Step-by-Step Solution
Key Concept: Case study on arithmetic progressions.
(a) Find the fixed yearly depreciation amount. [1 Mark]
Depreciation $=8{,}00{,}000-7{,}40{,}000=60{,}000$ per year. [1.0 Mark]
(b) Find the value of the car after 5 years. [1 Mark]
Value after $5$ years $=8{,}00{,}000-5(60{,}000)=8{,}00{,}000-3{,}00{,}000=5{,}00{,}000$. [1.0 Mark]
(c) In how many years will the car's value first drop to Rs 2,00,000 or below? [1 Mark]
$8{,}00{,}000-60{,}000n\leq2{,}00{,}000\Rightarrow60{,}000n\geq6{,}00{,}000\Rightarrow n\geq10$. So after $10$ years, the value first drops to Rs $2{,}00{,}000$ or below. [1.0 Mark]
(d) Is this depreciation model realistic for an indefinitely long time period? Briefly explain using the AP's behaviour. [1 Mark]
No — since the common difference is negative and constant, the AP eventually produces negative values (e.g. after more than $13.3$ years, the formula gives a negative 'value'), which is not realistic since a car's value cannot go below zero. In practice, depreciation models are usually not a simple AP over very long periods. [1.0 Mark]
Correct Answer: Depreciation $=8{,}00{,}000-7{,}40{,}000=60{,}000$ per year. [1.0 Mark] | Value after $5$ years $=8{,}00{,}000-5(60{,}000)=8{,}00{,}000-3{,}00{,}000=5{,}00{,}000$. [1.0 Mark] | $8{,}00{,}000-60{,}000n\leq2{,}00{,}000\Rightarrow60{,}000n\geq6{,}00{,}000\Rightarrow n\geq10$. So after $10$ years, the value first drops to Rs $2{,}00{,}000$ or below. [1.0 Mark] | No — since the common difference is negative and constant, the AP eventually produces negative values (e.g. after more than $13.3$ years, the formula gives a negative 'value'), which is not realistic since a car's value cannot go below zero. In practice, depreciation models are usually not a simple AP over very long periods. [1.0 Mark]