Building Expressions from Real Problems — Class 9 Mathematics Chapter 2

🖼 Image PendingA wide cinematic view of an Indian rural garden with a wooden picket fence on its long sides and a wire fence on its short sides; rows of vegetables and flowers stretch into a soft golden distance

AI Generation Prompt

Ultra-wide cinematic banner (16:5 ratio). A peaceful Indian rural garden at golden hour, viewed from a low angle. The garden is rectangular, with a tall wooden picket fence running along the two long sides and a thinner wire fence on the two short sides. Inside, neat rows of vegetables and yellow marigolds stretch into the soft golden distance. A small white farmhouse and a few trees sit in the background. The atmosphere conveys: every fence, every plot, every measurement is secretly an algebraic expression. Painterly cinematic illustration in the style of warm rural Indian art. Dark background. No text, no labels.

✦Think About It

A wire of length **20 cm** is bent into a rectangle. Different bends give different rectangles — long-and-thin, almost-square, etc. **Can you write a single expression that gives the area of *any* such rectangle, depending on how you bend the wire?**

If you call the length $x$, the width is forced — the four sides must add up to 20.

✦गणितसारसङ्ग्रह — महावीर (९वीं शताब्दी)

From Mahāvīra's Gaṇita-sāra-saṅgraha (c. 850 CE) — On Practical Calculation

क्षेत्रस्य परिमाणं वा मूल्यं वा निश्चयेन ज्ञेयम्

अक्षरैरेव संख्यया च — एतद् गणितस्य सार्थकता॥

(kṣetrasya parimāṇaṃ vā mūlyaṃ vā niścayena jñeyam / akṣarair eva saṃkhyayā ca — etad gaṇitasya sārthakatā)

'किसी ज़मीन का नाप हो या कीमत — दोनों को अक्षरों और संख्याओं से ठीक-ठीक जाना जा सकता है। यही गणित का मतलब है।'

'Whether it is the measurement of land or its price — both can be known precisely through letters and numbers. This is the meaning of mathematics.'

Mahāvīra, a Jain mathematician, wrote the Gaṇita-sāra-saṅgraha — one of the great Indian medieval algebra texts. He showed how everyday land-measurement, fencing, and pricing problems all become a few lines of algebra. The garden problem on this page is exactly the kind of thing his students worked through 1,200 years ago.

A garden of length $l$ and width $w$

A rectangular garden of length $l$ metres and width $w$ metres needs three things:

A wire fence along the two length sides — costs ₹100 per metre.
A wooden fence along the two width sides — costs ₹80 per metre.
Special seeds to be sown throughout the garden — costs ₹50 per square metre of area.

Let's compute the total cost step by step:

Length of wire fencing needed = $2l$ metres. Cost = $2l \times 100 = ₹200l$ .
Length of wooden fencing needed = $2w$ metres. Cost = $2w \times 80 = ₹160w$ .
Area to be sown = $l \times w$ square metres. Seed cost = $50 \times l \times w = ₹50lw$ .

Adding them up:

$\text{Total cost} = 200l + 160w + 50lw$

This expression has three terms: $200l$ , $160w$ , and $50lw$ . Two variables: $l$ and $w$ . No constant (because every bit of the cost depends on the size of the garden).

Notice the term $50lw$ — it has both variables multiplied together. We'll meet such terms again when you formally study quadratic and higher-degree polynomials in later chapters.

🖼 Image PendingA painted top-and-side view of a rectangular vegetable garden with a wooden picket fence on its long sides and a wire fence on its short sides. Length labelled 'l metres' along the long sides; width labelled 'w metres' along the short sides. Inside the garden are rows of marigolds and small huts in the background suggesting an Indian rural setting.

AI Generation Prompt

Painted illustration of a rectangular vegetable garden viewed slightly from above. Two long sides have a tall wooden picket fence; two short sides have a thinner wire mesh fence. Length of long sides labelled 'l metres' twice (once on each long side, with a thin red double-headed arrow); width of short sides labelled 'w metres' twice. Inside the garden: neat rows of yellow marigolds and small green vegetable plants. Background shows a small white farmhouse and trees, soft blue sky with white clouds. Style: warm hand-painted Indian textbook illustration. Dark background, orange accent labels, clean technical illustration style.

📸 Fig. 2.2 — A rectangular garden of length $l$ and width $w$, with both fences and seeds visible.

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Worked Example 1

SOLVED

For the cost expression $200l + 160w + 50lw$ , identify: (a) the number of terms, (b) the variables, (c) the coefficient of $lw$ , (d) is there a constant?

(a) Three terms: $200l$ , $160w$ , $50lw$ .

(b) Two variables: $l$ and $w$ .

(c) Coefficient of $lw$ : $50$ .

(d) No constant. Every term contains at least one variable — the cost is zero when the garden is zero-sized.

Compare with Page 1's expression $4x + 5y + 3$ : that one had a constant ( $3$ , the free pens). The garden expression doesn't, because no part of the cost is independent of the garden's dimensions.

From two variables to one — bending a wire

Now consider a different problem.

A wire of length 20 cm is bent into a rectangle. The wire goes around the rectangle's perimeter, so:

$2 \cdot \text{length} + 2 \cdot \text{width} = 20$

Dividing by 2: length + width = $10$ . So if the length is $x$ cm, the width must be $(10 - x)$ cm. The area is:

$\text{Area} = x \cdot (10 - x) = 10x - x^2 \text{ sq cm}$

Notice what just changed: this expression has only one variable, $x$ . The garden expression had two ( $l$ and $w$ ). The shop expression on Page 1 had two ( $x$ and $y$ ). This one has one.

Single-variable expressions are special enough to deserve their own name. We'll meet that name on the next page: polynomials.

Worked Example 2

SOLVED

For the area expression $10x - x^2$ , identify: (a) the number of terms, (b) the variable, (c) the coefficient of $x$ , (d) the coefficient of $x^2$ .

(a) Two terms: $10x$ and $-x^2$ .

(b) Variable: $x$ .

(c) Coefficient of $x$ : $10$ (from the term $10x$ ).

(d) Coefficient of $x^2$ : $-1$ (from the term $-x^2 = -1 \cdot x^2$ ). Don't miss the negative sign or the implicit 1.

Compare: the shop expression $4x + 5y + 3$ from Page 1 used two variables ( $x, y$ ). The area expression here uses one variable ( $x$ ) — this is the key feature that makes it a polynomial in one variable.

Worked Example 3

SOLVED

Use $\text{Area} = 10x - x^2$ to find the area of the rectangle when the length is $x = 3$ cm. Then find the area when $x = 7$ cm. What do you notice?

Step 1 — Substitute $x = 3$ .

$10(3) - (3)^2 = 30 - 9 = 21 \text{ sq cm}$

The rectangle is 3 cm × 7 cm — area 21 sq cm. ✓

Step 2 — Substitute $x = 7$ .

$10(7) - (7)^2 = 70 - 49 = 21 \text{ sq cm}$

The rectangle is 7 cm × 3 cm — area 21 sq cm. ✓

What do you notice? Both choices of $x$ give the same area! Why? Because the rectangles 3×7 and 7×3 are the same rectangle — just rotated 90°. The expression $10x - x^2$ is symmetric around $x = 5$ : as $x$ moves from 5, the area drops the same amount whether you go up or down.

Worked Example 4

SOLVED

Write an algebraic expression for the total cost of an auto-rickshaw ride: a base fare of ₹40 plus ₹15 per kilometre. Then evaluate the cost for a 7 km ride.

Step 1 — Define the variable.

Let $d$ = distance travelled in kilometres.

Step 2 — Write the expression.

Base fare is constant: ₹40. Per-km cost adds $15d$ . So:

$\text{Cost} = 15d + 40$

This is a one-variable expression (only $d$ ). The coefficient of $d$ is $15$ . The constant is $40$ .

Step 3 — Evaluate at $d = 7$ .

$15(7) + 40 = 105 + 40 = ₹145$

Answer: Cost = $15d + 40$ . For 7 km, the fare is ₹145.

Logical ReasoningLevel 2 · Application

Compare three expressions: (A) $4x + 5y + 3$ from the shop, (B) $200l + 160w + 50lw$ from the garden, (C) $10x - x^2$ from the bent wire. **Which one is fundamentally different from the other two, and why?**

Practice Yourself — Translate the Story into Algebra

For each scenario, write an algebraic expression. Cover the answers and try.

A taxi charges ₹50 base fare plus ₹12 per km. Distance = $d$ km.
A mobile data plan: ₹200 fixed + ₹2 per MB used. MB used = $m$ .
A rectangular room of length $l$ m and width $w$ m needs paint costing ₹15 per square metre.
A piggy bank has $x$ ₹2 coins and $y$ ₹5 coins. What is the total amount in rupees?
A train moves at 80 km/h for $t$ hours. Distance covered (km)?
A pen costs ₹15. You buy $p$ pens and pay ₹100, getting some change back. Express the change.
A square has side length $s$ cm. Express its perimeter.
A square has side length $s$ cm. Express its area. (Is this a polynomial in one variable?)

Answers: 1. $12d + 50$ · 2. $2m + 200$ · 3. $15lw$ · 4. $2x + 5y$ · 5. $80t$ · 6. $100 - 15p$ · 7. $4s$ · 8. $s^2$ — yes, a polynomial in one variable, but not linear (it has $s^2$ , so it's quadratic).

Quick Check

Q1.How many variables does the expression $7p + 11q - 4$ have?

AI Generation Prompt

✦Think About It

If you call the length $x$, the width is forced — the four sides must add up to 20.

✦गणितसारसङ्ग्रह — महावीर (९वीं शताब्दी)

From Mahāvīra's Gaṇita-sāra-saṅgraha (c. 850 CE) — On Practical Calculation

क्षेत्रस्य परिमाणं वा मूल्यं वा निश्चयेन ज्ञेयम्

अक्षरैरेव संख्यया च — एतद् गणितस्य सार्थकता॥

(kṣetrasya parimāṇaṃ vā mūlyaṃ vā niścayena jñeyam / akṣarair eva saṃkhyayā ca — etad gaṇitasya sārthakatā)

'Whether it is the measurement of land or its price — both can be known precisely through letters and numbers. This is the meaning of mathematics.'

A garden of length $l$ and width $w$

A rectangular garden of length $l$ metres and width $w$ metres needs three things:

A wire fence along the two length sides — costs ₹100 per metre.
A wooden fence along the two width sides — costs ₹80 per metre.
Special seeds to be sown throughout the garden — costs ₹50 per square metre of area.

Let's compute the total cost step by step:

Length of wire fencing needed = $2l$ metres. Cost = $2l \times 100 = ₹200l$ .
Length of wooden fencing needed = $2w$ metres. Cost = $2w \times 80 = ₹160w$ .
Area to be sown = $l \times w$ square metres. Seed cost = $50 \times l \times w = ₹50lw$ .

Adding them up:

$\text{Total cost} = 200l + 160w + 50lw$

This expression has three terms: $200l$ , $160w$ , and $50lw$ . Two variables: $l$ and $w$ . No constant (because every bit of the cost depends on the size of the garden).

Notice the term $50lw$ — it has both variables multiplied together. We'll meet such terms again when you formally study quadratic and higher-degree polynomials in later chapters.

AI Generation Prompt

📸 Fig. 2.2 — A rectangular garden of length $l$ and width $w$, with both fences and seeds visible.

Loading simulator…

Worked Example 1

SOLVED

For the cost expression $200l + 160w + 50lw$ , identify: (a) the number of terms, (b) the variables, (c) the coefficient of $lw$ , (d) is there a constant?

(a) Three terms: $200l$ , $160w$ , $50lw$ .

(b) Two variables: $l$ and $w$ .

(c) Coefficient of $lw$ : $50$ .

(d) No constant. Every term contains at least one variable — the cost is zero when the garden is zero-sized.

From two variables to one — bending a wire

Now consider a different problem.

A wire of length 20 cm is bent into a rectangle. The wire goes around the rectangle's perimeter, so:

$2 \cdot \text{length} + 2 \cdot \text{width} = 20$

Dividing by 2: length + width = $10$ . So if the length is $x$ cm, the width must be $(10 - x)$ cm. The area is:

$\text{Area} = x \cdot (10 - x) = 10x - x^2 \text{ sq cm}$

Notice what just changed: this expression has only one variable, $x$ . The garden expression had two ( $l$ and $w$ ). The shop expression on Page 1 had two ( $x$ and $y$ ). This one has one.

Single-variable expressions are special enough to deserve their own name. We'll meet that name on the next page: polynomials.

Worked Example 2

SOLVED

For the area expression $10x - x^2$ , identify: (a) the number of terms, (b) the variable, (c) the coefficient of $x$ , (d) the coefficient of $x^2$ .

(a) Two terms: $10x$ and $-x^2$ .

(b) Variable: $x$ .

(c) Coefficient of $x$ : $10$ (from the term $10x$ ).

(d) Coefficient of $x^2$ : $-1$ (from the term $-x^2 = -1 \cdot x^2$ ). Don't miss the negative sign or the implicit 1.

Worked Example 3

SOLVED

Use $\text{Area} = 10x - x^2$ to find the area of the rectangle when the length is $x = 3$ cm. Then find the area when $x = 7$ cm. What do you notice?

Step 1 — Substitute $x = 3$ .

$10(3) - (3)^2 = 30 - 9 = 21 \text{ sq cm}$

The rectangle is 3 cm × 7 cm — area 21 sq cm. ✓

Step 2 — Substitute $x = 7$ .

$10(7) - (7)^2 = 70 - 49 = 21 \text{ sq cm}$

The rectangle is 7 cm × 3 cm — area 21 sq cm. ✓

Worked Example 4

SOLVED

Write an algebraic expression for the total cost of an auto-rickshaw ride: a base fare of ₹40 plus ₹15 per kilometre. Then evaluate the cost for a 7 km ride.

Step 1 — Define the variable.

Let $d$ = distance travelled in kilometres.

Step 2 — Write the expression.

Base fare is constant: ₹40. Per-km cost adds $15d$ . So:

$\text{Cost} = 15d + 40$

This is a one-variable expression (only $d$ ). The coefficient of $d$ is $15$ . The constant is $40$ .

Step 3 — Evaluate at $d = 7$ .

$15(7) + 40 = 105 + 40 = ₹145$

Answer: Cost = $15d + 40$ . For 7 km, the fare is ₹145.

Logical ReasoningLevel 2 · Application

Quick Check

Q1.How many variables does the expression $7p + 11q - 4$ have?