NCERT Solutions for Class 9th Maths Chapter 2 Exercise Set 2.4 — Linear growth and linear decay

Book page 25–26 Updated on2026-09-08

Q1.

Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month. (i) Find the height after 7 months. (ii) Make a table of values for t varying from 0 to 10 months and show how the height, h, increases every month. (iii) Find an expression that relates h and t, and explain why it represents linear growth.

Answer

(i) Seven months of growth at 0.5 ft a month adds 3.5 ft.

h = 1.75 + 0.5 × 7
= 1.75 + 3.5
= 5.25 feet

(ii)

Month t	0	1	2	3	4	5	6	7	8	9	10
Height h (ft)	1.75	2.25	2.75	3.25	3.75	4.25	4.75	5.25	5.75	6.25	6.75

(iii)

h = 1.75 + 0.5t

Why this is linear growth: subtract two consecutive heights and see what survives.

h(t + 1) − h(t) = [1.75 + 0.5(t + 1)] − [1.75 + 0.5t]
= 0.5

The t cancels completely, so the monthly increase is the same fixed 0.5 ft at every stage — the plant does not grow faster as it gets taller. A fixed positive amount added over equal intervals is precisely the definition of linear growth. The expression has degree 1, the constant 1.75 is the height at t = 0, and 0.5 is the slope.

Check it yourself: in the table every step is +0.50 ft, and the 7-month entry 5.25 agrees with part (i).

Q2.

A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year. (i) Find the value of the phone after 3 years. (ii) Make a table of values for t varying from 0 to 8 years and show how the value of the phone, v, depreciates with time. (iii) Find an expression that relates v and t, and explain why it represents linear decay.

Answer

(i)

v = 10000 − 800 × 3
= 10000 − 2400
= ₹7600

(ii)

Year t	0	1	2	3	4	5	6	7	8
Value v (₹)	10000	9200	8400	7600	6800	6000	5200	4400	3600

(iii)

v = 10000 − 800t

Why this is linear decay: v(t + 1) − v(t) = −800 for every t — a fixed amount is subtracted over equal intervals, and the coefficient of t is negative. Note what this model assumes: the phone loses the same ₹800 in its eighth year as in its first. Real depreciation is often a fixed percentage instead, which would not be linear. The straight-line assumption is what makes the arithmetic here so simple.

Tip: the model runs out at 10000 − 800t = 0, i.e. t = 12.5 years. Beyond that the value would go negative, which no second-hand phone does.

Q3.

The initial population of a village is 750. Every year, 50 people move from a nearby city to the village. (i) Find the population of the village after 6 years. (ii) Make a table of values for t varying from 0 to 10 years and show how the population, P, increases every year. (iii) Find an expression that relates P and t, and explain why it represents linear growth.

Answer

(i)

P = 750 + 50 × 6
= 750 + 300
= 1050 people

(ii)

Year t	0	1	2	3	4	5	6	7	8	9	10
Population P	750	800	850	900	950	1000	1050	1100	1150	1200	1250

(iii)

P = 750 + 50t

Why this is linear growth: the yearly increase is P(t + 1) − P(t) = +50, the same number every year, so equal intervals of time bring equal increases in population. It is worth seeing why: the 50 arrivals come from the city, not from the village itself, so the number arriving does not depend on how big the village has become. If instead the village grew by births at a fixed rate per resident, the yearly increase would itself grow, and the pattern would not be linear.

Try This: when does the village first cross 2000? Solve 750 + 50t > 2000 to get t > 25, so in the 26th year.

Q4.

A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after the recharge. (i) Write an equation that models the remaining balance b(x) after using the scheme for x days. Explain why it represents linear decay. (ii) After how many days will the balance run out? (iii) Make a table of values for x varying from 1 to 10 days and show how the balance b(x), reduces with time.

Answer

(i)

b(x) = 600 − 15x

It is linear decay because b(x + 1) − b(x) = −15 for every x: a fixed amount is removed over equal intervals, and the coefficient of x is negative. The graph would be a straight line falling from (0, 600).

(ii) The balance runs out when b(x) = 0.

600 − 15x = 0
15x = 600
x = 40 days

(iii)

Day x	1	2	3	4	5	6	7	8	9	10
Balance b(x) (₹)	585	570	555	540	525	510	495	480	465	450

Why the zero of the polynomial answers (ii): “the balance runs out” means the output of the machine is 0, so we ask which input sends b to 0. For a linear polynomial ax + b with a ≠ 0 there is exactly one such input, x = −b/a — here −600/(−15) = 40. That is why questions of the form “when does it finish?” always have a single clean answer in a linear model.

Check it yourself: the table drops by exactly ₹15 each column, and continuing that fall from ₹450 on day 10 for another 30 days reaches ₹0 on day 40.

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