Growth & Decay in Action
A plant, a phone, and a village — three NCERT word problems modelled in three lines of algebra each
AI Generation Prompt
Ultra-wide cinematic banner (16:5 ratio) split into three softly-divided vertical panels. Left panel: a healthy young green plant pushing upward out of rich brown earth, with a faint upward arrow tracing its growth. Centre panel: a modern smartphone resting on a wooden surface, with a small price tag attached showing '₹10,000 → ₹9,200 → ₹8,400 …' — the numbers fading to suggest decreasing value. Right panel: a vibrant Indian village square at sunset with people walking, a few houses with terracotta roofs, and a faint upward arrow showing population growth. All three panels share the same warm golden-hour sunset light. The image conveys: linear models describe growth and decay everywhere — from a single plant to a depreciating gadget to a whole village. Painterly cinematic illustration. Dark background. No text labels other than what's described.
A village adds **50 people every year**. A phone loses **₹800 every year**. A plant grows **0.5 ft every month**. **They all change by a constant amount per unit time.** That tiny phrase — 'constant per unit time' — is the secret signature of every linear pattern in the world. **Where else can you spot it in your daily life?**
Think: monthly subscriptions, daily metro fares, weekly weight changes, hourly battery drain.
From the Subhāṣita-Ratna-Bhāṇḍāgāra
वृद्धिः क्षयश्च मानवानां जीवनस्य लक्षणे
सम-कालेन सम-गत्या निःशब्दं फलम् भवेत्॥
(vṛddhiḥ kṣayaś ca mānavānāṃ jīvanasya lakṣaṇe / sama-kālena sama-gatyā niḥśabdaṃ phalam bhavet)
'वृद्धि (बढ़ना) और क्षय (घटना) — दोनों मनुष्य के जीवन के चिन्ह हैं। बराबर समय में, बराबर गति से — चुपचाप फल मिलता है।'
'Vṛddhi (growth) and kṣaya (decay) — both are signs of human life. In equal time, at an equal pace, the result quietly comes.'
The Sanskrit terms vṛddhi (वृद्धि, growth) and kṣaya (क्षय, decay) are still used today in Indian medicine, agriculture, and economics — to describe exactly the kind of steady, equal-step change that linear polynomials capture mathematically. Algebra and Sanskrit agree: when something changes by an equal amount in equal time, you can predict its future.
NCERT Exercise Set 2.4 — three real-world models
Loading simulator…
A plant has height 1.75 feet and grows by 0.5 feet each month. Find its height after 7 months.
Step 1 — Set up the linear function.
Leading coefficient (the per-month growth rate); constant term (the starting height).
Step 2 — Substitute .
Answer: After 7 months, the plant is 5.25 ft tall.
Sense-check: the plant has grown by ft in 7 months. That's ft per month. ✓
Make a table of the plant's height for varying from 0 to 10 months.
Apply for each value of :
| (months) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| (ft) | 1.75 | 2.25 | 2.75 | 3.25 | 3.75 | 4.25 | 4.75 | 5.25 | 5.75 | 6.25 | 6.75 |
Notice the constant difference of 0.5 ft between consecutive months. That's the per-step rate, exactly equal to the leading coefficient.
Find an expression that relates and , and explain why it represents linear growth.
Expression: .
Why it represents linear growth:
- Linear because the highest power of is 1 (degree 1) and the successive differences are constant.
- Growth because the leading coefficient is positive — the height increases by 0.5 ft for every unit increase in .
Formally: a function represents linear growth iff . Here , so this is linear growth. ✓
A mobile phone is bought for ₹10{,}000. Its value decreases by ₹800 every year. Find the value after 3 years.
Step 1 — Set up the function.
Leading coefficient (negative because value is decreasing); constant term (the starting price).
Step 2 — Substitute .
Answer: After 3 years, the phone is worth ₹7,600 according to the linear model.
Make a table of the phone's value for varying from 0 to 8 years.
Apply :
| (years) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
| (₹) | 10000 | 9200 | 8400 | 7600 | 6800 | 6000 | 5200 | 4400 | 3600 |
Constant difference: −₹800 per year. ✓
Domain note: the formula gives at years. After that, it would give negative values — meaningless. The linear model is only realistic for the first ~12 years.
Find an expression that relates and , and explain why it represents linear decay.
Expression: .
Why it represents linear decay:
- Linear because the highest power of is 1 and successive differences are constant.
- Decay because the leading coefficient is negative — the value decreases by ₹800 for every unit increase in .
Formally: is linear decay iff . Here . ✓
The initial population of a village is 750. Every year, 50 people move from a nearby city to the village. Find the population after 6 years.
Step 1 — Set up the function.
Leading coefficient (positive; population grows); constant term (starting population).
Step 2 — Substitute .
Answer: After 6 years, the village has 1,050 people.
Lesson: this is linear growth (). The model assumes the per-year migration stays exactly constant — in reality, populations rarely grow this cleanly, but the linear model is an excellent first approximation.
The mobile-phone model says $v(t) = 10000 - 800t$. **At what value of $t$ would the model give zero value? What does this tell us about the *limits* of the linear model?**
Practice Yourself — More Growth/Decay Models
For each scenario, write the linear function , classify as growth/decay, and answer the question.
- Gym membership balance: starts at ₹6{,}000, decreases ₹500/month. When does it run out?
- Tree: 3 m tall, grows 0.4 m/year. Height after 8 years?
- Library overdue fine: ₹2 per day overdue (no fine before due date). Fine for 7 days overdue?
- Car odometer: reads 45{,}000 km, driven 80 km/day. Reading after 30 days?
- Ice block: 2 kg, melts 0.05 kg/min. When does it disappear?
- Daily revenue of a stall: ₹1{,}200 today, increases ₹40/day. Revenue on day 25?
- Battery: starts at 100% charge, drops 8% per hour during video playback. When does it die?
- Bank fixed deposit (simple interest, very simplified): ₹50{,}000 principal earning ₹250 interest per month. Total amount after 18 months?
Answers: 1. ; decay; runs out at months. 2. ; growth; m. 3. ; growth from zero; . 4. ; growth; km. 5. ; decay; gone at min. 6. ; growth; . 7. ; decay; dies at hr. 8. ; growth; .
Q1.A plant has height ft. Height after 10 months?
AI Generation Prompt
Ultra-wide cinematic banner (16:5 ratio) split into three softly-divided vertical panels. Left panel: a healthy young green plant pushing upward out of rich brown earth, with a faint upward arrow tracing its growth. Centre panel: a modern smartphone resting on a wooden surface, with a small price tag attached showing '₹10,000 → ₹9,200 → ₹8,400 …' — the numbers fading to suggest decreasing value. Right panel: a vibrant Indian village square at sunset with people walking, a few houses with terracotta roofs, and a faint upward arrow showing population growth. All three panels share the same warm golden-hour sunset light. The image conveys: linear models describe growth and decay everywhere — from a single plant to a depreciating gadget to a whole village. Painterly cinematic illustration. Dark background. No text labels other than what's described.
A village adds **50 people every year**. A phone loses **₹800 every year**. A plant grows **0.5 ft every month**. **They all change by a constant amount per unit time.** That tiny phrase — 'constant per unit time' — is the secret signature of every linear pattern in the world. **Where else can you spot it in your daily life?**
Think: monthly subscriptions, daily metro fares, weekly weight changes, hourly battery drain.
From the Subhāṣita-Ratna-Bhāṇḍāgāra
वृद्धिः क्षयश्च मानवानां जीवनस्य लक्षणे
सम-कालेन सम-गत्या निःशब्दं फलम् भवेत्॥
(vṛddhiḥ kṣayaś ca mānavānāṃ jīvanasya lakṣaṇe / sama-kālena sama-gatyā niḥśabdaṃ phalam bhavet)
'वृद्धि (बढ़ना) और क्षय (घटना) — दोनों मनुष्य के जीवन के चिन्ह हैं। बराबर समय में, बराबर गति से — चुपचाप फल मिलता है।'
'Vṛddhi (growth) and kṣaya (decay) — both are signs of human life. In equal time, at an equal pace, the result quietly comes.'
The Sanskrit terms vṛddhi (वृद्धि, growth) and kṣaya (क्षय, decay) are still used today in Indian medicine, agriculture, and economics — to describe exactly the kind of steady, equal-step change that linear polynomials capture mathematically. Algebra and Sanskrit agree: when something changes by an equal amount in equal time, you can predict its future.
NCERT Exercise Set 2.4 — three real-world models
Loading simulator…
A plant has height 1.75 feet and grows by 0.5 feet each month. Find its height after 7 months.
Step 1 — Set up the linear function.
Leading coefficient (the per-month growth rate); constant term (the starting height).
Step 2 — Substitute .
Answer: After 7 months, the plant is 5.25 ft tall.
Sense-check: the plant has grown by ft in 7 months. That's ft per month. ✓
Make a table of the plant's height for varying from 0 to 10 months.
Apply for each value of :
| (months) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| (ft) | 1.75 | 2.25 | 2.75 | 3.25 | 3.75 | 4.25 | 4.75 | 5.25 | 5.75 | 6.25 | 6.75 |
Notice the constant difference of 0.5 ft between consecutive months. That's the per-step rate, exactly equal to the leading coefficient.
Find an expression that relates and , and explain why it represents linear growth.
Expression: .
Why it represents linear growth:
- Linear because the highest power of is 1 (degree 1) and the successive differences are constant.
- Growth because the leading coefficient is positive — the height increases by 0.5 ft for every unit increase in .
Formally: a function represents linear growth iff . Here , so this is linear growth. ✓
A mobile phone is bought for ₹10{,}000. Its value decreases by ₹800 every year. Find the value after 3 years.
Step 1 — Set up the function.
Leading coefficient (negative because value is decreasing); constant term (the starting price).
Step 2 — Substitute .
Answer: After 3 years, the phone is worth ₹7,600 according to the linear model.
Make a table of the phone's value for varying from 0 to 8 years.
Apply :
| (years) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
| (₹) | 10000 | 9200 | 8400 | 7600 | 6800 | 6000 | 5200 | 4400 | 3600 |
Constant difference: −₹800 per year. ✓
Domain note: the formula gives at years. After that, it would give negative values — meaningless. The linear model is only realistic for the first ~12 years.
Find an expression that relates and , and explain why it represents linear decay.
Expression: .
Why it represents linear decay:
- Linear because the highest power of is 1 and successive differences are constant.
- Decay because the leading coefficient is negative — the value decreases by ₹800 for every unit increase in .
Formally: is linear decay iff . Here . ✓
The initial population of a village is 750. Every year, 50 people move from a nearby city to the village. Find the population after 6 years.
Step 1 — Set up the function.
Leading coefficient (positive; population grows); constant term (starting population).
Step 2 — Substitute .
Answer: After 6 years, the village has 1,050 people.
Lesson: this is linear growth (). The model assumes the per-year migration stays exactly constant — in reality, populations rarely grow this cleanly, but the linear model is an excellent first approximation.
The mobile-phone model says $v(t) = 10000 - 800t$. **At what value of $t$ would the model give zero value? What does this tell us about the *limits* of the linear model?**
Q1.A plant has height ft. Height after 10 months?