A vehicle was purchased in 2005 for Rs. 3,20,000. After ‘n’ years, its cost is Rs. 1 lakh. Assuming that every year it has been depreciating by Rs. 20 thousand, and it will attain its scrap value in ‘n + 4’ years, then what is its scrap value?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Rs. 20 thousand
With a fixed depreciation of Rs. 20,000 a year, the value after t years is 3,20,000 − 20,000t.
\( \displaystyle 320000-20000n=100000 \)
\( \displaystyle 20000n=220000 \)
\( \displaystyle n=11 \)
The scrap value is reached after n + 4 = 15 years:
\( \displaystyle \text{Scrap value}=320000-20000\times 15 \)
\( \displaystyle =320000-300000=20000 \)
So the scrap value is Rs. 20 thousand.
Shortcut: From the value Rs. 1 lakh, four more years of depreciation remove 4 × Rs. 20,000 = Rs. 80,000, leaving Rs. 20,000; n need not be found at all.