1. The population of a place increased to 54000 in 2003 at a rate of 5% per annum
(i) find the population in 2001
(ii) what would be its population in 2005?
Solution:
(i) It’s given that population in the year 2003 = 54,000
\(54,000 =\) (Population in 2001) \((1 + {5\over100})^2\)
\(54,000 =\) (Population in 2001) \(({105\over100})^2\)
Population in \(2001 = 54000 \times ({100\over105})^2\)
\(= 48979.59\)
Therefore, the population in the year 2001 was approximately 48,980
(ii) Population in \(2005 = 54000(1 + {5\over100})^2\)
\(= 54000({105\over100})^2\)
\(= 54000({21\over20})^2\)
\(= 59535\)
Therefore, the population in the year 2005 would be \(59,535\).
2. In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.
Solution:
The initial count of bacteria is given as \(5,06,000\)
Bacteria at the end of 2 hours \(= 506000(1 + {2.5\over100})^2\)
\(= 506000(1 + {1\over 40})^2\)
\(= 506000({41\over40})^2\)
\(= 531616.25\)
Therefore, the count of bacteria at the end of \(2\) hours will be \(5,31,616\) (approx.).
3. A scooter was bought at ₹ \(42,000\). Its value depreciated at the rate of 8% per annum. Find its value after one year.
Solution:
Principal = Cost price of the scooter = ₹ \(42,000\)
Depreciation \(= 8\%\) of ₹ \(42,000\) per year
= \({(P \times R \times T)\over100}\)
= \({(42000 \times 8 \times 1)\over 100}\)
= ₹ \(3360\)
Thus, the value after \(1\) year = ₹ \(42000\) − ₹ \(3360\) = ₹ \(38,640\).
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