Question

mr. durani bought a plot of land for ₹180000 and a car for ₹320000 at the same time. the valie 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 the land as well as the car after 3 years, what will be his profit or loss?

Answer 1

Solution :-

Purchase Price of the plot of land = Rs.180000

Appreciation in Price of land every year = 30 %

Price of the plot of land after 1st year = 180000 + (30*180000)/100

= 180000 + 54000

= Rs. 234000

Price of the plot of land after 2nd = 234000 + (234000*30)/100

= 234000 + 70200

= Rs. 304200

Price of the plot of land after 3rd year = 304200 + (304200*30)100

= 304200 + 91260

= Rs. 395460

So, after 3 years price of the plot of land will be Rs. 395460 (which is also its selling price).

Purchase Price of the car = Rs. 320000

Depreciation in the 1st year = 20 %

Price excluding depreciation = 320000 - (320000*20)/100

= 320000 - 64000

= Rs. 256000

Depreciation in 2nd year = 15 %

Price excluding depreciation = 256000 - (256000*15)/100

= 256000 - 38400

= Rs. 217600

Depreciation in the 3rd year = 15 %

Price excluding depreciation = 217600 - (217600*15)/100

= 217600 - 32640

= Rs. 184960

Reduced price of the car after 3 years (which is also its selling price) = Rs. 184960

Total money spent on plot of land and car = 180000 + 320000

= Rs. 500000

Selling Price of plot of land = Rs. 395460

Selling price of the car - Rs.184960

Selling price of both land and car = Rs. 395460 + 184960

= Rs. 580420

Profit = Selling - Cost price

⇒ 580420 - 500000

Profit = Rs. 80420

So, Mr. Durani earned Rs. 80420 as profit.