Solving Ratio, Proportion, and Variation Equations

Hello. In the previous lesson, you solved linear equations by preserving equality, clearing denominators with an LCD, and checking the final value in the original equation. Those skills are now used in word problems: the difficult part is often not the algebra, but deciding which equation the words require.

This lesson covers ratios, proportions, direct variation, and inverse variation. By the end, you should be able to turn common entrance-exam statements into equations, solve them efficiently, and distinguish relationships that look similar but behave differently.

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Ratios: parts, not necessarily actual quantities

A ratio compares quantities in a specified order. For example,

$$A:B=2:3$$

means that for every $2$ equal parts of $A$, there are $3$ equal parts of $B$. It does not mean $A=2$ and $B=3$. The actual values could be:

$$2:3,\quad 10:15,\quad 24:36$$

All have the same ratio because each is obtained by multiplying both parts by the same number.

The most dependable algebraic translation is:

$$A=2t,\qquad B=3t$$

where $t$ represents the value of one ratio-unit. This automatically preserves the relationship.

When a total is given

Suppose boys, girls, and teachers are in the ratio

$$2:3:5$$

and their total number is $200$.

Write each quantity using one common multiplier:

$$B=2t,\qquad G=3t,\qquad T=5t$$

Since the total is $200$,

$$2t+3t+5t=200$$ $$10t=200$$ $$t=20$$

Therefore,

$$B=40,\qquad G=60,\qquad T=100$$

The key shortcut is: add the ratio parts first.

$$2+3+5=10$$

So each part is

$$\frac{200}{10}=20.$$

When a difference is given

Suppose two quantities are in the ratio $5:3$, and their difference is $28$.

Let the quantities be $5t$ and $3t$. Their difference gives the equation:

$$5t-3t=28$$ $$2t=28$$ $$t=14$$

Thus the quantities are

$$5t=70,\qquad 3t=42.$$

For a ratio-and-difference question, use the difference of the ratio parts. For a ratio-and-total question, use the sum of the ratio parts.

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Proportions: equality of two ratios

A proportion says that two ratios are equal:

$$\frac{a}{b}=\frac{c}{d}$$

provided $b\neq0$ and $d\neq0$.

A ratio is a comparison, such as $2:5$. A proportion is an equation, such as

$$\frac{2}{5}=\frac{8}{20}.$$

To solve a proportion, multiply both sides by the product of the denominators. This produces the familiar cross-product rule:

$$\frac{a}{b}=\frac{c}{d}$$ $$ad=bc.$$

Cross-multiplication is therefore not a separate magic trick; it is simply the fraction-clearing method from the previous lesson.

Keep units in the same positions

Suppose a machine makes $14$ identical components in $6$ minutes at a constant rate. How many components, $x$, does it make in $15$ minutes?

Keep components in numerators and time in denominators:

$$\frac{14\text{ components}}{6\text{ min}} = \frac{x\text{ components}}{15\text{ min}}$$

Now cross-multiply:

$$6x=14(15)$$ $$6x=210$$ $$\boxed{x=35}$$

The equation is valid because both fractions mean the same thing: components per minute.

A frequent setup error is reversing only one ratio:

$$\frac{14}{6}=\frac{15}{x}$$

This compares components per minute on the left with minutes per component on the right, so it is invalid. Before solving, inspect the units: numerator units must match numerator units, and denominator units must match denominator units.

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Direct variation: a constant ratio

When $y$ varies directly with $x$, the relationship has the form

$$y=kx,$$

where $k$ is a fixed number called the constant of variation.

Equivalently,

$$\frac{y}{x}=k.$$

So in direct variation, the quotient $\frac{y}{x}$ stays constant. If $x$ doubles, $y$ also doubles; if $x$ becomes three times as large, $y$ becomes three times as large.

For example, suppose the cost $C$ of a material varies directly with the number of kilograms $q$. If $7$ kg costs $294$ rupees, find the equation and then the cost of $12.5$ kg.

Start with the direct-variation model:

$$C=kq$$

Use the known pair $C=294$, $q=7$:

$$294=7k$$ $$k=42.$$

Thus the complete equation is

$$\boxed{C=42q}.$$

For $q=12.5$,

$$C=42(12.5)$$ $$\boxed{C=525}.$$

Here, $k=42$ has a useful meaning: the material costs $42$ rupees per kilogram.

Not every straight-line equation is direct variation. Compare:

$$y=4x$$

and

$$y=4x+7.$$

The first is direct variation because it has the form $y=kx$ and passes through the origin. The second has a fixed extra amount of $7$, so the ratio $\frac{y}{x}$ is not constant. A taxi fare with a base charge plus a per-kilometre rate is usually linear, but not directly proportional.

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Inverse variation: a constant product

When $y$ varies inversely with $x$, the relationship is

$$y=\frac{k}{x}.$$

Multiplying by $x$ gives the form most useful in exams:

$$xy=k.$$

Thus, in inverse variation, the product $xy$ stays constant. If $x$ doubles, $y$ halves. If $x$ becomes three times as large, $y$ becomes one-third as large.

Suppose the time $t$ required to finish a fixed job varies inversely with the number $n$ of equally efficient workers. If $6$ workers take $15$ days, how long will $10$ workers take?

Because this is inverse variation,

$$tn=k.$$

Use the known values:

$$15(6)=k$$ $$k=90.$$

Therefore,

$$tn=90.$$

For $n=10$,

$$10t=90$$ $$\boxed{t=9}.$$

The answer is sensible: more workers should require less time. This model assumes all workers are equally productive and that the total job remains fixed.

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The fastest recognition test is:

Relationship	Constant quantity	Standard equation
Direct variation	$\frac{y}{x}$	$y=kx$
Inverse variation	$xy$	$y=\frac{k}{x}$
Neither necessarily	Neither is constant	Must inspect the equation

For instance, consider the pairs $(x,y)$:

$$(2,10),\qquad (4,20).$$

Their ratios are

$$\frac{10}{2}=5,\qquad \frac{20}{4}=5.$$

The ratio is constant, so this is direct variation:

$$y=5x.$$

Now consider:

$$(2,20),\qquad (4,10).$$

Their products are

$$2(20)=40,\qquad 4(10)=40.$$

The product is constant, so this is inverse variation:

$$y=\frac{40}{x}.$$

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Turning wording into the right equation

Certain phrases should trigger a particular model immediately:

Wording in the question	Algebraic model
“$y$ is directly proportional to $x$”	$y=kx$
“$y$ varies directly as $x$”	$y=kx$
“$y$ is inversely proportional to $x$”	$y=\frac{k}{x}$
“$y$ varies inversely as $x$”	$xy=k$
“$a:b=m:n$”	$a=mt,\ b=nt$
“For every $a$ units, there are $b$ units”	Set up equal ratios with matching units

A reliable four-step routine is:

1. Name the quantities and attach units where possible.
2. Identify the relationship: ratio, direct variation, inverse variation, or a general equation.
3. Write the general model before substituting numbers.
4. Check direction and units after solving.

The direction check catches many errors quickly:

• In direct variation, an increase in one positive quantity should produce an increase in the other.
• In inverse variation, an increase in one positive quantity should produce a decrease in the other.
• In a ratio-total question, the individual quantities must add to the stated total.
• In a ratio-difference question, the computed quantities must have the stated difference.

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An exam-speed decision routine

When a question presents two quantities, do not begin calculation immediately. Classify it first.

• If it gives a comparison such as $3:4$, introduce a common multiplier $t$.
• If it equates two rates or scale relationships, form a proportion with matching units.
• If it states “directly proportional,” use $y=kx$.
• If it states “inversely proportional,” use $xy=k$.
• If it gives an equation such as $y=5x+12$, do not force it into direct variation merely because $y$ rises when $x$ rises.

For IPMAT-style questions, this classification step usually takes only a few seconds, but it prevents spending a minute solving the wrong equation.

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Key takeaways

A ratio such as $a:b=m:n$ describes relative parts, best written as $a=mt$ and $b=nt$. Use the sum of ratio parts when a total is given, and their difference when a difference is given.

A proportion equates two ratios. Keep units in the same order, then clear fractions or cross-multiply. Direct variation has the model

$$y=kx,$$

so $\frac{y}{x}$ is constant. Inverse variation has the model

$$y=\frac{k}{x},$$

so $xy$ is constant.

Next, you will move into simultaneous linear equations, where two related equations determine two unknown quantities. The equation-formation discipline from this lesson will remain essential.