Proportions and Variations: Definition, Formula, Questions, Examples

"Proportions and variations are basic concepts in mathematics, important for understanding relationships between quantities. These concepts, including the difference in proportions and the difference between proportion and variation, are widely applied in the fields of physics, economics, and statistics. Understanding proportions helps in solving problems related to ratios, variations, and proportions, while variations are crucial for analyzing how one quantity changes with respect to another."

Concept of Proportion:

A proportion is an equation that states two ratios are equal. In other words, the proportion states the equality of the two fractions or the ratios and it is denoted by the ‘::’ symbol. For example, if we have the ratios $a:b$ and $c:d$, then they form a proportion if $\frac{a}{b} = \frac{c}{d}$ and it is written as $a:b::c:d$. Here, $a, b, c$, and $d$ are the terms of the proportions.

• First term: The first term is the numerator of the first ratio.

• Second term: The second term is the denominator of the first ratio.

• Third term: The third term is the numerator of the second ratio.

• Last term: The last term is the denominator of the last ratio.

Let's consider the proportion $4:5::12:15$ which can be written as $\frac{4}{5} = \frac{12}{15}$ Here, 4 is the first term, 5 is the second term, 12 is the third term and 15 is the last term.

Continued Proportion and Mean Term:

A continued proportion involves three quantities where the ratio of the first to the second is equal to the ratio of the second to the third. For example, $a, b, c$ will be a continued proportion if $\frac{a}{b} = \frac{b}{c}$ and it will be written as $a:b::b:c$, where b will be the mean term.

Let’s take another example where $2, 4, 8$ is in continued proportion as $\frac{2}{4}=\frac{4}{8}=\frac{1}{2}$, so we can say that $2:4::4:8$ and the mean term is 4.

What is Variation?

Variation describes how one quantity changes with respect to another. If two quantities are related in such a way that one quantity changes when the other changes, then they are said to be in a variation. It is denoted by the “∝” symbol. For example, If A and B are in direct variation then we can say that A ∝ B.

Direct Variation and Indirect Variation:

Variation can be classified into two types based on the pattern of change and the relationships between variables.

• Direct variation: When two quantities increase or decrease together, then they are said to be in a direct variation.

For example, if a car travels at a constant speed, the distance (s) travelled varies directly with time (t) and it is denoted by s ∝ t.

• Indirect or inverse variation: When one quantity increases while the other decreases and if one quantity decreases while the other increases, then they are said to be in an indirect or inverse variation.

For example, The time taken to complete a task varies inversely with the number of people working. If 4 people take 3 hours to complete a task, then 6 people would take $\frac{4×3}{6}$ = 2 hours.

Properties of Proportion

The important properties of proportion are:

• Addendo: If $a:b=c:d$ then $a+c:b+d$

• Subtrahendo: If $a:b=c:d$ then $a-c:b-d$

• Dividendo: If $a:b=c:d$ then $a-b:b=c-d:d$

• Componendo: If $a:b=c:d$ then $a+b:b=c+d:d$

• Alternendo: If $a:b=c:d$ then $a:c=b:d$

• Invertendo: If $a:b=c:d$ then $b:a=d:a$

• Componendo and dividendo: If $a:b=c:d$ then $a+b:a-b=c+d:c-d$

Properties of Variation

• If $x$ and $y$ are in direct variation, then the ratio $\frac{x}{y}$ is a constant.

• The graph of a direct variation is a straight line passing through the origin (0,0). This linear relationship indicates a consistent rate of change.

• In direct variation, both variables increase or decrease together. If one variable doubles, the other also doubles.

• If $x$ and $y$ are in indirect or inverse variation, then the product $xy$ is a constant.

• The graph of an inverse variation is a hyperbola. This non-linear relationship indicates that the rate of change varies with the values of the variables.

• In an indirect or inverse variation as one variable increases, the other decreases. If one variable doubles, the other is halved.

Summary of the formula used

• If $a, b, c, d$ are in proportion, then $a:b::c:d$ i.e. $\frac{a}{b} = \frac{c}{d}$

• If $a, b, c$ are in continued proportion, then $a:b::b:c$ i.e. $\frac{a}{b} = \frac{b}{c}$

• If $a$ and $b$ are in direct variation then $a ∝ b$ i.e. $a = kb$ or $\frac{a}{b}=k$, where $k$ is a constant.

• If $a$ and $b$ are in indirect or inverse variation then $a ∝ \frac{1}{b}$ i.e. $a = \frac{k}{b}$ or $ab=k$, where $k$ is a constant.

Tips and Tricks

• For problems related to proportions, use cross-multiplication and simplify to solve them.

• First, identify whether a problem involves direct or inverse variation then solve it accordingly.

• In direct variation, if one value doubles, the other also doubles.

• In inverse variation, if one value doubles, the other is halved.

• If $\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}$ and $a+b+c \neq 0$, then $a=b=c$.

Practice Questions/Solved Examples on Proportion and Variation

Q.1.

If $(2x-y)^{2}+(3y-2z)^{2}=0$, then the ratio $x:y:z$ is:

1. $1:3:2$

2. $1:2:3$

3. $3:1:2$

4. $3:2:1$

Hint: If the sum of squares of two terms is equal to zero, then the individual terms will also be zero.

Solution:

Given: $(2x - y)^2 + (3y - 2z)^2 = 0$

If the sum of squares of two terms is equal to zero, then the individual terms will also be zero.

Since, $(2x - y)^2 + (3y - 2z)^2 = 0$

$\therefore$ $2x - y = 0$ and $3y - 2z = 0$

$2x - y = 0$

$⇒ 2x = y$

$⇒ x:y = 1:2$

Also, $3y - 2z = 0$

$⇒ 3y = 2z$

$⇒ y:z = 2:3$

$\therefore$ $x:y:z = 1:2:3$

Hence, the correct answer is option (2).

Q.2.

The speeds of the three cars are in the ratio of $1:3:5$. The ratio of the time taken by these cars to travel the same distance is _____.

1. $3:5:15$

2. $15:3:5$

3. $15:5:3 $

4. $5:3:1$

Hint: Time is inversely proportional to the speed.

$\text{Time} \propto \frac{1}{\text{Speed}}$

Solution:

We know, $\text{Time} \propto \frac{1}{\text{Speed}}$

Required ratio of time = $1:\frac{1}{3}:\frac{1}{5}$

L.C.M. of 1, 3, and 5 is 15.

= $1 × 15 : \frac{1}{3}×15:\frac{1}{5}×15$

= $15:5:3$

Hence, the correct answer is option (3).

Q.3.

If a 7-storey building has a 28 m long shadow, the number of storeys of the building whose shadow is 48 m long is:

1. 14

2. 24

3. 16

4. 12

Hint: The length of the shadow is proportional to the number of storeys in the building.

Solution:

Length of shadow of a 7-storey building = 28 m

Let the number of stories of the building with a 48 m long shadow be $x$.

During the same time of the day, the sun shall cast proportional shadows on all the objects,

Hence, $\frac{7}{28}=\frac{x}{48}$

⇒ $x=\frac{48}{4}$

⇒ $x=12$.

Hence, the correct answer is option (4).

Q.4.

The force (in pound-force) needed to keep a car from skidding on a curve varies directly with the weight of the car (in pounds) and the square of its speed (in miles per hour [mph]) and inversely with the radius (in feet) of the curve. Suppose 6125 pound-force is required to keep a 2750 pound car, travelling at a speed of 35 mph, from skidding on a curve of radius 550 feet. How much pound-force is then required to keep a 3600-pound car, travelling at a speed of 50 mph, from skidding on a curve of radius 750 feet?

1. 11960

2. 12150

3. 12240

4. 12000

Hint: The force ($F$) needed to keep a car from skidding on a curve varies jointly as the weight of the car ($w$) and the square of the car's speed ($s$) and inversely as the radius of the curve ($r$).

$F = k[\frac{ws^2}{r}]$

Solution:

The force ($F$) needed to keep a car from skidding on a curve varies jointly as the weight of the car ($w$) and the square of the car's speed ($s$) and inversely as the radius of the curve ($r$).

$F = k[\frac{ws^2}{r}]$

Given: Force of 6125 pounds is needed to keep a 2750-pound car travelling at 35 mph from skidding on a curve of radius 550 feet.

⇒ $6125 = k[\frac{2750\times{35^2}}{550}]$

$k = 1$

Equation:

$F = 1[\frac{ws^2}{r}]$

⇒ $F = 1[\frac{3600\times{50^2}}{750}]$

⇒ $F = 12000$ pound-force

$\therefore$ 12000 pound-force is then required to keep a 3600-pound car, travelling at a speed of 50 mph, from skidding on a curve of radius 750 feet.

Hence, the correct answer is option (4).

Q.5.

If the cost of 120 m of cloth is INR 9,600, then what will be the cost of 147 m of that cloth?

1. INR 16,170

2. INR 11,670

3. INR 11,760

4. INR 17,160

Hint: Cost varies directly as the length of the cloth.

$\frac{\text{Cost}_1}{\text{Cost}_2} = \frac{\text{Length}_1}{\text{Length}_2}$

Solution:

Given: The cost of 120 m of cloth = INR 9,600

$\frac{\text{Cost}_1}{\text{Cost}_2} = \frac{\text{Length}_1}{\text{Length}_2}$

$⇒\frac{9600}{\text{Cost}_2}=\frac{120}{147}$

$\therefore$ The cost of 147m of cloth = 147 × 80 = INR 11760

Hence, the correct answer is option (3).

Q.6.

$A$ varies directly as $(B+18)$ and $A=108$ when $B=36$. Find the value of $A$ when $B=68$.

1. 75

2. 86

3. 127

4. 172

Hint: A varies directly as $(B+18)$

So, $\frac{A_1}{A_2} = \frac{B_1+18}{B_2+18}$

Solution:

A varies directly as $(B+18)$

So, $\frac{A_1}{A_2} = \frac{B_1+18}{B_2+18}$

⇒ $\frac{108}{A_2} = \frac{36+18}{68+18}$

⇒ $\frac{108}{A_2} = \frac{54}{86}$

⇒ $A_2 = 172$

Hence, the correct answer is option (4).

Q.7.

The third proportional of the following numbers $(x-y)^2, (x^2-y^2)^2$ is:

1. $(x+y)^3(x-y)^2$

2. $(x+y)^4(x-y)^2$

3. $(x+y)^2(x-y)^2$

4. $(x+y)^2(x-y)^3$

Hint: The formula to find the third proportional is $\frac{b^2}{a}$, where $a$ = first proportion and $b$ = second proportion.

Solution:

Given:

Two numbers are $(x-y)^2, (x^2-y^2)^2$.

So, let's take $a=(x-y)^2$ = 1st proportion,

$b=(x^2-y^2)^2$ = 2nd proportion

Now, $c$ = third proportion = $\frac{b^2}{a}=\frac{((x^2-y^2)^2)^2}{(x-y)^2}=\frac{(x-y)^4 (x+y)^4}{(x-y)^2}=(x+y)^4(x-y)^2$

Hence, the correct answer is $(x+y)^4(x-y)^2$.

Q.8.

What is the third proportional number to 10 and 20?

1. 30

2. 25

3. 50

4. 40

Hint: We know, if $x$ is the third proportion of $a$ and $b$, then $a:b::b:x$.

Solution:

Let $x$ be the third proportional of 10 and 20.

⇒ $10:20::20:x$

⇒ $\frac{10}{20}=\frac{20}{x}$

⇒ $x=\frac{400}{10}$ = 40

Hence, the correct answer is option (4).

Q.9.

What is the fourth proportional to 189, 273, and 153?

1. 117

2. 299

3. 221

4. 187

Hint: In the ratio $a:b::c:d$, $d$ is the fourth proportion.

Solution:

Let the fourth proportional be $a$.

According to the question,

⇒ $189:273=153:a$

⇒ $a=\frac{273×153}{189}$

⇒ $a=221$

Hence, the correct answer is option (3).

Q.10.

Find two numbers such that their mean proportion is 16 and the third proportion is 1024.

1. 4 and 32

2. 4 and 64

3. 8 and 64

4. 8 and 32

Hint: The mean proportion of two numbers $a$ and $b$ = $\sqrt{ab}$ and third proportion is $\frac{b^2}{a}$.

Solution:

Let the two numbers as $a$ and $b$. The mean proportion of two numbers is the square root of their product.

$\sqrt{ab} = 16$

Squaring both sides,

$ab = 256$.....................................(1)

The third proportional of two numbers $a$ and $b$ is the number which is to $b$ as $b$ is to $a$.

$\frac{b}{a} = \frac{1024}{b}$

⇒ $b^2 = 1024a$...............................(2)

We can substitute $a = \frac{256}{b}$ from equation (1) into equation (2).

⇒ $b^2 = 1024×\frac{256}{b}$

⇒ $b^3 = 1024×256$

⇒ $b$ = 64

Substituting $b = 64$ in equation (1)

⇒ $64a = 256$

⇒ $a$ = 4

So, the two numbers are 4 and 64.

Hence, the correct answer is option (2).