| Classwise Concept with Examples | ||||||
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| 6th | 7th | 8th | 9th | 10th | 11th | 12th |
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| Variation | Direct Variation | Inverse Variation |
| Time and Work | ||
Chapter 13 Direct and Inverse Proportions (Concepts)
Welcome to this fundamental chapter where we formally explore two crucial ways in which quantities can relate to and influence each other: Direct Proportion and Inverse Proportion. Often referred to as Direct Variation and Inverse Variation respectively, understanding these relationships is essential for modeling and solving a vast array of real-world problems. In countless situations, we observe that a change in one quantity brings about a predictable change in another. This chapter provides the mathematical framework to analyze, quantify, and predict these interconnected changes, moving beyond simple observation to precise calculation.
First, we delve into the concept of Direct Proportion. Two quantities, let's call them $x$ and $y$, are said to be in direct proportion if they change together in such a manner that their ratio remains constant. This means that if $x$ increases, $y$ increases proportionally, and if $x$ decreases, $y$ decreases proportionally. Mathematically, this constant relationship can be expressed as $\frac{x}{y} = k$, or more commonly rearranged as $y = kx$, where $k$ is a non-zero constant known as the constant of proportionality. Consider these scenarios exhibiting direct proportion:
- The total cost of buying apples is directly proportional to the number of apples purchased (assuming a fixed price per apple). If you double the apples, you double the cost ($\frac{Cost}{Number} = \text{Price per apple} = k$).
- The distance covered by a car travelling at a constant speed is directly proportional to the time it travels. More time means more distance ($\frac{Distance}{Time} = \text{Speed} = k$).
When solving problems involving direct proportion where we have two pairs of corresponding values $(x_1, y_1)$ and $(x_2, y_2)$, the constancy of the ratio gives us the powerful tool: $\frac{x_1}{y_1} = \frac{x_2}{y_2}$. This allows us to find an unknown value when the other three are known.
In contrast, we explore Inverse Proportion. Two quantities, $x$ and $y$, are in inverse proportion if an increase in one quantity leads to a proportional decrease in the other, and vice versa, such that their product remains constant. This constant product relationship is expressed mathematically as $x \times y = k$, where $k$ is again the constant of proportionality. Think about these examples:
- The time taken to travel a fixed distance is inversely proportional to the speed of travel. If you double the speed, you halve the time ($Speed \times Time = Distance = k$).
- The time required to complete a specific job is inversely proportional to the number of workers assigned to it (assuming they all work at the same constant rate). More workers mean less time ($Number \: of \: Workers \times Time = \text{Total Work} = k$).
For situations involving inverse proportion with corresponding value pairs $(x_1, y_1)$ and $(x_2, y_2)$, the constancy of the product leads to the relationship $x_1 y_1 = x_2 y_2$. This equation is used to solve for unknown quantities in inverse proportion scenarios.
A critical skill developed in this chapter is the ability to carefully read the context of a problem and accurately distinguish whether the quantities involved are varying directly or inversely. Misidentifying the type of proportion will lead to incorrect problem setup and solutions. We will practice this identification through various exercises, including analyzing tables of values to determine if $x/y$ or $x \times y$ remains constant, and interpreting word problems related to familiar contexts like time-work calculations, speed-distance-time scenarios, cost-quantity relationships, and resource allocation. Mastering direct and inverse proportion provides fundamental tools for mathematical modeling and practical problem-solving across numerous disciplines.
Variation
In mathematics and in the real world, we often observe how one quantity changes in relation to another. For instance, if you buy more pens of the same type, the total cost increases. If you travel at a faster speed, the time taken to cover a certain distance decreases. This relationship between two quantities, where a change in one affects the other, is called variation or proportion.
Understanding variation helps us to predict how one quantity will change if the other one changes, and it is a fundamental concept used in solving many problems in various fields like physics, economics, and engineering.
We primarily study two types of variations involving two quantities:
- Direct Variation (or Direct Proportion): Where both quantities increase or decrease together.
- Inverse Variation (or Inverse Proportion): Where an increase in one quantity leads to a decrease in the other, and vice versa.
These concepts allow us to model and solve problems involving proportional relationships between quantities.
Direct Variation
Two quantities are said to be in direct variation or direct proportion if they increase or decrease together in such a way that the ratio of their corresponding values remains constant. This means that if one quantity is doubled, the other is also doubled; if one is halved, the other is also halved, and so on.
If we say that a quantity $x$ is in direct variation with a quantity $y$, we can write this using the symbol of proportionality $\propto$ as $x \propto y$.
This proportionality can be expressed as an equation by introducing a constant. If $x \propto y$, then $x$ is equal to $y$ multiplied by a constant. This constant is called the constant of variation or constant of proportionality, usually denoted by $k$.
$x = ky$
... (i)
where $k$ is a non-zero constant.
From this equation, we can see that the ratio of $x$ to $y$ is always equal to this constant $k$ (provided $y \neq 0$):
$\frac{x}{y} = k$
... (ii)
This constant ratio is the key property of direct variation. If we have two pairs of corresponding values for $x$ and $y$, say $(x_1, y_1)$ and $(x_2, y_2)$, then their ratios must be equal:
$\frac{x_1}{y_1} = \frac{x_2}{y_2}$
... (iii)
This property is very useful for solving problems involving direct variation. If you know three of the four values in this proportion, you can find the fourth one.
Examples of Direct Variation:
- Cost and Number of Articles: If the price per article is constant, the total cost is directly proportional to the number of articles purchased. More articles mean higher cost.
- Distance and Time (at Constant Speed): If an object moves at a constant speed, the distance covered is directly proportional to the time taken. More time means greater distance. ($d = \text{speed} \times t$. Here, speed is the constant of proportionality if $d \propto t$).
- Amount of Work and Wages: If the wage rate per unit of work is constant, the total wages earned are directly proportional to the amount of work done. More work means higher wages.
- Amount of Substance and Weight: The weight of a substance is directly proportional to its amount (mass).
Example 1. If the cost of 5 metres of cloth is $\textsf{₹}210$, what would be the cost of 2 metres of cloth?
Answer:
Given:
- Length of cloth ($x_1$) = 5 metres, Cost ($y_1$) = $\textsf{₹}210$.
- We need to find the Cost ($y_2$) for a Length ($x_2$) = 2 metres.
Solution:
As the length of cloth increases, the cost increases proportionally (assuming the rate per metre is constant). This is a case of direct variation.
Let the length of the cloth be $x$ and the cost be $y$. We can say $x \propto y$, or $\frac{x}{y} = k$.
Using the property that the ratio of corresponding values is constant:
$\frac{x_1}{y_1} = \frac{x_2}{y_2}$
[Using Formula (iii)]
Substitute the given values $x_1=5, y_1=210, x_2=2$:
$\frac{5}{210} = \frac{2}{y_2}$
To solve for $y_2$, we can cross-multiply:
$5 \times y_2 = 2 \times 210$
$5y_2 = 420$
Divide both sides by 5:
$y_2 = \frac{420}{5}$
$y_2 = 84$
So, the cost of 2 metres of cloth would be $\textsf{₹}84$.
Alternate Method (Using Unitary Method):
In direct variation problems, the unitary method can often be used as an alternative approach.
Given: Cost of 5 metres of cloth = $\textsf{₹}210$.
Step 1: Find the cost of 1 unit (1 metre) of cloth.
Cost of 1 metre of cloth $= \frac{\text{Total Cost}}{\text{Total Length}} = \frac{\textsf{₹}210}{5 \text{ metres}} = \textsf{₹}42$ per metre.
Step 2: Use the cost of 1 unit to find the cost of the required quantity (2 metres).
Cost of 2 metres of cloth $= \text{Cost per metre} \times \text{Required length}
$= \textsf{₹}42/\text{m} \times 2 \text{ metres} = \textsf{₹}84$
This confirms the result obtained using the proportion formula.
Example 2. A car travels 40 km in 30 minutes. If the speed remains constant, how much distance will it cover in 1 hour 30 minutes?
Answer:
Given:
- Distance covered ($d_1$) = 40 km in Time ($t_1$) = 30 minutes.
- We need to find the Distance covered ($d_2$) in Time ($t_2$) = 1 hour 30 minutes.
- Speed is constant.
Solution:
At a constant speed, the distance covered is directly proportional to the time taken. As time increases, the distance covered increases proportionally.
Let the distance covered be $d$ and the time taken be $t$. We can say $d \propto t$, or $\frac{d}{t} = k$ (where $k$ is the constant speed).
First, ensure all time units are the same. Convert both time periods to minutes:
$t_1 = 30$ minutes.
$t_2 = 1$ hour 30 minutes $= (1 \times 60) \text{ minutes} + \ $$ 30 \text{ minutes} \ $$ = 60 + 30 = 90$ minutes.
Using the property that the ratio of corresponding values is constant:
$\frac{d_1}{t_1} = \frac{d_2}{t_2}$
[Using Formula (iii)]
Substitute the given values $d_1=40, t_1=30, t_2=90$:
$\frac{40}{30} = \frac{d_2}{90}$
Simplify the fraction on the left side:
$\frac{4}{3} = \frac{d_2}{90}$
To solve for $d_2$, multiply both sides by 90 (or cross-multiply):
$d_2 = \frac{4}{3} \times 90$
Simplify and calculate:
$d_2 = 4 \times \frac{\cancel{90}^{\normalsize 30}}{\cancel{3}_{\normalsize 1}} = 4 \times 30 = 120$
So, the car will cover 120 km in 1 hour 30 minutes.
Alternate Method (Using Unitary Method):
Given: Distance covered in 30 minutes = 40 km.
Step 1: Find the distance covered in 1 unit of time (1 minute).
Distance covered in 1 minute $= \frac{\text{Total Distance}}{\text{Total Time}} = \frac{40 \text{ km}}{30 \text{ minutes}} = \frac{4}{3}$ km per minute.
Step 2: Use the distance covered in 1 minute to find the distance covered in the required time (90 minutes).
Distance covered in 90 minutes
$= (\text{Distance per minute}) \times (\text{Required time})$
$= \frac{4}{3} \text{ km/minute} \times 90 \text{ minutes} = \frac{4}{3} \times 90 \text{ km}$
$= 4 \times 30 \text{ km} = 120 \text{ km}$
This confirms the result obtained using the proportion formula.
Inverse Variation
In contrast to direct variation, where quantities change in the same direction, inverse variation describes a relationship where an increase in one quantity leads to a decrease in the other, and vice versa, in a proportional manner. The relationship is such that the product of the corresponding values of the two quantities remains constant.
If we say that a quantity $x$ is in inverse variation with a quantity $y$, we can write this using the symbol of proportionality $\propto$ as $x \propto \frac{1}{y}$. This indicates that $x$ is directly proportional to the reciprocal of $y$.
This proportionality can be expressed as an equation by introducing a constant. If $x \propto \frac{1}{y}$, then $x$ is equal to $\frac{1}{y}$ multiplied by a constant. This constant is called the constant of variation or constant of proportionality, usually denoted by $k$.
$x = k \times \frac{1}{y}$
Multiplying both sides by $y$ (assuming $y \neq 0$), we get the key property of inverse variation:
$xy = k$
... (i)
where $k$ is a non-zero constant.
This means that for any pair of corresponding values of $x$ and $y$, their product is always the same constant $k$. If we have two pairs of corresponding values, say $(x_1, y_1)$ and $(x_2, y_2)$, then their products must be equal:
$x_1 y_1 = x_2 y_2$
... (ii)
This property is very useful for solving problems involving inverse variation. If you know three of the four values in this relationship, you can find the fourth one.
Examples of Inverse Variation:
- Speed and Time (for a Fixed Distance): If the distance to be covered is fixed, the speed of travel and the time taken are in inverse variation. Higher speed means less time. (Speed $\times$ Time = Distance. Here, Distance is the constant $k$ if Speed $\propto 1/$Time).
- Number of Workers and Time Taken (for Fixed Work): If the total amount of work is fixed and all workers work at the same rate, the number of workers and the time taken to complete the work are in inverse variation. More workers mean less time. (Number of workers $\times$ Time = Total work. Here, Total work is the constant $k$).
- Density and Volume (for a Fixed Mass): The density of an object and its volume are in inverse variation if the mass is constant. Higher density means smaller volume. (Density = Mass / Volume, so Mass = Density $\times$ Volume. Here, Mass is the constant $k$).
- Pressure and Volume (at Constant Temperature): For a fixed amount of gas at a constant temperature, the pressure and volume are in inverse variation (Boyle's Law). Higher pressure means smaller volume. (Pressure $\times$ Volume = Constant).
Example 1. If 10 workers can complete a piece of work in 5 days, how many days will 5 workers take to complete the same work? (Assume all workers work at the same rate).
Answer:
Given:
- Number of workers ($n_1$) = 10, Time taken ($d_1$) = 5 days.
- We need to find the Time taken ($d_2$) for Number of workers ($n_2$) = 5.
Solution:
Assuming all workers work at the same rate, the total amount of work required to complete the piece is constant. The number of workers and the time taken to complete the work are in inverse variation. As the number of workers decreases, the time taken increases proportionally.
Let the number of workers be $n$ and the time taken in days be $d$. We can say $n \propto \frac{1}{d}$, or $nd = k$ (where $k$ is the total amount of work in "worker-days").
Using the property that the product of corresponding values is constant:
$n_1 d_1 = n_2 d_2$
[Using Formula (ii)]
Substitute the given values $n_1=10, d_1=5, n_2=5$:
$10 \times 5 = 5 \times d_2$
Calculate the product on the left:
$50 = 5d_2$
Divide both sides by 5 to solve for $d_2$:
$d_2 = \frac{50}{5}$
$d_2 = 10$
So, 5 workers will take 10 days to complete the same work.
Alternate Method (Using Unitary Method - Work):
Given: 10 workers complete the work in 5 days.
The total amount of work can be measured in "worker-days".
Total Work $= \text{Number of workers} \times \text{Time taken}$
Total Work $= 10 \text{ workers} \times 5 \text{ days} = 50 \text{ worker-days}$
Now, we have 5 workers, and we want to find the number of days ($d$) they will take to complete this total work (50 worker-days).
$5 \text{ workers} \times d \text{ days} = 50 \text{ worker-days}$
$5d = 50$
$d = \frac{50}{5} = 10$ days
This confirms the result obtained using the inverse variation formula.
Example 2. A car takes 2 hours to reach a destination travelling at a speed of 60 km/h. How long will it take when it travels at a speed of 80 km/h? (Assume the distance is fixed).
Answer:
Given:
- Speed ($s_1$) = 60 km/h, Time taken ($t_1$) = 2 hours.
- We need to find the Time taken ($t_2$) for Speed ($s_2$) = 80 km/h.
- The distance to the destination is fixed.
Solution:
For a fixed distance, the speed of travel and the time taken are in inverse variation. As speed increases, the time taken decreases proportionally.
Let the speed be $s$ and the time taken be $t$. We know that Speed $\times$ Time = Distance. Since the distance is fixed, the product $st$ is constant. We can say $s \propto \frac{1}{t}$, or $st = k$ (where $k$ is the fixed distance).
Using the property that the product of corresponding values is constant:
$s_1 t_1 = s_2 t_2$
[Using Formula (ii)]
Substitute the given values $s_1=60, t_1=2, s_2=80$:
$60 \times 2 = 80 \times t_2$
Calculate the product on the left:
$120 = 80t_2$
Divide both sides by 80 to solve for $t_2$:
$t_2 = \frac{120}{80}$
Simplify the fraction:
$t_2 = \frac{12}{8} = \frac{3}{2}$
The time taken is $\frac{3}{2}$ hours, which is equal to $1.5$ hours or $1\frac{1}{2}$ hours or 1 hour 30 minutes.
So, the car will take 1.5 hours to reach the destination when travelling at 80 km/h.
Alternate Method (Using Unitary Method - Distance):
Given: Speed ($s_1$) = 60 km/h, Time ($t_1$) = 2 hours.
Step 1: Find the fixed distance to the destination.
Distance = Speed $\times$ Time
Distance $= 60 \text{ km/h} \times 2 \text{ hours} = 120 \text{ km}$
Step 2: Now, the car travels at a new speed ($s_2$ = 80 km/h) to cover the same distance (120 km). Use the formula Distance = Speed $\times$ Time to find the new time ($t_2$).
$120 \text{ km} = 80 \text{ km/h} \times t_2 \text{ hours}$
$120 = 80 t_2$
$t_2 = \frac{120}{80} = \frac{12}{8} = \frac{3}{2}$ hours
This confirms the result obtained using the inverse variation formula.
Time and Work
Problems involving the relationship between the time taken to complete a task and the number of individuals or machines doing the work are often categorised as Time and Work problems. These problems frequently involve the concept of inverse variation, especially when the total amount of work is constant.
Understanding Time and Work Relationship
The basic principle in time and work problems is that the total amount of work done is proportional to both the number of workers (or units doing the work) and the time taken, assuming a constant rate of work per worker. If we consider the total work as a fixed quantity, then the number of workers and the time taken are inversely related.
Key Relationship:
Total Work = Number of Workers $\times$ Time Taken
... (i)
If the "Total Work" is a fixed value (let's say $W$), then $W = \text{Number of Workers} \times \text{Time Taken}$. This means the product of the Number of Workers and the Time Taken is a constant ($W$). This is the definition of inverse variation. If the number of workers increases, the time taken decreases proportionally to keep the product (Total Work) constant.
We often assume that each worker works at the same constant rate unless specified otherwise. The work rate is the amount of work done per unit of time by a single worker.
Work Rate Concept
If a worker can complete a whole piece of work in $d$ days, it means that in each day, the worker completes a fraction of the total work. Assuming a constant work rate, the amount of work done by that worker in one day is the total work (1 whole piece) divided by the number of days.
Work done by one worker in 1 day $= \frac{1}{\text{Number of days the worker takes alone}}$
... (ii)
For example, if a worker takes 10 days to complete a job, their daily work rate is $\frac{1}{10}$ of the job per day. If they take 15 days, their daily work rate is $\frac{1}{15}$ of the job per day.
When multiple workers work together, their individual daily work rates are added to find the combined daily work rate, assuming they work independently and at the same rate.
Example 1. A alone can do a piece of work in 10 days and B alone can do it in 15 days. In how many days will they finish the work if they work together?
Answer:
Given:
- Time taken by A alone to complete the work = 10 days.
- Time taken by B alone to complete the work = 15 days.
To Find:
- Time taken by A and B together to complete the work.
Solution:
Step 1: Find the amount of work done by A and B individually in one day (their daily work rates).
Work done by A in 1 day $= \frac{1}{\text{Time taken by A alone}} = \frac{1}{10}$ of the total work.
[Using Formula (ii)]
Work done by B in 1 day $= \frac{1}{\text{Time taken by B alone}} = \frac{1}{15}$ of the total work.
[Using Formula (ii)]
Step 2: Find the combined amount of work done by A and B together in one day (their combined daily work rate). We add their individual daily work rates.
Work done by (A + B) together in 1 day $= (\text{Work by A in 1 day}) + (\text{Work by B in 1 day})$
$= \frac{1}{10} + \frac{1}{15}$
To add these fractions, find the LCM of the denominators 10 and 15. LCM(10, 15) = 30.
$= \frac{1 \times 3}{10 \times 3} + \frac{1 \times 2}{15 \times 2} = \frac{3}{30} + \frac{2}{30}$
$= \frac{3+2}{30} = \frac{5}{30} = \frac{1}{6}$ of the total work.
So, A and B together complete $\frac{1}{6}$ of the total work in one day.
Step 3: Find the total time taken for A and B together to complete the whole work. If they do $\frac{1}{6}$ of the work in 1 day, they will take the reciprocal of this fraction to complete the whole work (which is 1, or $\frac{6}{6}$).
Time taken by (A + B) together $= \frac{1}{\text{Combined work rate in 1 day}}$
$= \frac{1}{1/6} = 1 \div \frac{1}{6} = 1 \times 6 = 6$ days.
So, A and B will finish the work together in 6 days.
Example 2. A and B together can do a piece of work in 12 days. B alone can do it in 30 days. In how many days can A alone do the work?
Answer:
Given:
- Time taken by A and B together = 12 days.
- Time taken by B alone = 30 days.
To Find:
- Time taken by A alone to complete the work.
Solution:
Step 1: Find the work done in one day by A and B together, and by B alone.
Work done by (A + B) together in 1 day $= \frac{1}{\text{Time taken by A + B together}} = \frac{1}{12}$ of the total work.
Work done by B alone in 1 day $= \frac{1}{\text{Time taken by B alone}} = \frac{1}{30}$ of the total work.
Step 2: We know that the combined daily work rate is the sum of individual daily work rates. (Work by A in 1 day) + (Work by B in 1 day) = (Work by A + B in 1 day).
Let the time taken by A alone be $x$ days. So, work done by A alone in 1 day is $\frac{1}{x}$.
$\frac{1}{x} + \frac{1}{30} = \frac{1}{12}$
Step 3: Solve this equation for $\frac{1}{x}$ (which is A's daily work rate). Subtract $\frac{1}{30}$ from both sides:
$\frac{1}{x} = \frac{1}{12} - \frac{1}{30}$
To subtract the fractions, find the LCM of 12 and 30. LCM(12, 30) = 60.
$\frac{1}{x} = \frac{1 \times 5}{12 \times 5} - \frac{1 \times 2}{30 \times 2} = \frac{5}{60} - \frac{2}{60}$
$= \frac{5-2}{60} = \frac{3}{60}$
Simplify the fraction:
$\frac{1}{x} = \frac{\cancel{3}^{\normalsize 1}}{\cancel{60}_{\normalsize 20}} = \frac{1}{20}$
Step 4: Find the time taken by A alone. If A does $\frac{1}{20}$ of the work in 1 day, A will take the reciprocal of this fraction to complete the whole work.
Time taken by A alone ($x$) $= \frac{1}{\text{Work done by A in 1 day}}$
$x = \frac{1}{1/20} = 1 \div \frac{1}{20} = 1 \times 20 = 20$ days.
So, A alone can do the work in 20 days.