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ICSE • कक्षा 7 • Mathematics • अध्याय 6
अनुमानित समय: 45 Mins
प्रगति: अध्ययनरत

अनुपात और समानुपात (Ratio and Proportion)

In ICSE Class 7 Mathematics, "Ratio and Proportion" provides an authoritative, comparative master study guide analyzing the mathematical comparison of quantities of the same kind. This comprehensive chapter explores Concept of Ratio ($a : b = \frac{a}{b}$ where $b \neq 0$; Antecedent $a$ and Consequent $b$; Dimensionless nature; Ratio in simplest lowest terms; Comparison and ordering of ratios via LCM or cross-multiplication; Division of a given quantity in a given ratio), Concept of Proportion (Equality of two ratios $a : b = c : d$ or $a : b :: c : d$; Extremes $a, d$ and Means $b, c$; Cross-Product Rule: $\text{Product of Extremes} = \text{Product of Means}$ or $a \cdot d = b \cdot c$; Continued Proportion: $a : b :: b : c$ with Mean Proportional $b = \sqrt{ac}$ and Third Proportional $c = \frac{b^2}{a}$), The Unitary Method (Direct variation [both quantities increase or decrease in proportion] vs Inverse variation [one quantity increases as the other decreases]), and Real-World Applied Problems (Scale models, maps, speed-time trade-offs, and commercial cost calculations) aligned with the 2026–27 CISCE curriculum.

अध्याय CBT ऑनलाइन टेस्ट दें सभी अध्याय देखें
प्रस्तावना अध्याय अवधारणाएँ सूत्र एवं सर्वसमिकाएँ अध्याय सारांश स्व-मूल्यांकन अभ्यास टेस्ट दें

How Did Ancient Greek Sculptors Use a Single Magic Ratio to Carve Statues That Looked So Perfect People Believed They Were Living Gods?

Walk through the British Museum in London and examine the ancient marble sculptures carved by the master Athenian sculptor Phidias for the Parthenon in 447 BCE. Every curve, every muscle, every arch of the eyebrow radiates breathtaking aesthetic perfection. How did Phidias achieve this supernatural beauty? He did not guess; he used a divine mathematical proportion! Phidias discovered that human eyes perceive objects as most harmonious when dimensions follow the Golden Ratio: $\phi \approx 1.618 : 1$—the ratio where the ratio of the whole to the larger part equals the ratio of the larger part to the smaller part! From the spirals of nautilus seashells and the petals of sunflowers to the screen dimensions of your smartphone and the Mona Lisa of face, ratios govern the physical universe! What is the difference between Direct Variation and Inverse Variation? What is a Mean Proportional? How does the Unitary Method solve real-world problems? Let of master ratio and proportion.

यह अध्याय क्यों महत्वपूर्ण है

Ratio and proportion form the mathematical bedrock of chemistry (stoichiometric molar ratios, law of definite proportions), physics (velocity, density, pressure ratios), finance (currency exchange rates, profit-sharing in partnerships), culinary recipe scaling, and architectural map scale models. Mastery of these concepts is indispensable for ICSE mathematics and competitive exams.

अध्ययन से पूर्व (आवश्यक ज्ञान)

  • Multiplication and division of fractions and decimals.
  • Finding the HCF and LCM of natural numbers.
  • Solving simple linear equations.

इस अध्याय के लक्ष्य

  • Express and reduce ratios into simplest form with matching units.
  • Divide a given quantity into parts according to a specified ratio.
  • Verify whether four given numbers form a Proportion using the Extremes and Means rule ($ad = bc$).
  • Calculate the fourth proportional, third proportional, and mean proportional ($b = \sqrt{ac}$).
  • Differentiate between Direct Variation and Inverse Variation scenarios.
  • Apply the Unitary Method to solve multi-step commercial and rate problems.

अध्याय रूपरेखा एवं प्रगति

1 1. Concept of Ratio & Simplest Form
2 2. Proportion & The Cross-Product R...
3 3. Continued Proportion & Mean Prop...
4 4. The Unitary Method: Direct vs In...

सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन

1. Concept of Ratio & Simplest Form

Understand
A. What is a Ratio?

A Ratio is a mathematical comparison of two quantities of the same kind and in the same units by division:

$$\text{Ratio of } a \text{ to } b = a : b = \frac{a}{b} \quad (b \neq 0)$$
  • $a$ is called the first term or Antecedent.
  • $b$ is called the second term or Consequent.
  • Dimensionless Quantity: A ratio has no units (e.g., the ratio of $2\text{ m}$ to $5\text{ m}$ is $2 : 5$, NOT $2 : 5\text{ m}$).
  • Crucial Rule: Before forming a ratio, both quantities must be converted to the same unit! (e.g., ratio of $75\text{ paise}$ to $\text{Rs. } 3 = 75 : 300 = 1 : 4$).
B. Dividing a Quantity in a Given Ratio:

To divide a total quantity $Q$ into two parts in the ratio $m : n$:

$$\text{First Part} = \frac{m}{m + n} \times Q, \quad \text{Second Part} = \frac{n}{m + n} \times Q$$

2. Proportion & The Cross-Product Rule

Proportion Laws
A. Defining Proportion:

An equality of two ratios is called a Proportion. Four non-zero quantities $a, b, c, d$ are in proportion if:

$$a : b = c : d \quad \text{or} \quad a : b :: c : d$$
  • $a$ and $d$ are called the Extremes (outer terms).
  • $b$ and $c$ are called the Means (middle terms).
  • $d$ is called the Fourth Proportional to $a, b, c$.
B. The Fundamental Theorem of Proportion (Cross-Product Rule):
$$\frac{a}{b} = \frac{c}{d} \iff a \times d = b \times c$$ $$\mathbf{\text{Product of Extremes} = \text{Product of Means}}$$

3. Continued Proportion & Mean Proportional

Continued Proportion
A. Continued Proportion:

Three non-zero quantities $a, b, c$ are said to be in Continued Proportion if the ratio of the first to the second equals the ratio of the second to the third:

$$a : b :: b : c \iff \frac{a}{b} = \frac{b}{c}$$
  • Cross-multiplying gives: $b^2 = a \times c$.
  • Mean Proportional: The middle term $b$ is called the Mean Proportional between $a$ and $c$: $$b = \sqrt{a \times c}$$
  • Third Proportional: The term $c$ is called the Third Proportional to $a$ and $b$: $$c = \frac{b^2}{a}$$

4. The Unitary Method: Direct vs Inverse Variation

Unitary Method

The method of finding the value of a single unit first, and then finding the value of the required number of units:

A. Direct Variation (More $\to$ More, Less $\to$ Less):
  • Two quantities $x$ and $y$ vary directly if an increase in $x$ causes a proportionate increase in $y$ ($ rac{x}{y} = k$ is constant).
  • Rule: To find unit value $\to$ Divide; to find required value $\to$ Multiply. (e.g., 5 pens cost $\text{Rs. } 50 \implies 1\text{ pen costs } 50/5 = 10 \implies 8\text{ pens cost } 10 \times 8 = \text{Rs. } 80$).
B. Inverse Variation (More $\to$ Less, Less $\to$ More):
  • Two quantities vary inversely if an increase in $x$ causes a proportionate decrease in $y$ ($x \times y = k$ is constant).
  • Rule: To find unit value $\to$ Multiply; to find required value $\to$ Divide!
  • Example: 6 workers build a wall in 10 days. 1 worker takes $6 \times 10 = 60\text{ days}$. Therefore, 12 workers take $60 \div 12 = 5\text{ days}$!

महत्वपूर्ण सूत्र, सर्वसमिकाएँ एवं प्रमेय

Cross-Product Rule of Proportion
$$a : b :: c : d \iff a \cdot d = b \cdot c$$
Product of Extremes equals Product of Means.
Mean Proportional Formula
$$b = \sqrt{a \cdot c} \quad (\text{when } a : b :: b : c)$$
Geometric mean between two quantities.

Ratio, Proportion & Direct vs Inverse Variation

Ratio and Proportion: Extremes, Means & Variations RATIO (a : b) • a = Antecedent • b = Consequent • No Units! (Dimensionless) • Must match units before dividing:   75 paise : Rs 3 = 75 : 300 = 1 : 4 • Dividing Quantity Q (m : n):   Part 1 = [m / (m+n)] × Q   Part 2 = [n / (m+n)] × Q • Simplest form when gcd(a, b) = 1 PROPORTION (a : b :: c : d) Product of Extremes = Product of Means a × d = b × c • Continued Proportion (a:b::b:c):   b2 = a × c   Mean Proportional: b = √(ac)   Third Proportional: c = b2 / a • Fourth Proportional: d = (bc) / a UNITARY VARIATIONS • Direct Variation (x/y = k):   More items → More cost   1. Divide for 1 unit • 2. Multiply • Inverse Variation (x × y = k):   More workers → Less days   More speed → Less travel time   1. Multiply for 1 unit!   2. Divide for required units! • Classical work-time problems PROPORTION RULE: a × d = b × c • MEAN PROPORTIONAL b = √(ac)

अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष

मुख्य बिंदु 1
A ratio compares two quantities of the same kind by division; it has no units.
मुख्य बिंदु 2
Always convert quantities into identical units before finding their ratio.
मुख्य बिंदु 3
A proportion is an equality of two ratios: $a : b :: c : d$.
मुख्य बिंदु 4
Fundamental Theorem of Proportion: Product of Extremes = Product of Means ($ad = bc$).
मुख्य बिंदु 5
In continued proportion $a : b :: b : c$, the mean proportional is $b = \sqrt{ac}$.
मुख्य बिंदु 6
The third proportional $c$ to $a$ and $b$ is given by $c = b^2 / a$.
मुख्य बिंदु 7
Direct Variation: An increase in one quantity causes a proportional increase in the other ($x/y = k$).
मुख्य बिंदु 8
Inverse Variation: An increase in one quantity causes a proportional decrease in the other ($xy = k$).
मुख्य बिंदु 9
In Direct Variation, divide to find the unit value; in Inverse Variation, multiply to find the unit value.
मुख्य बिंदु 10
To divide $Q$ in ratio $m : n$, parts are $\frac{m}{m+n} Q$ and $\frac{n}{m+n} Q$.

स्व-मूल्यांकन अभ्यास (Check Your Understanding)

मूल वैचारिक स्पष्टता की जांच के लिए नैदानिक प्रश्न। पहले स्वयं हल करें, फिर उत्तर देखें।

1
Express the ratio of $45\text{ minutes}$ to $2\text{ hours } 15\text{ minutes}$ in simplest lowest terms.
उत्तर एवं व्याख्या देखें
उत्तर:

Step 1: Convert both quantities to the same unit (minutes):
First quantity $= 45\text{ minutes}$
Second quantity $= 2\text{ hours } 15\text{ minutes} = (2 \times 60) + 15 = 120 + 15 = 135\text{ minutes}$
Step 2: Express as a fraction and reduce by dividing by $\gcd(45, 135) = 45$:

$$\frac{45}{135} = \frac{45 \div 45}{135 \div 45} = \frac{1}{3}$$


The ratio in simplest form is $1 : 3$.


Convert 2 hours 15 minutes to 135 minutes, then reduce $45 / 135$ by dividing both by 45.
2
Divide $\text{Rs. } 7,200$ among $A$, $B$, and $C$ in the ratio $2 : 3 : 5$.
उत्तर एवं व्याख्या देखें
उत्तर: Step 1: Find the sum of the ratio terms:
$$\text{Sum of terms} = 2 + 3 + 5 = 10$$
Step 2: Calculate each share:
$$A\text{\ of share} = \frac{2}{10} \times 7,200 = 2 \times 720 = \mathbf{\text{Rs. } 1,440}$$
$$B\text{\ of share} = \frac{3}{10} \times 7,200 = 3 \times 720 = \mathbf{\text{Rs. } 2,160}$$
$$C\text{\ of share} = \frac{5}{10} \times 7,200 = 5 \times 720 = \mathbf{\text{Rs. } 3,600}$$.
Check: $1440 + 2160 + 3600 = 7,200$.
Sum of ratio terms $= 10$. Shares are $(2/10)$, $(3/10)$, and $(5/10)$ of 7,200.
3
Find the value of $x$ in the proportion: $16 : x :: 24 : 36$.
उत्तर एवं व्याख्या देखें
उत्तर:

By the Fundamental Theorem of Proportion:

$$\text{Product of Extremes} = \text{Product of Means}$$


$$16 \times 36 = x \times 24$$


$$24x = 576$$


$$x = \frac{576}{24} = \frac{16 \times 36}{24}$$


Cancel by 8: $\frac{2 \times 36}{3} = 2 \times 12 = \mathbf{24}$
The value of $x$ is $24$.


Apply $a \times d = b \times c$: $16 \times 36 = x \times 24$. Solve for $x$.
4
Find: (a) The Third Proportional to $9$ and $12$, (b) The Mean Proportional between $4$ and $36$.
उत्तर एवं व्याख्या देखें
उत्तर:

• (a) Third Proportional to $9$ and $12$:
Let the third proportional be $c$. Then $9, 12, c$ are in continued proportion:

$$9 : 12 :: 12 : c \implies \frac{9}{12} = \frac{12}{c}$$


$$9c = 12 \times 12 = 144 \implies c = \frac{144}{9} = \mathbf{16}$$

.
• (b) Mean Proportional between $4$ and $36$:
Let the mean proportional be $b$.

$$b = \sqrt{a \times c} = \sqrt{4 \times 36} = \sqrt{144} = \mathbf{12}$$

.


Third proportional is $c = b^2 / a = 144 / 9 = 16$. Mean proportional is $\sqrt{4 \times 36} = 12$.
5
If $15$ men can complete the construction of a canal in $24\text{ days}$, how many days will $20$ men take to complete the same work?
उत्तर एवं व्याख्या देखें
उत्तर:

This is an Inverse Variation problem (more men require fewer days):
15 men take $= 24\text{ days}$
1 man takes $= 15 \times 24 = 360\text{ days}$ (multiply to find unit value)
20 men take $= \frac{360}{20} = \mathbf{18\text{ days}}$ (divide to find required value).
20 men will take $18\text{ days}$.


Inverse variation: $15 \times 24 = 20 \times d \implies d = 360 / 20 = 18\text{ days}$.
6
What number must be added to each of the numbers $6, 15, 20, 43$ so that the resulting numbers are in proportion?
उत्तर एवं व्याख्या देखें
उत्तर: Let the required number be $x$. The numbers $(6+x), (15+x), (20+x), (43+x)$ are in proportion:
$$\frac{6 + x}{15 + x} = \frac{20 + x}{43 + x}$$
Cross-multiply:
$$(6 + x)(43 + x) = (15 + x)(20 + x)$$
$$258 + 49x + x^2 = 300 + 35x + x^2$$
Subtract $x^2$ from both sides:
$$49x - 35x = 300 - 258$$
$$14x = 42 \implies x = \frac{42}{14} = \mathbf{3}$$.
Check: $9 : 18 :: 23 : 46 \implies 1:2 = 1:2$. Correct!
Set $(6+x)(43+x) = (15+x)(20+x)$. The $x^2$ terms cancel out, giving $14x = 42$.
7
If $A : B = 3 : 4$ and $B : C = 8 : 9$, find the compounded ratio $A : C$ and the combined ratio $A : B : C$.
उत्तर एवं व्याख्या देखें
उत्तर:

• (a) Compounded Ratio $A : C$:

$$\frac{A}{C} = \frac{A}{B} \times \frac{B}{C} = \frac{3}{4} \times \frac{8}{9} = \frac{1 \times 2}{1 \times 3} = \frac{2}{3} \implies \mathbf{A : C = 2 : 3}$$


• (b) Combined Ratio $A : B : C$:
Make the common term $B$ identical in both ratios.
In $A : B$, $B = 4$. In $B : C$, $B = 8$.
Multiply the first ratio by $2$:

$$A : B = (3 \times 2) : (4 \times 2) = 6 : 8$$


$$B : C = 8 : 9$$


Since $B$ is now identical ($8$), combine:

$$\mathbf{A : B : C = 6 : 8 : 9}$$

.


Multiply $A : B$ by 2 so $B$ becomes 8 in both ratios, yielding $6 : 8 : 9$.
8
The scale of a geographical map is $1 : 25,000,000$. If two cities are $4.8\text{ cm}$ apart on the map, find the actual distance between them in kilometers.
उत्तर एवं व्याख्या देखें
उत्तर:

Map scale $= 1\text{ cm}$ represents $25,000,000\text{ cm}$ on ground.

$$\text{Actual Distance} = 4.8 \times 25,000,000\text{ cm} = 120,000,000\text{ cm}$$


Convert centimeters to kilometers ($1\text{ m} = 100\text{ cm}$, $1\text{ km} = 1,000\text{ m} \implies 1\text{ km} = 100,000\text{ cm}$):

$$\text{Distance in km} = \frac{120,000,000}{100,000} = \mathbf{1,200\text{ km}}$$

.
The actual distance between the two cities is $1,200\text{ kilometers}$.


Multiply $4.8$ by $25,000,000\text{ cm}$, then divide by $100,000$ to convert centimeters to kilometers.
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कक्षा 7 Mathematics के सभी अध्याय

अध्याय 1: पूर्णांक (Integers) अध्याय 2: परिमेय संख्याएँ (Rational Numbers) अध्याय 3: भिन्न (Fractions) अध्याय 4: घातांक (Exponents) अध्याय 5: समुच्चय (Sets) अध्याय 6: अनुपात और समानुपात (Ratio and Proportion) अध्याय 7: प्रतिशतता (Percentage) अध्याय 8: चाल, दूरी और समय (Speed, Distance and Time) अध्याय 9: बीजीय व्यंजक (Algebraic Expressions) अध्याय 10: एक चर वाले रैखिक समीकरण (Linear Equations) अध्याय 11: आकारों की समझ (Understanding Shapes) अध्याय 12: त्रिभुज (Triangles) अध्याय 13: सममिति (Symmetry) अध्याय 14: त्रिविमीय आकारों का द्विविमीय निरूपण (Representing 3D in 2D) अध्याय 15: सर्वांगसमता (Congruence) अध्याय 16: रचनाएँ (Constructions) अध्याय 17: क्षेत्रमिति (Mensuration) अध्याय 18: आँकड़ों का प्रबंधन (Data Handling) अध्याय 19: रैखिक असमिकाएँ (Linear Inequations)

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