ICSE Class 10 Maths Chapter 07 Ratio and Proportion Including Properties and Uses

Chapter 7: Ratio and Proportion

Including Properties and Uses

7.1 Introduction

Basic concepts of ratio and proportion have already been studied in earlier classes, especially in classes 8 and 9. In this chapter of Class 10, we shall study ratio and proportion in more detail.

7.2 Ratio

The ratio of two quantities of the same kind and in the same units is a comparison obtained by dividing the first quantity by the other.

If a and b are two quantities of the same kind and with the same units such that b ≠ 0; then the quotient \(\frac{a}{b}\) is called the ratio between a and b.

Remember

1. Ratio \(\frac{a}{b}\) has no unit and can be written as a : b (read as a is to b).

2. The quantities a and b are called terms of the ratio. The first quantity a is called the first term or the antecedent and the second quantity b is called the second term or the consequent of the ratio a : b.

The second term of a ratio cannot be zero.

i.e. (i) In the ratio a : b, the second term b cannot be zero (b ≠ 0).

(ii) In the ratio b : a, the second term a ≠ 0.

3. If both the terms of a ratio are multiplied or divided by the same non-zero number, the ratio remains unchanged.

4. A ratio must always be expressed in its lowest terms i.e. both the terms of the ratio must be co-prime.

The ratio is in its lowest terms, if the H.C.F. of its both terms is 1 (unity).

e.g. (i) The ratio 3 : 7 is in its lowest terms as the H.C.F. of its terms 3 and 7 is 1.

(ii) The ratio 4 : 20 is not in its lowest terms as the H.C.F. of its terms 4 and 20 is 4 and not 1.

5. Ratios a : b and b : a cannot be equal unless a = b

i.e., a : b ≠ b : a, unless a = b.

In other words, the order of the terms in a ratio is important.

Worked Example 1

(i) If 2x + 3y : 3x + 5y = 18 : 29, find x : y.

(ii) If x : y = 2 : 3, find the value of 3x + 2y : 2x + 5y.

Solution

(i) 2x + 3y : 3x + 5y = 18 : 29

\[\Rightarrow \frac{2x + 3y}{3x + 5y} = \frac{18}{29}\]

\[\Rightarrow 58x + 87y = 54x + 90y\]

\[\Rightarrow 4x = 3y\]

\[\Rightarrow \frac{x}{y} = \frac{3}{4} \text{ i.e. } x : y = 3 : 4\]

Ans.

(ii) x : y = 2 : 3 \[\Rightarrow \frac{x}{y} = \frac{2}{3}\]

Now, 3x + 2y : 2x + 5y = \[\frac{3x + 2y}{2x + 5y}\]

\[= \frac{3\left(\frac{x}{y}\right) + 2}{2\left(\frac{x}{y}\right) + 5}\]

[Dividing each term by y]

\[= \frac{3 \times \frac{2}{3} + 2}{2 \times \frac{2}{3} + 5}\]

\[\left[\because \frac{x}{y} = \frac{2}{3}\right]\]

\[= \frac{2 + 2}{\frac{4}{3} + 5}\]

\[= 12 : 19\]

Ans.

Alternative Method

\[\Rightarrow x : y = 2 : 3 \Rightarrow 3x = 2y \Rightarrow x = \frac{2y}{3}\]

\[\therefore \frac{3x + 2y}{2x + 5y} = \frac{3 \times \frac{2y}{3} + 2y}{2 \times \frac{2y}{3} + 5y} = \frac{4y}{19y} = \frac{4 \times 3}{19} = 12 : 19\]

Ans.

OR,

\[x : y = 2 : 3 \Rightarrow 3x = 2y \Rightarrow y = \frac{3x}{2}\]

\[\therefore \frac{3x + 2y}{2x + 5y} = \frac{3x + 2 \times \frac{3x}{2}}{2x + 5 \times \frac{3x}{2}} = \frac{6x}{19x} = \frac{6 \times 2}{19} = 12 : 19\]

Ans.

3rd Method

\[x : y = 2 : 3 \Rightarrow \text{if } x = 2k \text{ then } y = 3k\]

And, \[\frac{3x + 2y}{2x + 5y} = \frac{3 \times 2k + 2 \times 3k}{2 \times 2k + 5 \times 3k} = \frac{12k}{19k} = \frac{12}{19} = 12 : 19\]

Ans.

Precaution

For x : y = 2 : 3; if we take x = 2 and y = 3; then

\[\frac{3x + 2y}{2x + 5y} = \frac{3 \times 2 + 2 \times 3}{2 \times 2 + 5 \times 3} = \frac{12}{19} = 12 : 19;\]

which is the same as obtained in each solution given above. But this solution is absolutely wrong and for this solution, a student will score no marks.

Reason: Let the age of Mohit = 15 yrs. and the age of his elder brother Rahul = 24 yrs. The ratio between the ages of Mohit and Rahul = 15 yrs : 24 yrs = 5 : 8. Now read it otherwise, that the ratio between the ages of Mohit and Rahul is 5 : 8. What does it mean? Does it mean that Mohit's age is 5 years and Rahul's age is 8 years. The answer is simple, i.e. No.

In the same way, if x : y = 2 : 3, it does not mean x = 2 and y = 3.

Worked Example 2

If a : b = 5 : 3, find (5a + 8b) : (6a - 7b).

[2002]

Solution

Let a : b = 5 : 3 \[\Rightarrow\] if a = 5x, then b = 3x;

and

\[\frac{5a + 8b}{6a - 7b} = \frac{5 \times 5x + 8 \times 3x}{6 \times 5x - 7 \times 3x} = \frac{49x}{9x} = 49 : 9\]

Ans.

Worked Example 3

Two numbers are in the ratio 3 : 5. If 8 is added to each number, the ratio becomes 2 : 3. Find the numbers.

Solution

Since, the ratio between the numbers is 3 : 5

\[\Rightarrow\] if one number is 3x; the other number is 5x

Given:

\[\frac{3x + 8}{5x + 8} = \frac{2}{3} \Rightarrow 10x + 16 = 9x + 24\]

\[\Rightarrow x = 8\]

\[\therefore \text{Nos. are } 3x \text{ and } 5x = 3 \times 8 \text{ and } 5 \times 8 = 24 \text{ and } 40\]

Ans.

Worked Example 4

(i) What quantity must be added to each term of the ratio 8 : 15 so that it becomes equal to 3 : 5?

(ii) What quantity must be subtracted from each term of the ratio a : b so that it becomes c : d?

Solution

(i) Let x be added to each term of the ratio 8 : 15.

Given:

\[\frac{8 + x}{15 + x} = \frac{3}{5}\]

\[\Rightarrow 40 + 5x = 45 + 3x \Rightarrow x = 2\frac{1}{2}\]

Ans.

(ii) Let x be subtracted, then:

\[\frac{a - x}{b - x} = \frac{c}{d}\]

\[\Rightarrow ad - dx = bc - cx\]

\[\Rightarrow cx - dx = bc - ad\]

\[\Rightarrow x(c - d) = bc - ad \Rightarrow x = \frac{bc - ad}{c - d}\]

Ans.

Worked Example 5

The work done by (x - 3) men in (2x + 1) days and the work done by (2x + 1) men in (x + 4) days are in the ratio 3 : 10. Find the value of x.

[2003]

Solution

Assuming that all the men do the same amount of work in one day and one day work of each man = 1 unit; we get:

Amount of work done by (x - 3) men in (2x + 1) days

= amount of work done by (x - 3) (2x + 1) men in one day

= (x - 3) (2x + 1) units of work.

Similarly, amount of work done by (2x + 1) men in (x + 4) days.

= amount of work done by (2x + 1) (x + 4) men in one day.

= (2x + 1) (x + 4) units of work.

According to the given statement:

\[\frac{(x - 3)(2x + 1)}{(2x + 1)(x + 4)} = \frac{3}{10}\]

\[\Rightarrow \frac{2x^2 + x - 6x - 3}{2x^2 + 8x + x + 4} = \frac{3}{10}\]

i.e. \[\frac{2x^2 - 5x - 3}{2x^2 + 9x + 4} = \frac{3}{10}\]

\[\Rightarrow 20x^2 - 50x - 30 = 6x^2 + 27x + 12\]

\[\Rightarrow 14x^2 - 77x - 42 = 0\]

\[\Rightarrow 2x^2 - 11x - 6 = 0\]

\[\Rightarrow (x - 6)(2x + 1) = 0\]

[On factorising]

\[\Rightarrow x = 6, \text{ or } x = -\frac{1}{2}\]

x = -1/2 is not possible as it will make no. of men (x - 3) negative.

\[\therefore x = 6\]

Ans.

7.3 Increase (Or Decrease) In A Ratio

1. Let the price of an article increases from 20 to 24; we say that the price has increased in the ratio 20 : 24 = 5 : 6.

\[\Rightarrow\] The original price of the article : Its increased price = 5 : 6

2. Let the price of an article decreases from 24 to 20; we say that the price has decreased in the ratio 24 : 20 = 6 : 5.

\[\Rightarrow\] The original price of the article : Its decreased price = 6 : 5

In General

If a quantity increases or decreases in the ratio a : b.

\[\Rightarrow\] The new (resulting) quantity = \(\frac{b}{a}\) times of the original quantity.

Worked Example 6

When the fare of a certain journey by an airliner was increased in the ratio 5 : 7 the cost of the ticket for the journey became 1,421. Find the increase in the fare.

Solution

According to the given statement:

The original fair : Increased fair = 5 : 7

\[\Rightarrow 7 \times\] The original fare = 5 \(\times\) Increased fair

\[\Rightarrow 7 \times\] The original fare = 5 \(\times\) 1,421

\[\Rightarrow\] The original fare = \[\frac{5 \times 1,421}{7} =\] 1,015

\[\therefore\] Increase in the fare = (1,421 - 1,015) = 406

Ans.

Worked Example 7

In a regiment, the ratio of number of officers to the number of soldiers was 3 : 31 before a battle. In the battle 6 officers and 22 soldiers were killed. The ratio between the number of officers and the number of soldiers now is 1 : 13. Find the number of officers and soldiers in the regiment before the battle.

[1992]

Solution

Before the battle:

Let the number of officers be 3x

\[\Rightarrow\] the number of soldiers = 31x

After the battle:

The number of officers = 3x - 6

and, the number of soldiers = 31x - 22

Given: \[\frac{3x - 6}{31x - 22} = \frac{1}{13} \Rightarrow x = 7\]

[On solving]

\[\therefore\] The no. of officers before battle = 3x = 3 × 7 = 21

and the no. of soldiers before battle = 31x = 31 × 7 = 217

Ans.

Worked Example 8

If \[\frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{a + b}\] and a + b + c = 0; show that each given ratio is equal to -1.

Solution

Since,

a + b + c = 0 \[\Rightarrow\] a + b = -c, b + c = -a and c + a = -b

\[\therefore\] \[\frac{a}{b + c} = \frac{a}{-a} = -1;\] \[\frac{b}{c + a} = \frac{b}{-b} = -1\] and \[\frac{c}{a + b} = \frac{c}{-c} = -1\]

Hence, each of the given ratios is -1.

Ans.

Worked Example 9

If \[\frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{a + b}\] and a + b + c ≠ 0; show that each given ratio is equal to 1/2.

Solution

For any two or more equal ratios, each ratio is equal to the ratio between sum of their antecedents and sum of their consequents.

\[\therefore\] (i) \[\frac{a}{b} = \frac{c}{d}\] \[\Rightarrow\] \[\frac{a}{b} = \frac{c}{d} = \frac{a + c}{b + d}\]

(ii) \[\frac{a}{b} = \frac{c}{d} = \frac{e}{f}\] \[\Rightarrow\] \[\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = \frac{a + c + e}{b + d + f}\] and so on.

Given: \[\frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{a + b}\]

\[\Rightarrow\] \[\frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{a + b} = \frac{\text{sum of antecedents}}{\text{sum of consequents}}\]

\[= \frac{a + b + c}{(b + c) + (c + a) + (a + b)}\]

\[= \frac{a + b + c}{2a + 2b + 2c} = \frac{a + b + c}{2(a + b + c)}\]

\[= \frac{1}{2}\]

Ans.

7.4 Commensurable And Incommensurable Quantities

If the ratio between any two quantities of the same kind and having the same unit can be expressed exactly by the ratio between two integers; the quantities are said to be commensurable; otherwise incommensurable.

e.g. (i) The ratio between \[2\frac{1}{3}\] and \[3\frac{1}{2}\]

\[= \frac{7}{3} : \frac{7}{2} = \frac{7}{3} \times \frac{2}{7} = 2 : 3;\] which is the ratio between two integers 2 and 3.

Therefore, \[2\frac{1}{3}\] and \[3\frac{1}{2}\] are commensurable quantities.

(ii) The ratio between \[\sqrt{3}\] and 5 is \[\sqrt{3}\] : 5; which can never be expressed as the ratio between two integers; therefore \[\sqrt{3}\] and 5 are incommensurable quantities.

Teacher's Note

Understanding ratios helps in cooking recipes where ingredients must be mixed in specific proportions, like combining flour and sugar in a 2:1 ratio.

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