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NCERT Class 9 Mathematics Exercise Set 2.4 Solutions: Introduction to Linear Polynomials

Step-by-step working for every question of Exercise Set 2.4 in NCERT's Class 9 Mathematics book, chapter “Introduction to Linear Polynomials”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Ex 2.4 Q1

   A plant is 1.75 ft tall and grows 0.5 ft a month. (i) Height after 7 months; (ii) table of h for t = 0 to 10 months; (iii) the expression for h in terms of t and why it is linear growth.

   1. (i) The height of the plant after t months is given by h = 1.75 + 0.5t, where 1.75 is the initial height and 0.5 is the growth rate per month.
   2. (i) Substitute t = 7 into the expression to find the height after 7 months.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   3. (i) h = 1.75 + 0.5 × 7 = 1.75 + 3.5 = 5.25 feet.
   4. (ii) Calculate the height for each month from t = 0 to t = 10 using h = 1.75 + 0.5t.
   5. (ii) At t = 0, h = 1.75; at t = 1, h = 2.25; at t = 2, h = 2.75; at t = 3, h = 3.25; at t = 4, h = 3.75.
   6. (ii) At t = 5, h = 4.25; at t = 6, h = 4.75; at t = 7, h = 5.25; at t = 8, h = 5.75; at t = 9, h = 6.25.
   7. (ii) At t = 10, h = 1.75 + 0.5 × 10 = 6.75.
   8. (iii) The expression for height h in terms of time t is h = 1.75 + 0.5t.
   9. (iii) This is a linear growth because the height increases by a constant amount of 0.5 feet for every month, which is the definition of a linear relationship.

   Answer: (i) 5.25 feet; (ii) h = 1.75, 2.25, 2.75, …, 6.75 for t = 0, 1, 2, …, 10; (iii) h = 1.75 + 0.5t, linear growth (rises by 0.5 ft every month)

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2. Ex 2.4 Q2

   A phone bought for ₹10,000 loses ₹800 of value every year. (i) Value after 3 years; (ii) table of v for t = 0 to 8 years; (iii) the expression for v and why it is linear decay.

   1. (i) The initial value of the phone is ₹10,000.
   2. (i) The value decreases by ₹800 every year.
   3. (i) After 3 years, the total decrease is 3 × 800 = ₹2,400.
   4. (i) The value after 3 years is 10,000 − 2,400 = ₹7,600.
   5. (ii) At t = 0, v = 10,000.
   6. (ii) At t = 1, v = 10,000 − 800 = 9,200.
   7. (ii) At t = 2, v = 9,200 − 800 = 8,400.
   8. (ii) At t = 3, v = 8,400 − 800 = 7,600.
   9. (ii) At t = 4, v = 7,600 − 800 = 6,800.
   10. (ii) At t = 5, v = 6,800 − 800 = 6,000.
   11. (ii) At t = 6, v = 6,000 − 800 = 5,200.
   12. (ii) At t = 7, v = 5,200 − 800 = 4,400.
   13. (ii) At t = 8, v = 4,400 − 800 = 3,600.
   14. (iii) The value v after t years is given by v = 10000 − 800t.
   15. (iii) The value falls by the same ₹800 every year (v is a degree-1 expression in t with a negative coefficient of t), so this is linear decay.

   Answer: (i) ₹7,600; (ii) v = 10000, 9200, 8400, 7600, 6800, 6000, 5200, 4400, 3600 for t = 0 to 8; (iii) v = 10000 − 800t, linear decay

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3. Ex 2.4 Q3

   A village of 750 people gains 50 people every year. (i) Population after 6 years; (ii) table of P for t = 0 to 10 years; (iii) the expression for P and why it is linear growth.

   1. (i) The initial population is 750 and the yearly increase is 50.
   2. (i) Population after 6 years = 750 + 50 × 6.
   3. (i) Population after 6 years = 750 + 300 = 1050.
   4. (ii) For t = 0, P = 750 + 50(0) = 750.
   5. (ii) For t = 1, P = 750 + 50(1) = 800.
   6. (ii) For t = 2, P = 750 + 50(2) = 850.
   7. (ii) For t = 3, P = 750 + 50(3) = 900.
   8. (ii) For t = 4, P = 750 + 50(4) = 950.
   9. (ii) For t = 5, P = 750 + 50(5) = 1000.
   10. (ii) For t = 6, P = 750 + 50(6) = 1050.
   11. (ii) For t = 7, P = 750 + 50(7) = 1100.
   12. (ii) For t = 8, P = 750 + 50(8) = 1150.
   13. (ii) For t = 9, P = 750 + 50(9) = 1200.
   14. (ii) For t = 10, P = 750 + 50(10) = 1250.
   15. (iii) The expression for population P after t years is P = 750 + 50t.
   16. (iii) This is linear growth because the population increases by a constant amount of 50 every year.

   Answer: (i) 1050; (ii) P = 750, 800, 850, …, 1250 for t = 0, 1, 2, …, 10; (iii) P = 750 + 50t, linear growth (50 more people every year)

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4. Ex 2.4 Q4

   A ₹600 prepaid balance falls by ₹15 every day. (i) Equation for the balance b(x) after x days and why it is linear decay; (ii) when does it run out; (iii) table for x = 1 to 10 days.

   1. (i) The initial balance is ₹600 and it decreases by ₹15 for every day x.
   2. (i) The balance after x days is given by b(x) = 600 − 15x.
   3. (i) The balance goes down by the same ₹15 every day (b(x) is linear in x with a negative coefficient of x), so this is linear decay.
   4. (ii) To find when the balance runs out, set b(x) = 0.
   5. (ii) 600 − 15x = 0 implies 15x = 600.
   6. (ii) Solving for x gives x = 600 ÷ 15 = 40 days.
   7. (iii) For x = 1, b(1) = 600 − 15(1) = 585.
   8. (iii) For x = 2, b(2) = 600 − 15(2) = 570.
   9. (iii) For x = 3, b(3) = 600 − 15(3) = 555.
   10. (iii) For x = 4, b(4) = 600 − 15(4) = 540.
   11. (iii) For x = 5, b(5) = 600 − 15(5) = 525.
   12. (iii) For x = 6, b(6) = 600 − 15(6) = 510.
   13. (iii) For x = 7, b(7) = 600 − 15(7) = 495.
   14. (iii) For x = 8, b(8) = 600 − 15(8) = 480.
   15. (iii) For x = 9, b(9) = 600 − 15(9) = 465.
   16. (iii) For x = 10, b(10) = 600 − 15(10) = 450.

   Answer: (i) b(x) = 600 − 15x; (ii) 40 days; (iii) Table: x=1→585, x=2→570, x=3→555, x=4→540, x=5→525, x=6→510, x=7→495, x=8→480, x=9→465, x=10→450

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