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Introduction to Variation Concepts | Direct Variation and its Applications | Inverse Variation and its Applications |
Problems Combining Direct and Inverse Variation |
Variation: Direct and Inverse
Introduction to Variation Concepts
In mathematics and science, the concept of variation is used to describe how the value of one quantity changes in relation to the value of another quantity. When a change in one quantity consistently corresponds to a predictable change in another quantity, we say that there is a variation between them.
Understanding variation allows us to build mathematical models that represent relationships observed in the real world and solve problems based on these relationships.
The two fundamental types of variation are:
Types of Variation
- Direct Variation: This occurs when two quantities change in the same direction. If one quantity increases, the other also increases proportionally, and vice versa. The ratio of the two quantities remains constant.
- Inverse Variation: This occurs when two quantities change in opposite directions. If one quantity increases, the other decreases proportionally, and vice versa. The product of the two quantities remains constant.
In any relationship involving variation, there is a core constant that defines the specific link between the quantities. This constant is called the constant of variation or the constant of proportionality. It acts as a factor that translates the change in one quantity into the corresponding change in the other.
For example, the relationship between the distance covered by a car and the time taken, when the speed is constant, is a form of variation. Similarly, the relationship between the number of workers and the time taken to complete a job illustrates another type of variation.
Studying variation provides a powerful framework for setting up equations and solving problems across various domains, including physics (like force, mass, and acceleration), economics (like demand and price), engineering (like pressure and volume), and everyday calculations involving rates and quantities.
Direct Variation and its Applications
Direct variation describes a relationship between two quantities where one quantity is a direct multiple of the other. In simpler terms, if $y$ varies directly as $x$, then as $x$ increases, $y$ increases proportionally, and as $x$ decreases, $y$ decreases proportionally. The ratio of $y$ to $x$ remains constant throughout their change.
Definition and Formula
If a quantity $y$ varies directly as a quantity $x$, we denote this relationship using the proportionality symbol ($\propto$) as:
$\boldsymbol{y \propto x}$
To convert this proportionality into an equation, we introduce a non-zero constant, typically denoted by $k$. This constant is called the constant of direct variation or the constant of proportionality.
$\boldsymbol{y = kx}$
... (i)
Equation (i) is the defining equation for direct variation. From this equation, provided $x \neq 0$, we can find the constant $k$:
$\boldsymbol{k = \frac{y}{x}}$
[Constant of Direct Variation]
This formula emphasizes that in direct variation, the ratio of the two quantities is always equal to the constant $k$ for any corresponding pair of values (except when $x=0$).
Relationship between Two Pairs of Values in Direct Variation
Consider two different pairs of corresponding values for quantities $x$ and $y$ that are in direct variation. Let the first pair be $(x_1, y_1)$ and the second pair be $(x_2, y_2)$.
According to the definition of direct variation ($y = kx$), for the first pair $(x_1, y_1)$:
$\boldsymbol{y_1 = kx_1}$
... (ii)
And for the second pair $(x_2, y_2)$:
$\boldsymbol{y_2 = kx_2}$
... (iii)
From equation (ii), assuming $x_1 \neq 0$, we can express the constant $k$ as:
$\boldsymbol{k = \frac{y_1}{x_1}}$
Similarly, from equation (iii), assuming $x_2 \neq 0$, we can express the constant $k$ as:
$\boldsymbol{k = \frac{y_2}{x_2}}$
Since $k$ is the same constant for any corresponding pair of values in a direct variation, we can equate these two expressions for $k$:
$\boldsymbol{\frac{y_1}{x_1} = \frac{y_2}{x_2}}$
... (iv)
Equation (iv) is a fundamental relationship used to solve problems involving direct variation. If we are given three of the four quantities $(x_1, y_1, x_2, y_2)$, we can find the fourth one using this proportion.
This can also be written in the form of a proportion as $y_1:x_1 :: y_2:x_2$ or, more commonly, $x_1:y_1 :: x_2:y_2$ after applying Alternendo ($\frac{x_1}{y_1} = \frac{x_2}{y_2}$).
Graphical Representation
When the relationship $y = kx$ is plotted on a coordinate plane with $x$ on the horizontal axis and $y$ on the vertical axis, the graph is a straight line. Since $k$ is a non-zero constant, the equation $y=kx$ represents a linear equation of the form $y=mx+c$ where the y-intercept ($c$) is 0. Therefore, the graph of a direct variation is always a straight line that passes through the origin $(0,0)$. The slope of this line is equal to the constant of variation $k$.
Applications of Direct Variation in Real Life
Direct variation is a very common relationship observed in numerous situations:
1. Cost and Quantity: If you buy more quantity of an item at a fixed price per unit, the total cost increases proportionally. E.g., The total cost of apples is directly proportional to the number of apples purchased, assuming the price per apple is constant.
2. Distance and Time (at constant speed): If you travel at a constant speed, the distance you cover is directly proportional to the time you spend traveling. E.g., A car traveling at 60 km/h covers twice the distance in twice the time.
3. Work Done and Time (by fixed number of workers): If a fixed number of workers are working at a constant rate, the amount of work completed is directly proportional to the time they spend working. E.g., 5 workers can complete more tasks in 8 hours than in 4 hours.
4. Force and Acceleration (at constant mass): As per Newton's second law of motion ($F=ma$), if the mass of an object is kept constant, the force applied to it is directly proportional to the acceleration it experiences.
5. Simple Interest: Simple interest earned on a principal amount over a fixed period and at a fixed rate is directly proportional to the principal amount.
Example 1. If the cost of 8 notebooks is $\textsf{₹ } 192$, what would be the cost of 13 notebooks?
Answer:
Let the number of notebooks be $N$ and their total cost be $C$. Assuming the price per notebook is constant, the total cost $C$ is directly proportional to the number of notebooks $N$.
$\boldsymbol{C \propto N}$
This can be written as $C = kN$, where $k$ is the constant of proportionality (cost per notebook).
Given: Cost of 8 notebooks ($N_1 = 8$) is $\textsf{₹ } 192$ ($C_1 = 192$).
We can find the constant $k$ using the first pair of values:
From $C = kN$, $k = \frac{C}{N}$
$\boldsymbol{k = \frac{C_1}{N_1} = \frac{192}{8}}$
Performing the division:
$\frac{192}{8} = 24$
$\boldsymbol{k = 24}$
The constant of variation is 24, which represents that the cost of each notebook is $\textsf{₹ } 24$.
Now, we need to find the cost ($C_2$) of 13 notebooks ($N_2 = 13$). Using the relationship $C = kN$ with the value of $k$:
$\boldsymbol{C_2 = k \times N_2}$
$\boldsymbol{C_2 = 24 \times 13}$
Performing the multiplication:
$\begin{array}{cc}& & 2 & 4 \\ \times & & 1 & 3 \\ \hline & & 7 & 2 \\ & 2 & 4 & \times \\ \hline & 3 & 1 & 2 \\ \hline \end{array}$
$\boldsymbol{C_2 = 312}$
The cost of 13 notebooks is $\boldsymbol{\textsf{₹ } 312}$.
Alternate Solution using $\frac{y_1}{x_1} = \frac{y_2}{x_2}$ (or $\frac{C_1}{N_1} = \frac{C_2}{N_2}$) formula:
Let $N_1 = 8$ notebooks, $C_1 = \textsf{₹ } 192$.
Let $N_2 = 13$ notebooks, and $C_2$ be the required cost.
Since Cost ($C$) is directly proportional to the Number of Notebooks ($N$), we have:
$\boldsymbol{\frac{C_1}{N_1} = \frac{C_2}{N_2}}$
Substitute the given values into the formula:
$\frac{192}{8} = \frac{C_2}{13}$
Now, solve for $C_2$. Multiply both sides by 13:
$\boldsymbol{C_2 = \frac{192}{8} \times 13}$
Simplify the fraction $\frac{192}{8}$:
$\frac{192}{8} = 24$
$\boldsymbol{C_2 = 24 \times 13}$
Performing the multiplication:
$\begin{array}{cc}& & 2 & 4 \\ \times & & 1 & 3 \\ \hline & & 7 & 2 \\ & 2 & 4 & \times \\ \hline & 3 & 1 & 2 \\ \hline \end{array}$
$\boldsymbol{C_2 = 312}$
The cost of 13 notebooks is $\boldsymbol{\textsf{₹ } 312}$. Both methods give the same correct answer.
Inverse Variation and its Applications
Inverse variation, also known as inverse proportion, describes a relationship between two quantities where if one quantity increases, the other quantity decreases proportionally, and vice versa. The defining characteristic of inverse variation is that the product of the two related quantities remains constant.
Definition and Formula
If a quantity $y$ varies inversely as a quantity $x$, we express this relationship using the proportionality symbol ($\propto$) and the reciprocal of $x$:
$\boldsymbol{y \propto \frac{1}{x}}$
To convert this proportionality into an equation, we introduce a non-zero constant, $k$, called the constant of inverse variation or the constant of proportionality:
$\boldsymbol{y = \frac{k}{x}}$
... (i)
Here, $k$ is the constant. Provided $x \neq 0$ and $y \neq 0$, we can rearrange equation (i) by multiplying both sides by $x$ to find the constant $k$:
$\boldsymbol{xy = k}$
[Constant of Inverse Variation]
This formula shows that for any pair of corresponding non-zero values of $x$ and $y$ that are inversely related, their product is always equal to the same constant $k$.
Relationship between Two Pairs of Values in Inverse Variation
Consider two different pairs of corresponding values for quantities $x$ and $y$ that are related by inverse variation. Let the first pair be $(x_1, y_1)$ and the second pair be $(x_2, y_2)$.
According to the definition of inverse variation ($xy = k$), for the first pair $(x_1, y_1)$, their product equals the constant $k$:
$\boldsymbol{x_1 y_1 = k}$
... (ii)
Similarly, for the second pair $(x_2, y_2)$, their product also equals the same constant $k$:
$\boldsymbol{x_2 y_2 = k}$
... (iii)
Since both $x_1 y_1$ and $x_2 y_2$ are equal to the same constant $k$, we can equate them:
$\boldsymbol{x_1 y_1 = x_2 y_2}$
... (iv)
Equation (iv) is the fundamental equation used to solve problems involving inverse variation. If we are given any three of the four quantities $(x_1, y_1, x_2, y_2)$, we can use this equation to find the value of the fourth quantity.
This relationship can also be rearranged to form a proportion by dividing both sides by $x_2 y_1$ (assuming $x_2 \neq 0$ and $y_1 \neq 0$):
$\frac{x_1 y_1}{x_2 y_1} = \frac{x_2 y_2}{x_2 y_1}$
$\frac{x_1}{x_2} = \frac{y_2}{y_1}$
In ratio form, this is $x_1:x_2 :: y_2:y_1$. This shows that the ratio of the first quantity's values is proportional to the *inverse* ratio of the corresponding second quantity's values.
Graphical Representation
When the relationship $y = \frac{k}{x}$ (or $xy = k$) is plotted on a coordinate plane with $x$ on the horizontal axis and $y$ on the vertical axis, the graph is a curve called a hyperbola. For a positive constant $k$, the graph lies in the first and third quadrants. For a negative constant $k$, it lies in the second and fourth quadrants. The graph gets closer and closer to the x-axis and y-axis but never actually touches them. These axes are called asymptotes.
Applications of Inverse Variation in Real Life
Inverse variation relationships are frequently encountered in various fields:
1. Speed and Time (for a fixed distance): If the distance to be covered is constant, increasing the speed of travel reduces the time taken proportionally. For example, if you double your speed, you halve the time to reach the destination.
2. Number of Workers and Time (for a fixed amount of work): If a fixed amount of work needs to be completed, increasing the number of workers (assuming they all work at the same efficiency) decreases the time required. For example, if you double the number of workers, the job gets done in half the time.
3. Pressure and Volume of a Gas (at constant temperature): According to Boyle's Law, for a fixed amount of gas at a constant temperature, the pressure ($P$) is inversely proportional to its volume ($V$). This means $PV = k$, where $k$ is a constant.
4. Current and Resistance (at constant voltage): Ohm's Law states that for a constant voltage ($V$), the current ($I$) flowing through a circuit is inversely proportional to the resistance ($R$) ($I \propto \frac{1}{R}$ or $IR = V$).
5. Intensity of Light and Distance from Source: The intensity of light from a point source is inversely proportional to the square of the distance from the source. ($I \propto \frac{1}{d^2}$).
Example 1. 15 pipes of the same diameter can fill a tank in 12 minutes. How many minutes will 20 such pipes take to fill the same tank?
Answer:
Let the number of pipes be $N$ and the time taken to fill the tank be $T$ (in minutes). Since the tank capacity is fixed and the pipes have the same diameter (same filling rate), the number of pipes is inversely proportional to the time taken. (More pipes mean less time to fill the tank).
$\boldsymbol{T \propto \frac{1}{N}}$
This means their product is constant: $\boldsymbol{NT = k}$, where $k$ represents the total 'pipe-minutes' needed to fill the tank.
Given: 15 pipes ($N_1 = 15$) take 12 minutes ($T_1 = 12$).
We can find the constant $k$ using the first pair of values:
$\boldsymbol{k = N_1 \times T_1 = 15 \times 12}$
Performing the multiplication:
$\begin{array}{cc}& & 1 & 5 \\ \times & & 1 & 2 \\ \hline & & 3 & 0 \\ & 1 & 5 & \times \\ \hline & 1 & 8 & 0 \\ \hline \end{array}$
$\boldsymbol{k = 180}$
The constant of variation is 180, representing 180 pipe-minutes of work needed to fill the tank.
Now, we need to find the time ($T_2$) it takes for 20 pipes ($N_2 = 20$). Using the inverse variation relationship $NT = k$:
$\boldsymbol{N_2 \times T_2 = k}$
$\boldsymbol{20 \times T_2 = 180}$
... (v)
Divide both sides of equation (v) by 20 to solve for $T_2$:
$\boldsymbol{T_2 = \frac{180}{20}}$
Simplify the fraction:
$\frac{180}{20} = \frac{18}{2} = 9$
$\boldsymbol{T_2 = 9}$ minutes
So, it will take $\boldsymbol{9}$ minutes for 20 pipes to fill the same tank.
Alternate Solution using $x_1 y_1 = x_2 y_2$ (or $N_1 T_1 = N_2 T_2$) formula:
Let $N_1 = 15$ (pipes) and $T_1 = 12$ (minutes).
Let $N_2 = 20$ (pipes) and $T_2$ be the required time.
Since time ($T$) is inversely proportional to the number of pipes ($N$), we have $N_1 T_1 = N_2 T_2$.
Substitute the given values:
$\boldsymbol{15 \times 12 = 20 \times T_2}$
$\boldsymbol{180 = 20 T_2}$
Solve for $T_2$:
$\boldsymbol{T_2 = \frac{180}{20}}$
$\boldsymbol{T_2 = 9}$ minutes
It will take $\boldsymbol{9}$ minutes for 20 pipes to fill the tank.
Example 2. A garrison of 500 men has provisions for 24 days. If the garrison is joined by 300 more men, how long will the provisions last?
Answer:
Let the number of men be $M$ and the number of days the provisions last be $D$. Assuming the daily consumption per man is constant, the number of men is inversely proportional to the number of days the provisions last. (More men mean fewer days the provisions last).
$\boldsymbol{D \propto \frac{1}{M}}$
This means their product is constant: $\boldsymbol{MD = k}$, where $k$ represents the total 'man-days' of provisions.
Given: Initial number of men $M_1 = 500$, Provisions last for $D_1 = 24$ days.
Find the constant $k$ (total man-days):
$\boldsymbol{k = M_1 \times D_1 = 500 \times 24}$
Performing the multiplication:
$\begin{array}{cccc}& & & 2 & 4 \\ \times & & & 5 & 0 & 0 \\ \hline &&&&0&0 \\ &&&0&0&\times \\ &1&2&0&\times&\times \\ \hline &1&2&0&0&0 \\ \hline \end{array}$
$\boldsymbol{k = 12000}$ man-days
The total provisions are 12000 man-days.
Now, the garrison is joined by 300 more men.
New number of men $M_2 = 500 + 300 = 800$ men.
We need to find the number of days ($D_2$) the provisions will last for 800 men.
Using the inverse variation relationship $MD = k$:
$\boldsymbol{M_2 \times D_2 = k}$
$\boldsymbol{800 \times D_2 = 12000}$
... (vi)
Divide both sides of equation (vi) by 800 to solve for $D_2$:
$\boldsymbol{D_2 = \frac{12000}{800}}$
Simplify the fraction:
$\frac{12000}{800} = \frac{120}{8} = 15$
$\boldsymbol{D_2 = 15}$ days
The provisions will last for $\boldsymbol{15}$ days for 800 men.
Alternate Solution using $M_1 D_1 = M_2 D_2$ formula:
Let $M_1 = 500$ (men) and $D_1 = 24$ (days).
New number of men $M_2 = 500 + 300 = 800$ (men).
Let $D_2$ be the number of days the provisions last.
Since the number of days ($D$) is inversely proportional to the number of men ($M$), we have $M_1 D_1 = M_2 D_2$.
Substitute the values:
$\boldsymbol{500 \times 24 = 800 \times D_2}$
$\boldsymbol{12000 = 800 D_2}$
Solve for $D_2$:
$\boldsymbol{D_2 = \frac{12000}{800}}$
$\boldsymbol{D_2 = 15}$ days
The provisions will last for $\boldsymbol{15}$ days.
Competitive Exam Notes:
Inverse variation problems are common in Time and Work, Time Speed and Distance, and resource-based scenarios (like food stock). Recognizing the inverse relationship quickly is key.
- Key Relationship: The product of the two quantities is constant ($xy=k$). This leads to the fundamental equation $x_1 y_1 = x_2 y_2$.
- Identifying Inverse Variation: Look for scenarios where increasing one quantity leads to a decrease in the other for a fixed outcome (fixed distance, fixed amount of work, fixed amount of resource).
- Common Examples:
- Speed and Time (for fixed distance): $S_1 T_1 = S_2 T_2$
- Workers and Time (for fixed work): $W_1 D_1 = W_2 D_2$ (or $W_1 H_1 = W_2 H_2$ if time is in hours)
- Men and Provisions Duration (for fixed stock): $M_1 D_1 = M_2 D_2$
- Unit Consistency: Ensure that the units of the quantities are consistent across the two pairs of values (e.g., if $T_1$ is in hours, $T_2$ will be in hours).
- Comparison with Direct Variation: Remember that for direct variation, the ratio is constant ($\frac{y_1}{x_1} = \frac{y_2}{x_2}$), while for inverse variation, the product is constant ($x_1 y_1 = x_2 y_2$).
Problems Combining Direct and Inverse Variation
In real-world scenarios, the relationship between quantities is often more complex than simple direct or inverse variation. Many times, a quantity may depend on several other quantities, varying directly with some and inversely with others simultaneously. This type of relationship is known as combined variation.
Combined Variation
When a quantity varies directly as one or more quantities and inversely as one or more other quantities, it is a case of combined variation.
Mathematically, if a quantity $z$ varies directly as $x$ and inversely as $y$, we can write the combined proportionality as:
$\boldsymbol{z \propto \frac{x}{y}}$
To express this proportionality as an equation, we introduce a non-zero constant of variation, $k$:
$\boldsymbol{z = k \frac{x}{y}}$
... (i)
From equation (i), assuming $x \neq 0$ and $y \neq 0$, we can solve for the constant $k$ by multiplying both sides by $\frac{y}{x}$:
$\boldsymbol{k = \frac{zy}{x}}$
[Constant of Combined Variation]
This means that for any set of corresponding non-zero values $(x, y, z)$, the value of $\frac{zy}{x}$ will always be equal to the same constant $k$.
Relationship between Two Sets of Values in Combined Variation
Consider two different sets of corresponding values for quantities $x, y,$ and $z$ that are related by the combined variation $z \propto \frac{x}{y}$. Let the first set be $(x_1, y_1, z_1)$ and the second set be $(x_2, y_2, z_2)$.
Based on the definition $k = \frac{zy}{x}$, for the first set of values $(x_1, y_1, z_1)$, we have:
$\boldsymbol{\frac{z_1 y_1}{x_1} = k}$
... (ii)
And for the second set of values $(x_2, y_2, z_2)$, we have:
$\boldsymbol{\frac{z_2 y_2}{x_2} = k}$
... (iii)
Since both expressions are equal to the same constant $k$, we can equate them to get the relationship for solving problems:
$\boldsymbol{\frac{z_1 y_1}{x_1} = \frac{z_2 y_2}{x_2}}$
... (iv)
This equation is particularly useful when you are given one complete set of values $(x_1, y_1, z_1)$ and two values from a second set $(x_2, y_2 \text{ or } x_2, z_2 \text{ or } y_2, z_2)$, and you need to find the third unknown value.
Joint Variation
Joint variation is a special case of combined variation where a quantity varies directly as the product of two or more other quantities. There is no inverse variation involved in the basic definition of joint variation.
If a quantity $z$ varies jointly as $x$ and $y$, we write the proportionality as:
$\boldsymbol{z \propto xy}$
Introducing the constant of variation $k$, the equation becomes:
$\boldsymbol{z = kxy}$
... (v)
From equation (v), assuming $x \neq 0$ and $y \neq 0$, we can solve for the constant $k$ by dividing both sides by $xy$:
$\boldsymbol{k = \frac{z}{xy}}$
[Constant of Joint Variation]
This indicates that for any set of corresponding non-zero values $(x, y, z)$, the ratio $\frac{z}{xy}$ will always be equal to the same constant $k$.
Relationship between Two Sets of Values in Joint Variation
Consider two different sets of corresponding values $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ for quantities related by joint variation $z \propto xy$.
Based on the definition $k = \frac{z}{xy}$, for the first set of values $(x_1, y_1, z_1)$, we have:
$\boldsymbol{\frac{z_1}{x_1 y_1} = k}$
... (vi)
And for the second set of values $(x_2, y_2, z_2)$, we have:
$\boldsymbol{\frac{z_2}{x_2 y_2} = k}$
... (vii)
Equating the expressions for $k$ from equations (vi) and (vii), we get the relationship for solving problems:
$\boldsymbol{\frac{z_1}{x_1 y_1} = \frac{z_2}{x_2 y_2}}$
... (viii)
This equation is used when you are given one complete set of values $(x_1, y_1, z_1)$ and two values from a second set, and need to find the third unknown value in a joint variation scenario.
Applications of Combined and Joint Variation
These types of variations appear in many formulas and practical problems:
1. Work Done: The total work done often varies jointly with the number of workers, the time they work, and their efficiency. If work is $W$, number of workers is $N$, hours per day is $H$, and days worked are $D$, and efficiency is $E$, then $W \propto N \times H \times D \times E$. If efficiency is constant, $W \propto NHD$, which is a joint variation. If we compare two scenarios with different workers, hours, and days for the same amount of work ($W_1=W_2$), then $N_1 H_1 D_1 = N_2 H_2 D_2$ (a common application derived from $k = NHD/W$ where $W$ is fixed).
2. Area of a Rectangle: The area ($A$) of a rectangle varies jointly as its length ($L$) and breadth ($B$). $A \propto LB \implies A = kLB$. If $k=1$, $A=LB$.
3. Volume of a Cylinder: The volume ($V$) of a cylinder varies jointly as the square of its radius ($r$) and its height ($h$). $V \propto r^2 h \implies V = k r^2 h$. If $k=\pi$, $V = \pi r^2 h$.
4. Force of Attraction (Gravitation): The force of gravitational attraction ($F$) between two objects varies jointly as their masses ($m_1, m_2$) and inversely as the square of the distance ($r$) between their centers. $F \propto \frac{m_1 m_2}{r^2} \implies F = G \frac{m_1 m_2}{r^2}$, where $G$ is the gravitational constant. This is a classic example of combined variation.
5. Electrical Resistance: The resistance ($R$) of a wire varies directly as its length ($L$) and inversely as its cross-sectional area ($A$). $R \propto \frac{L}{A} \implies R = \rho \frac{L}{A}$, where $\rho$ is the resistivity of the material. This is another example of combined variation.
Example 1. If 8 workers can build a wall 18 metres long in 12 days, how many days will 12 workers take to build a wall 27 metres long?
Answer:
Let $W$ be the number of workers, $L$ be the length of the wall, and $D$ be the number of days.
The length of the wall built is directly proportional to the number of workers and the number of days they work (assuming constant work rate).
$\boldsymbol{L \propto W \times D}$
This is a case of joint variation. The equation is $L = kWD$, where $k$ is the constant related to the efficiency of the workers and the difficulty of the work.
From this, the constant $k$ is given by $k = \frac{L}{WD}$.
We have two scenarios:
Scenario 1: $W_1 = 8$ workers, $L_1 = 18$ metres, $D_1 = 12$ days.
Scenario 2: $W_2 = 12$ workers, $L_2 = 27$ metres, $D_2 = ?$ days.
Since $k$ is constant for both scenarios, we have $\frac{L_1}{W_1 D_1} = \frac{L_2}{W_2 D_2}$.
Substitute the given values:
$\boldsymbol{\frac{18}{8 \times 12} = \frac{27}{12 \times D_2}}$
... (ix)
Simplify the left side:
$\frac{18}{96} = \frac{\cancel{18}^3}{\cancel{96}^{16}} = \frac{3}{16}$
So, the equation becomes:
$\frac{3}{16} = \frac{27}{12 D_2}$
Now, solve for $D_2$ using cross-multiplication:
$\boldsymbol{3 \times (12 D_2) = 16 \times 27}$
$\boldsymbol{36 D_2 = 432}$
... (x)
Divide both sides of equation (x) by 36:
$\boldsymbol{D_2 = \frac{432}{36}}$
Performing the division:
$\frac{432}{36} = 12$
$\boldsymbol{D_2 = 12}$ days
It will take $\boldsymbol{12}$ days for 12 workers to build a wall 27 metres long.
Example 2. The time taken ($T$) to travel a certain distance varies directly as the distance ($D$) and inversely as the speed ($S$). If it takes 4 hours to travel 200 km at a speed of 50 km/h, how long will it take to travel 300 km at a speed of 60 km/h?
Answer:
Given that the time taken ($T$) varies directly as the distance ($D$) and inversely as the speed ($S$).
$\boldsymbol{T \propto \frac{D}{S}}$
This is a case of combined variation. The equation is $T = k \frac{D}{S}$, where $k$ is the constant of proportionality.
From this, the constant $k$ is given by $k = \frac{TS}{D}$. (Note: In this specific case, for consistent units, $k$ should actually be 1, as Time = Distance / Speed. However, we will solve it generally using $k$ as per the variation concept.)
We have two scenarios:
Scenario 1: $T_1 = 4$ hours, $D_1 = 200$ km, $S_1 = 50$ km/h.
Scenario 2: $T_2 = ?$ hours, $D_2 = 300$ km, $S_2 = 60$ km/h.
Since $k$ is constant for both scenarios, we have $\frac{T_1 S_1}{D_1} = \frac{T_2 S_2}{D_2}$.
Substitute the given values:
$\boldsymbol{\frac{4 \times 50}{200} = \frac{T_2 \times 60}{300}}$
... (xi)
Simplify both sides:
Left side: $\frac{4 \times 50}{200} = \frac{200}{200} = 1$
Right side: $\frac{T_2 \times 60}{300} = \frac{60 T_2}{300} = \frac{T_2}{5}$
So, the equation becomes:
$\boldsymbol{1 = \frac{T_2}{5}}$
Solve for $T_2$ by multiplying both sides by 5:
$\boldsymbol{T_2 = 1 \times 5}$
$\boldsymbol{T_2 = 5}$ hours
It will take $\boldsymbol{5}$ hours to travel 300 km at a speed of 60 km/h.
Alternate Method (Finding the constant $k$ first):
Using the first set of values to find $k = \frac{TS}{D}$:
$\boldsymbol{k = \frac{4 \times 50}{200} = \frac{200}{200} = 1}$
So the relationship is $T = 1 \times \frac{D}{S}$ or $T = \frac{D}{S}$ (as expected for Time, Distance, Speed).
Now use the second set of values to find $T_2$ when $D_2 = 300$ and $S_2 = 60$:
$\boldsymbol{T_2 = \frac{D_2}{S_2} = \frac{300}{60}}$
$\boldsymbol{T_2 = 5}$ hours
Both methods give the same result.
Example 3. The kinetic energy ($E$) of a moving object varies jointly as its mass ($m$) and the square of its velocity ($v$). If an object with mass 10 kg moving at 5 m/s has kinetic energy 125 Joules, what is the kinetic energy of an object with mass 20 kg moving at 3 m/s?
Answer:
Given that kinetic energy ($E$) varies jointly as mass ($m$) and the square of velocity ($v^2$).
$\boldsymbol{E \propto m v^2}$
This is a case of joint variation. The equation is $E = k m v^2$, where $k$ is the constant of proportionality.
From this, the constant $k$ is given by $k = \frac{E}{mv^2}$. (Note: The constant $k$ here is $\frac{1}{2}$ in the standard physics formula $E = \frac{1}{2}mv^2$. We will find this value).
We have two scenarios:
Scenario 1: $E_1 = 125$ J, $m_1 = 10$ kg, $v_1 = 5$ m/s.
Scenario 2: $E_2 = ?$ J, $m_2 = 20$ kg, $v_2 = 3$ m/s.
Since $k$ is constant for both scenarios, we have $\frac{E_1}{m_1 v_1^2} = \frac{E_2}{m_2 v_2^2}$.
Substitute the given values:
$\boldsymbol{\frac{125}{10 \times 5^2} = \frac{E_2}{20 \times 3^2}}$
... (xii)
Simplify both denominators:
$\frac{125}{10 \times 25} = \frac{E_2}{20 \times 9}$
$\frac{125}{250} = \frac{E_2}{180}$
Simplify the left side:
$\frac{125}{250} = \frac{1}{2}$
So, the equation becomes:
$\boldsymbol{\frac{1}{2} = \frac{E_2}{180}}$
Solve for $E_2$ by multiplying both sides by 180:
$\boldsymbol{E_2 = \frac{1}{2} \times 180}$
$\boldsymbol{E_2 = 90}$ Joules
The kinetic energy of the second object is $\boldsymbol{90}$ Joules.
Alternate Method (Finding the constant $k$ first):
Using the first set of values to find $k = \frac{E}{mv^2}$:
$\boldsymbol{k = \frac{125}{10 \times 5^2} = \frac{125}{10 \times 25} = \frac{125}{250} = \frac{1}{2}}$
The relationship is $E = \frac{1}{2} m v^2$.
Now use the second set of values to find $E_2$ when $m_2 = 20$ and $v_2 = 3$:
$\boldsymbol{E_2 = \frac{1}{2} \times m_2 \times v_2^2 = \frac{1}{2} \times 20 \times 3^2}$
$\boldsymbol{E_2 = \frac{1}{2} \times 20 \times 9 = 10 \times 9 = 90}$ Joules
Both methods give the same result and illustrate how the constant $k$ is $\frac{1}{2}$ in this physical formula.
Competitive Exam Notes:
Combined and Joint Variation problems require careful setup based on the problem statement. These are common in questions involving Work, Force, Energy, Rates, etc.
- Identifying Combined Variation: Look for phrases like "varies directly as ... and inversely as ...". Set up the proportionality like $z \propto \frac{x}{y}$ or $z \propto \frac{x_1 x_2}{y_1 y_2}$.
- Identifying Joint Variation: Look for phrases like "varies jointly as ... and ...". Set up the proportionality like $z \propto xy$ or $z \propto x_1 x_2 x_3$.
- Formulating the Equation: Convert the proportionality to an equation using a constant $k$: $z = k \frac{x}{y}$ or $z = kxy$.
- Using Two Sets of Values: Once the relationship is established ($k = \frac{zy}{x}$ or $k = \frac{z}{xy}$), equate the expression for $k$ for the two given scenarios: $\frac{z_1 y_1}{x_1} = \frac{z_2 y_2}{x_2}$ (for $z \propto x/y$) or $\frac{z_1}{x_1 y_1} = \frac{z_2}{x_2 y_2}$ (for $z \propto xy$). This is often the most direct way to solve problems without explicitly finding $k$ numerically first, though finding $k$ can be helpful for understanding the relationship.
- Work-Rate Problems: Many problems involving number of workers, days, hours, and amount of work fall under combined variation. A common template is $\frac{M_1 D_1 T_1}{W_1} = \frac{M_2 D_2 T_2}{W_2}$, where M=Men, D=Days, T=Time per day, W=Work done. Here, Work done ($W$) is directly proportional to Man ($M$), Days ($D$), and Time per day ($T$), i.e., $W \propto MDT$. So $\frac{W}{MDT} = k$, or $k = \frac{W}{MDT}$. Equating $k$ for two scenarios gives $\frac{W_1}{M_1 D_1 T_1} = \frac{W_2}{M_2 D_2 T_2}$, which rearranges to $\frac{M_1 D_1 T_1}{W_1} = \frac{M_2 D_2 T_2}{W_2}$ if $W_1, W_2 \neq 0$. If the work is the same ($W_1=W_2$), it simplifies to $M_1 D_1 T_1 = M_2 D_2 T_2$.
- Careful with Squares/Cubes: Pay attention if the variation involves the square or cube of a quantity (like $r^2$ or $v^2$ in the examples).