Advertisements
Advertisements
Question
A building worth Rs 1, 33,100 is constructed on a plot of land worth Rs 72,900. After how many years will the values of both be same, if the land appreciates at 10% p.a. and the building depreciates at 10% p.a.?
Advertisements
Solution
For the building:
`"V"_"n" = "V"_0 (1 - "r"/100)^"n"`
Vn = Rs 133100 `(1 - 10/100)^"n"`
Vn = Rs 133,100 x (0. 9)n
For the plot :
`"V"_"n" = "V"_0 (1 + "r"/100)^"n"`
Vn = Rs 72900 `(1 + 10/100)^"n"`
Vn = Rs 72900 × (1.1)n
Since, value becomes same:
1,33,100 x (0.9)n = 72, 900 x (1.1)n `lim_(x -> ∞)`
`(1.1)^"n"/(0.9)^"n" = 133100/72900`
`(1.1)^"n"/(0.9)^"n" = 1331 /729 = 11^3/9^3`
t = 3
Hence, after 3 years value of both will be same.
RELATED QUESTIONS
The simple interest and the compound interest on a certain sum of money for 2 years at the same rate of interest are Rs 8,000 and Rs 8,640 respectively. Calculate the rate of interest and the sum.
Find the difference between the compound interest and the simple interest in 3 years on Rs 15,000 at 8% p.a. compounded yearly.
Calculate the Amount and Cornpound Interest for the Following, when Cornpounded Annually:
Rs 12,000 for 3 years at 15 % p.a.
Find the difference betlween the compound interest compounded yearly and half-yearly for the following:
Rs 15,000 for `1 1/2` years at 12 % p.a.
The population of a town in the year 2005 was 4, 25,000. Find its population in the year 2007 if the rate of annual increase is 4% per year.
The population of a village increases at the rate of 50 per thousand. Its population after 2 years will be 22,050. Find the present population.
In a factory the production of scooters rose to 46,305 from 40,000 in 3 years. Find the annual rate of growth of the production of scooters.
A machine was purchased 2 years ago. Its value decreases by 10% every year. Its present value is Rs 19,083.60. For how much money was the machine purchased?
A new car is purchased for Rs 4, 00,000 .Its value depreciates at the rate of 10% per annum. What will be its value after 4 years?
A sum of money amounts to ₹ 2,240 at 4 % p.a., simple interest in 3 years. Find the interest on the same sum for 6 months at 3`(1)/(2)`% p.a.
