Growth and Depreciation
Growth and Depreciation
Sukhiram is the sarpanch of a village and is very good at mathematics. The current population of his village is 18550. From the record of past years, he observed that the population of the village is increasing at the rate of 4% per year. On the basis of the current growth rate, he wants to know the population of the village after 2 years. How can he do so?
Actually, it is a problem of growth (or appreciation). If he knows the formula, then he can easily find out what he wants to know.
Let us understand the formula that should be used here.
The growth per unit time is called the rate of growth.
Currently, the population of the village is increasing at the rate of 4% every year.
This means that in a year the population increases by 4% of the previous population. Hence, the rate of growth is 4% and is denoted by r.
\ r = 4%
If V0 is the current measure of quantity and V is the measure of quantity after n years, then
This is the relation between the current measure of a quantity and the measure of a quantity after n years.
On using this formula, we can find the population of the village after two years. Let us see how.
In the given example,
r = 4%
V0 = 18550
n = 2
On substituting these values in equation (i), we obtain
The population of the village cannot be in decimals. Hence,
V = 20064 (approximately)
In our daily lives, we come across several things that grow. For example: Population of a city or village, prices of goods, height or weight of children etc.
On the other hand, the value of machines or vehicles decreases with time.
The decrease in value per unit time is called rate of depreciation.
If the rate of depreciation is constant, then we have the following formula:
Here, r is the rate of depreciation per year; V0 is the current value and V is the value after n years.
When a quantity increases in the first year at the rate of, then decreases at the rate of in the second year and then increases at the rate of in the third year, then the formula is
Here, V0 is the initial measure and V is the quantity after 3 years.
Let us now solve some problems to understand these formulae better.
Example 1
Priya bought a diamond necklace worth Rs 300000. The value of the necklace appreciates by 6% every year. What will be the value of the necklace after 3 years?
Solution:
The value of the necklace after three years can be calculated by using the compound interest formula.
P = Rs 300000
Rate of appreciation, R = 6% p.a.
Therefore, the value of the necklace after 3 years
Thus, the value of the necklace after 3 years will be Rs 357304.80.
Example 2
There were 5000 students in a school in the year 2004. In the year 2005, the number of students increased by 5% of the number of students in the previous year. In the year 2006, the number of students decreased by 12% of the number of students in the previous year. How many students did the school have in the year 2006?
Solution:
V0 = 5000
Rate of appreciation in the year 2005, r1 = 5%
Rate of depreciation in the year 2006, r2 = 12%
Number of students in the year 2006 is given by
Therefore, the number of students in the year 2006
Thus, the number of students in the school in the year 2006 is 4620.
Actually, it is a problem of growth (or appreciation). If he knows the formula, then he can easily find out what he wants to know.
Let us understand the formula that should be used here.
The growth per unit time is called the rate of growth.
Currently, the population of the village is increasing at the rate of 4% every year.
This means that in a year the population increases by 4% of the previous population. Hence, the rate of growth is 4% and is denoted by r.
\ r = 4%
If V0 is the current measure of quantity and V is the measure of quantity after n years, then
This is the relation between the current measure of a quantity and the measure of a quantity after n years.
On using this formula, we can find the population of the village after two years. Let us see how.
In the given example,
r = 4%
V0 = 18550
n = 2
On substituting these values in equation (i), we obtain
The population of the village cannot be in decimals. Hence,
V = 20064 (approximately)
In our daily lives, we come across several things that grow. For example: Population of a city or village, prices of goods, height or weight of children etc.
On the other hand, the value of machines or vehicles decreases with time.
The decrease in value per unit time is called rate of depreciation.
If the rate of depreciation is constant, then we have the following formula:
Here, r is the rate of depreciation per year; V0 is the current value and V is the value after n years.
When a quantity increases in the first year at the rate of, then decreases at the rate of in the second year and then increases at the rate of in the third year, then the formula is
Here, V0 is the initial measure and V is the quantity after 3 years.
Let us now solve some problems to understand these formulae better.
Example 1
Priya bought a diamond necklace worth Rs 300000. The value of the necklace appreciates by 6% every year. What will be the value of the necklace after 3 years?
Solution:
The value of the necklace after three years can be calculated by using the compound interest formula.
P = Rs 300000
Rate of appreciation, R = 6% p.a.
Therefore, the value of the necklace after 3 years
Thus, the value of the necklace after 3 years will be Rs 357304.80.
Example 2
There were 5000 students in a school in the year 2004. In the year 2005, the number of students increased by 5% of the number of students in the previous year. In the year 2006, the number of students decreased by 12% of the number of students in the previous year. How many students did the school have in the year 2006?
Solution:
V0 = 5000
Rate of appreciation in the year 2005, r1 = 5%
Rate of depreciation in the year 2006, r2 = 12%
Number of students in the year 2006 is given by
Therefore, the number of students in the year 2006
Thus, the number of students in the school in the year 2006 is 4620.
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