The population of a town is 50,000. It grows at 4% every year. What will the population be after 2 years?

Papers Sign In Request access

Free previous-year papers & practice for JEE, NEET and CBSE Boards.

Papers Privacy Terms Contact

Papers Sign In Request access

Free previous-year papers & practice for JEE, NEET and CBSE Boards.

Papers Privacy Terms Contact

Rate of Growth and Depreciation Using Compound Interest | PadhAiPro

Question

MCQ (Single)Rate of Growth and Depreciation Using Compound Interest

Similar Questions

The value of a machine depreciates at 10% per annum. Its present value is ₹2,50,000. Find its value (in ₹) after 2 years.

NumericalRate of Growth and Depreciation Using Compound Interest

If a machine worth ₹P depreciates at R% per annum, its value after n years is P\left(1 + \frac{R}{100}\right)^n.

True/FalseRate of Growth and Depreciation Using Compound Interest

A printing machine costs ₹60,000. Its value depreciates at 15% per annum. Find its value (in ₹) after 2 years.

NumericalRate of Growth and Depreciation Using Compound Interest

In a lab experiment, a culture has 5,00,000 bacteria. The count increases at 2% per hour. How many bacteria will there be after 2 hours?

NumericalRate of Growth and Depreciation Using Compound Interest

The present population of a town is 1,35,200. It has been increasing at 4% per annum. What was the population of the town 2 years ago?

NumericalRate of Growth and Depreciation Using Compound Interest

The bar graph shows the value of Anil's scooter when he bought it and after 1 and 2 years. The value falls at the same rate every year. Find the rate of depreciation, and then the value of the scooter (in ₹) after 3 years.

NumericalRate of Growth and Depreciation Using Compound Interest