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.

 Compound Interest

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

Problems on Compound Interest

Variable Rate of Compound Interest

Difference of Compound Interest and Simple Interest

Practice Test on Compound Interest

Uniform Rate of Growth


 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 

From Uniform Rate of Depreciation to HOME PAGE


New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.



Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



Share this page: What’s this?

Recent Articles

  1. Multiplication of a Number by a 3-Digit Number |3-Digit Multiplication

    Mar 28, 24 06:33 PM

    Multiplying by 3-Digit Number
    In multiplication of a number by a 3-digit number are explained here step by step. Consider the following examples on multiplication of a number by a 3-digit number: 1. Find the product of 36 × 137

    Read More

  2. Multiply a Number by a 2-Digit Number | Multiplying 2-Digit by 2-Digit

    Mar 27, 24 05:21 PM

    Multiply 2-Digit Numbers by a 2-Digit Numbers
    How to multiply a number by a 2-digit number? We shall revise here to multiply 2-digit and 3-digit numbers by a 2-digit number (multiplier) as well as learn another procedure for the multiplication of…

    Read More

  3. Multiplication by 1-digit Number | Multiplying 1-Digit by 4-Digit

    Mar 26, 24 04:14 PM

    Multiplication by 1-digit Number
    How to Multiply by a 1-Digit Number We will learn how to multiply any number by a one-digit number. Multiply 2154 and 4. Solution: Step I: Arrange the numbers vertically. Step II: First multiply the d…

    Read More

  4. Multiplying 3-Digit Number by 1-Digit Number | Three-Digit Multiplicat

    Mar 25, 24 05:36 PM

    Multiplying 3-Digit Number by 1-Digit Number
    Here we will learn multiplying 3-digit number by 1-digit number. In two different ways we will learn to multiply a two-digit number by a one-digit number. 1. Multiply 201 by 3 Step I: Arrange the numb…

    Read More

  5. Multiplying 2-Digit Number by 1-Digit Number | Multiply Two-Digit Numb

    Mar 25, 24 04:18 PM

    Multiplying 2-Digit Number by 1-Digit Number
    Here we will learn multiplying 2-digit number by 1-digit number. In two different ways we will learn to multiply a two-digit number by a one-digit number. Examples of multiplying 2-digit number by

    Read More