Advertisements
Advertisements
Question
The annual rate of growth in population of a town is 10%. If its present population is 26620, then the population 3 years ago was _________
Advertisements
Solution
₹ 20,000
Explanation;
Hint:
Rate of growth of population r = 10%
Present population = 26620
Let population 3 years ago be x
∴ Applying the formula for population growth which is similar to compound interest,
26620 = `x(1 + "r"/100)^3`
∴ 26620 = `x(1 + 10/100)^3`
= `x(110/100)^3`
∴ x = `26620 xx (100/110)^3`
= ₹ 20,000
The population 3 years ago was ₹ 20,000
APPEARS IN
RELATED QUESTIONS
Calculate the amount and compound interest on Rs 18000 for `2 1/2` years at 10% per annum compounded annually.
Find the compound interest at the rate of 5% per annum for 3 years on that principal which in 3 years at the rate of 5% per annum gives Rs 1200 as simple interest.
In how much time would Rs 5000 amount to Rs 6655 at 10% per annum compound interest?
A sum of Rs. 8,000 is invested for 2 years at 10% per annum compound interest. Calculate:
(i) interest for the first year.
(ii) principal for the second year.
(iii) interest for the second year.
(iv) the final amount at the end of the second year
(v) compound interest earned in 2 years.
Calculate the compound interest for the second year on Rs. 15000 invested for 5 years at 6% per annum.
The compound interest on ₹ 5000 at 12% p.a for 2 years, compounded annually is ___________
If the present population of a city is P and it increases at the rate of r% p.a, then the population n years ago would be `"P"(1 + "r"/100)^"n"`
The present value of a machine is ₹ 16800. It depreciates at 25% p.a. Its worth after 2 years is ₹ 9450
Find the C.I. on ₹ 15000 for 3 years if the rates of interest are 15%, 20% and 25% for the I, II and III years respectively
