Published 2026-10-28
Chapter: Ratio and Proportion

Ratio and Proportion - Understanding ratios, equivalent ratios, checking proportions, and applying the unitary method to solve practical problems

In mathematics, comparing quantities is a fundamental skill used to analyze relationships between numbers, measurements, and real-world scenarios. While subtraction allows us to compare quantities by finding their difference, it does not always provide a meaningful comparison when quantities vary significantly in scale. To solve this, we use division to compare quantities. This method of comparison by division forms the foundation of Ratio and Proportion.

Mastering ratios, equivalent ratios, proportions, and the unitary method allows us to solve everyday practical problems, such as scaling recipes, determining speed and distance relationships, adjusting map scales, and calculating costs during shopping.


1. Concept of Ratio

1.1 Why Comparison by Division?

Consider two trees: Tree A is 3 metres3 \text{ metres} tall, and Tree B is 12 metres12 \text{ metres} tall.

  • Comparison by Difference: Height of Tree B−Height of Tree A=12 m−3 m=9 m\text{Height of Tree B} - \text{Height of Tree A} = 12\text{ m} - 3\text{ m} = 9\text{ m}. We can say Tree B is 9 metres9\text{ metres} taller than Tree A.
  • Comparison by Division: Height of Tree BHeight of Tree A=123=4\frac{\text{Height of Tree B}}{\text{Height of Tree A}} = \frac{12}{3} = 4. We can say Tree B is 44 times as tall as Tree A.

Now consider an insect of length 1 cm1\text{ cm} and a snake of length 100 cm100\text{ cm}. Comparing them by difference gives 99 cm99\text{ cm}, which gives little context regarding scale. Comparing them by division shows that the snake is 100100 times longer than the insect. Therefore, comparison by division provides a clearer understanding of relative magnitude.

1.2 Definition of Ratio

A ratio is a relationship between two quantities of the same kind, expressed as the quotient of one divided by the other. It shows how many times one quantity contains another.

If aa and bb are two non-zero quantities of the same unit, the ratio of aa to bb is written as:

a:boraba : b \quad \text{or} \quad \frac{a}{b}

The symbol :: is read as "is to". Thus, a:ba : b is read as "aa is to bb".

  • The first term (aa) is called the antecedent (or first term).
  • The second term (bb) is called the consequent (or second term).

1.3 Essential Rules of Ratios

  1. Same Units Requirement: The quantities being compared must be expressed in the same unit of measurement. If they are not, they must first be converted to a common unit.
    • Example: To find the ratio of 50 paise50\text{ paise} to ₹ 2\text{₹ } 2, convert ₹ 2\text{₹ } 2 into paise (200 paise200\text{ paise}) before making the comparison.
  2. Order Matters: The order of terms in a ratio is strict. a:ba : b is not equal to b:ab : a unless a=ba = b.
    • Example: If a class has 1515 boys and 2020 girls, the ratio of boys to girls is 15:2015 : 20, whereas the ratio of girls to boys is 20:1520 : 15.
  3. No Units in the Final Ratio: A ratio is a pure number and has no unit.
  4. Simplest Form (Lowest Terms): A ratio a:ba : b is in its simplest form if the highest common factor (HCF) of aa and bb is 11.
              ┌─────────────────────────────────────────┐
              │           Quantities a and b            │
              └────────────────────┬────────────────────┘
                                   │
                         Convert to Same Units
                                   │
                                   ▼
              ┌─────────────────────────────────────────┐
              │         Write as Fraction (a/b)         │
              └────────────────────┬────────────────────┘
                                   │
                     Divide by HCF of (a, b)
                                   │
                                   ▼
              ┌─────────────────────────────────────────┐
              │      Simplest Form Ratio (m : n)        │
              └─────────────────────────────────────────┘

2. Equivalent Ratios

2.1 Definition

Two ratios are said to be equivalent if their corresponding simplified fractions are equal.

Just like equivalent fractions, an equivalent ratio is obtained by multiplying or dividing both the antecedent (first term) and the consequent (second term) by the same non-zero number.

Given a ratio a:ba : b: Equivalent Ratio (by multiplication)=a×kb×k=(a⋅k):(b⋅k)[k≠0]\text{Equivalent Ratio (by multiplication)} = \frac{a \times k}{b \times k} = (a \cdot k) : (b \cdot k) \quad [k \neq 0]

Equivalent Ratio (by division)=a÷mb÷m=(a÷m):(b÷m)[m≠0]\text{Equivalent Ratio (by division)} = \frac{a \div m}{b \div m} = (a \div m) : (b \div m) \quad [m \neq 0]

2.2 Finding Equivalent Ratios

Let us find three equivalent ratios for 2:32 : 3:

2×23×2=46  ⟹  4:6\frac{2 \times 2}{3 \times 2} = \frac{4}{6} \implies 4 : 6

2×33×3=69  ⟹  6:9\frac{2 \times 3}{3 \times 3} = \frac{6}{9} \implies 6 : 9

2×43×4=812  ⟹  8:12\frac{2 \times 4}{3 \times 4} = \frac{8}{12} \implies 8 : 12

Thus, 2:3=4:6=6:9=8:122 : 3 = 4 : 6 = 6 : 9 = 8 : 12.

2.3 Comparing Ratios

To compare two ratios to determine which is larger, convert them into fractions with a common denominator (using the Least Common Multiple, or LCM).

Ratio 1Ratio 2Fractional FormLCM of DenominatorsEqualized FractionsComparison
2:32 : 33:53 : 523\frac{2}{3} and 35\frac{3}{5}LCM(3,5)=15\text{LCM}(3, 5) = 152×53×5=1015\frac{2 \times 5}{3 \times 5} = \frac{10}{15}, 3×35×3=915\frac{3 \times 3}{5 \times 3} = \frac{9}{15}1015>915  ⟹  2:3>3:5\frac{10}{15} > \frac{9}{15} \implies 2 : 3 > 3 : 5
4:74 : 75:85 : 847\frac{4}{7} and 58\frac{5}{8}LCM(7,8)=56\text{LCM}(7, 8) = 564×87×8=3256\frac{4 \times 8}{7 \times 8} = \frac{32}{56}, 5×78×7=3556\frac{5 \times 7}{8 \times 7} = \frac{35}{56}3256<3556  ⟹  4:7<5:8\frac{32}{56} < \frac{35}{56} \implies 4 : 7 < 5 : 8

3. Proportion

3.1 Understanding Proportion

When two ratios are equal, they are said to be in proportion.

If the ratio a:ba : b is equal to the ratio c:dc : d, then the four quantities aa, bb, cc, and dd are in proportion. This is written as:

a:b=c:dora:b::c:da : b = c : d \quad \text{or} \quad a : b :: c : d

The double colon symbol :::: stands for "is as". The statement is read as: "aa is to bb as cc is to dd".

3.2 Terms of a Proportion

In the proportion a:b::c:da : b :: c : d:

  • a,b,c,da, b, c, d are called the terms of the proportion.
  • aa is the first term, bb is the second term, cc is the third term, and dd is the fourth term.
  • The first and fourth terms (aa and dd) are located at the ends, so they are called the Extreme Terms (or Extremes).
  • The second and third terms (bb and cc) are located in the middle, so they are called the Middle Terms (or Means).

a⏟Extreme:b:c⏞Means:d⏟Extreme\Large \underbrace{a}_{\text{Extreme}} : \overbrace{b : c}^{\text{Means}} : \underbrace{d}_{\text{Extreme}}

3.3 The Fundamental Property of Proportion

For four numbers to be in proportion, the product of the extremes must equal the product of the means.

a:b::c:d  ⟹  ab=cda : b :: c : d \implies \frac{a}{b} = \frac{c}{d}

Cross-multiplying gives:

Product of Extremes=Product of Means\text{Product of Extremes} = \text{Product of Means} a×d=b×ca \times d = b \times c

  • Rule to Check Proportion: If a×d=b×ca \times d = b \times c, then a,b,c,da, b, c, d are in proportion. If a×d≠b×ca \times d \neq b \times c, they are not in proportion.

4. The Unitary Method

4.1 Definition

The Unitary Method is a problem-solving technique in which we first find the value of a single unit (one item) by division, and then find the value of the required number of units by multiplication.

                  ┌─────────────────────────────────────┐
                  │      Value of 'N' Given Units       │
                  └──────────────────┬──────────────────┘
                                     │
                         Divide by N (Division)
                                     │
                                     ▼
                  ┌─────────────────────────────────────┐
                  │        Value of 1 Single Unit       │
                  └──────────────────┬──────────────────┘
                                     │
                       Multiply by M (Multiplication)
                                     │
                                     ▼
                  ┌─────────────────────────────────────┐
                  │    Value of 'M' Required Units      │
                  └─────────────────────────────────────┘

4.2 Two Primary Steps of the Unitary Method

  1. Step 1: Find the value of 1 unit (Unit Value) Value of 1 unit=Total value of given itemsNumber of given items\text{Value of 1 unit} = \frac{\text{Total value of given items}}{\text{Number of given items}}
  2. Step 2: Find the value of required items Value of M units=(Value of 1 unit)×M\text{Value of } M \text{ units} = (\text{Value of 1 unit}) \times M

5. Real-World Applications

Application 1: Scaling Recipes in Cooking

When cooking for different numbers of guests, ingredients must maintain a constant ratio to preserve taste and texture. If a pancake recipe for 44 people requires 200 grams200\text{ grams} of flour, the ratio of flour to people is 200:4=50:1200 : 4 = 50 : 1. For 1010 people, using equivalent ratios or the unitary method shows that we need 50×10=500 grams50 \times 10 = 500\text{ grams} of flour.

Application 2: Map Reading and Blueprint Design

Maps represent massive geographical areas on a small sheet of paper. Map legends use scale ratios such as 1:100,0001 : 100,000, which means 1 cm1\text{ cm} on the map represents 100,000 cm100,000\text{ cm} (1 km1\text{ km}) on the ground. Architects use proportional ratios to scale buildings up or down accurately.

Application 3: Smart Comparison Shopping

Supermarkets sell items in varying pack sizes. To find the better deal, shoppers calculate the unit price using the unitary method. If Pack A gives 500 g500\text{ g} for ₹ 100\text{₹ } 100 (₹ 0.20/g\text{₹ } 0.20/\text{g}) and Pack B gives 800 g800\text{ g} for ₹ 144\text{₹ } 144 (₹ 0.18/g\text{₹ } 0.18/\text{g}), comparing unit prices shows that Pack B is the more economical choice.


6. Step-by-Step Solved Textbook Examples

Example 1: Simplifying Ratios with Different Units

Problem: Find the ratio of 75 cm75\text{ cm} to 1.5 metres1.5\text{ metres} in the simplest form.

Solution:

  • Step 1: Identify the given quantities. Quantity 1=75 cm\text{Quantity 1} = 75\text{ cm} Quantity 2=1.5 m\text{Quantity 2} = 1.5\text{ m}

  • Step 2: Convert both quantities to the same unit. Since 1 m=100 cm1\text{ m} = 100\text{ cm}: Quantity 2=1.5×100 cm=150 cm\text{Quantity 2} = 1.5 \times 100\text{ cm} = 150\text{ cm}

  • Step 3: Write as a ratio and express in fractional form. Ratio=75150\text{Ratio} = \frac{75}{150}

  • Step 4: Simplify by dividing both numerator and denominator by their HCF. HCF(75,150)=75\text{HCF}(75, 150) = 75 Ratio in simplest form=75÷75150÷75=12\text{Ratio in simplest form} = \frac{75 \div 75}{150 \div 75} = \frac{1}{2}

Final Answer: The required ratio is 1:2.\text{The required ratio is } 1 : 2.


Example 2: Dividing a Quantity in a Given Ratio

Problem: Divide ₹ 1,200\text{₹ } 1,200 between Rahul and Ananya in the ratio 3:53 : 5.

Solution:

  • Step 1: Add the terms of the ratio to find the total number of equal parts. Total parts=3+5=8 parts\text{Total parts} = 3 + 5 = 8\text{ parts}

  • Step 2: Express each person's share as a fraction of the total amount. Rahul’s share=38 of ₹ 1,200\text{Rahul's share} = \frac{3}{8} \text{ of } \text{₹ } 1,200 Ananya’s share=58 of ₹ 1,200\text{Ananya's share} = \frac{5}{8} \text{ of } \text{₹ } 1,200

  • Step 3: Calculate each share. Rahul’s share=38×1200=3×150=₹ 450\text{Rahul's share} = \frac{3}{8} \times 1200 = 3 \times 150 = \text{₹ } 450 Ananya’s share=58×1200=5×150=₹ 750\text{Ananya's share} = \frac{5}{8} \times 1200 = 5 \times 150 = \text{₹ } 750

  • Step 4: Verify that the sum equals the original total. ₹ 450+₹ 750=₹ 1,200(Correct)\text{₹ } 450 + \text{₹ } 750 = \text{₹ } 1,200 \quad \text{(Correct)}

Final Answer: Rahul receives ₹ 450 and Ananya receives ₹ 750.\text{Rahul receives } \text{₹ } 450 \text{ and Ananya receives } \text{₹ } 750.


Example 3: Checking for Proportion

Problem: Determine whether the numbers 25,30,40,25, 30, 40, and 4848 are in proportion.

Solution:

  • Step 1: Identify the terms. a=25,b=30,c=40,d=48a = 25, \quad b = 30, \quad c = 40, \quad d = 48

  • Step 2: Calculate the Product of Extremes (a×da \times d). Product of Extremes=25×48=1,200\text{Product of Extremes} = 25 \times 48 = 1,200

  • Step 3: Calculate the Product of Means (b×cb \times c). Product of Means=30×40=1,200\text{Product of Means} = 30 \times 40 = 1,200

  • Step 4: Compare the products. Product of Extremes=Product of Means=1,200\text{Product of Extremes} = \text{Product of Means} = 1,200

Final Answer: Since 25×48=30×40, the numbers 25,30,40, and 48 are in proportion (25:30::40:48).\text{Since } 25 \times 48 = 30 \times 40, \text{ the numbers } 25, 30, 40, \text{ and } 48 \text{ are in proportion } (25 : 30 :: 40 : 48).


Example 4: Real-World Unitary Method Application

Problem: A motorbike travels 210 km210\text{ km} using 6 litres6\text{ litres} of petrol. How far will it travel using 15 litres15\text{ litres} of petrol?

Solution:

  • Step 1: State the given information. Distance traveled with 6 litres of petrol=210 km\text{Distance traveled with } 6\text{ litres of petrol} = 210\text{ km}

  • Step 2: Apply the Unitary Method to find the distance covered using 1 litre1\text{ litre} of petrol (Division). Distance traveled with 1 litre of petrol=2106=35 km\text{Distance traveled with } 1\text{ litre of petrol} = \frac{210}{6} = 35\text{ km}

  • Step 3: Calculate the distance covered using 15 litres15\text{ litres} of petrol (Multiplication). Distance traveled with 15 litres of petrol=35×15=525 km\text{Distance traveled with } 15\text{ litres of petrol} = 35 \times 15 = 525\text{ km}

Final Answer: The motorbike will travel 525 km on 15 litres of petrol.\text{The motorbike will travel } 525\text{ km} \text{ on } 15\text{ litres of petrol.}


7. Common Student Mistakes to Avoid

Common ErrorIncorrect ApproachCorrect ApproachWhy It Happens
Comparing Different UnitsFinding the ratio of 500 m500\text{ m} to 2 km2\text{ km} as 500:2=250:1500 : 2 = 250 : 1.Convert 2 km2\text{ km} to 2,000 m2,000\text{ m}. Ratio = 500:2000=1:4500 : 2000 = 1 : 4.Forgetting that ratios require all quantities to share the exact same units before simplifying.
Reversing the TermsWriting the ratio of 10 boys10\text{ boys} to 15 girls15\text{ girls} as 15:1015 : 10 (3:23 : 2).First term must match the first named quantity: Boys : Girls = 10:15=2:310 : 15 = 2 : 3.Assuming order does not matter in ratios, treating a:ba : b as equivalent to b:ab : a.
Attaching Units to RatiosWriting the simplified ratio of 10 kg10\text{ kg} to 20 kg20\text{ kg} as 1:2 kg1 : 2\text{ kg}.Write the final ratio purely as 1:21 : 2.Misunderstanding that units cancel out during division, making ratios unitless numbers.
Incorrect Step Order in Unitary MethodMultiplying first, then dividing without finding the unit value.Divide first to find the value of 11 unit, then multiply by the target quantity.Skipping the basic intermediate step of finding the value of a single unit.

8. Practice Questions for Self-Assessment

Question 1

Problem: Give two equivalent ratios for 4:94 : 9, and compare 4:94 : 9 with 5:125 : 12 to determine which is larger.

Solution:

  1. Finding Equivalent Ratios:

    • Multiply by 22: 4×29×2=818  ⟹  8:18\frac{4 \times 2}{9 \times 2} = \frac{8}{18} \implies 8 : 18
    • Multiply by 33: 4×39×3=1227  ⟹  12:27\frac{4 \times 3}{9 \times 3} = \frac{12}{27} \implies 12 : 27
    • Two equivalent ratios are 8:188 : 18 and 12:2712 : 27.
  2. Comparing 4:94 : 9 and 5:125 : 12:

    • Convert to fractions: 49\frac{4}{9} and 512\frac{5}{12}.
    • Find LCM of denominators 99 and 1212: LCM(9,12)=36\text{LCM}(9, 12) = 36.
    • Convert to like fractions: 49=4×49×4=1636\frac{4}{9} = \frac{4 \times 4}{9 \times 4} = \frac{16}{36} 512=5×312×3=1536\frac{5}{12} = \frac{5 \times 3}{12 \times 3} = \frac{15}{36}
    • Compare numerators: Since 16>1516 > 15, we have 1636>1536\frac{16}{36} > \frac{15}{36}.
    • Therefore, 4:9>5:124 : 9 > 5 : 12.

Question 2

Problem: Are the ratios 12 cm:3 m12\text{ cm} : 3\text{ m} and 8 seconds:3 minutes8\text{ seconds} : 3\text{ minutes} in proportion? Show complete working.

Solution:

  1. Simplify Ratio 1 (12 cm:3 m12\text{ cm} : 3\text{ m}):

    • Convert metres to centimetres: 3 m=300 cm3\text{ m} = 300\text{ cm}.
    • Ratio 1=12300=12÷12300÷12=125  ⟹  1:25\text{Ratio 1} = \frac{12}{300} = \frac{12 \div 12}{300 \div 12} = \frac{1}{25} \implies 1 : 25.
  2. Simplify Ratio 2 (8 seconds:3 minutes8\text{ seconds} : 3\text{ minutes}):

    • Convert minutes to seconds: 3 minutes=3×60=180 seconds3\text{ minutes} = 3 \times 60 = 180\text{ seconds}.
    • Ratio 2=8180=8÷4180÷4=245  ⟹  2:45\text{Ratio 2} = \frac{8}{180} = \frac{8 \div 4}{180 \div 4} = \frac{2}{45} \implies 2 : 45.
  3. Compare simplified ratios:

    • 125≠245\frac{1}{25} \neq \frac{2}{45} (Cross multiplication: 1×45=451 \times 45 = 45, while 25×2=5025 \times 2 = 50).
    • Since the simplified ratios are not equal, the given ratios are NOT in proportion.

Question 3

Problem: The cost of 1212 notebooks is ₹ 480\text{₹ } 480.

  1. Find the cost of 77 notebooks.
  2. How many notebooks can be purchased for ₹ 320\text{₹ } 320?

Solution:

Part 1: Find the cost of 7 notebooks

  • Cost of 1212 notebooks = ₹ 480\text{₹ } 480
  • Cost of 11 notebook = 48012=₹ 40\frac{480}{12} = \text{₹ } 40
  • Cost of 77 notebooks = 40×7=₹ 28040 \times 7 = \text{₹ } 280

Part 2: Find notebooks purchased for ₹ 320\text{₹ } 320

  • Number of notebooks bought for ₹ 40\text{₹ } 40 = 11
  • Number of notebooks bought for ₹ 320\text{₹ } 320 = 32040=8 notebooks\frac{320}{40} = 8\text{ notebooks}

Final Answers:

  1. The cost of 77 notebooks is ₹ 280\text{₹ } 280.
  2. 88 notebooks can be purchased for ₹ 320\text{₹ } 320.

9. Exam Revision & FAQs

Question A: What is the main difference between ratio and proportion?

Answer:

  • A ratio is used to compare two quantities of the same unit by division (e.g., a:ba : b).
  • A proportion is a mathematical statement showing that two ratios are equal (e.g., a:b=c:da : b = c : d). A ratio involves two terms, whereas a proportion involves four terms across two ratios.

Question B: If x:12::5:20x : 12 :: 5 : 20, find the value of xx.

Answer: Using the property Product of Extremes = Product of Means: x×20=12×5x \times 20 = 12 \times 5 20x=6020x = 60 x=6020=3x = \frac{60}{20} = 3

Value of xx is 33.


Question C: Can a ratio be formed between 5 litres5\text{ litres} of milk and 20 kilograms20\text{ kilograms} of rice? Explain.

Answer: No, a ratio cannot be formed between 5 litres5\text{ litres} of milk and 20 kg20\text{ kg} of rice. Ratios can only compare quantities of the same kind that can be expressed in the same units. Litres measure volume, while kilograms measure mass; since these physical quantities are different, they cannot be converted into a common unit.


Question D: How do we divide a total quantity QQ into three parts according to a ratio a:b:ca : b : c?

Answer:

  1. Find the sum of all ratio terms: Total Parts=a+b+c\text{Total Parts} = a + b + c.
  2. Compute each individual share: First Share=aa+b+c×Q\text{First Share} = \frac{a}{a + b + c} \times Q Second Share=ba+b+c×Q\text{Second Share} = \frac{b}{a + b + c} \times Q Third Share=ca+b+c×Q\text{Third Share} = \frac{c}{a + b + c} \times Q

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