Linear Patterns in Daily Life — Class 9 Mathematics Chapter 2

🖼 Image PendingA wide cinematic split-frame image — on the left, a girl removes a small ₹5 note from a clay piggy bank; on the right, a green-and-yellow Indian auto-rickshaw moves through a sunlit street with its meter visible

AI Generation Prompt

Ultra-wide cinematic banner (16:5 ratio) with a soft visual split. Left half: a young girl in school uniform reaches into a clay piggy bank on a wooden table and removes a small ₹5 currency note; on the table beside the bank, a small chalkboard reads '₹100 → ₹95 → ₹90 → ₹85'. Right half: a green-and-yellow Indian auto-rickshaw moves through a sunlit Indian street, with its meter visible at the front showing '₹40' — and a faint paper sign on the side reading '₹25 base + ₹15/km'. Both halves are unified by a warm late-afternoon golden light. The image conveys: linear patterns are not abstract mathematics — they are everywhere in daily Indian life. Painterly cinematic illustration. Dark background. No additional text labels other than what's described.

✦Think About It

Bela starts the month with **₹100** of pocket money. She spends **₹5 each day**. After how many days will she have only **₹40** left? **Notice:** her money *decreases* by a constant amount each day. Is this still a 'linear pattern' — even though it's going *down*, not up?

Linear patterns can go up *or* down — only the *step size* needs to be constant. A constant negative step is just as 'linear' as a constant positive step.

✦A Sanskrit Subhāṣita on Steady Habits

From the Subhāṣita-Ratna-Bhāṇḍāgāra (a classical Sanskrit anthology)

अल्पम् अल्पम् त्यजेद् वित्तं पुण्यार्थम् बुद्धि-पूर्वकम्

अल्पम् अल्पम् च गृह्णीयाद् अध्ययनं नित्य-नूतनम्॥

(alpam alpam tyajed vittaṃ puṇyārtham buddhi-pūrvakam / alpam alpam ca gṛhṇīyād adhyayanam nitya-nūtanam)

'थोड़ा-थोड़ा करके धन को (अच्छे काम के लिए) सोच-समझकर खर्च करो; और थोड़ा-थोड़ा करके हर दिन कुछ नया सीखो।'

'Spend wealth in small amounts, mindfully, for good purposes. Take in learning in small amounts — but every single day.'

The Sanskrit verse celebrates steady, equal-step change — exactly what makes a pattern linear. ₹5 spent each day, one new fact learnt each day — these are the rhythms of slow, reliable change. The mathematics of this page is the algebra of that quiet daily discipline.

Bela's pocket money — a decreasing linear pattern

Worked Example 1Bela starts with ₹100, spends ₹5/day (NCERT Ex 7)

NCERT

Bela has ₹100 for pocket money. She spends ₹5 every day. After how many days will she be left with ₹40?

Step 1 — Set up a table.

Day $n$	0	1	2	3	4	…
Amount left (₹)	100	$100 - 5 \times 1 = 95$	$100 - 5 \times 2 = 90$	$100 - 5 \times 3 = 85$	$100 - 5 \times 4 = 80$	…

Step 2 — General formula. On day $n$ :

$\text{Amount left} = 100 - 5n \text{ rupees}$

This is a linear polynomial in $n$ — leading coefficient $-5$ (note the negative sign, because money is decreasing), constant term $100$ .

Step 3 — Set the formula equal to ₹40 and solve.

$100 - 5n = 40$

Subtract 100: $-5n = -60$ . Divide by $-5$ : $n = 12$ .

Answer: Bela will have ₹40 left after 12 days. (Check: $100 - 5 \times 12 = 100 - 60 = 40$ ✓.)

Worked Example 2Bela on day 15, and when she runs out (NCERT Think-Reflect)

NCERT

(a) What amount will Bela have on the 15th day? (b) On which day will she have spent everything?

(a) Day 15. Substitute $n = 15$ :

$100 - 5 \times 15 = 100 - 75 = ₹25$

Bela has ₹25 left on day 15.

(b) When does she run out? Set the amount to 0:

$100 - 5n = 0 \Rightarrow n = 20$

On day 20, Bela will have spent her entire ₹100. (After that, the formula gives negative values — which would mean she'd be in debt; the model breaks down beyond the natural domain.)

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The auto-rickshaw fare — an increasing linear pattern

Worked Example 3Auto-rickshaw fare (NCERT Ex 8)

NCERT

An auto-rickshaw fare starts at ₹25 for the first 2 km. After that, it increases by ₹15 per kilometre. What is the fare for a 10-km journey? Find a formula for the fare for any $n$ km, when $n \ge 2$ .

Step 1 — Build the table for 1-6 km.

Km	1	2	3	4	5	6
Fare (₹)	25	25	$25 + 1(15) = 40$	$25 + 2(15) = 55$	$25 + 3(15) = 70$	$25 + 4(15) = 85$

Step 2 — Direct calculation for 10 km.

For a 10-km journey, the first 2 km cost ₹25 (flat). The remaining $10 - 2 = 8$ km cost $8 \times 15 = ₹120$ . So total $= 25 + 120 = ₹145$ .

Step 3 — General formula for $n$ km, $n \ge 2$ .

Fare $= 25 + 15(n - 2) = 25 + 15n - 30 = 15n - 5$ .

$\text{Fare}(n) = 15n - 5 \quad (n \ge 2)$

Check: $n = 10 \Rightarrow 15(10) - 5 = 150 - 5 = 145$ ✓.

Answer: Fare for 10 km is ₹145. General formula: $15n - 5$ rupees for $n \ge 2$ km.

Worked Example 4Auto fare = ₹130, find the distance (NCERT Think-Reflect)

NCERT

For how many kilometres is the auto-rickshaw fare ₹130?

Set the formula equal to 130:

$15n - 5 = 130$

Add 5: $15n = 135$ . Divide by 15: $n = 9$ .

Answer: Fare is ₹130 for a 9 km journey. (Check: $15(9) - 5 = 135 - 5 = 130$ ✓.)

Linear pattern — the formal definition

After two examples — Bela's decreasing pocket money and the increasing auto fare — we can now write down the formal definition.

A linear pattern is a sequence of numbers in which the difference between two consecutive terms is constant.

If the constant difference is $d$ and the first term is $f(1) = a + d$ (i.e., the underlying linear polynomial is $f(n) = dn + a$ ), then:

$d > 0$ → linear growth (each term is larger than the previous one)
$d < 0$ → linear decay (each term is smaller than the previous one)
$d = 0$ → constant pattern (technically not 'linear' since the polynomial reduces to a constant, but the test still works)

Linear patterns will return as 'arithmetic progressions' (APs) in a later chapter on Sequences and Progressions. What you're learning here — the constant-difference signature, the nth-term formula — is the algebraic core of arithmetic progressions.

Exercise Set 2.3 — five practice scenarios

Q1Student savings account (NCERT Ex Set 2.3, Q1)

NCERT

A student has ₹500 in her savings bank account. She gets ₹150 every month as pocket money. How much money will she have at the end of every month from the second month onwards? Find a linear expression for the amount she will have in the nth month.

Pattern setup. At the end of month 1: $500 + 150 = ₹650$ . End of month 2: $650 + 150 = ₹800$ . End of month 3: ₹950. Etc.

End of month $n$	1	2	3	4	5
Amount (₹)	650	800	950	1100	1250

Constant difference: $150$ . So $f(n) = 150n + b$ . To match $f(1) = 650$ : $150 + b = 650$ , so $b = 500$ .

Linear expression: $f(n) = 150n + 500$ rupees.

Answer: Amount in nth month = ₹ $(150n + 500)$ . (Check: $f(2) = 800$ ✓.)

Q2Rally with members dropping out (NCERT Q2)

NCERT

A rally starts with 120 members. Each hour, 9 members drop out. How many members will remain after 1, 2, 3, … hours? Find a linear expression for the number of members at the end of the nth hour.

Pattern setup. Hour 1: $120 - 9 = 111$ . Hour 2: $111 - 9 = 102$ . Hour 3: 93. Etc.

Constant difference: $-9$ . So $f(n) = -9n + b$ . To match $f(1) = 111$ : $-9 + b = 111$ , so $b = 120$ .

Linear expression: $f(n) = 120 - 9n$ members.

Domain note: this formula gives 0 members when $n = \frac{120}{9} \approx 13.33$ , so realistically the rally is over after about 13 hours (the 14th hour would give a negative count, which is meaningless).

Q3Rectangle with fixed length 13 (NCERT Q3)

NCERT

Suppose the length of a rectangle is 13 cm. Find the area if the breadth is (i) 12 cm, (ii) 10 cm, (iii) 8 cm. Find the linear pattern representing the area as a function of breadth.

Areas:

(i) $13 \times 12 = 156$ sq cm
(ii) $13 \times 10 = 130$ sq cm
(iii) $13 \times 8 = 104$ sq cm

Linear pattern: Area is $A = 13b$ where $b$ is the breadth.

Why is this linear? The length 13 is fixed, so $A = 13b$ is a linear polynomial in $b$ (leading coefficient 13, constant 0). If both length and breadth varied (as $A = lb$ ), it would not be a polynomial in one variable at all.

Q4Box with fixed length and breadth (NCERT Q4)

NCERT

Suppose a rectangular box has length 7 cm and breadth 11 cm. Find the volume if the height is (i) 5 cm, (ii) 9 cm, (iii) 13 cm. Find the linear pattern for the volume.

Volumes:

(i) $7 \times 11 \times 5 = 385$ cu cm
(ii) $7 \times 11 \times 9 = 693$ cu cm
(iii) $7 \times 11 \times 13 = 1001$ cu cm

Linear pattern: Volume is $V = 77h$ where $h$ is the height.

Leading coefficient is $77 = 7 \times 11$ — the area of the fixed base. Since the base is fixed, volume scales linearly with height.

Q5Sarita reading a 500-page book (NCERT Q5)

NCERT

Sarita is reading a book of 500 pages. She reads 20 pages every day. How many pages will be left after 15 days? Express this as a linear pattern.

Pattern. After day 1: $500 - 20 = 480$ pages left. Day 2: $460$ . Etc.

Linear expression: Pages left after day $n$ : $f(n) = 500 - 20n$ .

After 15 days: $f(15) = 500 - 20 \times 15 = 500 - 300 = 200$ pages left.

Domain: Sarita finishes the book when $f(n) = 0$ , i.e. $n = 25$ days. After that, the formula would give negative pages — meaningless.

Logical ReasoningLevel 4 · Evaluation

In Q3 above, the rectangle has length **fixed** at 13 cm. The area is $A = 13b$ — linear in the breadth $b$. If you let the **length and breadth both vary**, the area becomes $A = lb$ — a product of two variables. **Is $lb$ a polynomial in one variable? Why or why not?**

Practice Yourself — Real-World Linear Patterns

Find the linear expression $f(n)$ and answer the question.

A water tank starts with 2000 L and loses 150 L per day to evaporation. How much water is left after 10 days? When does it run dry?
A library charges ₹50 for a membership and ₹5 per book borrowed. Cost for 12 books?
A bus has 44 seats and 6 passengers get on at each stop (it starts empty). At which stop does the bus first become full?
A weight-loss programme: a person starts at 80 kg, loses 0.5 kg per week. Their weight after 16 weeks?
A rope of 30 m is cut into pieces of 2 m each. Length remaining after $n$ cuts?
A plant is 15 cm tall and grows 2 cm per week. Its height after $n$ weeks?

Answers: 1. $f(n) = 2000 - 150n$ ; after 10 days ${} = 500$ L; runs dry on day $\frac{2000}{150} \approx 13.3$ , so day 14. 2. $f(b) = 5b + 50$ ; for 12 books ${} = ₹110$ . 3. $f(s) = 6s$ ; full when $6s = 44 \Rightarrow s \approx 7.3$ , so the 8th stop. 4. $f(w) = 80 - 0.5w$ ; at $w = 16$ , weight = 72 kg. 5. $f(n) = 30 - 2n$ m. 6. $f(n) = 2n + 15$ cm.

Quick Check

Q1.Bela starts with ₹100 and spends ₹5 a day. The amount left on day $n$ is $100 - 5n$ . What is the leading coefficient of this linear polynomial, and what does its sign tell you?

AI Generation Prompt

✦Think About It

Linear patterns can go up *or* down — only the *step size* needs to be constant. A constant negative step is just as 'linear' as a constant positive step.

✦A Sanskrit Subhāṣita on Steady Habits

From the Subhāṣita-Ratna-Bhāṇḍāgāra (a classical Sanskrit anthology)

अल्पम् अल्पम् त्यजेद् वित्तं पुण्यार्थम् बुद्धि-पूर्वकम्

अल्पम् अल्पम् च गृह्णीयाद् अध्ययनं नित्य-नूतनम्॥

(alpam alpam tyajed vittaṃ puṇyārtham buddhi-pūrvakam / alpam alpam ca gṛhṇīyād adhyayanam nitya-nūtanam)

'Spend wealth in small amounts, mindfully, for good purposes. Take in learning in small amounts — but every single day.'

Bela's pocket money — a decreasing linear pattern

Worked Example 1Bela starts with ₹100, spends ₹5/day (NCERT Ex 7)

NCERT

Bela has ₹100 for pocket money. She spends ₹5 every day. After how many days will she be left with ₹40?

Step 1 — Set up a table.

Day $n$	0	1	2	3	4	…
Amount left (₹)	100	$100 - 5 \times 1 = 95$	$100 - 5 \times 2 = 90$	$100 - 5 \times 3 = 85$	$100 - 5 \times 4 = 80$	…

Step 2 — General formula. On day $n$ :

$\text{Amount left} = 100 - 5n \text{ rupees}$

This is a linear polynomial in $n$ — leading coefficient $-5$ (note the negative sign, because money is decreasing), constant term $100$ .

Step 3 — Set the formula equal to ₹40 and solve.

$100 - 5n = 40$

Subtract 100: $-5n = -60$ . Divide by $-5$ : $n = 12$ .

Answer: Bela will have ₹40 left after 12 days. (Check: $100 - 5 \times 12 = 100 - 60 = 40$ ✓.)

Worked Example 2Bela on day 15, and when she runs out (NCERT Think-Reflect)

NCERT

(a) What amount will Bela have on the 15th day? (b) On which day will she have spent everything?

(a) Day 15. Substitute $n = 15$ :

$100 - 5 \times 15 = 100 - 75 = ₹25$

Bela has ₹25 left on day 15.

(b) When does she run out? Set the amount to 0:

$100 - 5n = 0 \Rightarrow n = 20$

On day 20, Bela will have spent her entire ₹100. (After that, the formula gives negative values — which would mean she'd be in debt; the model breaks down beyond the natural domain.)

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The auto-rickshaw fare — an increasing linear pattern

Worked Example 3Auto-rickshaw fare (NCERT Ex 8)

NCERT

Step 1 — Build the table for 1-6 km.

Km	1	2	3	4	5	6
Fare (₹)	25	25	$25 + 1(15) = 40$	$25 + 2(15) = 55$	$25 + 3(15) = 70$	$25 + 4(15) = 85$

Step 2 — Direct calculation for 10 km.

For a 10-km journey, the first 2 km cost ₹25 (flat). The remaining $10 - 2 = 8$ km cost $8 \times 15 = ₹120$ . So total $= 25 + 120 = ₹145$ .

Step 3 — General formula for $n$ km, $n \ge 2$ .

Fare $= 25 + 15(n - 2) = 25 + 15n - 30 = 15n - 5$ .

$\text{Fare}(n) = 15n - 5 \quad (n \ge 2)$

Check: $n = 10 \Rightarrow 15(10) - 5 = 150 - 5 = 145$ ✓.

Answer: Fare for 10 km is ₹145. General formula: $15n - 5$ rupees for $n \ge 2$ km.

Worked Example 4Auto fare = ₹130, find the distance (NCERT Think-Reflect)

NCERT

For how many kilometres is the auto-rickshaw fare ₹130?

Set the formula equal to 130:

$15n - 5 = 130$

Add 5: $15n = 135$ . Divide by 15: $n = 9$ .

Answer: Fare is ₹130 for a 9 km journey. (Check: $15(9) - 5 = 135 - 5 = 130$ ✓.)

Linear pattern — the formal definition

After two examples — Bela's decreasing pocket money and the increasing auto fare — we can now write down the formal definition.

A linear pattern is a sequence of numbers in which the difference between two consecutive terms is constant.

If the constant difference is $d$ and the first term is $f(1) = a + d$ (i.e., the underlying linear polynomial is $f(n) = dn + a$ ), then:

$d > 0$ → linear growth (each term is larger than the previous one)
$d < 0$ → linear decay (each term is smaller than the previous one)
$d = 0$ → constant pattern (technically not 'linear' since the polynomial reduces to a constant, but the test still works)

Exercise Set 2.3 — five practice scenarios

Q1Student savings account (NCERT Ex Set 2.3, Q1)

NCERT

Pattern setup. At the end of month 1: $500 + 150 = ₹650$ . End of month 2: $650 + 150 = ₹800$ . End of month 3: ₹950. Etc.

End of month $n$	1	2	3	4	5
Amount (₹)	650	800	950	1100	1250

Constant difference: $150$ . So $f(n) = 150n + b$ . To match $f(1) = 650$ : $150 + b = 650$ , so $b = 500$ .

Linear expression: $f(n) = 150n + 500$ rupees.

Answer: Amount in nth month = ₹ $(150n + 500)$ . (Check: $f(2) = 800$ ✓.)

Q2Rally with members dropping out (NCERT Q2)

NCERT

Pattern setup. Hour 1: $120 - 9 = 111$ . Hour 2: $111 - 9 = 102$ . Hour 3: 93. Etc.

Constant difference: $-9$ . So $f(n) = -9n + b$ . To match $f(1) = 111$ : $-9 + b = 111$ , so $b = 120$ .

Linear expression: $f(n) = 120 - 9n$ members.

Q3Rectangle with fixed length 13 (NCERT Q3)

NCERT

Suppose the length of a rectangle is 13 cm. Find the area if the breadth is (i) 12 cm, (ii) 10 cm, (iii) 8 cm. Find the linear pattern representing the area as a function of breadth.

Areas:

(i) $13 \times 12 = 156$ sq cm
(ii) $13 \times 10 = 130$ sq cm
(iii) $13 \times 8 = 104$ sq cm

Linear pattern: Area is $A = 13b$ where $b$ is the breadth.

Q4Box with fixed length and breadth (NCERT Q4)

NCERT

Suppose a rectangular box has length 7 cm and breadth 11 cm. Find the volume if the height is (i) 5 cm, (ii) 9 cm, (iii) 13 cm. Find the linear pattern for the volume.

Volumes:

(i) $7 \times 11 \times 5 = 385$ cu cm
(ii) $7 \times 11 \times 9 = 693$ cu cm
(iii) $7 \times 11 \times 13 = 1001$ cu cm

Linear pattern: Volume is $V = 77h$ where $h$ is the height.

Leading coefficient is $77 = 7 \times 11$ — the area of the fixed base. Since the base is fixed, volume scales linearly with height.

Q5Sarita reading a 500-page book (NCERT Q5)

NCERT

Sarita is reading a book of 500 pages. She reads 20 pages every day. How many pages will be left after 15 days? Express this as a linear pattern.

Pattern. After day 1: $500 - 20 = 480$ pages left. Day 2: $460$ . Etc.

Linear expression: Pages left after day $n$ : $f(n) = 500 - 20n$ .

After 15 days: $f(15) = 500 - 20 \times 15 = 500 - 300 = 200$ pages left.

Domain: Sarita finishes the book when $f(n) = 0$ , i.e. $n = 25$ days. After that, the formula would give negative pages — meaningless.

Logical ReasoningLevel 4 · Evaluation

Quick Check

Q1.Bela starts with ₹100 and spends ₹5 a day. The amount left on day $n$ is $100 - 5n$ . What is the leading coefficient of this linear polynomial, and what does its sign tell you?