COMPARING QUANTITIES
Category : 7th Class
Learning Objectives:
RATIO
Ratio is comparison of two or more quantities of the same kind using division. It shown as a: b. The first term a is called antecedent and term b is called consequent.
IMPORTANT FACTS ABOUT RATIO:
\[e.g.\,\,4:7=\frac{4}{7}=\frac{4\times 3}{7\times 3}=\frac{12}{21}\]
\[\Rightarrow \,\,4:7=12:21\]
(i) Ratio cannot exist between height and weight.
(ii) We cannot write a ratio between the age of a student and marks obtained by him.
(b) Since ratio is a number, it has no units.
(c) To find the ratio of quantities of same kind, quantities should be in same unit.
e.g. To compare 3 : 5 and 2 : 3
\[\frac{3}{5}\]and\[\frac{2}{3}\]
LCM of 5 and 3 = 15
\[\frac{3}{5}=\frac{3}{5}\times \frac{3}{3}=\frac{9}{15}\]and \[\frac{2}{3}\times \frac{5}{5}=\frac{10}{15}\]
Since \[\frac{10}{15}>\frac{9}{15}\Rightarrow \frac{2}{3}>\frac{3}{5}\]
i.e. \[2:3>3:5\]
e.g. To convert the 15 : 35 in its lowest term
\[15=3\times 5\]
\[35=7\times 5\]
H.C.F.\[=5\]
Dividing both the terms by HCF
\[\frac{15\div 5}{35\div 5}=\frac{3}{7}=3:7\]
If a given quantity increases or decreases in the ratio\[a:b,\]then new quantity\[=\frac{b}{a}\]of the original quantity.
The fraction by which the original quantity is multiplied to get the new (increased) quantity is called the multiplying ratio (or factor).
\[\frac{\text{New}\left( \text{increased} \right)\text{quantity}}{\text{Original quantity}}=\text{Multiplying factor}\]
e.g. (i) To increase 24 kg in the ratio\[2:3.\]
New weight after increase\[=\frac{3}{2}\]of \[24=\frac{3}{2}\times 24=36\,kg.\]
(ii) To decrease Rs. 104 in the ratio \[8:5\]
New amount after decrease \[=\frac{5}{8}\times 104=Rs.\,\,65.\]
Note: Whenever we say that two numbers are in the ratio \[2:3,\] we can write them as \[2x\] and \[3x,\] similarly, three numbers in the ratio \[4:5:8\] can be written as \[4x,\,\,5x\] and \[8x.\]
PROPORTION:
A proportion is an equation that states that two ratios are equal.
Four numbers are said to be in proportion if the ratio of the first two is equal to the ratio of the last two i.e. a, b, c and d are said to be in proportion if a : b = c : d.
This is expressed as a : b :: c : d. The first and fourth terms are called the extremes and second and third terms are called the means.
Product of extremes = Product of means
Three quantities, a b and c (of the same kind) are said to be in continued proportion
if \[a:b::b:c\]
\[\frac{a}{b}=\frac{b}{c}\Rightarrow {{b}^{2}}=ac\]
Here, b is called the mean proportional a and c are known as the first proportional and third proportional respectively.
Mean proportion between\[a\]and\[c=\sqrt{ac}\]
Note: In ratio both the terms should be of the same kind, but in proportion, the first two should be of the same kind and the last two should be of the same kind.
Example:
(i) Determine the following numbers are in proportion or not 1.2, 2.7, 0.4, 0.9
Solution: Product of extremes\[=1.2\times 0.9=1.08\]
Product of means\[=2.7\times 0.4=1.08.\]
Product of extremes = Product of means
So these numbers are in proportion.
(ii) Find the fourth proportional to 4.8, 1.6, and 5.4.
Solution: Let the fourth proportion be \[x.\] Then 4.8, 1.6, 5.4, \[x\] are in proportion
\[\therefore \] Product of extremes = Product of means
i.e., \[4.8\times x=1.6\times 5.4\]
\[x=\frac{1.6\times 5.4}{4.8}\Rightarrow 1.8\]
(iii) Find the mean proportional between \[\frac{1}{4}\] and \[\frac{1}{25}\]
Solution: Let\[x\]be the mean proportional between \[\frac{1}{4}\] and \[\frac{1}{25}.\]
then \[{{x}^{2}}=\frac{1}{4}\times \frac{1}{25}=\frac{1}{100}\]
\[x=\sqrt{\frac{1}{100}}=\frac{1}{10}\]
\[\therefore \] \[\frac{1}{10}\] is the mean proportional between \[\frac{1}{4}\] and\[\frac{1}{25}.\]
(iv) Find the third proportional to i.e. 3.6, 1.8
Solution: Let the third proportional is\[x.\]Then 3.6 and 1.8 are in continued proportion.
\[\therefore \] \[\frac{3.6}{1.8}=\frac{1.8}{x}\]
\[3.6x=1.8\times 1.8\]
\[x=\frac{1.8\times 1.8}{3.6\times 10}\]
\[x=0.9\]
UNITARY METHOD
The method of finding the value of the required number of quantity by first finding the value of the unit quantity is called unitary method.
Example: A scooter consumes 28 liters of petrol in covering 2100 km. How much petrol will be needed to cover a distance of 3600 km?
Solution: 2100 km can be covered in 28 liters of petrol
\[\therefore \] 1 km can be covered in\[\left( 28\times \frac{1}{2100} \right)\]liters
\[\therefore \] 3600 km can be covered in \[\frac{28}{2100}\times 3600\] liters = 48 liters.
PERCENTAGE
The word 'per cent' is an abbreviation of the Latin phrase 'per centum' which means per hundred or hundredths.
Thus, the term per cent means per hundred or for every hundred.
By a certain per cent we mean that many hundredths.
IMPORTANT FORMULAE
\[\text{Percentage}\,\text{increase}=\left( \frac{\text{Increase}\,\text{in}\,\text{quantity}}{\text{Original}\,\text{quantity}}\times 100 \right)%\]
\[\text{Percentage}\,\text{decrease}=\left( \frac{\text{Decrease}\,\text{in}\,\text{quantity}}{\text{Original}\,\text{quantity}}\times 100 \right)%\]
PROFIT AND LOSS
The cost price is abbreviated as C.P.
The selling price is abbreviated as S.P.
IMPORTANT FORMULAE
Example: By selling a washing machine for Rs. 7200, Rajesh loses 10%. Find the Cost Price of the washing machine.
Solution: Let cost price of washing machine be\[Rs.\,\,'x'\]
Given,
SP = Rs. 7200, % Loss = 10%
As % loss given, so CP > SP
\[\Rightarrow \,\,\text{SP}=\text{CP}-\text{Loss}\]
\[\Rightarrow \,\,7200=x-10%\,\,of\,x\]
\[\Rightarrow \,\,x-\frac{10}{100}x=7200\]
\[\Rightarrow \,\,\frac{9x}{10}=7200\]
\[\Rightarrow \,\,x=\frac{7200\times 10}{9}=8000\]
\[\therefore \] Cost price of washing machine = Rs. 8000
IMPORTANT FORMULAE
Discount: Sometimes to increase the sales or dispose of the old stock, dealer or seller offers his goods at reduced prices. This reduction in price offered by dealer/seller is called discount.
Marked Price (M.P.): The printed price or the tagged price of an article is called its marked price (M.P).
It is also called list price.
Discount is always calculated on M.P. of the article.
S.P. = Marked price\[-\]Discount.
For example, two discounts of 15% and 4% are equivalent to a single discount of
\[\left( 15+4-\frac{15\times 4}{100} \right)=19-\frac{60}{1000}=19-\frac{3}{5}=19-0.6=18.4%\]
SIMPLE INTEREST
For example, a rate of '10% per annum', means Rs. 10 on Rs. 100 for 1 year.
IMPORTANT FORMULAE
S.I. = Simple interest, P = Principal amount, R = Rate of interest, T = Time
Example: At what rate percent by simple interest, will a sum of money double itself in 5 years 4 months?
Solution: Let P = Rs.\[x\]
Amount A = Rs.\[2x\]
\[\therefore \] S.I. = A\[-\]P = Rs.\[2x\]\[-\]Rs.\[x\] = Rs.\[x\]
\[T=5\]years 4 months = \[5\frac{4}{12}\]years = \[5\frac{1}{3}\] years = \[\frac{16}{3}\] years
Let R be the rate percent per annum.
Using \[R=\frac{S.I.\times 100}{P\times T},\]
We get \[R=\frac{x\times 100}{x\times \frac{16}{3}}=\frac{300}{16}=18.75.\]
Hence required rate = 18.75% p.a.
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