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📐 Chapter 2: Introduction to Linear Polynomials • Exercise Set 2.4

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Exercise Set 2.1 5Exercise Set 2.2 7Exercise Set 2.3 5Exercise Set 2.4 4Exercise Set 2.5 3Exercise Set 2.6 1End-of-Chapter Exercises 14All Exercises 39
Q1

Exercise Set 2.4

1. 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.

  • (i) Height after 7 months: 1.75 + (7 × 0.5) = 1.75 + 3.5 = 5.25 feet.
  • (ii) Table of values:
    Month (t)012345678910
    Height (h)1.752.252.753.253.754.254.755.255.756.256.75
  • (iii) Expression: h = 1.75 + 0.5t. It represents linear growth because the quantity (height) increases by a constant positive amount (0.5 feet) over equal intervals (each month).
Q2

Exercise Set 2.4

2. 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.

  • (i) Value after 3 years: 10000 - (3 × 800) = 10000 - 2400 = ₹7600.
  • (ii) Table of values:
    Year (t)012345678
    Value (v)1000092008400760068006000520044003600
  • (iii) Expression: v = 10000 - 800t. It represents linear decay because the value decreases by a constant fixed amount (₹800) over equal intervals (each year).
Q3

Exercise Set 2.4

3. 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.

  • (i) Population after 6 years: 750 + (6 × 50) = 750 + 300 = 1050.
  • (ii) Table of values:
    Year (t)012345678910
    Population (P)750800850900950100010501100115012001250
  • (iii) Expression: P = 750 + 50t. It represents linear growth because the population increases by a constant amount (50 people) over equal intervals (each year).
Q4

Exercise Set 2.4

4. 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.

  • (i) Equation: b(x) = 600 - 15x. It represents linear decay because the balance decreases by a fixed amount (₹15) at constant intervals (every day).
  • (ii) Balance runs out: Set b(x) = 0.
    600 - 15x = 0 ⇒ 15x = 600 ⇒ x = 40 days.
  • (iii) Table of values:
    Days (x)12345678910
    Balance b(x)585570555540525510495480465450