Maharashtra Board Class 9 Maths Part 1 Algebra Chapter 4 Ratio and Proportion Set 4.3 Solutions

Question 1. If \( \frac{a}{b} = \frac{7}{3} \), then find the values of the following ratios.
(i) \( \frac{5a+3b}{5a-3b} \)
(ii) \( \frac{2a^2+3b^2}{2a^2-3b^2} \)
(iii) \( \frac{a^3-b^3}{b^3} \)
(iv) \( \frac{7a+9b}{7a-9b} \)
Answer:
(i)
\( \frac{a}{b} = \frac{7}{3} \) ...[Given]
\( \frac{a}{b} \times \frac{5}{3} = \frac{7}{3} \times \frac{5}{3} \)

...[Multiplying both sides by \( \frac{5}{3} \)]
\( \frac{5a}{3b} = \frac{35}{9} \)
\( \frac{5a+3b}{5a-3b} = \frac{35+9}{35-9} \)

...[By componendo - dividendo]
\( \frac{5a+3b}{5a-3b} = \frac{44}{26} \)
\( = \frac{2 \times 22}{2 \times 13} = \frac{22}{13} \)
\( \therefore \frac{5a+3b}{5a-3b} = 22:13 \)
Alternate method:
\( \frac{a}{b} = \frac{7}{3} \) ...[Given]
Let the common multiple be m.
then, \( a = 7m \) and \( b = 3m \)
\( \frac{5a+3b}{5a-3b} = \frac{5(7m)+3(3m)}{5(7m)-3(3m)} \)
\( = \frac{35m+9m}{35m-9m} = \frac{44m}{26m} = \frac{44}{26} \)
\( = \frac{2 \times 22}{2 \times 13} = \frac{22}{13} \)
\( \therefore \frac{5a+3b}{5a-3b} = 22:13 \)

(ii)
\( \frac{a}{b} = \frac{7}{3} \) ...[Given]
\( \therefore (\frac{a}{b})^2 = (\frac{7}{3})^2 \)

...[Squaring both sides]
\( \frac{a^2}{b^2} = \frac{49}{9} \)
\( \therefore \frac{a^2}{b^2} \times \frac{2}{3} = \frac{49}{9} \times \frac{2}{3} \)

...[Multiplying both sides by \( \frac{2}{3} \)]
\( \frac{2a^2}{3b^2} = \frac{98}{27} \)
\( \therefore \frac{2a^2+3b^2}{2a^2-3b^2} = \frac{98+27}{98-27} \)

...[By componendo - dividendo]
\( \frac{2a^2+3b^2}{2a^2-3b^2} = \frac{125}{71} \)
\( \therefore \frac{2a^2+3b^2}{2a^2-3b^2} = 125:71 \)

(iii)
\( \frac{a}{b} = \frac{7}{3} \) ...[Given]
\( \therefore (\frac{a}{b})^3 = (\frac{7}{3})^3 \)

...[Cubing both sides]
\( \frac{a^3}{b^3} = \frac{343}{27} \)
\( \therefore \frac{a^3-b^3}{b^3} = \frac{343-27}{27} \)

...[By dividendo]
\( \therefore \frac{a^3-b^3}{b^3} = \frac{316}{27} \)
\( \therefore \frac{a^3-b^3}{b^3} = 316:27 \)

(iv)
\( \frac{a}{b} = \frac{7}{3} \) ...[Given]
\( \therefore \frac{a}{b} \times \frac{7}{9} = \frac{7}{3} \times \frac{7}{9} \)

...[Multiplying both sides by \( \frac{7}{9} \)]
\( \therefore \frac{7a}{9b} = \frac{49}{27} \)
\( \therefore \frac{7a+9b}{7a-9b} = \frac{49+27}{49-27} \)

...[By componendo - dividendo]
\( \therefore \frac{7a+9b}{7a-9b} = \frac{76}{22} = \frac{2 \times 38}{2 \times 11} = \frac{38}{11} \)
\( \therefore \frac{7a+9b}{7a-9b} = 38:11 \)
In simple words: This question involves applying various properties of ratios and proportions such as componendo-dividendo, squaring, cubing, and invertendo to find the values of derived ratios based on an initial given ratio. The core idea is to manipulate the initial ratio to match the structure of the target ratios.

🎯 Exam Tip: Master the properties of ratios (alternendo, invertendo, componendo, dividendo, componendo-dividendo) as they are frequently tested. Show each step clearly for full marks in derivation problems.

Question 2. If \( \frac{15a^2+4b^2}{15a^2-4b^2} = \frac{47}{7} \), then find the value of the following ratios.
(i) \( \frac{a}{b} \)
(ii) \( \frac{7a-3b}{7a+3b} \)
(iii) \( \frac{b^2-2a^2}{b^2+2a^2} \)
(iv) \( \frac{b^3-2a^3}{b^3+2a^3} \)
Answer:
(i)
\( \frac{15a^2+4b^2}{15a^2-4b^2} = \frac{47}{7} \) ...[Given]
\( \therefore \frac{15a^2+4b^2+(15a^2-4b^2)}{15a^2+4b^2-(15a^2-4b^2)} = \frac{47+7}{47-7} \)

...[By componendo - dividendo]
\( \therefore \frac{15a^2+4b^2+15a^2-4b^2}{15a^2+4b^2-15a^2+4b^2} = \frac{54}{40} \)
\( \therefore \frac{30a^2}{8b^2} = \frac{54}{40} \)
\( \therefore \frac{a^2}{b^2} = \frac{54 \times 8}{40 \times 30} \)
\( \therefore \frac{a^2}{b^2} = \frac{9}{25} \)
\( \therefore \frac{a}{b} = \frac{3}{5} \)

...[Taking positive square root of both sides]
\( \therefore \frac{a}{b} = 3:5 \)

(ii)
\( \frac{a}{b} = \frac{3}{5} \)
\( \therefore \frac{a}{b} \times \frac{7}{3} = \frac{3}{5} \times \frac{7}{3} \)

...[Multiplying both sides by \( \frac{7}{3} \)]
\( \therefore \frac{7a}{3b} = \frac{7}{5} \)
\( \therefore \frac{7a+3b}{7a-3b} = \frac{7+5}{7-5} \)

...[By componendo - dividendo]
\( \therefore \frac{7a+3b}{7a-3b} = \frac{12}{2} \)
\( \therefore \frac{7a+3b}{7a-3b} = 6 \)
\( \therefore \frac{7a-3b}{7a+3b} = \frac{1}{6} \)

...[By invertendo]
\( \therefore \frac{7a-3b}{7a+3b} = 1:6 \)

(iii)
\( \frac{a}{b} = \frac{3}{5} \)
\( \therefore (\frac{a}{b})^2 = (\frac{3}{5})^2 \)

...[Squaring both sides]
\( \frac{a^2}{b^2} = \frac{9}{25} \)
\( \therefore \frac{b^2}{a^2} = \frac{25}{9} \)

...[By invertendo]
\( \therefore \frac{b^2}{a^2} \times \frac{1}{2} = \frac{25}{9} \times \frac{1}{2} \)

...[Multiplying both sides by \( \frac{1}{2} \)]
\( \therefore \frac{b^2}{2a^2} = \frac{25}{18} \)
\( \therefore \frac{b^2+2a^2}{b^2-2a^2} = \frac{25+18}{25-18} \)

...[By componendo - dividendo]
\( \therefore \frac{b^2+2a^2}{b^2-2a^2} = \frac{43}{7} \)
\( \therefore \frac{b^2-2a^2}{b^2+2a^2} = \frac{7}{43} \)

...[By invertendo]
\( \therefore \frac{b^2-2a^2}{b^2+2a^2} = 7:43 \)

(iv)
\( \frac{a}{b} = \frac{3}{5} \)
\( \therefore (\frac{a}{b})^3 = (\frac{3}{5})^3 \)

...[Cubing both sides]
\( \therefore \frac{a^3}{b^3} = \frac{27}{125} \)
\( \therefore \frac{b^3}{a^3} = \frac{125}{27} \)

...[By invertendo]
\( \therefore \frac{b^3}{a^3} \times \frac{1}{2} = \frac{125}{27} \times \frac{1}{2} \)

...[Multiplying both sides by \( \frac{1}{2} \)]
\( \therefore \frac{b^3}{2a^3} = \frac{125}{54} \)
\( \therefore \frac{b^3+2a^3}{b^3-2a^3} = \frac{125+54}{125-54} \)

...[By componendo - dividendo]
\( \therefore \frac{b^3+2a^3}{b^3-2a^3} = \frac{179}{71} \)
\( \therefore \frac{b^3-2a^3}{b^3+2a^3} = \frac{71}{179} \)

...[By invertendo]
\( \therefore \frac{b^3-2a^3}{b^3+2a^3} = 71:179 \)
In simple words: This question starts with a complex ratio of squared terms and requires finding simpler ratios and ratios involving cubes. The strategy involves using componendo-dividendo to simplify the initial ratio, then finding \( \frac{a}{b} \), and finally applying powers and other ratio properties to derive the required expressions.

🎯 Exam Tip: When given a ratio involving squared or cubed terms, use componendo-dividendo first to simplify it into a basic ratio like \( \frac{a}{b} \), which makes subsequent calculations easier. Remember to show the property used for each step.

Question 3. If \( \frac{3a+7b}{3a-7b} = \frac{4}{3} \) then find the value of the ratio \( \frac{3a^2-7b^2}{3a^2+7b^2} \).
Answer::
\( \frac{3a+7b}{3a-7b} = \frac{4}{3} \) ...[Given]
\( \therefore \frac{3a+7b+(3a-7b)}{3a+7b-(3a-7b)} = \frac{4+3}{4-3} \)

...[By componendo - dividendo]
\( \therefore \frac{3a+7b+3a-7b}{3a+7b-3a+7b} = \frac{7}{1} \)
\( \therefore \frac{6a}{14b} = \frac{7}{1} \)
\( \therefore \frac{a}{b} = \frac{7 \times 14}{6} \)
\( \therefore \frac{a}{b} = \frac{49}{3} \)
\( \therefore (\frac{a}{b})^2 = (\frac{49}{3})^2 \)

...[Squaring both sides]
\( \therefore \frac{a^2}{b^2} = \frac{2401}{9} \)
\( \therefore \frac{a^2}{b^2} \times \frac{3}{7} = \frac{2401}{9} \times \frac{3}{7} \)

...[Multiplying both sides by \( \frac{3}{7} \)]
\( \therefore \frac{3a^2}{7b^2} = \frac{343}{3} \)
\( \therefore \frac{3a^2+7b^2}{3a^2-7b^2} = \frac{343+3}{343-3} \)

...[By componendo - dividendo]
\( \therefore \frac{3a^2+7b^2}{3a^2-7b^2} = \frac{346}{340} \)
\( \therefore \frac{3a^2+7b^2}{3a^2-7b^2} = \frac{2 \times 173}{2 \times 170} = \frac{173}{170} \)
\( \therefore \frac{3a^2-7b^2}{3a^2+7b^2} = \frac{170}{173} \)

...[By invertendo]
\( \therefore \frac{3a^2-7b^2}{3a^2+7b^2} = 170:173 \)
In simple words: This problem asks to find a ratio of squared terms, starting from a ratio of linear terms. The solution involves applying componendo-dividendo to the initial ratio to find \( \frac{a}{b} \), then squaring it, adjusting with a constant multiplier, and finally using componendo-dividendo and invertendo again to get the desired ratio.

🎯 Exam Tip: Pay close attention to the target ratio required (e.g., \( \frac{X-Y}{X+Y} \) vs. \( \frac{X+Y}{X-Y} \)) to correctly apply invertendo if needed as the last step. Errors here are common.

Question 4. Solve the following equations.

(i) \( \frac{x^2+12x-20}{3x-5} = \frac{x^2+8x+12}{2x+3} \)
Answer:
\( \frac{x^2+12x-20}{3x-5} = \frac{x^2+8x+12}{2x+3} \)
\( \therefore \frac{x^2+12x-20}{4(3x-5)} = \frac{x^2+8x+12}{4(2x+3)} \)

...[Multiplying both sides by \( \frac{1}{4} \)]
\( \therefore \frac{x^2+12x-20}{12x-20} = \frac{x^2+8x+12}{8x+12} \)
\( \therefore \frac{(x^2+12x-20)-(12x-20)}{12x-20} = \frac{(x^2+8x+12)-(8x+12)}{8x+12} \)

...[By dividendo]
\( \therefore \frac{x^2+12x-20-12x+20}{12x-20} = \frac{x^2+8x+12-8x-12}{8x+12} \)
\( \therefore \frac{x^2}{12x-20} = \frac{x^2}{8x+12} \)
This equation is true for \( x = 0 \)
\( \therefore x = 0 \) is one of the solutions.
If \( x \neq 0 \), then \( x^2 \neq 0 \)
\( \therefore \frac{1}{12x-20} = \frac{1}{8x+12} \) ...[Dividing both sides by \( x^2 \)]
\( \therefore 8x+12 = 12x-20 \)
\( \therefore 12+20 = 12x-8x \)
\( \therefore 32 = 4x \)
\( \therefore x = 8 \)
\( \therefore x = 0 \) or \( x = 8 \) are the solutions of the given equation.
In simple words: This equation is solved by manipulating ratios to isolate \( x^2 \) on both sides using the dividendo property, leading to one solution \( x=0 \). If \( x \neq 0 \), dividing by \( x^2 \) simplifies the equation to a linear one, yielding the second solution.

🎯 Exam Tip: Always consider \( x=0 \) as a potential solution when you simplify by dividing by a term containing \( x \), to avoid losing a valid root.

(ii) \( \frac{10x^2+15x+63}{5x^2-25x+12} = \frac{2x+3}{x-5} \)
Answer:
If \( x = 0 \) then
L.H.S. \( = \frac{10x^2+15x+63}{5x^2-25x+12} = \frac{10(0)^2+15(0)+63}{5(0)^2-25(0)+12} = \frac{63}{12} = \frac{21}{4} \)
R.H.S. \( = \frac{2x+3}{x-5} = \frac{2(0)+3}{0-5} = \frac{3}{-5} = -\frac{3}{5} \)
\( \frac{21}{4} = -\frac{3}{5} \) which is a contradiction.
\( \therefore x \neq 0 \)
Now, \( \frac{10x^2+15x+63}{5x^2-25x+12} = \frac{2x+3}{x-5} \)
\( \therefore \frac{10x^2+15x+63}{2x+3} = \frac{5x^2-25x+12}{x-5} \)

...[By alternendo]
\( \therefore \frac{10x^2+15x+63}{5x(2x+3)} = \frac{5x^2-25x+12}{5x(x-5)} \)

...[Multiplying both sides by \( \frac{1}{5x} \) as \( x \neq 0 \)]
\( \therefore \frac{10x^2+15x+63}{10x^2+15x} = \frac{5x^2-25x+12}{5x^2-25x} \)
\( \therefore \frac{(10x^2+15x+63)-(10x^2+15x)}{10x^2+15x} = \frac{(5x^2-25x+12)-(5x^2-25x)}{5x^2-25x} \)

...[By dividendo]
\( \therefore \frac{10x^2+15x+63-10x^2-15x}{10x^2+15x} = \frac{5x^2-25x+12-5x^2+25x}{5x^2-25x} \)
\( \therefore \frac{63}{10x^2+15x} = \frac{12}{5x^2-25x} \)
\( \therefore \frac{63}{5x(2x+3)} = \frac{12}{5x(x-5)} \)
\( \therefore \frac{21}{2x+3} = \frac{4}{x-5} \)

...[Multiplying both sides by \( \frac{5x}{3} \) as \( x \neq 0 \)]
\( \therefore 21(x-5) = 4(2x+3) \)
\( \therefore 21x-105 = 8x+12 \)
\( \therefore 21x-8x = 12+105 \)
\( \therefore 13x = 117 \)
\( \therefore x = 9 \)
\( \therefore x = 9 \) is the solution of the given equation.
In simple words: This problem is solved by first testing if \( x=0 \) is a solution, then using the alternendo property to rearrange terms, followed by a clever multiplication by \( \frac{1}{5x} \) and applying dividendo to simplify the equation significantly, leading to a linear equation.

🎯 Exam Tip: When equations have complex polynomial ratios, try to simplify them using ratio properties (like alternendo and dividendo) to reduce the degree of polynomials and make the equation solvable.

(iii) \( \frac{(2x+1)^2+(2x-1)^2}{(2x+1)^2-(2x-1)^2} = \frac{17}{8} \)
Answer:
\( \frac{(2x+1)^2+(2x-1)^2}{(2x+1)^2-(2x-1)^2} = \frac{17}{8} \)
\( \therefore \frac{[(2x+1)^2+(2x-1)^2]+[(2x+1)^2-(2x-1)^2]}{[(2x+1)^2+(2x-1)^2]-[(2x+1)^2-(2x-1)^2]} = \frac{17+8}{17-8} \)

...[By componendo - dividendo]
\( \therefore \frac{2(2x+1)^2}{2(2x-1)^2} = \frac{25}{9} \)
\( \therefore \frac{(2x+1)^2}{(2x-1)^2} = \frac{25}{9} \)
\( \therefore \frac{2x+1}{2x-1} = \pm \frac{5}{3} \)

...[Taking square root of both sides]
Case 1: \( \frac{2x+1}{2x-1} = \frac{5}{3} \)
\( \therefore 3(2x+1) = 5(2x-1) \)
\( \therefore 6x+3 = 10x-5 \)
\( \therefore 3+5 = 10x-6x \)
\( \therefore 8 = 4x \)
\( \therefore x = 2 \)
Case 2: \( \frac{2x+1}{2x-1} = -\frac{5}{3} \)
\( \therefore 3(2x+1) = -5(2x-1) \)
\( \therefore 6x+3 = -10x+5 \)
\( \therefore 6x+10x = 5-3 \)
\( \therefore 16x = 2 \)
\( \therefore x = \frac{2}{16} = \frac{1}{8} \)
\( \therefore x = 2 \) or \( x = \frac{1}{8} \) is the solution of the given equation.
In simple words: This equation is solved by applying the componendo-dividendo property to simplify the ratio of squared terms, then taking the square root of both sides. This leads to two separate linear equations, each yielding a distinct solution for \( x \).

🎯 Exam Tip: Remember to consider both positive and negative roots when taking the square root of both sides of an equation to avoid missing potential solutions.

(iv) \( \frac{\sqrt{4x+1}+\sqrt{x+3}}{\sqrt{4x+1}-\sqrt{x+3}} = \frac{4}{1} \)
Answer:
\( \frac{\sqrt{4x+1}+\sqrt{x+3}}{\sqrt{4x+1}-\sqrt{x+3}} = \frac{4}{1} \)
\( \therefore \frac{(\sqrt{4x+1}+\sqrt{x+3})+(\sqrt{4x+1}-\sqrt{x+3})}{(\sqrt{4x+1}+\sqrt{x+3})-(\sqrt{4x+1}-\sqrt{x+3})} = \frac{4+1}{4-1} \)

...[By componendo - dividendo]
\( \therefore \frac{\sqrt{4x+1}+\sqrt{x+3}+\sqrt{4x+1}-\sqrt{x+3}}{\sqrt{4x+1}+\sqrt{x+3}-\sqrt{4x+1}+\sqrt{x+3}} = \frac{5}{3} \)
\( \therefore \frac{2\sqrt{4x+1}}{2\sqrt{x+3}} = \frac{5}{3} \)
\( \therefore \frac{\sqrt{4x+1}}{\sqrt{x+3}} = \frac{5}{3} \)
\( \therefore \frac{4x+1}{x+3} = \frac{25}{9} \)

...[Squaring both sides]
\( \therefore 9(4x+1) = 25(x+3) \)
\( \therefore 36x+9 = 25x+75 \)
\( \therefore 36x-25x = 75-9 \)
\( \therefore 11x = 66 \)
\( \therefore x = 6 \)
\( \therefore x = 6 \) is the solution of the given equation.
In simple words: This equation with square roots is simplified by applying componendo-dividendo to remove the nested structure of roots. Then, squaring both sides converts it into a linear equation, which is straightforward to solve for \( x \).

🎯 Exam Tip: Equations involving square roots often require squaring to eliminate the roots. Before squaring, try to simplify the expression using ratio properties or by isolating a single square root term.

(v) \( \frac{(4x+1)^2+(2x+3)^2}{4x^2+12x+9} = \frac{61}{36} \)
Answer:
\( \frac{(4x+1)^2+(2x+3)^2}{4x^2+12x+9} = \frac{61}{36} \)
\( \therefore \frac{(4x+1)^2+(2x+3)^2}{(2x)^2+2 \times 2x \times 3+(3)^2} = \frac{61}{36} \)
\( \therefore \frac{(4x+1)^2+(2x+3)^2}{(2x+3)^2} = \frac{61}{36} \)

...[ \( a^2+2ab+b^2 = (a+b)^2 \) ]
\( \therefore \frac{[(4x+1)^2+(2x+3)^2]-(2x+3)^2}{(2x+3)^2} = \frac{61-36}{36} \)

...[By dividendo]
\( \therefore \frac{(4x+1)^2}{(2x+3)^2} = \frac{25}{36} \)
\( \therefore \frac{4x+1}{2x+3} = \pm \frac{5}{6} \)

...[Taking square root of both sides]
Case 1: \( \frac{4x+1}{2x+3} = \frac{5}{6} \)
\( \therefore 6(4x+1) = 5(2x+3) \)
\( \therefore 24x+6 = 10x+15 \)
\( \therefore 24x-10x = 15-6 \)
\( \therefore 14x = 9 \)
\( \therefore x = \frac{9}{14} \)
Case 2: \( \frac{4x+1}{2x+3} = -\frac{5}{6} \)
\( \therefore 6(4x+1) = -5(2x+3) \)
\( \therefore 24x+6 = -10x-15 \)
\( \therefore 24x+10x = -15-6 \)
\( \therefore 34x = -21 \)
\( \therefore x = -\frac{21}{34} \)
\( \therefore x = \frac{9}{14} \) or \( x = -\frac{21}{34} \) is the solution of the given equation.
In simple words: This equation is simplified by recognizing the denominator as a perfect square and then applying the dividendo property. This isolates the squared term of \( (4x+1) \), allowing us to take the square root and solve the resulting two linear equations.

🎯 Exam Tip: Always look for perfect square trinomials (e.g., \( a^2+2ab+b^2 \)) to simplify denominators or numerators. This significantly reduces the complexity before applying ratio properties.

(vi) \( \frac{(3x-4)^3-(x+1)^3}{(3x-4)^3+(x+1)^3} = \frac{61}{189} \)
Answer:
\( \frac{(3x-4)^3-(x+1)^3}{(3x-4)^3+(x+1)^3} = \frac{61}{189} \)
\( \therefore \frac{(3x-4)^3+(x+1)^3}{(3x-4)^3-(x+1)^3} = \frac{189}{61} \)

...[By invertendo]
\( \therefore \frac{[(3x-4)^3+(x+1)^3]+[(3x-4)^3-(x+1)^3]}{[(3x-4)^3+(x+1)^3]-[(3x-4)^3-(x+1)^3]} = \frac{189+61}{189-61} \)

...[By componendo - dividendo]
\( \therefore \frac{2(3x-4)^3}{2(x+1)^3} = \frac{250}{128} \)
\( \therefore \frac{(3x-4)^3}{(x+1)^3} = \frac{125}{64} \)
\( \therefore \frac{3x-4}{x+1} = \frac{5}{4} \)

...[Taking cube root of both sides]
\( \therefore 4(3x-4) = 5(x+1) \)
\( \therefore 12x-16 = 5x+5 \)
\( \therefore 12x-5x = 5+16 \)
\( \therefore 7x = 21 \)
\( \therefore x = 3 \)
\( \therefore x = 3 \) is the solution of the given equation.
In simple words: This equation, involving cubes, is solved by first using invertendo and then componendo-dividendo to simplify the complex ratio. This isolates the cubic terms, allowing us to take the cube root of both sides, which transforms the equation into a simple linear form for \( x \).

🎯 Exam Tip: For equations with ratios of cubed terms, employ invertendo and componendo-dividendo to simplify them into a ratio of linear terms, then take the cube root to solve efficiently.