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Linear growth and decay · 4 marks

A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year.

  1. (i) Find the value of the phone after 3 years.
  2. (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.
  3. (iii) Find an expression that relates v and t, and explain why it represents linear decay.
Answer: (i) ₹7,600 (ii) ₹10,000, 9,200, 8,400, …, 3,600 for t = 0 to 8 (iii) v = 10000 − 800t; linear decay because the value falls by the same ₹800 every year.

Step-by-step solution

Idea: Start value ₹10,000; it loses a fixed ₹800 each year, so after t years it has lost 800t rupees.

(i) Find the value of the phone after 3 years.

  1. Decrease in 3 years = 800 × 3 = ₹2,400.½ mark
  2. Value = 10,000 − 2,400 = ₹7,600.½ mark
₹7,600

(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.

  1. Subtract ₹800 for each year:
    Year, tValue, v (₹)
    010,000
    19,200
    28,400
    37,600
    46,800
    56,000
    65,200
    74,400
    83,600
    1 mark
  2. The value goes down by the same ₹800 every year.½ mark
10,000, 9,200, 8,400, 7,600, 6,800, 6,000, 5,200, 4,400, 3,600

(iii) Find an expression that relates v and t, and explain why it represents linear decay.

  1. v = 10000 − 800t1 mark
  2. It is linear in t with a negative coefficient (−800): the value decreases by a constant ₹800 over each equal interval of one year. So it represents linear decay.½ mark
v = 10000 − 800t; linear decay (constant decrease of ₹800 per year).
(i) ₹7,600 (ii) 10,000, 9,200, 8,400, 7,600, 6,800, 6,000, 5,200, 4,400, 3,600 rupees (iii) v = 10000 − 800t, linear decay since the value drops by a constant ₹800 each year.

Check: v(8) = 10000 − 6400 = 3600, the last entry of the table ✓.

Answer to write in the exam

(i)

Decrease in 3 years = 800 × 3 = ₹2,400

Value = 10,000 − 2,400

∴ Value after 3 years = ₹7,600

(ii)

t: 0, 1, 2, 3, 4, 5, 6, 7, 8

v (₹): 10000, 9200, 8400, 7600, 6800, 6000, 5200, 4400, 3600

∴ v decreases by ₹800 every year

(iii)

v = 10000 − 800t

v decreases by a constant ₹800 in each year (equal intervals)

∴ It represents linear decay

Common mistakes that cost marks

  • Taking off 800 only once: giving ₹9,200 for 3 years.
  • Writing v = 800t − 10000, which is negative at the start.
  • Using percentages (8% a year). The question says a fixed amount, ₹800, each year.

How this can come in the exam

MCQ (1 mark)

A scooter bought for ₹60,000 loses ₹4,500 in value each year. Its value after 4 years is

  1. ₹42,000
  2. ₹55,500
  3. ₹18,000
  4. ₹45,000
Show answer

(A) ₹42,000
60000 − 4500 × 4 = 60000 − 18000 = ₹42,000.

Short answer (2 marks)

A laptop bought for ₹45,000 loses ₹5,000 in value every year. Write its value v after t years and find when it will be worth ₹20,000.

Show answerv = 45000 − 5000t (1 mark). 5000t = 25000 ⇒ t = 5 years (1 mark).

Try one yourself

For the same phone, after how many years will the value be ₹2,000?

Show answer

10000 − 800t = 2000 ⇒ 800t = 8000 ⇒ t = 10 years.

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