ICSE Class 8 Maths Chapter 06 Ratio and Proportion

Chapter 6: Ratio And Proportion

(Including Proportion Parts)

6.1 Review

Ratio

A ratio is the relationship between two quantities which expresses how many times one quantity is of the other quantity of the same kind and in the same unit.

Remember: The two quantities must be of the same kind and in the same unit.

There can be a ratio between ₹ 15 and ₹ 20, but there can be no ratio between ₹ 15 and 20 oranges.

1. The ratio between two quantities is obtained by dividing the first quantity by the second.

e.g. if A = 36 and B = 24; then ratio between A and B = \(\frac{36}{24} = \frac{3}{2} = 3 : 2\) and ratio between B and A = \(\frac{24}{36} = \frac{2}{3} = 2 : 3\).

[3 : 2 is read as '3 is to 2' and 2 : 3 is read as '2 is to 3']

2. The two quantities (numbers) in a ratio are called its terms. The first term is called the antecedent (means 'that which goes before') and the second term is called the consequent (means 'that which goes after').

3. A ratio is a pure number and has no unit.

4. A ratio should always be expressed in lowest terms.

e.g. ratio between 20 and 32 = \(\frac{20}{32} = \frac{20 + 4}{32 + 4} = \frac{5}{8} = 5 : 8\)

Clearly, the ratio between two quantities is equivalent to the fraction that one quantity is of the other.

e.g. the ratio between A and B is 3 : 2 \(\Rightarrow\) A = \(\frac{3}{2}\) times of B.

Test Yourself

1. Since, 2 kg = ......... g

The ratio between 800 g and 2 kg = .......................... = ..............

2. Since, 2 km and 2 kg are not of the ...........

and not in the ...................; they do not form ........... .

Teacher's Note

Ratios help us compare quantities in everyday life, such as recipe ingredients or comparing prices of items at different stores.

6.2 Comparing Ratios

Example 1:

Which ratio is greater 3 : 7 or 10 : 21 ?

Solution:

First method: (By converting each ratio into a decimal fraction).

\(3 : 7 = \frac{3}{7} = 0.4285\) .... and \(10 : 21 = \frac{10}{21} = 0.4761\) ....

As 0.4761 .... is greater than 0.4285 ....; the ratio 10 : 21 is greater.

Second method: The given ratios are 3 : 7 and 10 : 21

i.e. \(\frac{3}{7}\) and \(\frac{10}{21}\)

or, \(\frac{3 \times 3}{7 \times 3}\) and \(\frac{10}{21}\)

(Making the denominators equal)

or, \(\frac{9}{21}\) and \(\frac{10}{21}\)

Since, \(\frac{10}{21} > \frac{9}{21}\)

\(\therefore 10 : 21\) is greater

Third method:

For any two ratios a : b and c : d, if:

(i) a \(\times\) d > b \(\times\) c \(\Rightarrow\) a : b is greater than c : d.

(ii) a \(\times\) d < b \(\times\) c \(\Rightarrow\) a : b is less than c : d.

(iii) a \(\times\) d = b \(\times\) c \(\Rightarrow\) a : b is equal to c : d.

For given ratios 3 : 7 and 10 : 21

3 \(\times\) 21 = 63 and 7 \(\times\) 10 = 70

Since, 3 \(\times\) 21 < 7 \(\times\) 10 \(\Rightarrow\) 3 : 7 is less than 10 : 21

\(\Rightarrow\) 10 : 21 is greater

Teacher's Note

Comparing ratios is useful when shopping, like determining which store offers better value by comparing price-to-quantity ratios.

6.3 To Divide A Given Quantity In A Given Ratio

Example 2:

Divide ₹ 832 into two parts in the ratio 4 : 9.

Solution:

Since, 4 + 9 = 13

\(\therefore\) 1st part = \(\frac{4}{13}\) of the whole = \(\frac{4}{13} \times\) ₹ 832 = ₹ 256

and, 2nd part = \(\frac{9}{13}\) of the whole = \(\frac{9}{13} \times\) ₹ 832 = ₹ 576

Alternative method

Since, the two parts are in the ratio 4 : 9

Let the parts be ₹ 4x and ₹ 9x

\(\Rightarrow\) 4x + 9x = 832

\(\Rightarrow\) 13x = 832 and x = 64

\(\therefore\) 1st part = ₹ 4x = ₹ 4 \(\times\) 64 = ₹ 256

and, 2nd part = ₹ 9x = ₹ 9 \(\times\) 64 = ₹ 576

Example 3:

Two numbers are in the ratio 5 : 4. If 3 is subtracted from the first and 2 is subtracted from the second, they become in the ratio 6 : 5. Find the numbers.

Solution:

Since, the numbers are in the ratio 5 : 4.

Let the numbers be 5x and 4x.

Given: \(\frac{5x - 3}{4x - 2} = \frac{6}{5}\) \(\Rightarrow\) 25x - 15 = 24x - 12 \(\Rightarrow\) x = 3

\(\therefore\) Numbers = 5x and 4x = 5 \(\times\) 3 and 4 \(\times\) 3 = 15 and 12

Example 4:

The ratio of the number of boys to the number of girls in a school of 450 pupil is 4 : 5. When some new boys and girls are admitted, the number of boys increases by 25 and ratio of boys to girls changes to 9 : 13. Calculate the number of new girls admitted.

Solution:

No. of boys in the school = \(\frac{4}{9}\) \(\times\) 450 = 200 [\(\therefore\) 4 + 5 = 9]

and, no. of girls in the school = \(\frac{5}{9}\) \(\times\) 450 = 250

Let the number of new girls admitted be x, then new no. of boys = 200 + 25 = 225 and new no. of girls = 250 + x.

Given: \(\frac{225}{250 + x} = \frac{9}{13}\) \(\Rightarrow\) 2250 + 9x = 2925 [By cross multiplication]

\(\Rightarrow\) 9x = 2925 - 2250 = 675

\(\Rightarrow\) x = \(\frac{675}{9}\) = 75

\(\therefore\) 75 new girls are admitted.

Teacher's Note

Dividing quantities in given ratios applies to real situations like splitting expenses among friends or distributing resources in schools.

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