We will discuss here how to apply the principle of compound interest in the problems of uniform rate of depreciation.
If the rate of decrease is uniform, we denote this as uniform decrease or depreciation.
If the present value P of a quantity decreases at the rate of r% per unit of time then the value Q of the quantity after n units of time is given by
Q = P(1 - \(\frac{r}{100}\))\(^{n}\) and depreciation in value = P - Q = P{1 – (1 - \(\frac{r}{100}\))\(^{n}\)}
If the present population of a car = P, rate of depreciation = r% per annum then the price of the car after n years is Q, where
Q = P(1 - \(\frac{r}{100}\))\(^{n}\) and depreciation = P - Q = P{1 – (1 - \(\frac{r}{100}\))\(^{n}\)}
The fall of efficiency of a machine due to
constant use, decrease in valuations of old buildings and furniture, decrease
in valuations of the movable properties of the transports, decrease in the
number of diseases as a result of alertness come under uniform decrease or
depreciation.
Solved examples on the principle of compound interest in the
uniform rate of depreciation:
1. The price of a machine depreciates by 10% every year. If the machine is bought for $ 18000 and sold after 3 years, what price will it fetch?
Solution:
The present price of the machine, P = $ 18000, r = 10, n = 3
Q = P(1 - \(\frac{r}{100}\))\(^{n}\)
⟹ Q = 18000(1 - \(\frac{10}{100}\))\(^{3}\)
⟹ Q = 18000(1 - \(\frac{1}{10}\))\(^{3}\)
⟹ Q = 18000(\(\frac{9}{10}\))\(^{3}\)
⟹ Q = 18000 × (\(\frac{9}{10}\)) × (\(\frac{9}{10}\)) × (\(\frac{9}{10}\))
⟹ Q = 18000 × (\(\frac{9 × 9 × 9}{10 × 10 × 10}\))
⟹ Q = 18 × 81 × 9
= 13122
Therefore, the machine will fetch 13122 after 3 years.
2. The value of a machine in a factory depreciates at 10% of its value at the beginning of the year. If its present value be $ 60,000, what will be its estimated value after 3 years?
Solution:
Let the present value of the machine (P) = Rs. 10000, r = 10, n = 3
Q = P(1 - \(\frac{r}{100}\))\(^{n}\)
⟹ Q = 60,000(1 - \(\frac{10}{100}\))\(^{3}\)
⟹ Q = 60,000(1 - \(\frac{1}{10}\))\(^{3}\)
⟹ Q = 60,000(\(\frac{9}{10}\))\(^{3}\)
⟹ Q = 60,000 × (\(\frac{9}{10}\)) × (\(\frac{9}{10}\)) × (\(\frac{9}{10}\))
⟹ Q = 60,000 × (\(\frac{9 × 9 × 9}{10 × 10 × 10}\))
⟹ Q = 43,740
Therefore, the value of the machine will be$ 43,740 after 3 years.
3. The price of a car depreciates by 20% every year. By what percent will the price of the car reduce after 3 years?
Solution:
Let the present price of the car be P. Here, r = 20 and n = 3
Q = P(1 - \(\frac{r}{100}\))\(^{n}\)
⟹ Q = P(1 - \(\frac{20}{100}\))\(^{3}\)
⟹ Q = P(1 - \(\frac{1}{5}\))\(^{3}\)
⟹ Q = P(\(\frac{4}{5}\))\(^{3}\)
⟹ Q = P × (\(\frac{4}{5}\)) × (\(\frac{4}{5}\)) × (\(\frac{4}{5}\))
⟹ Q = (\(\frac{64P}{125}\))
Therefore, the reduced price = (\(\frac{64P}{125}\)); so reduction in price = P - (\(\frac{64P}{125}\)) = (\(\frac{61P}{125}\))
Therefore, the percent reduction in price = (\(\frac{\frac{61P}{125}}{P}\)) × 100% = \(\frac{61}{125}\) × 100% = 48.8%
4. The cost of a school bus depreciates by 10% every year. If its present worth is $ 18,000; what will be its value after three years?
Solution:
The present population P = 18,000,
Rate (r) = 10
Unit of time being year (n) = 3
Now applying the formula of depreciation we get:
Q = P(1 - \(\frac{r}{100}\))\(^{n}\)
⟹ Q = $18,000(1 - \(\frac{10}{100}\))\(^{3}\)
⟹ Q = $18,000(1 - \(\frac{1}{10}\))\(^{3}\)
⟹ Q = $18,000(\(\frac{9}{10}\))\(^{3}\)
⟹ Q = $18,000 × (\(\frac{9}{10}\)) × (\(\frac{9}{10}\)) × (\(\frac{9}{10}\))
⟹ Q = $18,000 × (\(\frac{9 × 9 × 9}{10 × 10 × 10}\))
⟹ Q = $18 × 81 × 9
= $13,122
Therefore, the value of the school bus will be $ 13,122 after 3 years.
● Compound Interest
Compound Interest with Growing Principal
Compound Interest with Periodic Deductions
Compound Interest by Using Formula
Compound Interest when Interest is Compounded Yearly
Compound Interest when Interest is Compounded Half-Yearly
Compound Interest when Interest is Compounded Quarterly
Variable Rate of Compound Interest
Difference of Compound Interest and Simple Interest
Practice Test on Compound Interest
● Compound Interest - Worksheet
Worksheet on Compound Interest
Worksheet on Compound Interest when Interest is Compounded Half-Yearly
Worksheet on Compound Interest with Growing Principal
Worksheet on Compound Interest with Periodic Deductions
Worksheet on Variable Rate of Compound Interest
Worksheet on Difference of Compound Interest and Simple Interest
8th Grade Math Practice
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