Growth
Growth refers to an increase in the value of a quantity over time, such as population, investment, or business revenue. It is often expressed as a percentage.
1. Population Growth Formula (Exponential Growth):
\(P_T = P_0(1 + \frac{R}{100})^T\)
Where:
- \( P_T \) = Final population after growthÂ
- \( P_0 \) = Initial Population Value (starting amount)Â
- \( R \) = Growth rate
- \( T\) = Time periodÂ
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2. Formula for \(R_1 %\), \(R_2 %\), \(R_3 %\) Population Growth Rate Every year:
\(P_T = P_0(1 + \frac{R_1}{100})(1 + \frac{R_2}{100})(1 + \frac{R_3}{100})\)
Note: It’s every year so, \(T = 1\).Â
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3. Formula for Migrated in \(M_{in}\), Migrated out \(M_{out}\), Dead \(D\), Born \(B\):
Let us consider, in a particular year \(M_{in}\) and\(M_{out}\) people are migrated in and migrated out respectively. Also, \(D\) people are dead and \(B\) people are born. Then the population at the end of \(T\) years \(P_T\) is, given by,
\(P_T = P_0(1 + \frac{R}{100})^T + M_{in} – M_{out} + B - D\)
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Depreciation
Depreciation is the decrease in the value of an asset over time due to wear and tear, obsolescence, or usage.
1. Value Depreciation Formula (Exponential /Declining Balance Method)
\(V_T = V_0(1 - \frac{R}{100})^T\)
Where:
- \( V_T\) = Value of asset after depreciationÂ
- \(V_0\) = Initial value (original cost)Â
- \( R\) = Depreciation rateÂ
- \( T\) = Time periodÂ
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2. Formula for \(R_1 %\), \(R_2 %\), \(R_3 %\) Value Depreciation Rate Every year:
\(V_T = V_0(1 - \frac{R_1}{100})(1 - \frac{R_2}{100})(1 - \frac{R_3}{100})\)
Note: It’s every year so, \(T = 1\).Â
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Attempt these questions independently to reinforce your learning:
Q1.
The population of a village is 10,000. The population grows by 2% annually in the village. [SEE O80 SP]
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a. Write the formula used to find the population after T years. [1K]
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b. After how many years the population of the village will be 10,404? Find it. [2U]
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c. If the population increases at the rate of 4% per annum, by what number will the population of that village be increased in 2 years? Find it. [1A]
Q2.
A machinery good bought for Rs. 1,00,000 depreciates by 10% in the first 2 years and then by 5% in the next 1 year. [SEE MODEL]
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a. Find the price after 1 year. [1U]
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b. What will be the price after 3 years? [2A]
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c. If the depreciation rate of two years and that of the next one year are exchanged, then what would be the difference in the price after 3 years? [2HA]
Class 10: Mathematics
