# 6th Class Mathematics Commercial Mathematics COMMERCIAL MATHEMATICS

COMMERCIAL MATHEMATICS

Category : 6th Class

Learning Objective

• To understand the term percentage and value of percentage.
• To understand the terms cost price, selling price, profit and loss.
• To learn how to calculate profit, loss, profit percent, loss percent, selling price cost price.
• To understand the terms simple interest and amount.
• To learn how to calculate simple interest and amount.

PERCENTAGE

The term percent means "for every hundred". A fraction whose denominator is 100 is called percentage and the numerator of the fraction is called the rate percent. Thus, when we say a man made a profit of 20 percent we mean to say that he gained Rs 20 for every hundred rupees he invested in the business, i.e., 20/100 rupees for each Rupee. The abbreviation of percent is p.c. and it is generally denoted by %.

VALUE OF PERCENTAGE

Value of percentage always depends on the quantity to which it refers consider the statement:

“65% of the students in this class are boys". From the context, it is understood that boys form 65% of the total number of students in the class. To know the value of 65%, the value of the total number of student should be mown.

If the total number of students is 200, then, the number of boys $=\frac{200\times 65}{100}=130;$ It can also be written as $(200)\times (0.65)=130$.

Note that the expressions 6%, 63%, 72%, 155% etc. do not have any value intrinsic to themselves. Their values depend on the quantities to which they refer.

COMPARING PERCENTAGES

Which of the three 25%, 5% and 125% is the largest?

If should be remembered that no comparison can be made about the above percentages, because they do not have intrinsic values. If 25% refers to 25% of 10,000 then its value is 0.25 x 10,000 = 2,500 and if 75% of 100, its value is 0.75 x 100 =75.

And so we can conclude that 25% of 10,000 > 75% of 100.

Note: Percentages can be compared only when the quantities they refer to are known.

IMPORTANT RESULTS

1. To express a percentage as a fraction divide it by $100\Rightarrow$ a percentage = 1/100.

Example:

Express the following as fraction

(a) 25%

(b) $33\frac{1}{3}%$

Sol.

(a) $25%\,=\frac{25}{10}\left[ \text{since}\,\text{ }\!\!%\!\!\text{ }\,\text{means}\,\frac{1}{100} \right]=\frac{1}{4}$

(b)  $33\frac{1}{3}%\,=\frac{100}{3}%\,=\frac{100}{3\times 100}=\frac{1}{3}$

2. To express a fraction as a percent multiply it by $100\,\frac{a}{b}\,=\left[ \left( \frac{a}{b}\times 100 \right) \right]\,%$

Example:

Express $\frac{1}{8}$ as a percentage.

Sol.

$\frac{1}{8}=\frac{1}{8}\times 100%$

$=\frac{100}{8}%=\frac{25}{2}%\,=12\frac{1}{2}%$

3. To express percentage as a decimal we remove the symbol% and shift the decimal point by two places the left.

Example:

Express $6\frac{1}{2}%$ as a decimal.

Sol.

$6\frac{1}{2}%=\frac{13}{2}%\,=6.5%\,=\frac{65}{100}=0.065$

4. To express decimal as a percentage we shift the decimal point by two places to the right and write number obtained with the symbol % or simply we multiply the decimal with 100.

Example:

0.345 as a percentage.

Sol. $0.345\times 100%$

= 34.5%

If A is R% of a given number N, then $N=\frac{A\times 100}{R}$

Example:

25% of a number is 80. What is the number?

Sol.

Let the number be X.

According to the given condition

$\frac{25}{100}\,\times \,X=80\Rightarrow \,X=\frac{80\times 100}{25}=320$.

PROFIT AND LOSS

COST PRICE (CP)

The price for which an article is bought is called its cost price.

SELLING PRICE (SP)

The price at which an article is sold is called its selling price.

PROFIT (GAIN)

The difference between the selling price and cost price is called the profit. For profit, selling price should be greater than cost price.

LOSS

The difference between the cost price and the selling price is called the loss. When cost price is greater than the selling price there is a loss.

Wit and loss is generally represented as a percent of the cost price, unless otherwise stated.

FORMULAE

• Profit = S.P – C.P, if S.P > C.P
• Loss C.P – S.P, if C.P > S.P
• Profit%$\frac{\text{Profit}\,\times 100}{C.P}$
• $Loss%\frac{Loss\,\times 100}{C.P}$
• $S.P=\left( \frac{100+\text{Profit}%}{100} \right)\times C.P$
• $S.P=\left( \frac{100+Loss%}{100} \right)\times C.P$
• $C.P=\left( \frac{100}{100+\text{Profit}%} \right)\times S.P$
• $C.P=\left( \frac{100}{100-\text{Profit}%} \right)\,\times S.P$

Example:

A man buys a radio for Rs 600 and sells it at a gain of 25%, what will be the selling price for him?

Sol.

C.P = Rs 600 and Gain (Profit) % = 25%

$S.P=\left( \frac{100+\text{Profit}%}{100} \right)\times C.P$

$=\left( \frac{100+25}{100} \right)$

$=\frac{125}{100}\times 600$

= Rs 70

Example:

Find the cost price of good, sells at Rs 160, having Loss% = 20%

Sol.

S.P = Rs 160% = 20%

$C.P=\frac{S.P\times 100}{100-Loss%}$

$=\frac{160\times 100}{80}$

= Rs 200

Cost price of goods = Rs 200.

SIMPLE INTEREST

Suppose Vishal borrows money from a bank for his higher studies, then at the end of the specified period, he has to pay the money he borrowed and some additional money for the privilege of having used the bank money

Now, we can define the following terms:

The money borrowed is called the Principal ‘p’;

The additional money paid is called the Interest 'SI;

The total money paid is called the Amount 'A’

Formula

1. Amount = Principal + Interests A =P+S.I.

2. Simple Interest

$=\frac{\text{Principal}\,\text{ }\!\!\times\!\!\text{ Time}\,\text{ }\!\!\times\!\!\text{ Rate}}{\text{100}}\,\Rightarrow \,S.I.=\frac{P\times T\times R}{100}$

3. Principal

$=\frac{\text{100 }\!\!\times\!\!\text{ Simple}\,\text{Interest}}{\text{Time }\!\!\times\!\!\text{ Rate}}\,\Rightarrow \,P=\frac{100\times S.I.}{T\times R}$

4. Rate $=\frac{\text{100 }\!\!\times\!\!\text{ Simple}\,\text{Interest}}{\text{Principle}\,\times \text{Time}}\,\Rightarrow \,R=\frac{100\times S.I.}{P\times T}$

5. Time $=\frac{\text{100 }\!\!\times\!\!\text{ Simple}\,\text{Interest}}{\text{Principle}\,\text{ }\!\!\times\!\!\text{ Rate}}\,\Rightarrow \,T=\frac{100\times S.I.}{P\times R}$

Example

Find (i) the interest, (ii) the amount on Rs 5000 at 5% per annum S.I. for 4 years.

Sol.

(i)  Given P = Rs 5000,     T = 4 years, R = 5%

We know $S.I.=\frac{P\times T\times R}{100}$$\therefore \,\,S.I.\,=\frac{5000\times 4\times 5}{100}$

Hence, S.I. =Rs 1000.

(ii) Amount = P + S.I.

$\therefore$ Amount = Rs 5000 + Rs 1000 = Rs 6000.

Example

What sum of money will amount to Rs 1400 at the rate of 8% per annum S.I. in 5 years?

Sol. Let the principal be Rs 100

100x5x8

$\therefore$ S.I. on Rs 100 for 5 years at 8% $=\frac{100\times 5\times 8}{100}=5\times 8=\text{Rs}40$

$\therefore$ Amount = P + S.I.

=Rs (100+40)=Rs 140.

If the amount is Rs 140, then Principal = Rs 100

If the amount is Rs 1400, then Principal

$=\frac{100\times 1400}{140}$

$\therefore$ Principal = Rs 1000.

Example

In what time will Rs 450 amount to Rs 540 at 5% per annum S.I.?

Sol.        Given A = Rs 540, P = Rs 450, R = 5%, T = Rs

$\therefore$ S.I. = A - P = Rs 540 - Rs 450 == Rs 90

$\therefore$  $T=\frac{100\times S.I.}{P\times R}$

$\therefore$  $T=\frac{100\times 90}{450\times 5}=4$

$\therefore$ Time = 4 years.

Example

In how many years will a sum of money double itself at 10% per annum S.I.?

Sol. Let the principle be Rs 'x’

$\therefore$  Amount == 2x

$\therefore$ S.I. = A – P

$\therefore$ S.I. $=2x-x=x$

$T=\frac{100\times S.I.}{P\times R}$

$\therefore$  $T=\frac{100\times x}{x\times 10}\,=10$

$\therefore$ Time = 10 years.

Note:

1. If a sum of money doubles, means S.I. = 2P - P = P

2. If a sum of money triple, means S.I. = 3P - P = 2P

Example

If Rs 800 amounts to Rs 960 in 4 years, find the rate percent per annum S.I.

Sol.        Given: P = Rs 800, A, = Rs 960

$\therefore$ S.I. = Rs 960 – Rs 800 = Rs 160

$R=\frac{100\times S.I.}{P\times T}\,=\frac{100\times 160}{800\times }=5$

$\therefore$ Rate == 5%.

AVERAGE OR MEAN

If ${{X}_{1}},{{X}_{2}},\,{{X}_{3}}....{{X}_{n}}$ be n observations then their arithmetic mean is given by:

$\overline{X}=\frac{{{X}_{1}}+{{X}_{2}}+......{{X}_{n}}}{n}$

Example

Find the mean height if the heights of 5 persons are 144cm, 153 cm, 150cm, 158 cm and 155 cm respectively

Sol.

Mean Height $=\frac{144+153+150+158+155}{5}$

$=\frac{760}{5}\,=152\,cm$

Example

Find the mean of the first 10 odd numbers.

Sol.

First ten odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19

So mean

$(\overline{x})=\frac{1+3+5+9+11+13+15+17+19}{10}$

$=\frac{100}{10}=10$

SPEED

The speed of a body is defined as the distance covered by it in unit time.

Speed $=\frac{\text{Distance}}{\text{Time}}$ distance is constant

Time $=\frac{\text{Distance}}{\text{Speed}};$ time is constant

Distance = Speed x Time; Speed is constant

UNITS OF MEASUREMENT

• Time is measure in seconds (sec), minutes (min) or hours (hr)
• Distance is usually measured in metres (m), kilometres (Km), miles, yards or feet.
• Speed is usually measured in metre/sec (m/s), kilometre/hour (Km/hr), miles/hr.

AVERAGE SPEED

Total distance travelled

Average speed $=\frac{\text{Total}\,\text{distance}\,\text{travelled}}{\text{Total}\,\text{time}\,\text{take}}$

Example

Find the Speed of train travels 90 km in 2 hours.

Sol.

$Speed\,=\frac{\text{Distance}}{\text{Time}}$

Speed $=\frac{90}{2}$

Speed = 45 Km/hr

Example

Find the average speed of a person if he goes 5 km from point A to B and then goes 3 km from point B to C, he takes 2 hours to cover the distance from point A to C.

Sol.

Total distance covered by man = (5 + 3) km = 8 km.

Average speed

$=\frac{\text{Total}\,\text{distance}}{\text{Total}\,\text{time}\,\text{take}}=\frac{8}{2\,}=4\,km/hr$

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