# Uniform Rate of Depreciation

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.

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