CBSE Class 11 Applied Mathematics Commercial Mathematics Notes Set 01

Profit and Loss

Introduction

In our everyday life and in the business world, we encounter transactions involving sales and purchases. Every time such a transaction occurs, it may be observed that there is a seller and a buyer involved. The seller sells some goods for a certain amount paid by the buyer. The seller eventually makes some profit or loss in the transaction.

Cost Price and Selling Price

Cost price (CP): The price at which an article is purchased is called its cost price.
Selling price (SP): The price at which an article is sold is called its selling price.

Profit

If the selling price of an article is greater than its cost price, we say that there is a profit or gain.
Profit = Selling price − Cost price. The percentage of profit is always calculated on the cost price of an article.
When SP > CP:
1. Profit = SP − CP
2. Profit percentage = \( \frac{\text{Profit}}{\text{CP}} \times 100\% \)
3. Profit = \( \frac{\text{Profit percentage}}{100} \times \text{CP} \)
4. When CP and profit percentage are given, \( \text{SP} = \text{CP}\left(\frac{100 + \text{profit percentage}}{100}\right) \)
5. When SP and profit percentage are given, \( \text{CP} = \frac{100 \times \text{SP}}{100 + \text{profit percentage}} \)

Loss

If the selling price of an article is less than its cost price, we say that there is a loss.
Loss = Cost price − Selling price. The percentage of loss is always calculated on the cost price of the article.
When SP < CP:
1. Loss = CP − SP
2. Loss percentage = \( \frac{\text{Loss}}{\text{CP}} \times 100\% \)
3. Loss = \( \frac{\text{Loss percentage}}{100} \times \text{CP} \)
4. When CP and loss percentage are given, \( \text{SP} = \text{CP}\left(\frac{100 - \text{loss percentage}}{100}\right) \)
5. When SP and loss percentage are given, \( \text{CP} = \frac{100 \times \text{SP}}{100 - \text{loss percentage}} \)

Overheads

All the expenditure incurred on transportation, repairs, etc. (if any) is categorised as overheads. These overheads are always included in the CP of the article.
Note: When there are two articles having the same cost price, and one article is sold at a% profit and the other is sold at the same loss percent, then effectively neither profit nor loss is made. If there are two articles having the same selling price, and one is sold at x% profit and the other at x% loss, then effectively there is always a loss, and the loss percentage is \( \left(\frac{x}{10}\right)^2\% \).

Example 1. Rakesh purchased a TV for Rs. 5000 and paid Rs. 250 for its transportation. If he sold the TV for Rs. 5075, then find his profit or loss percentage.
Answer:
Price at which the TV was bought = Rs. 5000
Overheads in the form of transportation = Rs. 250
∴ The total cost price of the TV = Rs. (5000 + 250) = Rs. 5250
Selling price of the TV = Rs. 5075
SP < CP ⇒ there is a loss.
The amount of loss = CP − SP = 5250 − 5075 = Rs. 175
∴ Loss percentage = \( \frac{\text{Loss}}{\text{CP}} \times 100\% = \frac{175}{5250} \times 100\% = \frac{10}{3}\% \) = 3.33%
∴ Rakesh incurred a loss of 3.33%.

Example 2. By selling 24 pens, Kranthi lost an amount equal to the CP of 3 pens. Find his loss percentage.
Answer:
Let us assume that the cost price of each pen is Rs. 1.
⇒ CP of 24 pens = Rs. 24
Loss = CP of 3 pens = 3 × 1 = Rs. 3
⇒ Loss percentage = \( \frac{\text{Loss}}{\text{CP}} \times 100\% = \frac{3}{24} \times 100\% \) = 12.5%
∴ Kranthi’s loss is 12.5%.

Example 3. Naresh sold two books for Rs. 600 each, thereby gaining 20% on one book and losing 20% on the other book. Find his overall loss or gain percent.
Answer:
Selling price of the first book = Rs. 600, profit = 20%
⇒ CP = \( \frac{100 \times \text{SP}}{100 + \text{profit percentage}} = \frac{100 \times 600}{100 + 20} \) = Rs. 500
Selling price of the second book = Rs. 600, loss = 20%
⇒ CP = \( \frac{100 \times \text{SP}}{100 - \text{loss percentage}} = \frac{100 \times 600}{100 - 20} \) = Rs. 750
So, the total cost price of the books = Rs. 500 + Rs. 750 = Rs. 1250
The total selling price of the books = 2 × 600 = Rs. 1200
As the total selling price of the books < the total cost price of the books, there is a loss.
Loss = CP − SP = Rs. 1250 − Rs. 1200 = Rs. 50
⇒ Loss percentage = \( \frac{\text{Loss}}{\text{CP}} \times 100\% = \frac{50}{1250} \times 100\% \) = 4%
∴ Naresh’s loss is 4%. (Check: \( \left(\frac{20}{10}\right)^2\% \) = 4%.)

Simple Interest and Compound Interest

Interest

Interest is paid to the lender by the borrower for using money for a specified period of time. The various terms and their general representation are as follows:
1. Principal: The original sum borrowed. It is denoted by P.
2. Time: The time for which the money is borrowed. It is denoted by n.
3. Rate: The rate at which interest is calculated on the original sum. It is denoted by r.
4. Amount: The sum of the principal and the interest. It is denoted by A.

Simple Interest

When interest is calculated every year on the original principal, such interest is called simple interest. Here, year after year, the interest earned gets accumulated and is due to the lender, but this accumulated interest is not taken into account for the purpose of calculating interest for later years.
Simple interest = \( \frac{PNR}{100} \)
Total amount \( A = P + \frac{PNR}{100} = P\left(1 + \frac{NR}{100}\right) \)

Compound Interest

In compound interest, the interest is added to the principal at the end of each period to arrive at the new principal for the next period. In other words, the amount at the end of the first year becomes the principal for the second year, the amount at the end of the second year becomes the principal for the third year, and so on.
If P denotes the principal at the beginning of period 1, then
P at the beginning of year 2 = \( P\left(1 + \frac{r}{100}\right) = PR \)
P at the beginning of year 3 = \( P\left(1 + \frac{r}{100}\right)^2 = PR^2 \)
P at the beginning of year (n + 1) = \( P\left(1 + \frac{r}{100}\right)^n = PR^n \), where \( R = 1 + \frac{r}{100} \)
Hence, the amount after n years \( A = PR^n \)
Interest \( I = A - P = P(R^n - 1) \)

Example 4. If Rs. 5000 becomes Rs. 5700 in a year’s time at simple interest, what will Rs. 7000 become at the end of 5 years at the same rate of interest?
Answer:
Amount = Principal + Interest; Principal = Rs. 5000
Therefore, 5700 = 5000 + Interest ⇒ Interest = Rs. 700
\( I = \frac{PNR}{100} \Rightarrow 700 = \frac{5000 \times R \times 1}{100} \Rightarrow R = 14\% \)
∴ The simple interest on Rs. 7000 at the rate of 14% for a period of 5 years is \( \frac{7000 \times 14 \times 5}{100} \) = Rs. 4900
Therefore, at the end of 5 years, Rs. 7000 amounts to Rs. 7000 + Rs. 4900 = Rs. 11,900.

Example 5. The cost of an electronic device reduces at the rate of 5% per annum. If its present cost is Rs. 32,490, what was its cost two years ago?
Answer:
\( 32490 = P\left(1 - \frac{R}{100}\right)^2 = P\left(1 - \frac{5}{100}\right)^2 \)
\( \Rightarrow 32490 = P\left(\frac{20 - 1}{20}\right)^2 = P \times \frac{19}{20} \times \frac{19}{20} \)
∴ P = Rs. 36,000

Nominal and Effective Rate of Interest

In the formula \( A = P\left(1 + \frac{r}{100}\right)^n \), when compounding is done more than once a year, the rate (r) will not be for one year but for the time period over which compounding is done, and the power to which the term inside the bracket is raised will not be the number of years but the number of years multiplied by the number of times compounding is done per year.
When compounding is done more than once a year, the rate of interest given in the problem is called the nominal rate of interest.
We can also calculate the rate of interest which will yield simple interest in one year equal to the interest obtained under compound interest at the given nominal rate of interest. The rate of interest so calculated is called the effective rate of interest.

Present Value, Future Value and Annuity

Present Value and Future Value

If we deposit a sum of money with present value PV in a bank that pays interest at the rate r (per annum, as a decimal), then after one year it will become PV(1 + r). Let us call this amount its future value FV. We may write it as
FV = PV(1 + r)
We may also think of (1 + r) as a growth factor. Continuing this process for another year, compounding the interest annually, the future value will become
\( FV = [PV(1 + r)](1 + r) = PV(1 + r)^2 \)
This gives the future value after two years. If we continue this compounding for n years, the future value becomes
\( FV = PV(1 + r)^n \) …(i)
The above expression is valid for annual compounding. If we do the compounding quarterly, the interest credited will be only at the rate \( \frac{r}{4} \), but there will also be 4n compounding periods in n years, so \( FV = PV\left(1 + \frac{r}{4}\right)^{4n} \). Similarly, for monthly compounding the interest rate is \( \frac{r}{12} \) per month and the compounding occurs 12n times in n years. Thus the above equation becomes
\( FV = PV\left(1 + \frac{r}{12}\right)^{12n} \)
From equation (i), we get
\( PV = \frac{FV}{(1 + r)^n} \) …(ii)
This gives the present value of a future payment. Discounting is the procedure to convert the future value of a sum of money to its present value. Discounting is a very important concept in finance because it allows us to compare the present values of different future payments.

Annuity

In many financial situations, we have to deal with a stream of payments, such as rent receipts or monthly pay cheques. An annuity represents such a series of cash payments, e.g. monthly or weekly payments. Another example of an annuity is a loan that you take out and then pay back in monthly instalments. Many insurance companies give the proceeds of a life insurance policy either as a lump sum or in the form of an annuity.
If there is a cash flow C at the end of the first, second, third, … period, then the sum of the discounted cash flows is given by
\( S = \frac{C}{1 + r} + \frac{C}{(1 + r)^2} + \ldots \) up to n terms …(i)
Here S represents the present value of all future cash flows. We compare it with the standard form of a geometric series
\( S = a + ax + \ldots + ax^{n-1} \) …(ii)
We notice that the first term is \( a = \frac{C}{1 + r} \) and the ratio between the terms is \( x = \frac{1}{1 + r} \). We know its sum is
\( S_n = \frac{a(1 - x^n)}{1 - x} \) …(iii)
This gives \( S = \frac{\frac{C}{1 + r}\left(1 - \frac{1}{(1 + r)^n}\right)}{1 - \frac{1}{1 + r}} \Rightarrow S = \frac{C[1 - (1 + r)^{-n}]}{r} \) …(iv)
Using sigma notation for the summation, we may write (i) as \( S = \sum_{i=1}^{n} \frac{C}{(1 + r)^i} \). Thus we get
\( \sum_{i=1}^{n} \frac{C}{(1 + r)^i} = \frac{C[1 - (1 + r)^{-n}]}{r} \)
We can use formula (iii) to find the future value of an annuity as well.

Example 6. Present value, interest rate: You expect to receive Rs. 10,000 as a bonus after 5 years on the job. You have calculated the present value of this bonus and the answer is Rs. 8000. What discount rate did you use in your calculation?
Answer:
To find the PV of a future sum of money, we use \( PV = \frac{FV}{(1 + r)^n} \)
This gives \( 8000 = \frac{10000}{(1 + r)^5} \)
or \( (1 + r)^5 = \frac{10000}{8000} = 1.25 \)
\( 1 + r = (1.25)^{1/5} = 1.0456 \)
And thus r = 4.56%

Depreciation

Depreciation

Depreciation refers to two aspects of the same concept: first, the actual decrease in the fair value of an asset, such as the decrease in value of factory equipment each year as it is used and wears out; and second, the allocation in the accounting statements of the original cost of the asset to the periods in which the asset is used.
Depreciation is thus the decrease in the value of assets and the method used to reallocate the cost of a tangible asset over its useful life span. Generally the cost is allocated as depreciation expense among the periods in which the asset is expected to be used. Depreciation has been defined as the diminution in the utility or value of an asset, and is a non-cash expense.

Methods of Providing or Allocating Depreciation

Different methods are suitable for different assets depending upon the nature and type of the asset.
Straight line method: This method is also termed the ‘original cost method’ because under this method depreciation is charged at a fixed percentage on the original cost of the asset. The amount of depreciation remains equal from year to year and as such the method is also known as the ‘equal instalment method’ or ‘fixed instalment method’. Under this method, the amount of depreciation is calculated by deducting the scrap value from the original cost of the asset and then dividing the remaining balance by the number of years of its estimated life. The depreciation so calculated and charged annually will reduce the original cost of the asset to zero, or to its scrap value, as the case may be, at the end of its estimated life.
Under this method, the amount of depreciation is calculated by the following formula:
\( \text{Yearly depreciation} = \frac{\text{Original cost of the asset} - \text{Estimated scrap value}}{\text{Estimated life of the asset}} \)
For example, if the original cost of an asset is Rs. 1,00,000 and its scrap value is likely to be Rs. 15,000 after its estimated life of 10 years, the depreciation written off will be \( \frac{100000 - 15000}{10} \) = Rs. 8500 every year.

Special Features or Characteristics of Depreciation

1. Depreciation is a decline in the value of fixed assets (except land).
2. Such a fall is of a permanent nature. Once the value of an asset is reduced due to depreciation, it cannot be restored to its original cost.
3. Depreciation is a gradual and continuing process, because the value of the assets declines either by their constant use or by obsolescence due to expiry of time.
4. Depreciation is not a process of valuation of an asset but a process of allocation of the cost of an asset over its effective span of life.
5. It affects only the book value of the asset, not the market value.
6. It is a non-cash expense. It does not involve any cash outflow.

Causes of Depreciation

(1) By constant use: Due to the constant use of fixed assets in business operations, wear and tear arises in them, which results in the reduction of their values.
(2) By expiry of time: The value of the majority of assets decreases with the passage of time even if they are not being put to use in the business. Natural forces such as rain, winds etc. contribute to the deterioration of their values.
(3) By accident: Sometimes a machine may be destroyed due to fire, earthquake, flood etc., or a vehicle may be damaged due to an accident.
(4) By permanent fall in market price: Fluctuations in the market value of fixed assets are not recorded, because such assets are not meant for resale but for use in the business. Sometimes, however, a fall in the value of certain fixed assets is treated as depreciation, such as a permanent fall in the value of investments.