Growth and Decay
The formula A = P\((1+{ \frac { R }{ 100 } ) }^{ n} \) is called the compound interest law, and applies to any quantity which increases or decreases so that the amount at the end of each period of constant length bears a constant ratio to amount at the beginning of that period. This ratio is called the growth factor, if it is greater than 1, and the decay factor, if less than 1.
For example, if the population of a town increases steadily by 2% p.a. of the population at the beginning each year, the yearly growth factor is \( (1+\frac { 2 }{ 100 } ) \) i.e., 1.02, and the population after n years is \( { (1.02) }^{ n } \) times population at the beginning of that period. If the population decreases by 2%, then the yearly decay factor is \( (1-\frac { 2 }{ 100 } ) \) , i.e., 0.98.
Illustrative Examples
Example 1: If the population of a town decreases 2.5% annually and the present population is 3,26,40,000, find its population after 3 years.
Solution. Required population = P\((1+{ \frac { R }{ 100 } ) }^{ n} \), Here d = 2.5
Therefore, Population after 3 years
= 32640000 \((1+{ \frac { R }{ 100 } ) }^{ n} \)
= 32640000 \((1+{ \frac { 2.5 }{ 100 } ) }^{ 2} \)
= 32640000 \((1+{ \frac { 25 }{ 1000 } ) }^{ 2} \)
= 32640000 x \( (\frac { 15 }{ 16 } ) \) x \( (\frac { 15 }{ 16 } ) \) x \( (\frac { 15 }{ 16 } ) \)
= 31028400
Example 2: The population of a certain city is 1,25,000. If the annual birth rate is 3.3% and the annual death rate is 1.3%, calculate the population after 3 years.
Solution. Present population of the city (P) = 125000
Time (n) = 3 years, Rate of birth \( ({ R }_{ 1 } ) \) = 3.3%,
Rate of death \( ({ R }_{ 2 } ) \) = 1.3%.
So the net rate of increase (\( { R }_{ 1 } \) — \( { R }_{ 2 } \))
= 3.3 — 1.3 = 2%
Therefore, Population after 3 years
= 1250001\((1+{ \frac { 2}{ 100 } ) }^{ 3} \)
= 125000 x \( (\frac { 51 }{ 50 } ) \) x \( (\frac { 51 }{ 50 } ) \) x \( (\frac { 51 }{ 50 } ) \)
= 51 x 51 x 51
= 132651.
Appreciation and Depreciation
When the value of an article increases with the passage of time, the article is said to appreciate. When the value of an article decreases with the passage of time, the article is said to depreciate.
For example, if a man buys a car and uses it for two years, it is obvious that the car will not be worth the e as a new one. The car will thus have depreciated in value.
On the other hand, if a man buys a piece of land, he will probably find that in a few years he will be able to get a better price for it than the price he paid for it. The value of the land will thus have appreciated. When things are difficult to obtain, they have a rarity value and appreciate.
Illustrative Examples
Example 1: In a certain experiment the count of bacteria was increasing at the rate of 2.5% per hour. Initially, the count was 512000. Find the bacteria at the end of 2 hours.
Solution. Using the formula P\((1+{ \frac { R }{ 100 } ) }^{ n} \)
Bacteria at the end of 2 hours
= { 512000 x \((1+{ \frac { 5 }{ 2 \times 100 } ) }^{ 2} \) }
= (512000 x \( (\frac { 41 }{ 40 } ) \) x \( (\frac { 41 }{ 40 } ) \))
= 537920
Hence, the bacteria at the end of 2 hours = 537920.
Example 2: The value of a residential flat constructed at a cost of Rs 1,00,000 is appreciating at the rate 10% per year annum. What will be its value 3 years after construction?
Solution.
Present value of the flat (P) = Rs 100000, rate of appreciation = 10%, Time (n) = 3 years.
Therefore, Value of the flat after 3 years
= Rs 100000 \((1+{ \frac { 10 }{ 100 } ) }^{ 3} \)
= Rs 100000 \((1+{ \frac { 1 }{ 10 } ) }^{ 3} \)
= l00000 x \( (\frac { 11}{ 10 } ) \) x \( (\frac { 11}{ 10 } ) \) x \( (\frac { 11}{ 10 } ) \)
= Rs 1,33,100.
Example 3: A motorcycle Is bought at Rs 160000. Its value depreciates at the rate of 10% per annum. Find its value after 2 years.
Solution.
Value of the motorcycle after 2 year
= Rs { 160000 x \((1-{ \frac { 10 }{ 100 } ) }^{ 2} \) }
= Rs { 160000 x \((1-{ \frac { 1 }{ 10 } ) }^{ 2} \) }
= Rs (160000 x \( (\frac { 9 }{ 10 } ) \) x \( (\frac { 9 }{ 10 } ) \) )
= Rs 129600
Therefore, value after 1 year = Rs 129600
Example 4: Raghu purchased a boat for Rs 16,000. If the cost of the boat is depreciating at the rate of 5% annum, calculate its value after 2 years.
Solution.
Present value of the boat (P) = Rs 16,000, Rate (r) of depreciation = — 5%, Time (n) = 2 years.
Therefore, The value of the boat after 2 years
= 16000 x \( (1-{ \frac { 5 }{ 100 } ) }^{ 2} \)
= Rs 16000 x \( (1-{ \frac { 1 }{ 20 } ) }^{ 2} \)
= Rs 16000 x \( (\frac { 19 }{ 20 } ) \) x \( (\frac { 19 }{ 20 } ) \)
= Rs 14,440.
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