COMMERCIAL MATHEMATICS
Category : 8th Class
Learning Objective
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
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
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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