Advertisements
Advertisements
प्रश्न
The population of a town increases at the rate of 40 per thousand annually. If the present population be 175760, what was the population three years ago.
Advertisements
उत्तर
Population after three years = P \[\left( 1 + \frac{R}{100} \right)^3 \]
\[175, 760 = P \left( 1 + \frac{40}{1000} \right)^3 \]
\[175, 760 = P \left( 1 . 04 \right)^3 \]
\[P = \frac{175, 760}{1 . 124864}\]
\[ = 156, 250\]
Thus, the population three years ago was 156, 250.
APPEARS IN
संबंधित प्रश्न
Compute the amount and the compound interest in the following by using the formulae when:
Principal = Rs 12800, Rate = \[7\frac{1}{2} %\], Time = 3 years
The compound interest on Rs 1800 at 10% per annum for a certain period of time is Rs 378. Find the time in years.
The present population of a town is 25000. It grows at 4%, 5% and 8% during first year, second year and third year respectively. Find its population after 3 years.
The population of a town increases at the rate of 50 per thousand. Its population after 2 years will be 22050. Find its present population.
Mohan purchased a house for Rs 30000 and its value is depreciating at the rate of 25% per year. Find the value of the house after 3 years.
The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago. If its present value is Rs 43740, find its purchase price.
On the construction work of a flyover bridge there were 320 workers initially. The number of workers were increased by 25% every year. Find the number of workers after 2 years.
A loan of ₹ 15000 was taken on compound interest. If the rate of compound interest is 12 p.c.p.a. find the amount to settle the loan after 3 years.
The cost of a sewing machine is Rs 7,000. Its value depreciates at 8% p.a. Then the value of the machine after 2 years is Rs 5,924.80.
