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Maths > Comparing Quantities > Growth and Depreciation
Comparing Quantities

Growth and Depreciation

In our day to day life we observe that there are things or entities like the population of a city, the value of the property, the height of a tree, weight, and height of a child, the number of bacteria, etc. which increase in magnitude over a period of time. The relative increase in a quantity or entity is called growth and growth per unit of time is called the rate of growth.

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The Formula for Population Growth

  • Formula 1: Let P be the population of a city at the beginning of the certain year and the population grows at a constant rate R% per annum, then: $$Population\quad after\quad n\quad years=P\left( 1+\frac { R }{ 100 } \right) ^{ n }$$
  • Formula 2:  Let P be the population of a city or a town at the beginning of a certain year. If the population grows at the rate of R1% during the first year and R2% during the second year, then: $$Population\quad after\quad 2\quad years=P\left( 1+\frac { R_{ 1 } }{ 100 } \right) \times \left( 1+\frac { R_{ 2 } }{ 100 } \right) $$
  • Formula 3: Let P be the population a city or a town at the beginning of a certain year. If the population decrease at the rate of R% per annum, then: $$Population\quad after\quad n\quad years=P\left( 1-\frac { R }{ 100 } \right) ^{ n }$$

growth

                                                                              Learn Uses of Percentages here. 

Depreciation

It is a well-known fact that the constant use of any machine or any other article causes wear and tear due to which its value decreases with time. The relative decreases in the value of a machine over a period of time is its depreciation. Depreciation per unit of time gives the rate of depreciation. The value at any time is called the depreciated value.

Formula for Depreciation

  • First Formula: If V0 is the value of an article at a certain time and R% per annum is the rate of depreciation, then the value of Vn at the end of a year is: $$V_{ n } = V_{ 0 } \left( 1-\frac { R }{ 100 } \right) ^{ n }$$
  • Second Formula: If V0 is the value of an article at a certain time and the rate of depreciation is R1% for the first n1 years, R2% for next n2 years and so on and Rk% for the last nk years the value at the end of n1 + n2 +………..+ nk years is: $${ V }={ V }_{ 0 }\left( 1+\frac { R_{ 1 } }{ 100 } \right) ^{ { n }_{ 1 } }\times \left( 1+\frac { R_{ 2 } }{ 100 } \right) ^{ { n }_{ 2 } }……\times \left( 1+\frac { R_{ 2 } }{ 100 } \right) ^{ { n }_{ k } }$$

                                                              What are Profit and Loss?? Learn here. 

Solved Example For You

Q 1: The value of a residential flat constructed at a cost of Rs100000 is depreciating at the rate of 10% per annum. What will be its value 3 years after construction?

Solution: We have: V0= Initial value= Rs 100000, R= Rate of depreciation = 10% per annum, n = 3 years. Therefore: Value after n years, $$V_{ n } = V_{ 0 } \left( 1-\frac { R }{ 100 } \right) ^{ n }$$ after putting value of n and R, we get: $$V=100000\left( 1-\frac { 10 }{ 100 } \right) ^{ 3 }\\ V=100000\left( 1-\frac { 1 }{ 10 } \right) ^{ 3 }\\ V=100000\left( \frac { 9 }{ 10 } \right) ^{ 3 }\\ V=100\times 9\times 9\times 9\\ V=Rs72900\\ $$

Q 2: The population of a town is increasing at the rate of 5% per annum. What will be the population of the town on this basis after two years, if the present population is 16000?

Solution: Here, P =Initial population = 16000, R =Rate of growth of population =5% per annum, n Number of years =2; as we know: $$Population\quad after\quad n\quad years=P\left( 1+\frac { R }{ 100 } \right) ^{ n }$$ Therefore, after putting vaules of P,R and n: $$Population\quad after\quad 2\quad years=16000\times \left( 1+\frac { 5 }{ 100 } \right) ^{ 2 }\\ Population\quad after\quad 2\quad years=16000\times \left( 1+\frac { 1 }{ 20 } \right) ^{ 2 }\\ Population\quad after\quad 2\quad years=16000\times \left( \frac { 21 }{ 20 } \right) ^{ 2 }\\ Population\quad after\quad 2\quad years=16000\times \left( \frac { 21 }{ 20 } \right) \times \left( \frac { 21 }{ 20 } \right) \\ Population\quad after\quad 2\quad years=40\times 21\times 21\\ Population\quad after\quad 2\quad years=17640\\ $$

Question. How can one calculate growth?

Answer. To calculate the growth rate, one must, first of all, subtract the past value from the current value. Afterwards, one must divide that number by the past value. Finally, one must multiply the answer by 100 so that it can be expressed as a percentage. For example, if the value a company’s value was $100 and now its $200, then one would subtract 100 from 200 and get 100.

Question. What does growth rate mean?

Answer. Growth rates refer to the percentage change of a specific variable. Moreover, this happens in a particular context and within a specific time period.

Question. Explain what it means to grow exponentially?

Answer. Exponential growth refers to a specific way that a particular quantity can escalate or increase over time. Exponential growth takes place when the rate of growth is proportional to the current amount. It does not simply mean to grow quickly or faster.

Question. Explain the formula of exponential growth?

Answer. The earlier original exponential formula was y = abx. Now, in the new functions, the b value (growth factor) has been replaced either by (1 – r) or (1 + r).

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