ICSE Class 10 Maths Chapter 09 Proportion Reference Content

Chapter 9

Proportion

Exercise 9 (A)

Q. 1. Find x, when :

(i) \(3 : 4 : : 2-4 : x\)

(ii) \(1 : 3 : : x : 7\)

(iii) \(x : 1-5 : : 3 : 5\)

Sol.

(i) \(∴ 3 : 4 : : 2-4 : x\)

\(∴ 3 × x = 4 × 2-4\)

\(x = \frac{4 × 2-4}{3} = 3-2\) Ans.

(ii) \(∴ 1 : 3 : : x : 7\)

\(∴ 3 × x = 1 × 7\)

\(x = \frac{1 × 7}{3} = \frac{7}{3} = 2\frac{1}{3}\) Ans.

(iii) \(∴ x : 1-5 : : 3 : 5\)

\(∴ x × 5 = 1-5 × 3\)

\(x = \frac{1-5 × 3}{5} = 0-9\) Ans.

Q. 2. Find the fourth proportional to :

(i) 3, 8 and 21

(ii) 1-4, 3-2 and 7

(iii) 1-5, 4-5 and 3-6

(iv) \(a^2\), \(ab\) and \(b^2\)

(v) \((a^2 - ab + b^2)\), \((a^3 + b^3)\) and \((a - b)\)

Sol.

Let \(x\) be the fourth proportional then,

(i) \(∴ 3 : 8 : : 21 : x\)

\(⇒ 3 × x = 8 × 21\)

\(⇒ x = \frac{8 × 21}{3} = 56\)

\(∴\) Fourth proportional \(= 56\) Ans.

(ii) \(∴ 1-4 : 3-2 : : 7 : x\)

\(⇒ 1-4 × x = 3-2 × 7\)

\(⇒ x = \frac{3-2 × 7}{1-4} = 16\)

\(∴\) Fourth proportional \(= 16\) Ans.

(iii) \(∴ 1-5 : 4-5 : : 3-6 : x\)

\(⇒ 1-5 × x = 4-5 × 3-6\)

\(⇒ x = \frac{4-5 × 3-6}{1-5} = 10-8\)

\(∴\) Fourth proportional \(= 10-8\) Ans.

(iv) \(∴ a^2 : ab : : b^2 : x\)

\(⇒ a^2 × x = ab × b^2\)

\(⇒ x = \frac{ab × b^2}{a^2} = \frac{b^3}{a}\)

\(∴\) Fourth proportional \(= \frac{b^3}{a}\) Ans.

Q. 3. Find the third proportional to :

(i) 9 and 6

(ii) \(2\frac{2}{3}\) and 4

(iii) 1-6 and 2-4

(iv) \((2 + \sqrt{3})\) and \((5 + 4\sqrt{3})\)

(v) \(\left(\frac{a}{b} + \frac{b}{a}\right)\) and \(\sqrt{a^2 + b^2}\)

Sol.

(i) Let \(x\) be the third proportional to 9 and 6

then \(9 : 6 : : 6 : x\)

\(⇒ 9 × x = 6 × 6\)

\(⇒ x = \frac{6 × 6}{9} = 4\)

\(∴\) Third proportional \(= 4\) Ans.

(ii) Let \(x\) be the third proportional to \(2\frac{2}{3}\) and 4

then \(2\frac{2}{3} : 4 : : 4 : x\)

\(⇒ \frac{8}{3} : 4 : : 4 : x\) \(⇒ \frac{8}{3} × x = 4 × 4\)

\(⇒ x = \frac{4 × 4 × 3}{8} ⇒ x = 6\)

\(∴\) Third proportional \(= 6\) Ans.

(iii) Let \(x\) be the third proportional to 1-6 and 2-4

then \(1-6 : 2-4 : : 2-4 : x\) \(⇒ 1-6 × x = 2-4 × 2-4\)

\(⇒ x = \frac{2-4 × 2-4}{1-6} = 3-6\)

\(∴\) Third proportional \(= 3-6\) Ans.

(iv) Let \(x\) be the third proportional to \((2 + \sqrt{3})\) and \((5 + 4\sqrt{3})\)

then \((2 + \sqrt{3}) : (5 + 4\sqrt{3}) : : (5 + 4\sqrt{3}) : x\)

\(⇒ (2 + \sqrt{3}) × x = (5 + 4\sqrt{3})(5 + 4\sqrt{3})\)

\(⇒ x = \frac{(5 + 4\sqrt{3})(5 + 4\sqrt{3})}{2 + \sqrt{3}}\)

\(= \frac{(5)^2 + 2 × 5 × 4\sqrt{3} + (4\sqrt{3})^2}{2 + \sqrt{3}}\)

\(= \frac{25 + 40\sqrt{3} + 48}{2 + \sqrt{3}} = \frac{73 + 40\sqrt{3}}{2 + \sqrt{3}}\)

\(= \frac{(73 + 40\sqrt{3})(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})}\)

[Rationalizing the denominator]

\(⇒ x = \frac{(73 + 40\sqrt{3})(2 - \sqrt{3})}{4 - 3}\)

\(= \frac{(73 + 40\sqrt{3})(2 - \sqrt{3})}{1}\)

\(= 146 + 80\sqrt{3} - 73\sqrt{3} - 120\)

\(= (26 + 7\sqrt{3})\)

\(∴\) Third proportional \(= (26 + 7\sqrt{3})\) Ans.

(v) Let \(x\) be the third proportional to \(\left(\frac{a}{b} + \frac{b}{a}\right)\) and \(\sqrt{a^2 + b^2}\), then

\(\left(\frac{a}{b} + \frac{b}{a}\right) : \sqrt{a^2 + b^2} : : \sqrt{a^2 + b^2} : x\)

\(\left(\frac{a}{b} + \frac{b}{a}\right) × x = \sqrt{a^2 + b^2} × \sqrt{a^2 + b^2}\)

\(= (a^2 + b^2)\)

\(x = \frac{(a^2 + b^2)}{ab}\)

Hence, third proportional \(= ab\). Ans.

Teacher's Note

Understanding proportions helps in solving real-world problems like recipe scaling in cooking or adjusting ratios in construction projects.

Q. 4. Find the mean proportional between :

and 63

(ii) 2-5 and 0-9

(iii) 6-25 and 1-6

(iv) \((\sqrt{26} - \sqrt{17})\) and \((\sqrt{26} + \sqrt{17})\)

(v) \((6 + 3\sqrt{3})\) and \((8 - 4\sqrt{3})\)

Sol.

(i) Let \(x\) be the mean proportional between 28 and 63, then \(28 : x : : x : 63\)

\(= 28 × 63 ⇒ x = \sqrt{28 × 63} = \sqrt{1764} = 42\)

\(∴\) Mean proportional \(= 42\) Ans.

(ii) Let \(x\) be the mean proportional between 2-5 and 0-9, then \(2-5 : x : : x : 0-9\)

\(= 2-5 × 0-9 ⇒ x = \sqrt{2-25 × 0-9} = \sqrt{2-25} = 1-5\)

Hence, Mean proportional \(= 1-5\) Ans.

(iii) Let \(x\) be the mean proportional between 6-25 and 1-6, then

\(6-25 : x : : x : 1-6\) \(⇒ x^2 = 6-25 × 1-6\)

\(\sqrt{6-25 × 1-6} = \sqrt{10-000} = \sqrt{10}\)

Hence, Mean proportional \(= \sqrt{10}\) Ans.

(iv) Let \(x\) be the mean proportional between \((\sqrt{26} - \sqrt{17})\) and \((\sqrt{26} + \sqrt{17})\), then

\((\sqrt{26} - \sqrt{17}) : x : : x : (\sqrt{26} + \sqrt{17})\)

\(= (\sqrt{26} - \sqrt{17})(\sqrt{26} + \sqrt{17})\)

\(= 26 - 17\) \(∴\) (a + b)(a - b) = \(a^2 - b^2\)]

\(= 9 = (3)^2\) \(∴ x = 3\)

Hence, Mean proportional \(= 3\) Ans.

(v) Let \(x\) be the mean proportional between \((6 + 3\sqrt{3})\) and \((8 - 4\sqrt{3})\), then

\((6 + 3\sqrt{3}) : x : : x : (8 - 4\sqrt{3})\)

\(⇒ x^2 = (6 + 3\sqrt{3})(8 - 4\sqrt{3})\)

\(= 48 - 24\sqrt{3} + 24\sqrt{3} - 36 = 12\)

\(∴ x = \sqrt{12} = \sqrt{4 × 3} = 2\sqrt{3}\)

Hence, Mean proportional \(= 2\sqrt{3}\) Ans.

Q. 5. (i) What must be added to each of the numbers 6, 15, 20 and 43 so that the resulting numbers are in proportion ?

(ii) What least number must be added to each of the numbers 5, 11, 19 and 37 so that they are in proportion ? (2009)

Sol.

(i) Let \(x\) be added to each of the numbers

6, 15, 20 and 43, then

\(6 + x, 15 + x, 20 + x\)

and \(43 + x\) are proportional

then \(6 + x : 15 + x : : 20 + x : 43 + x\)

\(⇒ (6 + x)(43 + x) = (15 + x)(20 + x)\)

\(⇒ 258 + 6x + 43x + x^2 = 300 + 15x + 20x + x^2\)

\(⇒ 258 + 49x + x^2 = 300 + 35x + x^2\)

\(⇒ x^2 + 49x - x^2 - 35x = 300 - 258\)

\(⇒ 14x = 42 ⇒ x = \frac{42}{14} = 3\)

\(∴\) The required number to be added \(= 3\) Ans.

(ii) Let \(x\) be added to 5, 11, 19 and 37, that the remainders are in proportion

\(∴ \frac{5 + x}{11 + x} = \frac{19 + x}{37 + x}\)

\(⇒ (5 + x)(37 + x) = (19 + x)(11 + x)\)

\(⇒ 185 + 5x + 37x + x^2 = 209 + 19x + 11x + x^2\)

\(⇒ 185 + 42x + x^2 = 209 + 30x + x^2\)

\(⇒ 42x + x^2 - 30x - x^2 = 209 - 185\)

\(⇒ 12x = 24 ⇒ x = \frac{24}{12} = 2\)

\(∴\) 2 is to be added.

Q. 6. What must be subtracted from each of 23, 30, 57 and 78 so that the remainders are in proportion ?

(2004)

Sol.

Let the number to be subtracted \(= x\)

Then, \(23 - x, 30 - x, 57 - x\) and \(78 - x\) are in proportions.

\(∴ \frac{23 - x}{30 - x} = \frac{57 - x}{78 - x}\)

By cross multiplication

\((23 - x)(78 - x) = (57 - x)(30 - x)\)

\(⇒ 1794 - 23 x - 78 x + x^2\)

\(= 1710 - 57 x - 30 x + x^2\)

\(⇒ 1794 - 101x + x^2 = 1710 - 87x + x^2\)

\(⇒ 1794 - 1710 = - 87 x + 101 x\)

\(⇒ 84 = 14 x ⇒ x = \frac{84}{14} = 6\)

Hence, required number \(= 6\) Ans.

Q. 7. If (x - 2), (x + 2), (2x + 1) and (2x + 19) are in proportion, find the value of x.

Teacher's Note

Proportions are essential in scaling recipes and blueprints, helping us maintain proper ratios in everyday situations.

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