ICSE Class 8 Maths Commercial Arithmetic Chapter 12 Ratio and Proportion PDF Download

Ratio and Proportion

Overview

Ratio

Proportion

Direct and Inverse Proportion

Ratio

A ratio represents a relation between two quantities. If a and b are two quantities, their ratio is written as a : b (read as 'a is to' b).

If Rahul weighs 48 kg and Rohan weighs 50 kg, the relation between their weights can be represented in many ways.

If the ratio between the weights of silver and gold in an alloy is given as 8 : 5, it means that for every 5 mg of gold in the alloy, there is 8 mg of silver. It also means that for every 5 kg of gold in the alloy, there is 8 kg of silver. Thus, we conclude that:

1. A ratio does not indicate the exact measure of two quantities but is simply the relation between two measures. A ratio has no units.

2. For ratios to be used for comparing two measures, the measures need to be represented in the same units.

3. The first term of the ratio is known as its antecedent while the second term is known as its consequent. If the antecedent becomes the consequent, then the meaning of the ratio is reversed. The ratio of silver and gold in an alloy being 5 : 8 would mean that for every 8 mg of gold, the alloy would have 5 mg silver, or there would be more gold than silver.

Methods of Comparison

Example 1: Out of 45 girls in a class, 18 girls join a choir. Compare the number of girls in the choir to the number of girls in the class using a fraction, a decimal, a percentage, and a ratio.

Fraction

Fraction of girls in the choir = \[\frac{18}{45} = \frac{2}{5}\]

Thus, \[\frac{2}{5}\] of all girls are members of the choir.

Fraction of girls not in the choir = \[1 - \frac{18}{45} = 1 - \frac{2}{5} = \frac{3}{5}\]

Thus, \[\frac{3}{5}\] of all girls are not in the choir.

Decimals

Girls in the choir = \[\frac{2}{5} = 0.4\]

Thus, 0.4 of all girls join the choir.

Girls not in the choir = 1 - 0.4 = 0.6.

Thus, 0.6 of all girls are not in the choir.

Percentage

Percentage of girls in the choir = \[\frac{2}{5} \times 100 = 40\%\]

Thus 40% of the girls are in the choir.

Percentage of girls not in the choir = 100 - 40 = 60%.

Thus, 60% of the girls are not in the choir.

Ratio

\[\frac{2}{5}\] = 2 : 5. That means 2 out of every 5 girls in the class join the choir.

\[\frac{3}{5}\] = 3 : 5. That means 3 out of every 5 girls do not join the choir.

If we divide the above ratios, we get \[\frac{2}{3} = 2 : 3\], or for every 2 girls who joined the choir, 3 girls in the class did not join the choir.

Teacher's Note

Understanding ratios helps us compare prices at the grocery store - knowing that a 2 kg bag costs the same as two 1 kg bags allows us to make smarter shopping decisions.

Example 2: Arrange the Following Ratios in Ascending Order

6 : 7, 11 : 13, 5 : 4, 3 : 4, 17 : 19

Converting the given ratios into fractions we have

\[\frac{6}{7}, \frac{11}{13}, \frac{5}{4}, \frac{3}{4}, \frac{17}{19}\]

Converting the fractions into decimals, correct up to 2 places, we have 0.86, 0.85, 1.25, 0.75, 0.89

Now 0.75 < 0.85 < 0.86 < 0.89 < 1.25

Thus 3 : 4 - 11 : 13 < 6 : 7 < 17 : 19 < 5 : 4

Example 3: Reduce the Ratio Between 7 min 12 s and 8 min 6 s to its Simplest Form

7 min 12 s = 432 s

8 min 6 s = 486 s

Thus ratio = 432 : 486

Dividing both terms by their HCF we have

\[\frac{432}{54} : \frac{486}{54} = 8 : 9\]

Example 4: The Length and Breadth of a Parallelogram are in the Ratio 11 : 9

If the perimeter of the parallelogram is 50 cm, find its length and breadth.

The number of parts the perimeter has to be divided into = 11 + 9 = 20, of which length is 11 parts and breadth is 9 parts.

Given: The sum of the length and breadth

Perimeter of the parallelogram = 50 cm

Here, 2(l + b) = 50 cm

\[\frac{\text{Perimeter}}{2} = \frac{50}{2} = 25 \text{ cm}\]

Thus, length = \[\frac{11}{20} \times 25 = 13.75 \text{ cm}\]

and breadth = \[\frac{9}{20} \times 25 = 11.25 \text{ cm}\]

CHECK: 2(13.75 + 11.25) = 2 × 25 = 50 cm

Teacher's Note

Architects use ratios to scale building dimensions from blueprints to actual construction, ensuring all parts of a building are proportionally correct.

Example 5: The Weights of Anksh and his Sister Anju are in the Ratio 11 : 7

Find their weights if the sum of their weights is given as 93.6 kg.

11 : 7 can be written as \[\frac{11}{7}\]

\[\text{or} \quad \frac{\text{Anksh's weight}}{\text{Anju's weight}} = \frac{11}{7}\]

Now 11 + 7 = 18, but Anksh's weight + Anju's weight is given as 93.6 kg.

If 18 parts = 93.6 kg, one part = \[\frac{93.6}{18} = 5.2 \text{ kg}\]

Thus converting the ratios into quantities (with units), we have \[\frac{11 \times 5.2 \text{ kg}}{7 \times 5.2 \text{ kg}} = \frac{57.2 \text{ kg}}{36.4 \text{ kg}}\]

Thus, the weights of Anksh and Anju are 57.2 kg and 36.4 kg, respectively.

CHECK: 57.2 + 36.4 = 93.6 kg

Example 6: The Weights of a Father and a Mother are in the Ratio 6 : 4

The weights of the mother and the son are in the ratio 6 : 5. If the three family members weigh 160 kg in all, how much does each weigh?

Given ratios are father : mother = 6 : 4 and mother : son = 6 : 5

Mother's weight, being the common term, is expressed as 4 and 6 in the two ratios.

Find the equivalent ratios such that the mother's weight is represented by the same number.

LCM of 4 and 6 = 12

6 : 4 = 6 × 3 : 4 × 3 = 18 : 12

6 : 5 = 6 × 2 : 5 × 2 = 12 : 10

Now that the mother's weight is represented by the same number, the ratios 6 : 4 and 6 : 5 can be written as 18 : 12 and 12 : 10

or the weights of the father : mother : son = 18 : 12 : 10

Now 18 + 12 + 10 = 40

But given weights of father + mother + son = 160 kg

Thus father's weight = \[\frac{18}{40} \times 160 = 72 \text{ kg}\]

mother's weight = \[\frac{12}{40} \times 160 = 48 \text{ kg}\]

son's weight = \[\frac{10}{40} \times 160 = 40 \text{ kg}\]

Teacher's Note

When recipes are scaled for different numbers of people, ingredients are adjusted using ratios - doubling a recipe means doubling each ingredient's ratio to water or flour.

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