Mr. Durani bought a plot of land for ₹ 180000 and a car for ₹ 320000 at the same time. The value of the plot of land grows uniformly at the rate of 30% p.a.., while the value of the car depreciates by 20% in the first year and by 15% p.a. thereafter. If he sells the plot of land as well as the car after 3 years, what will be his profit or loss?
Solution:
It is given that
Price of plot of land = ₹ 180000
Growth rate = 30% p.a.
Period (n) = 3 years
We know that
Amount after 3 years = P (1 + R/100)n
Substituting the values
= 180000 (1 + 30/100)3
By further calculation
= 180000 × (13/10)3
It can be written as
180000 × 13/10 × 13/10 × 13/10
= ₹ 395460
Here
Price of car = ₹ 320000
Rate of depreciation = 20% for the first year and 15% for next period
Period (n) = 3 years
We know that
Amount after 3 years = A (1 – R1/100)n × (1 – R2/100)2
Substituting the values
= 320000 (1 – 20/100) (1 – 15/100)2
By further calculation
= 320000 × 4/5 × (17/20)2
So we get
= 320000 × 4/5 × 17/20 × 17/20
= ₹ 184960
Here
Total cost of plot and car = 180000 + 320000 = ₹ 500000
Total sale price of plot and car = 395460 + 184960 = ₹ 580420
We know that
Profit = S.P. – C.P.
Substituting the values
= 580420 – 500000
= ₹ 80420
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