7th Class Mathematics Comparing Quantities COMPARING QUANTITIES

COMPARING QUANTITIES

Category : 7th Class

Learning Objectives:

To understand the concept of ratio and proportion.
To learn how to find the value of required quantity by using unitary method.
To understand percentage and some important formulae related to percentage.
To understand the terms cost price, selling price, profit, loss, discount, marked price etc,
To learn some formulae which are useful to calculate CP, SP, profit, loss, discount, marked price etc.
To understand the concept of simple interest and learn how to calculate it.

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:

In a ratio, the order of terms is very important, i.e., the ratio \[2:3\] is different from the ratio \[3:2.\]
Since ratio is a fraction, the ratio will remain unchanged if each term of the ratio is multiplied or divided by the same non-zero number.

\[e.g.\,\,4:7=\frac{4}{7}=\frac{4\times 3}{7\times 3}=\frac{12}{21}\]

\[\Rightarrow \,\,4:7=12:21\]

(a) Ratio exists between quantities of the same kind, i.e.

(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.

To compare two ratios, we either convert them into equivalent like fractions (fractions with same denominator by finding the LCM of the denominators) or convert them to decimal form.

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. \[2:3>3:5\]

A ratio\[a:b=\frac{a}{b}\]is in its lowest terms if the HCF of a and b is 1.

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

\[\frac{15\div 5}{35\div 5}=\frac{3}{7}=3:7\]

Increase or decrease in a given ratio \[a:b\]

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

To convert a given percentage to a fraction or decimal, divide it by 100 and remove the sign%.
To convert a given fraction or decimal into percentage, multiply it by 100 and put the sign%.
\[\frac{a}{b}=\left( \frac{a}{b}\times 100 \right)%\]

\[\text{Percentage}\,\text{increase}=\left( \frac{\text{Increase}\,\text{in}\,\text{quantity}}{\text{Original}\,\text{quantity}}\times 100 \right)%\]

\[\text{Percentage}\,\text{decrease}=\left( \frac{\text{Decrease}\,\text{in}\,\text{quantity}}{\text{Original}\,\text{quantity}}\times 100 \right)%\]

If A's income is \[x%\] more than that of B, then B's income is less than that of A by \[\left\{ \frac{x}{(100+x)}\times 100 \right\}%\]
If A's income is \[x%\] less than that of B, then B's income is more than that of A by \[\left\{ \frac{x}{(100-x)}\times 100 \right\}%\]
If the price of an item is increased by \[r%,\]then the reduction in consumption, so that expenditure is not increased, \[=\left\{ \left( \frac{r}{r+100} \right)\times 100 \right\}%\]
If the price of a commodity decreases by \[r%,\] then the increase in consumption, so that expenditure remains the same, \[=\left\{ \left( \frac{r}{100-r} \right)\times 100 \right\}%\]

PROFIT AND LOSS

Cost Price: The price at which an article is purchased is known as its Cost price.

The cost price is abbreviated as C.P.

Selling Price: The price at which an article is sold is known as its Selling price.

The selling price is abbreviated as S.P.

Profit: If the selling price (S.P.) of an article is greater than the cost price (C.R), then the difference between the selling price and cost price is called Profit.
Loss: If the selling price (S.P.) of an article is less than the cost price (C.P), the difference between the cost price (C.P.) and the selling price (S.P.) is called Loss.

IMPORTANT FORMULAE

Gain = S.P. - C.P, if S.P. > C.P.
Loss = C.P. - S.P, if C.P. > S.P.
\[\text{Gain}%=\frac{\text{Gain}\times 100}{\text{C}\text{.P}\text{.}},Loss%=\frac{Loss\times 100}{\text{C}\text{.P}\text{.}}\]
\[\text{S}\text{.P}\text{.}=\frac{100+\text{gain}%}{100}\times \text{C}.P.\]
\[\text{S}\text{.P}\text{.}=\frac{100-\text{loss}%}{100}\times \text{C}.P.\]
When the selling price and gain per cent are given,
\[\text{C}\text{.P}\text{.}=\frac{100}{100-\text{gain}%}\times \text{S}.P.\]
When the selling price and loss per cent are given
\[\text{C}\text{.P}\text{.}=\frac{100}{100-\text{loss}%}\times \text{S}.P.\]

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.

When discount is offered on an article, then we calculate the selling price (S.P.) as:

S.P. = Marked price\[-\]Discount.

Discount = M.P.\[-\]S.P. = Marked price\[-\]Selling price
\[\text{Discount }\!\!%\!\!\text{ }=\frac{\text{Discount}}{\text{M}\text{.P}\text{.}}\times 100.\]
\[\text{S}\text{.P}\text{.}=M.P-Discount=\text{M}\text{.P}\text{.}-\frac{\text{Discount }\!\!%\!\!\text{ }\times \text{M}\text{.P}\text{.}}{100}\]
\[=\text{M}\text{.P}\text{.}\times \left\{ 1-\frac{\text{Discount }\!\!%\!\!\text{ }}{100} \right\}\]
\[=\text{M}\text{.P}\text{.}\times \left\{ \frac{100-\text{Discount }\!\!%\!\!\text{ }}{100} \right\}\]
\[\text{M}\text{.P}\text{.=}\frac{100\times \text{S}.P.}{100-\text{Discount }\!\!%\!\!\text{ }}\]
Two successive discounts of \[x%\] and \[y%\]allowed on an item are equivalent to a single discount of \[\left( x+y-\frac{xy}{100} \right)%\] which is less than the sum of individual discounts.

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

Whenever we borrow money from some lending source such as a bank or a financial institution, we have to pay some extra money to the service provider, which depends upon the sum borrowed by us and the period of time for which we wish to borrow it. This extra money is called Interest. The borrowed money is called Principal.
On the other hand, when we deposit money in a bank for safekeeping, we earn interest. Interest is calculated according to an agreement which specifies the rate of interest. Generally the rate of interest is taken as "per cent per annum" which means "per Rs. 100 per year."

For example, a rate of '10% per annum', means Rs. 10 on Rs. 100 for 1 year.

When interest is calculated simply on the original principal, it is known as simple interest. When the interest for a specific period is added to the principal, then the sum is called the Amount.

IMPORTANT FORMULAE

\[S.I.=\frac{P\times R\times T}{100}.\]
\[P=\frac{100\times S.I.}{R\times T}\]
\[R=\frac{100\times S.I.}{P\times T},\]
\[T=\frac{100\times S.I.}{R\times P}\]

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.

7th Class Mathematics Comparing Quantities COMPARING QUANTITIES

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Notes - Comparing Quantities