Ex 2.4, 2

Transcript

Ex 2.4, 2 A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year. 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. Given that a mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year. Thus, Initial Value = ₹ 10,000 Value after 1 year = 10,000 – 800 = ₹ 9,200 Value after 2 years = 10,000 – 800 × 2 = 10,000 – 1,600 = ₹ 8,400 Similarly, we can write Value after 3 years = 10,000 – 800 × 3 = 10,000 – 2,400 = ₹ 7,600 Now, our (ii) question is (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. Finding value after each year Initial Value = Value after 0 years = ₹ 10,000 Value after 1 year = 10,000 – 800 = ₹ 9,200 Value after 2 years = 10,000 – 800 × 2 = 10,000 – 1,600 = ₹ 8,400 Value after 3 years = 10,000 – 800 × 3 = 10,000 – 2,400 = ₹ 7,600 Value after 4 years = 10,000 – 800 × 4 = 10,000 – 3,200 = ₹ 6,800 Value after 5 years = 10,000 – 800 × 5 = 10,000 – 4,000 = ₹ 6,000 Value after 6 years = 10,000 – 800 × 6 = 10,000 – 4,800 = ₹ 5,200 Value after 7 years = 10,000 – 800 × 7 = 10,000 – 5,600 = ₹ 4,400 Value after 8 years = 10,000 – 800 × 8 = 10,000 – 6,400 = ₹ 3,600 Thus, value decreases by ₹ 800 every month Writing it in a table Now, our (iii) question is (iii) Find an expression that relates v and t, and explain why it represents linear decay. Let Time in months = t And, value of plant (in ₹) after t years be v(t) Since value v starts at 10,000 and subtracts 800 every year t. Our Expression is v(t) = 10,000 – 800t Explanation of Linear Decay It represents linear decay because the value decreases by a constant amount (₹800) over equal intervals of time (each year). Now, our (iii) question is (iii) Find an expression that relates v and t, and explain why it represents linear decay. Let Time in months = t And, value of plant (in ₹) after t years be v(t) Since value v starts at 10,000 and subtracts 800 every year t. Our Expression is v(t) = 10,000 – 800t Explanation of Linear Decay It represents linear decay because the value decreases by a constant amount (₹800) over equal intervals of time (each year).