} = \pi r^2 h$$

  $$\text{Curved Surface Area (CSA)} = (\text{Circumference of base}) \times \text{height} = 2 \pi r h$$

  $$\text{Total Surface Area (TSA)} = \text{CSA} + 2 \times (\text{Area of base}) = 2 \pi r h + 2 \pi r^2 = 2 \pi r (h + r)$$

• Cone (Right Circular Cone): A solid with a circular base and a curved surface tapering to a single point (apex). If 'r' is the radius of the base, 'h' is the perpendicular height, and 's' is the slant height (distance from apex to a point on the edge of the base circle):

  The slant height can be found using the Pythagorean theorem: $s = \sqrt{r^2 + h^2}$.

  $$\text{Volume} = \frac{1}{3} \times (\text{Area of base}) \times \text{height} = \frac{1}{3} \pi r^2 h$$

  $$\text{Curved Surface Area (CSA)} = \pi \times \text{radius} \times \text{slant height} = \pi r s$$

  $$\text{Total Surface Area (TSA)} = \text{CSA} + \text{Area of base} = \pi r s + \pi r^2 = \pi r (s + r)$$

• Sphere: A perfectly round geometrical object in three-dimensional space that is the surface of a completely round ball. If 'r' is the radius:

  $$\text{Volume} = \frac{4}{3} \pi r^3$$

  $$\text{Surface Area} = 4 \pi r^2$$

• Hemisphere: Exactly half of a sphere. If 'r' is the radius:

  $$\text{Volume} = \frac{1}{2} \times (\text{Volume of Sphere}) = \frac{1}{2} \times \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3$$

  $$\text{Curved Surface Area (CSA)} = \frac{1}{2} \times (\text{Surface Area of Sphere}) = \frac{1}{2} \times 4 \pi r^2 = 2 \pi r^2$$

  The total surface area includes the curved surface and the circular base:

  $$\text{Total Surface Area (TSA)} = \text{CSA} + (\text{Area of circular base}) = 2 \pi r^2 + \pi r^2 = 3 \pi r^2$$

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Seating Arrangement

Seating arrangement problems are a common type of logical reasoning puzzle where you are given a set of conditions about the relative positions of individuals or objects and are asked to deduce their arrangement. These problems require careful reading, systematic thinking, and the ability to handle multiple pieces of information simultaneously. The most frequently encountered types of arrangements are linear (in a line), circular (around a table), and sometimes rectangular or square.

Types of Arrangements

1. Linear Arrangement

In linear arrangement problems, individuals are seated or items are placed in a straight line. The line can be arranged horizontally or vertically. The direction in which the individuals are facing is very important as it determines their left and right sides.

Key aspects and terminologies:

• Total Positions: Identify the total number of positions available in the line.
• Facing Direction: All individuals might be facing the same direction (e.g., all facing North, all facing South) or different directions. This dictates which side is their "left" and "right".
  • If facing North: Your left is their left, your right is their right.
  • If facing South: Your left is their right, your right is their left.
  • If facing East: Left is towards North, Right is towards South.
  • If facing West: Left is towards South, Right is towards North.
• Absolute Positions: Clues that fix a person's position irrespective of others. Examples: "A is sitting at the extreme left end", "B is exactly in the middle of the row", "C is at the second position from the right end".
• Relative Positions: Clues that describe the position of one person relative to another. Examples: "B is to the right of C", "D is between E and F", "G is third to the left of H".
• Immediate vs. General Position: Pay close attention to words like "immediate left" or "immediately to the right" (meaning adjacent positions) versus "to the left" or "to the right" (meaning somewhere on that side, not necessarily adjacent).

For example: Seven friends A, B, C, D, E, F, G are sitting in a row facing North. "C is to the immediate right of B" means they are next to each other in the order B C. "D is to the right of C" means D is somewhere after C in the row (C ... D).

2. Circular Arrangement

In circular arrangement problems, individuals are seated around a circular table or arranged in a circle.

Key aspects and terminologies:

• Number of Arrangements:
  • If the positions around the circle are distinct (e.g., seats are numbered, or there's a fixed point of reference like a door), then for $n$ individuals, the number of arrangements is $n!$.
  • If the positions are not distinct and only the relative arrangement of individuals matters (which is typical for people around a table), the number of arrangements of $n$ distinct individuals is $(n-1)!$. This is because fixing the position of one person makes the remaining $n-1$ positions relative to the first person linear.
  • If the arrangement is considered the same when viewed clockwise or anti-clockwise (e.g., arranging beads in a necklace), the number of arrangements is $\frac{(n-1)!}{2}$. However, for seating people, left and right usually matter, so the relative (n-1)! is the standard.
• Facing Direction: Individuals can be facing the centre of the circle or facing away from the centre. This is crucial for determining left and right.
  • If facing the centre: "Right" means clockwise movement around the circle, and "Left" means anti-clockwise movement.
  • If facing away from the centre: "Right" means anti-clockwise movement around the circle, and "Left" means clockwise movement.
• Relative Positions: Clues are given about positions relative to others (e.g., "P is third to the right of Q", "R is immediately to the left of S", "T is sitting opposite to U"). The concept of "opposite" is usually applicable when there is an even number of positions.
• Neighbours: Pay attention to conditions like "A and B are neighbours" or "C is not a neighbour of D". Neighbours are persons sitting immediately next to each other.

Example: Eight persons P, Q, R, S, T, U, V, W are sitting around a round table facing the centre. "P is third to the right of Q". Starting from Q, move 1st right, 2nd right, 3rd right (clockwise) to place P.

3. Rectangular/Square Arrangement

These are similar to circular arrangements but involve a rectangular or square table. People might be seated at the corners, along the sides, or both. Their facing direction (towards or away from the centre) is again important.

Key aspects and terminologies:

• Positions: Clearly distinguish between individuals sitting at the corners and those sitting in the middle of the sides. The corner positions might be different from side positions in terms of number of neighbours or opposite positions.
• Facing Direction: Usually, individuals are all facing the centre or all facing away from the centre. Left/right is determined based on this direction, similar to circular arrangements.
• Relative Positions: Clues can involve positions relative to others (e.g., "A is third to the right of B"), positions relative to the table structure ("C is at a corner"), or relationships between corner and side occupants ("The person at the corner is immediately to the right of the person in the middle of the side").
• Opposite: In a square or rectangular arrangement with an even number of seats, persons diagonally opposite (at corners) or directly across (in the middle of opposite sides) are considered opposite.

Example: Four people are at the corners and four are in the middle of the sides of a square table, all facing the centre. "A is at a corner". "B is in the middle of a side". "C is second to the right of the person at a corner".

Strategy for Solving Seating Arrangement Problems

Solving seating arrangement problems systematically is key to avoiding errors and finding the correct arrangement. Here's a general approach:

1. Understand the Setup: Read the problem carefully to determine the type of arrangement (linear, circular, etc.), the total number of individuals, and the direction they are facing (if applicable). Visualize the layout.

2. Draw a Diagram: Always draw a diagram representing the seating arrangement. For a linear arrangement, draw a line with marked positions. For a circular or rectangular arrangement, draw the shape and mark the positions. This helps in visualizing the relationships.

3. Identify Definite Information: Look for clues that specify the absolute position of one or more individuals. Start placing these individuals in your diagram. Absolute information is the most reliable starting point.

4. Use Connecting Information: After placing individuals based on absolute clues, look for relative clues that connect directly to the individuals you have already placed. For example, if you placed 'A', look for clues like "B is next to A" or "C is opposite A".

5. Handle Relative Information (Left/Right): Use the relative position clues (left, right, second to the left, third to the right) carefully, keeping the facing direction in mind. If facing the centre of a circle, 'right' is clockwise; if facing away, 'right' is anti-clockwise. For linear, it depends on the direction faced.

6. Look for Constraints and Negative Information: Note down conditions that specify who is NOT sitting next to whom, or who is NOT in a particular position. These negative constraints are important for eliminating possibilities later.

7. Deal with Conditional Information: Some clues might be conditional or provide multiple possibilities (e.g., "A is to the left of B, but not necessarily immediately"). If you exhaust definite and connecting clues, you might have to consider different cases arising from such conditional information. Draw separate diagrams for each potential case.

8. Combine Clues: Continuously try to combine different pieces of information. For example, if "A is next to B" and "B is next to C", then A, B, and C must be sitting together in the sequence A-B-C or C-B-A.

9. Eliminate Possibilities: Use negative information and later clues to eliminate cases that contradict the given conditions. If a case leads to a contradiction at any point, discard it.

10. Fill in Remaining Positions: Once you have used most of the clues, fill in the remaining positions with the remaining individuals.

11. Verify the Final Arrangement: Before answering the questions, re-read all the original conditions one by one and check if your final arrangement satisfies every single condition. If even one condition is not met, your arrangement is incorrect.

12. Answer the Questions: Once the arrangement is verified, answer the specific questions asked based on your final diagram.

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Answer:

This is a circular arrangement problem with 5 persons sitting around a table. They are all facing the centre. For those facing the centre, 'right' is clockwise, and 'left' is anti-clockwise.

Draw a circle with 5 positions marked around it.

Condition (i): R is to the immediate right of P. Let's arbitrarily place P at one position. Since R is to P's immediate right and they are facing the centre, R must be in the position immediately clockwise to P.

Arrangement snippet: P, then R clockwise.

Condition (iii): T is sitting between P and S. This means P, T, and S are sitting in sequence. The sequence could be P-T-S or S-T-P. From condition (i), we know R is immediate right of P, so the position immediately left of P is not R. Since T is between P and S, and R is immediately right of P, T and S must be on P's left side relative to R.

So the sequence P, T, S must be in the anti-clockwise direction from P. Let's place T immediately anti-clockwise to P, and then S immediately anti-clockwise to T.

The arrangement so far (starting from P and going anti-clockwise) is P, T, S. Following clockwise from P, we placed R immediately after P. So the arrangement looks like ... S - T - P - R ... around the circle.

We have placed P, R, S, and T. The only person remaining is Q, and there is one empty position left in the circle, which is between R and S.

Let's place Q in the remaining spot between R and S.

The arrangement (clockwise from P) is now P, R, Q, S, T.

Now, let's verify this complete arrangement with all the given conditions:

(i) R is to the immediate right of P: Starting at P, moving immediately clockwise leads to R. (Satisfied)

(iii) T is sitting between P and S: Moving anti-clockwise from P, we find T and then S. (Satisfied)

(ii) S is to the immediate left of Q: Find Q. Since Q is facing the centre, Q's immediate left is the position immediately anti-clockwise to Q. This position is occupied by S. (Satisfied)

All conditions are satisfied. The final arrangement (clockwise from P) is P - R - Q - S - T - (back to P).

The question asks: Who is sitting to the immediate left of R?

Find R in the arrangement. Since they are facing the centre, R's immediate left is the position immediately anti-clockwise to R.

Looking at the diagram, the person immediately anti-clockwise to R is Q.

Therefore, Q is sitting to the immediate left of R.