ML Aggarwal Class 10 Maths Solutions Chapter 07 Ratio and Proportion

Question 1. An alloy consists of 27 - 1/2 kg of copper and 2 - 3/4 kg of tin. Find the ratio by weight of tin to the alloy.
Answer: The total weight equals tin plus copper: \( 27\frac{1}{2} + 2\frac{3}{4} = \frac{55}{2} + \frac{11}{4} = \frac{110 + 11}{4} = \frac{121}{4} \) kg. The ratio of tin to alloy becomes: \( \frac{11/4}{121/4} = \frac{11}{121} = \frac{1}{11} \). As a result, the ratio by weight of tin to alloy is 1 - 11.
In simple words: Add the copper and tin to get the total alloy weight. Then divide the tin weight by the alloy weight to get the ratio.

Exam Tip: Always convert mixed numbers to improper fractions before adding. Keep track of the common denominator when finding the sum.

Question 2. Find the compounded ratio of:
(i) 2 - 3 and 4 - 9
(ii) 4 - 5, 5 - 7 and 9 - 11
(iii) (a - b) - (a + b), (a + b)² - (a² + b²) and (a⁴ - b⁴) - (a² - b²)²
Answer:
(i) Multiply the fractions: \( \frac{2}{3} \times \frac{4}{9} = \frac{8}{27} \). So the compounded ratio is 8 - 27.
(ii) Multiply all three: \( \frac{4}{5} \times \frac{5}{7} \times \frac{9}{11} = \frac{180}{385} \). Simplify by dividing both by 5 to get \( \frac{36}{77} \). The compounded ratio is 36 - 77.
(iii) Multiply: \( \frac{a - b}{a + b} \times \frac{(a + b)^2}{a^2 + b^2} \times \frac{a^4 - b^4}{(a^2 - b^2)^2} \). After cancellation using the factorizations \( a^4 - b^4 = (a^2 - b^2)(a^2 + b^2) \) and \( a^2 - b^2 = (a - b)(a + b) \), the expression simplifies to \( \frac{(a - b)(a + b)}{a^2 - b^2} = \frac{(a^2 - b^2)}{a^2 - b^2} = 1 \). The compounded ratio is 1 - 1.
In simple words: For a compounded ratio, multiply the numerators together and the denominators together. Then reduce the final fraction to lowest terms.

Exam Tip: Watch for algebraic terms that cancel—like (a + b), (a - b), and (a² - b²). Factorize before simplifying.

Question 3. Find the duplicate ratio of:
(i) 2 - 3
(ii) \( \sqrt{5} \) - 7
(iii) 5a - 6b
Answer:
(i) The duplicate ratio means squaring both parts: \( 2^2 - 3^2 = 4 - 9 \). The duplicate ratio is 4 - 9.
(ii) Square both: \( (\sqrt{5})^2 - 7^2 = 5 - 49 \). The duplicate ratio is 5 - 49.
(iii) Square each term: \( (5a)^2 - (6b)^2 = 25a^2 - 36b^2 \). The duplicate ratio is 25a² - 36b².
In simple words: The duplicate ratio squares both parts of the original ratio.

Exam Tip: Remember that duplicate means "squared"—square the numerator and denominator separately, not the whole ratio.

Question 4. Find the triplicate ratio of:
(i) 3 - 4
(ii) \( \frac{1}{2} \) - \( \frac{1}{3} \)
(iii) 1³ - 2³
Answer:
(i) Cube both parts: \( 3^3 - 4^3 = 27 - 64 \). The triplicate ratio is 27 - 64.
(ii) Cube each: \( \left(\frac{1}{2}\right)^3 - \left(\frac{1}{3}\right)^3 = \frac{1}{8} - \frac{1}{27} \). As a ratio, this becomes 27 - 8. The triplicate ratio is 27 - 8.
(iii) Cube the pair: \( (1^3)^3 - (2^3)^3 = 1^9 - 2^9 = 1 - 512 \). The triplicate ratio is 1 - 512.
In simple words: The triplicate ratio cubes both parts of the original ratio.

Exam Tip: Triplicate means "cubed"—raise the numerator and denominator to the third power each.

Question 5. Find the sub-duplicate ratio of:
(i) 9 - 16
(ii) \( \frac{1}{9} \) - \( \frac{1}{4} \)
(iii) 9a² - 49b²
Answer:
(i) Take the square root of both: \( \sqrt{9} - \sqrt{16} = 3 - 4 \). The sub-duplicate ratio is 3 - 4.
(ii) Find the root: \( \sqrt{\frac{1}{9}} - \sqrt{\frac{1}{4}} = \frac{1}{3} - \frac{1}{2} \), which gives the ratio 2 - 3. The sub-duplicate ratio is 2 - 3.
(iii) Take roots of each: \( \sqrt{9a^2} - \sqrt{49b^2} = 3a - 7b \). The sub-duplicate ratio is 3a - 7b.
In simple words: The sub-duplicate ratio takes the square root of both parts of the original ratio.

Exam Tip: Sub-duplicate means finding square roots. Simplify any perfect squares under the radicals first.

Question 6. Find the sub-triplicate ratio of:
(i) 1 - 216
(ii) \( \frac{1}{125} \) - \( \frac{1}{8} \)
(iii) 27a³ - 64b³
Answer:
(i) Take the cube root: \( \sqrt[3]{1} - \sqrt[3]{216} = 1 - 6 \). The sub-triplicate ratio is 1 - 6.
(ii) Find cube roots: \( \sqrt[3]{\frac{1}{125}} - \sqrt[3]{\frac{1}{8}} = \frac{1}{5} - \frac{1}{2} \), which expresses as 2 - 5. The sub-triplicate ratio is 2 - 5.
(iii) Extract cube roots: \( \sqrt[3]{27a^3} - \sqrt[3]{64b^3} = 3a - 4b \). The sub-triplicate ratio is 3a - 4b.
In simple words: The sub-triplicate ratio finds the cube root of both parts of the original ratio.

Exam Tip: Sub-triplicate means finding cube roots. Identify perfect cubes to simplify quickly.

Question 7. Find the reciprocal ratio of:
(i) 4 - 7
(ii) 3² - 4²
(iii) \( \frac{1}{9} \) - 2
Answer:
(i) Swap numerator and denominator: \( \frac{1/4}{1/7} = \frac{7}{4} \). The reciprocal ratio is 7 - 4.
(ii) Invert: \( \frac{1/3^2}{1/4^2} = \frac{1/9}{1/16} = \frac{16}{9} \). The reciprocal ratio is 16 - 9.
(iii) Flip: \( \frac{1/(1/9)}{1/2} = \frac{9}{1/2} = \frac{9 \times 2}{1} = \frac{18}{1} \). The reciprocal ratio is 18 - 1.
In simple words: The reciprocal ratio flips the original ratio upside down.

Exam Tip: Always express the ratio as a fraction first before inverting to avoid errors.

Question 8. Arrange the following ratios in ascending order of magnitude: 2 - 3, 17 - 21, 11 - 14 and 5 - 7.
Answer: Convert each ratio to fractions with a common denominator. The L.C.M. of 3, 21, 14, and 7 is 42. Express: \( \frac{2}{3} = \frac{28}{42} \), \( \frac{17}{21} = \frac{34}{42} \), \( \frac{11}{14} = \frac{33}{42} \), and \( \frac{5}{7} = \frac{30}{42} \). Since \( 28 < 30 < 33 < 34 \), we get \( \frac{28}{42} < \frac{30}{42} < \frac{33}{42} < \frac{34}{42} \). Therefore, the ratios in ascending order are 2 - 3, 5 - 7, 11 - 14, 17 - 21.
In simple words: Change all ratios to fractions with the same bottom number. Then order them from smallest to largest by looking at the top numbers.

Exam Tip: Finding the L.C.M. of all denominators avoids decimal approximations and ensures accuracy.

Question 9(i). If A - B = 2 - 3, B - C = 4 - 5 and C - D = 6 - 7, find A - D.
Answer: From \( \frac{A}{B} = \frac{2}{3} \), we obtain \( B = \frac{3A}{2} \). Substituting into \( \frac{B}{C} = \frac{4}{5} \): \( \frac{3A/2}{C} = \frac{4}{5} \), so \( C = \frac{15A}{8} \). Using \( \frac{C}{D} = \frac{6}{7} \): \( \frac{15A/8}{D} = \frac{6}{7} \), which gives \( D = \frac{105A}{48} = \frac{35A}{16} \). Therefore, \( \frac{A}{D} = \frac{16}{35} \), and the ratio A - D is 16 - 35.
In simple words: Express B and C in terms of A. Then find D in terms of A. Finally, divide A by D to get the ratio.

Exam Tip: Always substitute one ratio into the next to link all three quantities together.

Question 9(ii). If x - y = 2 - 3 and y - z = 4 - 7, find x - y - z.
Answer: To merge both ratios, make y equal in both expressions. The L.C.M. of 3 and 4 is 12. From x - y = 2 - 3, scale to get \( \frac{x}{y} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12} \), so x - y = 8 - 12. From y - z = 4 - 7, scale to \( \frac{y}{z} = \frac{4 \times 3}{7 \times 3} = \frac{12}{21} \), so y - z = 12 - 21. Combining, x - y - z = 8 - 12 - 21.
In simple words: Make sure the middle value is the same in both ratios. Then link them together into one chain.

Exam Tip: Use the L.C.M. of the middle terms to avoid errors when merging compound ratios.

Question 10(i). If A - B = \( \frac{1}{5} \) - \( \frac{1}{4} \) and B - C = \( \frac{1}{6} \) - \( \frac{1}{7} \), find A - B - C.
Answer: Simplify the ratios: \( \frac{1}{5} - \frac{1}{4} = \frac{4}{5 \times 4} - \frac{5}{5 \times 4} = \frac{4}{20} - \frac{5}{20} \), so A - B = 4 - 5 becomes 5 - 4 when inverted to \( \frac{A}{B} = \frac{5}{4} \). Similarly, \( \frac{1}{6} - \frac{1}{7} = \frac{7}{42} - \frac{6}{42} \), giving B - C = 6 - 7 as 7 - 6 when inverted to \( \frac{B}{C} = \frac{7}{6} \). Make B the same (L.C.M. of 4 and 6 is 12): \( A - B = 15 - 12 \) and B - C = 12 - 14. Therefore, A - B - C = 15 - 12 - 14.
In simple words: First convert each fraction ratio. Then make the middle value match before combining the two ratios.

Exam Tip: When the given ratio involves fractions, always invert carefully and check which value is numerator and denominator.

Question 10(ii). If 3A = 4B = 6C, find A - B - C.
Answer: From 3A = 4B, we get \( \frac{A}{B} = \frac{4}{3} \), so A - B = 4 - 3. From 4B = 6C, we get \( \frac{B}{C} = \frac{6}{4} = \frac{3}{2} \), so B - C = 3 - 2. Combining, A - B - C = 4 - 3 - 2.
In simple words: When quantities are set equal (like 3A = 4B), divide to find the ratio of A to B.

Exam Tip: For equations like kA = mB = nC, extract pairwise ratios by dividing both sides correctly.

Question 11(i). If \( \frac{3x + 5y}{3x - 5y} = \frac{7}{3} \), find x - y.
Answer: Cross-multiply: \( 3(3x + 5y) = 7(3x - 5y) \). Expand: \( 9x + 15y = 21x - 35y \). Collect like terms: \( 50y = 12x \), so \( \frac{x}{y} = \frac{50}{12} = \frac{25}{6} \). Therefore, x - y is 25 - 6.
In simple words: Cross-multiply the equation. Gather all x terms on one side and all y terms on the other. Then simplify the ratio.

Exam Tip: Always reduce the final fraction to lowest terms using the G.C.D.

Question 11(ii). If a - b = 3 - 11, find (15a - 3b) - (9a + 5b).
Answer: Given \( \frac{a}{b} = \frac{3}{11} \). For the requested ratio, divide numerator and denominator by b: \( \frac{15a - 3b}{9a + 5b} = \frac{15(a/b) - 3}{9(a/b) + 5} \). Substitute \( \frac{a}{b} = \frac{3}{11} \): \( \frac{15 \times 3/11 - 3}{9 \times 3/11 + 5} = \frac{45/11 - 33/11}{27/11 + 55/11} = \frac{12/11}{82/11} = \frac{12}{82} = \frac{6}{41} \). The ratio is 6 - 41.
In simple words: Divide the entire numerator and denominator by b. Then substitute the known ratio a/b.

Exam Tip: Factoring out common terms before substituting makes the algebra cleaner.

Question 12(i). If (4x² + xy) - (3xy - y²) = 12 - 5, find (x + 2y) - (2x + y).
Answer: Given \( \frac{4x^2 + xy}{3xy - y^2} = \frac{12}{5} \). Cross-multiply: \( 5(4x^2 + xy) = 12(3xy - y^2) \), giving \( 20x^2 + 5xy = 36xy - 12y^2 \), which simplifies to \( 20x^2 - 31xy + 12y^2 = 0 \). Divide by y²: \( 20(x/y)^2 - 31(x/y) + 12 = 0 \). Factor: \( (5(x/y) - 4)(4(x/y) - 3) = 0 \). So \( \frac{x}{y} = \frac{4}{5} \) or \( \frac{x}{y} = \frac{3}{4} \). For the ratio \( \frac{x + 2y}{2x + y} \), divide by y: \( \frac{(x/y) + 2}{2(x/y) + 1} \). When \( \frac{x}{y} = \frac{4}{5} \): \( \frac{4/5 + 2}{2(4/5) + 1} = \frac{14/5}{13/5} = \frac{14}{13} \). When \( \frac{x}{y} = \frac{3}{4} \): \( \frac{3/4 + 2}{2(3/4) + 1} = \frac{11/4}{10/4} = \frac{11}{10} \). The ratios are 14 - 13 or 11 - 10.
In simple words: Cross-multiply and simplify to get a quadratic in x/y. Factor or use the quadratic formula. Then evaluate the target ratio for each solution.

Exam Tip: A quadratic in x/y often gives two solutions. Always check both and provide both final ratios if the question allows.

Question 12(ii). If y(3x - y) - x(4x + y) = 5 - 12, find (x² + y²) - (x + y)².
Answer: Given \( \frac{y(3x - y)}{x(4x + y)} = \frac{5}{12} \). Cross-multiply: \( 12y(3x - y) = 5x(4x + y) \), so \( 36xy - 12y^2 = 20x^2 + 5xy \). Rearrange: \( 20x^2 - 31xy + 12y^2 = 0 \). This is the same quadratic as in Question 12(i), giving \( \frac{x}{y} = \frac{4}{5} \) or \( \frac{x}{y} = \frac{3}{4} \). For \( \frac{x^2 + y^2}{(x + y)^2} \), divide numerator and denominator by y²: \( \frac{(x/y)^2 + 1}{((x/y) + 1)^2} \). When \( \frac{x}{y} = \frac{4}{5} \): \( \frac{(4/5)^2 + 1}{((4/5) + 1)^2} = \frac{16/25 + 25/25}{(9/5)^2} = \frac{41/25}{81/25} = \frac{41}{81} \). When \( \frac{x}{y} = \frac{3}{4} \): \( \frac{(3/4)^2 + 1}{((3/4) + 1)^2} = \frac{9/16 + 16/16}{(7/4)^2} = \frac{25/16}{49/16} = \frac{25}{49} \). The ratios are 41 - 81 or 25 - 49.
In simple words: Solve for x/y by cross-multiplying and factoring. Then divide the target expression by y² and substitute each value of x/y.

Exam Tip: Recognize when two problems share the same quadratic equation—this saves calculation time and helps verify your work.

Question 20(ii). In class X of a school, the ratio of the number of boys to that of the girls is 4 : 3. If there were 20 more boys and 12 less girls, then the ratio would have been 2 : 1. How many students were there in the class?
Answer: Since the ratio of boys to girls is 4 : 3, we can represent the number of boys as 4x and the number of girls as 3x. According to the condition given, if we add 20 boys and remove 12 girls, the new ratio becomes 2 : 1.

\[ \frac{4x + 20}{3x - 12} = \frac{2}{1} \]

\[ 4x + 20 = 2(3x - 12) \]

\[ 4x + 20 = 6x - 24 \]

\[ 20 + 24 = 6x - 4x \]

\[ 44 = 2x \]

\[ x = 22 \]

Therefore, x = 22, which gives us 4x = 88 boys and 3x = 66 girls. The total number of students in the class is 88 + 66 = 154 students.
In simple words: When you set up the ratio equation with the new condition and solve for x, you find there are 88 boys and 66 girls originally, making 154 students total in the class.

Exam Tip: Always set up the ratio equation carefully with the changed values, and cross-multiply correctly to avoid sign errors when collecting terms.

Question 21. In an examination, the ratio of passes to failures was 4 : 1. If 30 less had appeared and 20 less passed, the ratio of passes to failures would have been 5 : 1. How many students appeared for the examination?
Answer: Let the number of students who passed be 4x and those who failed be x. Then, the total students who appeared = 5x.

In the second situation, students appearing = 5x - 30, students passing = 4x - 20, and students failing = (5x - 30) - (4x - 20) = x - 10.

Given that the new ratio of passes to failures is 5 : 1, we write: \( \frac{4x - 20}{x - 10} = \frac{5}{1} \)

Cross-multiplying: 4x - 20 = 5(x - 10)

\( \implies 4x - 20 = 5x - 50 \)

\( \implies -x = -30 \)

\( \implies x = 30 \)

Wait, let me recalculate. From 4x - 20 = 5x - 50, we get 4x - 5x = -50 + 20, so -x = -30, thus x = 30. But checking: if x = 30, then 5x = 150. Let me verify with the original ratio: passes = 4(30) = 120, fails = 30, ratio = 120:30 = 4:1 ✓. In second case: passes = 120 - 20 = 100, appears = 150 - 30 = 120, fails = 120 - 100 = 20, ratio = 100:20 = 5:1 ✓. So total students appeared = 150.

In simple words: We set up two ratios - one for the original situation and one after removing students. By solving the equation formed from these ratios, we find that 150 students appeared for the exam.

Exam Tip: Always define your variable clearly and write both conditions separately before forming the proportion equation. Cross-multiply carefully and verify your answer by substituting back into both given conditions.

Exercise 6.2

Question 1. Find the value of x in the following proportions:
(i) 10 : 35 = x : 42
(ii) 3 : x = 24 : 2
(iii) 2.5 : 1.5 = x : 3
(iv) x : 50 :: 3 : 2
Answer:
(i) From 10 : 35 = x : 42, we get \( \frac{10}{35} = \frac{x}{42} \)

\( \implies x = \frac{10}{35} \times 42 = \frac{420}{35} = 12 \)

(ii) From 3 : x = 24 : 2, we get \( \frac{3}{x} = \frac{24}{2} \)

\( \implies x = \frac{3 \times 2}{24} = \frac{6}{24} = \frac{1}{4} \)

(iii) From 2.5 : 1.5 = x : 3, we get \( \frac{2.5}{1.5} = \frac{x}{3} \)

\( \implies x = \frac{2.5 \times 3}{1.5} = \frac{7.5}{1.5} = 5 \)

(iv) From x : 50 :: 3 : 2, we get \( \frac{x}{50} = \frac{3}{2} \)

\( \implies x = \frac{3 \times 50}{2} = \frac{150}{2} = 75 \)
In simple words: In each proportion, cross-multiply the known numbers and divide to find x. The product of the outer terms always equals the product of the inner terms.

Exam Tip: Remember the property: if a : b = c : d, then ad = bc. Use this to set up your equation quickly and avoid calculation errors.

Question 2. Find the fourth proportional to:
(i) 3, 12, 15
(ii) \( \frac{1}{3}, \frac{1}{4}, \frac{1}{5} \)
(iii) 1.5, 2.5, 4.5
(iv) 9.6 kg, 7.2 kg, 28.8 kg
Answer:
(i) Let the fourth proportional be x. Then 3, 12, 15 and x are in proportion.

\( \implies 3 : 12 :: 15 : x \)

\( \implies \frac{3}{12} = \frac{15}{x} \)

\( \implies x = \frac{15 \times 12}{3} = \frac{180}{3} = 60 \)

(ii) Let the fourth proportional be x. Then \( \frac{1}{3}, \frac{1}{4}, \frac{1}{5} \) and x are in proportion.

\( \implies \frac{1}{3} : \frac{1}{4} :: \frac{1}{5} : x \)

\( \implies \frac{\frac{1}{3}}{\frac{1}{4}} = \frac{\frac{1}{5}}{x} \)

\( \implies \frac{4}{3} = \frac{1}{5x} \)

\( \implies x = \frac{3}{5 \times 4} = \frac{3}{20} \)

(iii) Let the fourth proportional be x. Then 1.5, 2.5, 4.5 and x are in proportion.

\( \implies 1.5 : 2.5 :: 4.5 : x \)

\( \implies \frac{1.5}{2.5} = \frac{4.5}{x} \)

\( \implies x = \frac{4.5 \times 2.5}{1.5} = \frac{11.25}{1.5} = 7.5 \)

(iv) Let the fourth proportional be x. Then 9.6 kg, 7.2 kg, 28.8 kg and x are in proportion.

\( \implies 9.6 : 7.2 :: 28.8 : x \)

\( \implies \frac{9.6}{7.2} = \frac{28.8}{x} \)

\( \implies x = \frac{28.8 \times 7.2}{9.6} = \frac{207.36}{9.6} = 21.6 \text{ kg} \)
In simple words: To find the fourth proportional, set up the proportion and cross-multiply. The fourth number is always found by multiplying the second and third numbers, then dividing by the first.

Exam Tip: Always write the proportion clearly and check that your ratio simplifies correctly. For decimal or fractional answers, verify by checking the ratio.

Question 3. Find the third proportional to:
(i) 5, 10
(ii) 0.24, 0.6
(iii) Rs 3, Rs 12
(iv) \( 5\frac{1}{4} \) and 7
Answer:
(i) Let the third proportional be x. Then 5, 10, x are in continued proportion.

\( \implies \frac{5}{10} = \frac{10}{x} \)

\( \implies x = \frac{10 \times 10}{5} = \frac{100}{5} = 20 \)

(ii) Let the third proportional be x. Then 0.24, 0.6, x are in continued proportion.

\( \implies \frac{0.24}{0.6} = \frac{0.6}{x} \)

\( \implies x = \frac{0.6 \times 0.6}{0.24} = \frac{0.36}{0.24} = 1.5 \)

(iii) Let the third proportional be x. Then 3, 12, x are in continued proportion.

\( \implies \frac{3}{12} = \frac{12}{x} \)

\( \implies x = \frac{12 \times 12}{3} = \frac{144}{3} = 48 \)

So the third proportional is Rs 48.

(iv) Let the third proportional be x. Then \( 5\frac{1}{4}, 7, x \) are in continued proportion.

Converting: \( 5\frac{1}{4} = \frac{21}{4} \)

\( \implies \frac{\frac{21}{4}}{7} = \frac{7}{x} \)

\( \implies x = \frac{7 \times 7}{\frac{21}{4}} = \frac{49 \times 4}{21} = \frac{196}{21} = \frac{28}{3} = 9\frac{1}{3} \)
In simple words: In continued proportion, the middle number appears twice. Multiply the two numbers and divide by the first to get the third.

Exam Tip: For continued proportion, remember that if a, b, x are in continued proportion, then a : b = b : x, which gives b² = ax. This makes finding x straightforward.

Question 4. Find the mean proportion of:
(i) 5 and 80
(ii) \( \frac{1}{12} \) and \( \frac{1}{75} \)
(iii) 8.1 and 2.5
(iv) (a - b) and (a³ - a²b), a > b
Answer:
(i) Let the mean proportion be x.

\( \therefore \frac{5}{x} = \frac{x}{80} \)

\( \implies x^2 = 5 \times 80 = 400 \)

\( \implies x = \sqrt{400} = 20 \)

(ii) Let the mean proportion be x.

\( \therefore \frac{\frac{1}{12}}{x} = \frac{x}{\frac{1}{75}} \)

\( \implies x^2 = \frac{1}{12} \times \frac{1}{75} = \frac{1}{900} \)

\( \implies x = \sqrt{\frac{1}{900}} = \frac{1}{30} \)

(iii) Let the mean proportion be x.

\( \therefore \frac{8.1}{x} = \frac{x}{2.5} \)

\( \implies x^2 = 8.1 \times 2.5 = 20.25 \)

\( \implies x = \sqrt{20.25} = 4.5 \)

(iv) Let the mean proportion be x.

\( \therefore \frac{(a-b)}{x} = \frac{x}{(a^3 - a^2b)} \)

\( \implies x^2 = (a-b) \times (a^3 - a^2b) \)

\( \implies x^2 = (a-b) \times a^2(a - b) = a^2(a-b)^2 \)

\( \implies x = \sqrt{a^2(a-b)^2} = a(a-b) \)
In simple words: The mean proportion is always the square root of the product of the two numbers. This number sits exactly in the middle of the two given values in a geometric sense.

Exam Tip: For mean proportion, always square-root the product. Watch your signs carefully, especially with algebraic expressions - ensure the final answer makes sense in the context given.

Question 5. If a, 12, 16 and b are in continued proportion, find a and b.
Answer: Since a, 12, 16 and b are in continued proportion:

\( \therefore \frac{a}{12} = \frac{12}{16} = \frac{16}{b} \)

From \( \frac{a}{12} = \frac{12}{16} \):

\( \implies a = \frac{12 \times 12}{16} = \frac{144}{16} = 9 \)

From \( \frac{16}{b} = \frac{12}{16} \):

\( \implies b = \frac{16 \times 16}{12} = \frac{256}{12} = \frac{64}{3} \)

Therefore, a = 9 and b = \( \frac{64}{3} \).
In simple words: When four numbers are in continued proportion, the ratio of the first to the second equals the ratio of the second to the third and the ratio of the third to the fourth. Use this property to find the missing values.

Exam Tip: In continued proportion, each pair of adjacent terms has the same ratio. Set up equations using this property and solve step by step.

Question 6. What number must be added to each of the numbers 5, 11, 19 and 37 so that they are in proportion?
Answer: Let the number to be added be x. Then the new numbers are 5 + x, 11 + x, 19 + x, and 37 + x.

Since these numbers are in proportion:

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

Cross-multiplying: (5 + x)(37 + x) = (19 + x)(11 + x)

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

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

\( \implies 42x - 30x = 209 - 185 \)

\( \implies 12x = 24 \)

\( \implies x = 2 \)

Therefore, the number that must be added is 2.
In simple words: We add the same number to all four values and set up a proportion. By cross-multiplying and solving, we find what number makes them proportional.

Exam Tip: Always expand both sides carefully when cross-multiplying. The x² terms will cancel out, leaving a simple linear equation to solve.

Question 7. What numbers should be subtracted from each of the numbers 23, 30, 57 and 78 so that remainders are in proportion?
Answer: Let the number to be subtracted be x. Then the new numbers are 23 - x, 30 - x, 57 - x, and 78 - x.

Since these numbers are in proportion:

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

Cross-multiplying: (23 - x)(78 - x) = (57 - x)(30 - x)

\( \implies 1794 - 23x - 78x + x^2 = 1710 - 57x - 30x + x^2 \)

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

\( \implies 1794 - 101x = 1710 - 87x \)

\( \implies -101x + 87x = 1710 - 1794 \)

\( \implies -14x = -84 \)

\( \implies x = 6 \)

Therefore, the number that must be subtracted is 6.
In simple words: We subtract the same number from all four values and set up a proportion. Solving the equation formed by cross-multiplying gives us the answer.

Exam Tip: Be careful with negative signs when expanding (a - x)(b - x). Check your arithmetic by verifying that the four remainders actually form a proportion when x = 6.

Question 8. If k + 3, k + 2, 3k - 7 and 2k - 3 are in proportion, find k.
Answer: Since k + 3, k + 2, 3k - 7 and 2k - 3 are in proportion:

\( \therefore \frac{k+3}{k+2} = \frac{3k-7}{2k-3} \)

Cross-multiplying: (k + 3)(2k - 3) = (3k - 7)(k + 2)

\( \implies 2k^2 - 3k + 6k - 9 = 3k^2 + 6k - 7k - 14 \)

\( \implies 2k^2 + 3k - 9 = 3k^2 - k - 14 \)

\( \implies 2k^2 - 3k^2 + 3k + k - 9 + 14 = 0 \)

\( \implies -k^2 + 4k + 5 = 0 \)

Multiplying by -1: \( k^2 - 4k - 5 = 0 \)

\( \implies k^2 - 5k + k - 5 = 0 \)

\( \implies k(k-5) + 1(k-5) = 0 \)

\( \implies (k+1)(k-5) = 0 \)

\( \implies k = -1 \text{ or } k = 5 \)

Therefore, k = -1 or k = 5.
In simple words: We cross-multiply the proportion, expand carefully, and rearrange into a quadratic equation. Factoring gives us two possible values for k.

Exam Tip: When you get a quadratic, always try factoring first by grouping terms. Remember to check both solutions to ensure they don't make any denominator zero in the original proportion.

Question 9. If (x + 5) is the mean proportion between x + 2 and x + 9, find the value of x.
Answer: Since (x + 5) is the mean proportion between x + 2 and x + 9:

\( \therefore \frac{x+2}{x+5} = \frac{x+5}{x+9} \)

\( \implies (x+5)^2 = (x+2)(x+9) \)

\( \implies x^2 + 10x + 25 = x^2 + 9x + 2x + 18 \)

\( \implies x^2 + 10x + 25 = x^2 + 11x + 18 \)

\( \implies 10x - 11x = 18 - 25 \)

\( \implies -x = -7 \)

\( \implies x = 7 \)

Therefore, the value of x is 7.
In simple words: When a number is the mean proportion between two others, its square equals their product. We use this relationship to form an equation and solve for x.

Exam Tip: Always expand (x + a)² as x² + 2ax + a² to avoid sign errors. Check your answer by substituting back to verify the mean proportion relationship.

Question 10. What numbers must be added to each of the numbers 16, 26 and 40 so that the resulting numbers must be in continued proportion?
Answer: Let the number to be added to each number be x. Then the new numbers are 16 + x, 26 + x, and 40 + x.

Since the new numbers are in continued proportion:

\( \therefore \frac{16+x}{26+x} = \frac{26+x}{40+x} \)

Cross-multiplying: (16 + x)(40 + x) = (26 + x)(26 + x)

\( \implies 640 + 16x + 40x + x^2 = 676 + 26x + 26x + x^2 \)

\( \implies 640 + 56x + x^2 = 676 + 52x + x^2 \)

\( \implies 56x - 52x = 676 - 640 \)

\( \implies 4x = 36 \)

\( \implies x = 9 \)

Therefore, the number to be added to each number is 9.
In simple words: When three numbers are in continued proportion, the middle number squared equals the product of the first and third. We use this to set up and solve the equation.

Exam Tip: In continued proportion with three numbers, the right side of the equation is a perfect square (26 + x)². This often simplifies the algebra when you expand and cancel.

Question 11. Find two numbers such that the mean proportional between them is 28 and the third proportional to them is 224.
Answer: Let the two numbers be x and y.

Given that 28 is the mean proportional between x and y:

\( \therefore \frac{x}{28} = \frac{28}{y} \)

\( \implies xy = 28 \times 28 = 784 \)

\( \implies x = \frac{784}{y} \quad \text{...(Eq 1)} \)

Given that 224 is the third proportional to x and y:

\( \therefore \frac{x}{y} = \frac{y}{224} \)

\( \implies y^2 = 224x \)

Substituting the value of x from Eq 1:

\( \implies y^2 = 224 \times \frac{784}{y} \)

\( \implies y^3 = 224 \times 784 = 175616 \)

\( \implies y = \sqrt[3]{175616} = \sqrt[3]{56^3} = 56 \)

From Eq 1: \( x = \frac{784}{56} = 14 \)

Therefore, the two numbers are 14 and 56.
In simple words: We use the mean and third proportional relationships to write two equations. Substituting one into the other gives us a cubic equation which we solve to find both numbers.

Exam Tip: Always check your answer by verifying both conditions: that 28 is the mean proportion of 14 and 56 (since \( \sqrt{14 \times 56} = \sqrt{784} = 28 \)), and that 224 is the third proportional (since \( \frac{14}{56} = \frac{56}{224} \)).

Question 12. If b is the mean proportional between a and c, prove that a, c, a² + b² and b² + c² are proportional.
Answer: Given that b is the mean proportional between a and c:

\( b^2 = ac \quad \text{...(Eq 1)} \)

To prove that a, c, a² + b² and b² + c² are proportional, we need to show:

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

Or equivalently: \( a(b^2 + c^2) = c(a^2 + b^2) \)

Solving L.H.S:

\( a(b^2 + c^2) = a(ac + c^2) \quad \text{[Substituting } b^2 = ac \text{ from Eq 1]} \)

\( = ac(a + c) \)

Solving R.H.S:

\( c(a^2 + b^2) = c(a^2 + ac) \quad \text{[Substituting } b^2 = ac \text{ from Eq 1]} \)

\( = ac(a + c) \)

Since L.H.S = R.H.S, we have proved that a, c, a² + b² and b² + c² are proportional. Hence proved.
In simple words: We substitute the given relationship b² = ac into both sides of the equality we need to prove. When both sides simplify to the same expression, the proportionality is confirmed.

Exam Tip: Always start a proof by writing down what is given and what needs to be proved. Substitute the given condition strategically to simplify the expressions on both sides, and show they are equal.

Question 13. If b is the mean proportional between a and c, prove that (ab + bc) is the mean proportional between (a² + b²) and (b² + c²).
Answer: Given that b is the mean proportional between a and c, we have:
\( b^2 = ac \) [....Eq 1]

To show (ab + bc) is the mean proportional between (a² + b²) and (b² + c²), we need to verify:
\( (ab + bc)^2 = (a^2 + b^2)(b^2 + c^2) \)

Working on the left side:
\( (ab + bc)^2 = a^2b^2 + 2ab^2c + b^2c^2 \)

Substituting \( b^2 = ac \) from Eq 1:
\( = a^2(ac) + 2ac(ac) + c^2(ac) \)
\( = a^3c + 2a^2c^2 + ac^3 \)
\( = ac(a^2 + 2ac + c^2) \)
\( = ac(a + c)^2 \)

Working on the right side:
\( (a^2 + b^2)(b^2 + c^2) = a^2b^2 + a^2c^2 + b^4 + b^2c^2 \)

Substituting \( b^2 = ac \):
\( = a^2(ac) + a^2c^2 + (ac)^2 + (ac)c^2 \)
\( = a^3c + a^2c^2 + a^2c^2 + ac^3 \)
\( = a^3c + 2a^2c^2 + ac^3 \)
\( = ac(a^2 + 2ac + c^2) \)
\( = ac(a + c)^2 \)

Since L.H.S. = R.H.S., the result is proved.
In simple words: When we expand both sides of the equation and replace \( b^2 \) with its equivalent value ac, both the left and right sides become equal to ac(a + c)², confirming the relationship.

Exam Tip: This is a standard proportionality proof - always substitute the given condition early and work systematically through both sides, factoring completely to show equality.

Question 14. If y is the mean proportional between x and z, prove that xyz(x + y + z)³ = (xy + yz + zx)³.
Answer: Given that y is the mean proportional between x and z, we have:
\( y^2 = xz \) [....Eq 1]

We need to prove:
\( xyz(x + y + z)^3 = (xy + yz + zx)^3 \)

Starting with the left side:
\( xyz(x + y + z)^3 = xz \cdot y(x + y + z)^3 \)

Substituting \( xz = y^2 \) from Eq 1:
\( = y^2 \cdot y(x + y + z)^3 \)
\( = y^3(x + y + z)^3 \)
\( = (y(x + y + z))^3 \)
\( = (xy + y^2 + yz)^3 \)

Substituting \( y^2 = xz \) again:
\( = (xy + xz + yz)^3 \)
\( = (xy + yz + zx)^3 \) = R.H.S.

Since L.H.S. = R.H.S., the result is hence proved.
In simple words: By replacing xz with y² and factoring out y³ from both the coefficient and the bracket, the left side transforms into the cube of (xy + yz + zx), which equals the right side.

Exam Tip: Recognize that substituting the mean proportional condition early allows you to express both sides in terms of cubes, making the proof clearer and more direct.

Question 15. If a + c = mb and \( \frac{1}{b} + \frac{1}{d} = \frac{m}{c} \), prove that a, b, c and d are in proportion.
Answer: Given:
\( a + c = mb \) and \( \frac{1}{b} + \frac{1}{d} = \frac{m}{c} \)

From the first condition, dividing by b:
\( \frac{a}{b} + \frac{c}{b} = m \) [....Eq 1]

From the second condition, multiplying by c:
\( \frac{c}{b} + \frac{c}{d} = m \)

Substituting the value of m from Eq 1:
\( \frac{c}{b} + \frac{c}{d} = \frac{a}{b} + \frac{c}{b} \)

Simplifying:
\( \frac{c}{d} = \frac{a}{b} \)

Therefore:
\( \frac{a}{b} = \frac{c}{d} \)

This shows that a, b, c, and d are in proportion.
In simple words: By using the two given equations and substituting values systematically, we arrive at the relationship \( \frac{a}{b} = \frac{c}{d} \), which is the definition of four numbers being in proportion.

Exam Tip: Always manipulate the given conditions carefully - dividing or multiplying by strategic terms helps isolate the ratios you need to prove equality.

Question 16. If \( \frac{x}{a} = \frac{y}{b} = \frac{z}{c} \), prove that
(i) \( \frac{x^3}{a^2} + \frac{y^3}{b^2} + \frac{z^3}{c^2} = \frac{(x + y + z)^3}{(a + b + c)^2} \)
(ii) \( \left(\frac{a^2x^2 + b^2y^2 + c^2z^2}{a^3x + b^3y + c^3z}\right)^3 = \frac{xyz}{abc} \)
(iii) \( \frac{ax - by}{(a + b)(x - y)} + \frac{by - cz}{(b + c)(y - z)} + \frac{cz - ax}{(c + a)(z - x)} = 3 \)
Answer:

(i) Let \( \frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k \)

Therefore: \( x = ak, y = bk, z = ck \)

L.H.S. = \( \frac{x^3}{a^2} + \frac{y^3}{b^2} + \frac{z^3}{c^2} \)
\( = \frac{(ak)^3}{a^2} + \frac{(bk)^3}{b^2} + \frac{(ck)^3}{c^2} \)
\( = \frac{a^3k^3}{a^2} + \frac{b^3k^3}{b^2} + \frac{c^3k^3}{c^2} \)
\( = ak^3 + bk^3 + ck^3 \)
\( = k^3(a + b + c) \)

R.H.S. = \( \frac{(x + y + z)^3}{(a + b + c)^2} \)
\( = \frac{(ak + bk + ck)^3}{(a + b + c)^2} \)
\( = \frac{k^3(a + b + c)^3}{(a + b + c)^2} \)
\( = k^3(a + b + c) \)

Since L.H.S. = R.H.S., hence proved that \( \frac{x^3}{a^2} + \frac{y^3}{b^2} + \frac{z^3}{c^2} = \frac{(x + y + z)^3}{(a + b + c)^2} \).

(ii) Let \( \frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k \)

Therefore: \( x = ak, y = bk, z = ck \)

L.H.S. = \( \left(\frac{a^2x^2 + b^2y^2 + c^2z^2}{a^3x + b^3y + c^3z}\right)^3 \)
\( = \left(\frac{a^2(ak)^2 + b^2(bk)^2 + c^2(ck)^2}{a^3(ak) + b^3(bk) + c^3(ck)}\right)^3 \)
\( = \left(\frac{a^4k^2 + b^4k^2 + c^4k^2}{a^4k + b^4k + c^4k}\right)^3 \)
\( = \left(\frac{k^2(a^4 + b^4 + c^4)}{k(a^4 + b^4 + c^4)}\right)^3 \)
\( = (k)^3 = k^3 \)
\( = k \times k \times k \)
\( = \frac{x}{a} \times \frac{y}{b} \times \frac{z}{c} \)
\( = \frac{xyz}{abc} \) = R.H.S.

Hence proved that \( \left(\frac{a^2x^2 + b^2y^2 + c^2z^2}{a^3x + b^3y + c^3z}\right)^3 = \frac{xyz}{abc} \).

(iii) Let \( \frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k \)

Therefore: \( x = ak, y = bk, z = ck \)

L.H.S. = \( \frac{ax - by}{(a + b)(x - y)} + \frac{by - cz}{(b + c)(y - z)} + \frac{cz - ax}{(c + a)(z - x)} \)

\( = \frac{a(ak) - b(bk)}{(a + b)((ak) - (bk))} + \frac{b(bk) - c(ck)}{(b + c)((bk) - (ck))} + \frac{c(ck) - a(ak)}{(c + a)((ck) - (ak))} \)

\( = \frac{a^2k - b^2k}{(a + b)(ak - bk)} + \frac{b^2k - c^2k}{(b + c)(bk - ck)} + \frac{c^2k - a^2k}{(c + a)(ck - ak)} \)

\( = \frac{k(a^2 - b^2)}{k(a + b)(a - b)} + \frac{k(b^2 - c^2)}{k(b + c)(b - c)} + \frac{k(c^2 - a^2)}{k(c + a)(c - a)} \)

\( = \frac{(a - b)(a + b)}{(a + b)(a - b)} + \frac{(b - c)(b + c)}{(b + c)(b - c)} + \frac{(c - a)(c + a)}{(c + a)(c - a)} \)

\( = 1 + 1 + 1 = 3 \) = R.H.S.

Hence proved that \( \frac{ax - by}{(a + b)(x - y)} + \frac{by - cz}{(b + c)(y - z)} + \frac{cz - ax}{(c + a)(z - x)} = 3 \).
In simple words: Using the common ratio k, we replace x, y, z with their equivalents ak, bk, ck. This substitution allows each complex fraction to simplify to 1, making the sum equal to 3.

Exam Tip: Always introduce the common ratio k when given equal ratios - this technique transforms complicated algebraic expressions into manageable forms that reveal hidden cancellations.

Question 17. If \( \frac{a}{b} = \frac{c}{d} = \frac{e}{f} \), prove that
(i) \( (b^2 + d^2 + f^2)(a^2 + c^2 + e^2) = (ab + cd + ef)^2 \)
(ii) \( \frac{(a^3 + c^3)^2}{(b^3 + d^3)^2} = \frac{e^6}{f^6} \)
(iii) \( \frac{a^2}{b^2} + \frac{c^2}{d^2} + \frac{e^2}{f^2} = \frac{ac}{bd} + \frac{ce}{df} + \frac{ae}{bf} \)
(iv) \( bdf\left(\frac{a + b}{b} + \frac{c + d}{d} + \frac{e + f}{f}\right)^3 = 27(a + b)(c + d)(e + f) \)
Answer:

(i) Let \( \frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k \)

Therefore: \( a = bk, c = dk, e = fk \)

L.H.S. = \( (b^2 + d^2 + f^2)(a^2 + c^2 + e^2) \)
\( = (b^2 + d^2 + f^2)((bk)^2 + (dk)^2 + (fk)^2) \)
\( = (b^2 + d^2 + f^2)(b^2k^2 + d^2k^2 + f^2k^2) \)
\( = k^2(b^2 + d^2 + f^2)(b^2 + d^2 + f^2) \)
\( = k^2(b^2 + d^2 + f^2)^2 \)

R.H.S. = \( (ab + cd + ef)^2 \)
\( = ((bk) \cdot b + (dk) \cdot d + (fk) \cdot f)^2 \)
\( = (b^2k + d^2k + f^2k)^2 \)
\( = (k(b^2 + d^2 + f^2))^2 \)
\( = k^2(b^2 + d^2 + f^2)^2 \)

Since L.H.S. = R.H.S., hence proved that \( (b^2 + d^2 + f^2)(a^2 + c^2 + e^2) = (ab + cd + ef)^2 \).

(ii) Let \( \frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k \)

Therefore: \( a = bk, c = dk, e = fk \)

L.H.S. = \( \frac{(a^3 + c^3)^2}{(b^3 + d^3)^2} \)
\( = \frac{((bk)^3 + (dk)^3)^2}{(b^3 + d^3)^2} \)
\( = \frac{(b^3k^3 + d^3k^3)^2}{(b^3 + d^3)^2} \)
\( = \frac{(k^3(b^3 + d^3))^2}{(b^3 + d^3)^2} \)
\( = \frac{k^6(b^3 + d^3)^2}{(b^3 + d^3)^2} \)
\( = k^6 \)

R.H.S. = \( \frac{e^6}{f^6} \)
\( = \frac{(fk)^6}{f^6} \)
\( = \frac{f^6k^6}{f^6} \)
\( = k^6 \)

Since L.H.S. = R.H.S., hence proved.

(iii) Let \( \frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k \)

Therefore: \( a = bk, c = dk, e = fk \)

L.H.S. = \( \frac{a^2}{b^2} + \frac{c^2}{d^2} + \frac{e^2}{f^2} \)
\( = \frac{(bk)^2}{b^2} + \frac{(dk)^2}{d^2} + \frac{(fk)^2}{f^2} \)
\( = \frac{b^2k^2}{b^2} + \frac{d^2k^2}{d^2} + \frac{f^2k^2}{f^2} \)
\( = k^2 + k^2 + k^2 = 3k^2 \)

R.H.S. = \( \frac{ac}{bd} + \frac{ce}{df} + \frac{ae}{bf} \)
\( = \frac{(bk)(dk)}{bd} + \frac{(dk)(fk)}{df} + \frac{(bk)(fk)}{bf} \)
\( = \frac{bdk^2}{bd} + \frac{dfk^2}{df} + \frac{bfk^2}{bf} \)
\( = k^2 + k^2 + k^2 = 3k^2 \)

Since L.H.S. = R.H.S., hence proved.

(iv) Let \( \frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k \)

Therefore: \( a = bk, c = dk, e = fk \)

L.H.S. = \( bdf\left(\frac{a + b}{b} + \frac{c + d}{d} + \frac{e + f}{f}\right)^3 \)
\( = bdf\left(\frac{bk + b}{b} + \frac{dk + d}{d} + \frac{fk + f}{f}\right)^3 \)
\( = bdf(k + 1 + k + 1 + k + 1)^3 \)
\( = bdf(3k + 3)^3 \)
\( = bdf(3)^3(k + 1)^3 \)
\( = 27bdf(k + 1)^3 \)

R.H.S. = \( 27(a + b)(c + d)(e + f) \)
\( = 27(bk + b)(dk + d)(fk + f) \)
\( = 27b(k + 1) \cdot d(k + 1) \cdot f(k + 1) \)
\( = 27bdf(k + 1)^3 \)

Since L.H.S. = R.H.S., hence proved.
In simple words: By setting all three ratios equal to a common constant k, we express each numerator in terms of its denominator. Substituting these expressions simplifies each complex statement to an identity.

Exam Tip: The common ratio method is powerful for proportion problems - once you set k equal to all ratios, subsequent algebra becomes straightforward manipulation without needing clever insights.

Question 18. If ax = by = cz, prove that \( \frac{x^2}{yz} + \frac{y^2}{zx} + \frac{z^2}{xy} = \frac{bc}{a^2} + \frac{ca}{b^2} + \frac{ab}{c^2} \).
Answer: Let \( ax = by = cz = k \)

Then: \( x = \frac{k}{a}, y = \frac{k}{b}, z = \frac{k}{c} \)

L.H.S. = \( \frac{x^2}{yz} + \frac{y^2}{zx} + \frac{z^2}{xy} \)

\( = \frac{\left(\frac{k}{a}\right)^2}{\frac{k}{b} \times \frac{k}{c}} + \frac{\left(\frac{k}{b}\right)^2}{\frac{k}{c} \times \frac{k}{a}} + \frac{\left(\frac{k}{c}\right)^2}{\frac{k}{a} \times \frac{k}{b}} \)

\( = \frac{\frac{k^2}{a^2}}{\frac{k^2}{bc}} + \frac{\frac{k^2}{b^2}}{\frac{k^2}{ac}} + \frac{\frac{k^2}{c^2}}{\frac{k^2}{ab}} \)

\( = \frac{k^2 \times bc}{a^2 \times k^2} + \frac{k^2 \times ac}{b^2 \times k^2} + \frac{k^2 \times ab}{c^2 \times k^2} \)

\( = \frac{bc}{a^2} + \frac{ac}{b^2} + \frac{ab}{c^2} = \frac{bc}{a^2} + \frac{ca}{b^2} + \frac{ab}{c^2} \) = R.H.S.

Since L.H.S. = R.H.S., hence proved.
In simple words: By expressing x, y, and z in terms of k and the given constants, the fractions simplify to show both sides are identical.

Exam Tip: When given equations like ax = by = cz, setting this common value equal to k allows you to express variables as fractions - this method often reveals hidden symmetries in algebraic expressions.

Question 19. If a, b, c, d are in proportion, prove that:
(i) \( \frac{(5a + 7b)(2c - 3d)}{(5c + 7d)(2a - 3b)} = 1 \)
(ii) \( (ma + nb) : b = (mc + nd) : d \)
(iii) \( (a^4 + c^4) : (b^4 + d^4) = a^2c^2 : b^2d^2 \)
(iv) \( \frac{a^2 + ab}{c^2 + cd} = \frac{b^2 - 2ab}{d^2 - 2cd} \)
(v) \( \frac{(a + c)^3}{(b + d)^3} = \frac{a(a - c)^2}{b(b - d)^2} \)
(vi) \( \frac{a^2 + ab + b^2}{a^2 - ab + b^2} = \frac{c^2 + cd + d^2}{c^2 - cd + d^2} \)
(vii) \( \frac{a^2 + b^2}{c^2 + d^2} = \frac{ab + ad - bc}{bc + cd - ad} \)
(viii) \( abcd\left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{d^2}\right) = a^2 + b^2 + c^2 + d^2 \)
Answer:

(i) Since a, b, c, d are in proportion:
\( \frac{a}{b} = \frac{c}{d} = k \)

Therefore: \( a = bk, c = dk \)

L.H.S. = \( (5a + 7b)(2c - 3d) \)
\( = (5bk + 7b)(2dk - 3d) \)
\( = b(5k + 7) \cdot d(2k - 3) \)
\( = bd(5k + 7)(2k - 3) \)

R.H.S. = \( (5c + 7d)(2a - 3b) \)
\( = (5dk + 7d)(2bk - 3b) \)
\( = d(5k + 7) \cdot b(2k - 3) \)
\( = bd(5k + 7)(2k - 3) \)

Since L.H.S. = R.H.S., hence proved.

(ii) Since a, b, c, d are in proportion:
\( \frac{a}{b} = \frac{c}{d} = k \)

Therefore: \( a = bk, c = dk \)

L.H.S. = \( (ma + nb) : b \)
\( = \frac{ma + nb}{b} \)
\( = \frac{m(bk) + nb}{b} \)
\( = mk + n \)

R.H.S. = \( (mc + nd) : d \)
\( = \frac{mc + nd}{d} \)
\( = \frac{m(dk) + nd}{d} \)
\( = mk + n \)

Since L.H.S. = R.H.S., hence proved.

(iii) Since a, b, c, d are in proportion:
\( \frac{a}{b} = \frac{c}{d} = k \)

Therefore: \( a = bk, c = dk \)

L.H.S. = \( (a^4 + c^4) : (b^4 + d^4) \)
\( = \frac{a^4 + c^4}{b^4 + d^4} \)
\( = \frac{(bk)^4 + (dk)^4}{b^4 + d^4} \)
\( = \frac{k^4b^4 + k^4d^4}{b^4 + d^4} \)
\( = k^4 \)

R.H.S. = \( a^2c^2 : b^2d^2 \)
\( = \frac{a^2c^2}{b^2d^2} \)
\( = \frac{(bk)^2(dk)^2}{b^2d^2} \)
\( = \frac{k^4b^2d^2}{b^2d^2} \)
\( = k^4 \)

Since L.H.S. = R.H.S., hence proved.

(iv) Since a, b, c, d are in proportion:
\( \frac{a}{b} = \frac{c}{d} = k \)

Therefore: \( a = bk, c = dk \)

L.H.S. = \( \frac{a^2 + ab}{c^2 + cd} \)
\( = \frac{(bk)^2 + (bk)b}{(dk)^2 + (dk)d} \)
\( = \frac{b^2k^2 + b^2k}{d^2k^2 + d^2k} \)
\( = \frac{b^2k(k + 1)}{d^2k(k + 1)} \)
\( = \frac{b^2}{d^2} \)

R.H.S. = \( \frac{b^2 - 2ab}{d^2 - 2cd} \)
\( = \frac{b^2 - 2(bk)b}{d^2 - 2(dk)d} \)
\( = \frac{b^2 - 2b^2k}{d^2 - 2d^2k} \)
\( = \frac{b^2(1 - 2k)}{d^2(1 - 2k)} \)
\( = \frac{b^2}{d^2} \)

Since L.H.S. = R.H.S., hence proved.

(v) Since a, b, c, d are in proportion:
\( \frac{a}{b} = \frac{c}{d} = k \)

Therefore: \( a = bk, c = dk \)

L.H.S. = \( \frac{(a + c)^3}{(b + d)^3} \)
\( = \frac{(bk + dk)^3}{(b + d)^3} \)
\( = \frac{(k(b + d))^3}{(b + d)^3} \)
\( = k^3 \)

R.H.S. = \( \frac{a(a - c)^2}{b(b - d)^2} \)
\( = \frac{(bk)(bk - dk)^2}{b(b - d)^2} \)
\( = \frac{(bk)(k(b - d))^2}{b(b - d)^2} \)
\( = \frac{bk \cdot k^2(b - d)^2}{b(b - d)^2} \)
\( = k^3 \)

Since L.H.S. = R.H.S., hence proved.

(vi) Since a, b, c, d are in proportion:
\( \frac{a}{b} = \frac{c}{d} = k \)

Therefore: \( a = bk, c = dk \)

L.H.S. = \( \frac{a^2 + ab + b^2}{a^2 - ab + b^2} \)
\( = \frac{(bk)^2 + (bk)b + b^2}{(bk)^2 - (bk)b + b^2} \)
\( = \frac{b^2k^2 + b^2k + b^2}{b^2k^2 - b^2k + b^2} \)
\( = \frac{b^2(k^2 + k + 1)}{b^2(k^2 - k + 1)} \)
\( = \frac{k^2 + k + 1}{k^2 - k + 1} \)

R.H.S. = \( \frac{c^2 + cd + d^2}{c^2 - cd + d^2} \)
\( = \frac{(dk)^2 + (dk)d + d^2}{(dk)^2 - (dk)d + d^2} \)
\( = \frac{d^2(k^2 + k + 1)}{d^2(k^2 - k + 1)} \)
\( = \frac{k^2 + k + 1}{k^2 - k + 1} \)

Since L.H.S. = R.H.S., hence proved.

(vii) Since a, b, c, d are in proportion:
\( \frac{a}{b} = \frac{c}{d} = k \)

Therefore: \( a = bk, c = dk \)

L.H.S. = \( \frac{a^2 + b^2}{c^2 + d^2} \)
\( = \frac{(bk)^2 + b^2}{(dk)^2 + d^2} \)
\( = \frac{b^2k^2 + b^2}{d^2k^2 + d^2} \)
\( = \frac{b^2(k^2 + 1)}{d^2(k^2 + 1)} \)
\( = \frac{b^2}{d^2} \)

R.H.S. = \( \frac{ab + ad - bc}{bc + cd - ad} \)
\( = \frac{(bk)b + (bk)d - b(dk)}{b(dk) + (dk)d - (bk)d} \)
\( = \frac{b^2k + bkd - bdk}{bdk + d^2k - bkd} \)
\( = \frac{b^2k + bkd - bdk}{bdk + d^2k - bkd} \)
\( = \frac{b^2k}{d^2k} \)
\( = \frac{b^2}{d^2} \)

Since L.H.S. = R.H.S., hence proved.

(viii) Since a, b, c, d are in proportion:
\( \frac{a}{b} = \frac{c}{d} = k \)

Therefore: \( a = bk, c = dk \)

L.H.S. = \( abcd\left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{d^2}\right) \)
\( = (bk) \cdot b \cdot (dk) \cdot d \left(\frac{1}{(bk)^2} + \frac{1}{b^2} + \frac{1}{(dk)^2} + \frac{1}{d^2}\right) \)
\( = b^2d^2k^2 \left(\frac{1}{b^2k^2} + \frac{1}{b^2} + \frac{1}{d^2k^2} + \frac{1}{d^2}\right) \)
\( = b^2d^2k^2 \cdot \frac{1}{b^2k^2} + b^2d^2k^2 \cdot \frac{1}{b^2} + b^2d^2k^2 \cdot \frac{1}{d^2k^2} + b^2d^2k^2 \cdot \frac{1}{d^2} \)
\( = d^2 + d^2k^2 + b^2 + b^2k^2 \)
\( = b^2(1 + k^2) + d^2(1 + k^2) \)
\( = (b^2 + d^2)(1 + k^2) \)

Let me recalculate this more carefully:
L.H.S. = \( abcd\left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{d^2}\right) \)
\( = \frac{abcd}{a^2} + \frac{abcd}{b^2} + \frac{abcd}{c^2} + \frac{abcd}{d^2} \)
\( = \frac{bcd}{a} + \frac{acd}{b} + \frac{abd}{c} + \frac{abc}{d} \)
\( = \frac{bd}{k} + \frac{kd}{1} + \frac{kb}{1} + \frac{b^2}{k} \)
\( = b^2 + d^2 + b^2k^2 + d^2k^2 \)
\( = b^2(1 + k^2) + d^2(1 + k^2) \)

R.H.S. = \( a^2 + b^2 + c^2 + d^2 \)
\( = (bk)^2 + b^2 + (dk)^2 + d^2 \)
\( = b^2k^2 + b^2 + d^2k^2 + d^2 \)
\( = b^2(1 + k^2) + d^2(1 + k^2) \)
\( = (b^2 + d^2)(1 + k^2) \)

Since L.H.S. = R.H.S., hence proved.
In simple words: When numbers are in proportion, setting their common ratio equal to k allows us to rewrite each number as a multiple of one of them. Substituting these expressions into even the most complex algebraic relationships shows they reduce to identical forms.

Exam Tip: For proportion problems with multiple parts, always start by identifying the common ratio k - this single step unlocks most algebraic manipulations that follow, making even eight different proof statements manageable using the same substitution technique.

Question 20. If x, y, z are in continued proportion, prove that: \( \frac{(x + y)^2}{(y + z)^2} = \frac{x}{z} \)
Answer: Since x, y, z are in continued proportion:
\( \therefore \frac{x}{y} = \frac{y}{z} = k \)
\( \implies y = zk \text{ and } x = yk = zk^2 \)
L.H.S. \( = \frac{(x + y)^2}{(y + z)^2} = \frac{(zk^2 + zk)^2}{(zk + z)^2} = \frac{z^2k^2(k + 1)^2}{z^2(k + 1)^2} = k^2 \)
R.H.S. \( = \frac{x}{z} = \frac{zk^2}{z} = k^2 \)
Since, L.H.S. = R.H.S., hence proved.
In simple words: When three numbers are in continued proportion, you can express the first and second in terms of the third and a constant k. Substitute these into the given expression and simplify to get the same result on both sides.

Exam Tip: Always establish the relationship between variables using the continued proportion condition first, then substitute into L.H.S. and R.H.S. separately to avoid algebraic errors.

Question 21. If a, b, c are in continued proportion, prove that: \( \frac{pa^2 + qab + rb^2}{pb^2 + qbc + rc^2} = \frac{a}{c} \)
Answer: Since a, b, c are in continued proportion:
\( \therefore \frac{a}{b} = \frac{b}{c} = k \)
\( \implies b = ck \text{ and } a = bk = ck^2 \)
L.H.S. \( = \frac{pa^2 + qab + rb^2}{pb^2 + qbc + rc^2} = \frac{p(ck^2)^2 + q(ck^2)(ck) + r(ck)^2}{p(ck)^2 + q(ck)c + rc^2} \)
\( = \frac{pc^2k^4 + qc^2k^3 + rc^2k^2}{pc^2k^2 + qc^2k + rc^2} = \frac{c^2k^2(pk^2 + qk + r)}{c^2(pk^2 + qk + r)} = k^2 \)
R.H.S. \( = \frac{a}{c} = \frac{ck^2}{c} = k^2 \)
Since, L.H.S. = R.H.S., hence proved that \( \frac{pa^2 + qab + rb^2}{pb^2 + qbc + rc^2} = \frac{a}{c} \)
In simple words: Express a and b in terms of c and k using the continued proportion. Substitute these into both sides of the equation. The same factor cancels out from numerator and denominator on the left side, leaving the same result as the right side.

Exam Tip: Factor out common terms from numerator and denominator carefully. The key is recognizing that the coefficients p, q, r combine with the powers of k to produce a factorable expression.

Question 22. If a, b, c are in continued proportion prove that:
(i) \( \frac{a + b}{b + c} = \frac{a^2(b - c)}{b^2(a - b)} \)
(ii) \( \frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3} = \frac{a}{b^2c^2} + \frac{b}{c^2a^2} + \frac{c}{a^2b^2} \)
(iii) \( a : c = (a^2 + b^2) : (b^2 + c^2) \)
(iv) \( a^2b^2c^2(a^{-4} + b^{-4} + c^{-4}) = b^{-2}(a^4 + b^4 + c^4) \)
(v) \( abc(a + b + c)^3 = (ab + bc + ca)^3 \)
(vi) \( (a + b + c)(a - b + c) = a^2 + b^2 + c^2 \)
Answer:
(i) Since a, b, c are in continued proportion:
\( \therefore \frac{a}{b} = \frac{b}{c} = k \implies b = ck \text{ and } a = bk = ck^2 \)
L.H.S. \( = \frac{a + b}{b + c} = \frac{ck^2 + ck}{ck + c} = \frac{c(k^2 + k)}{c(k + 1)} = \frac{ck(k + 1)}{c(k + 1)} = k \)
R.H.S. \( = \frac{a^2(b - c)}{b^2(a - b)} = \frac{(ck^2)^2(ck - c)}{(ck)^2(ck^2 - ck)} = \frac{c^2k^4 \cdot c(k - 1)}{c^2k^2 \cdot ck(k - 1)} = \frac{c^3k^4(k - 1)}{c^3k^3(k - 1)} = k \)
Since, L.H.S. = R.H.S., hence proved that \( \frac{a + b}{b + c} = \frac{a^2(b - c)}{b^2(a - b)} \)
(ii) Since a, b, c are in continued proportion:
\( \therefore \frac{a}{b} = \frac{b}{c} = k \implies b = ck \text{ and } a = bk = ck^2 \)
L.H.S. \( = \frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3} = \frac{1}{(ck^2)^3} + \frac{1}{(ck)^3} + \frac{1}{c^3} = \frac{1}{c^3k^6} + \frac{1}{c^3k^3} + \frac{1}{c^3} \)
\( = \frac{1}{c^3}\left(\frac{1}{k^6} + \frac{1}{k^3} + 1\right) \)
R.H.S. \( = \frac{a}{b^2c^2} + \frac{b}{c^2a^2} + \frac{c}{a^2b^2} = \frac{ck^2}{(ck)^2c^2} + \frac{ck}{c^2(ck^2)^2} + \frac{c}{(ck^2)^2(ck)^2} \)
\( = \frac{ck^2}{c^3k^2} + \frac{ck}{c^3k^4} + \frac{c}{c^4k^6} = \frac{1}{c^3} + \frac{1}{c^3k^3} + \frac{1}{c^3k^6} = \frac{1}{c^3}\left(1 + \frac{1}{k^3} + \frac{1}{k^6}\right) \)
Since, L.H.S. = R.H.S., hence proved that \( \frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3} = \frac{a}{b^2c^2} + \frac{b}{c^2a^2} + \frac{c}{a^2b^2} \)
(iii) Since a, b, c are in continued proportion:
\( \therefore \frac{a}{b} = \frac{b}{c} = k \implies b = ck \text{ and } a = bk = ck^2 \)
L.H.S. \( = a : c = \frac{a}{c} = \frac{ck^2}{c} = k^2 \)
R.H.S. \( = (a^2 + b^2) : (b^2 + c^2) = \frac{a^2 + b^2}{b^2 + c^2} = \frac{(ck^2)^2 + (ck)^2}{(ck)^2 + c^2} = \frac{c^2k^4 + c^2k^2}{c^2k^2 + c^2} \)
\( = \frac{c^2(k^4 + k^2)}{c^2(k^2 + 1)} = \frac{k^2(k^2 + 1)}{k^2 + 1} = k^2 \)
Since, L.H.S. = R.H.S., hence proved that \( a : c = (a^2 + b^2) : (b^2 + c^2) \)
(iv) Since a, b, c are in continued proportion:
\( \therefore \frac{a}{b} = \frac{b}{c} = k \implies b = ck \text{ and } a = bk = ck^2 \)
L.H.S. \( = a^2b^2c^2(a^{-4} + b^{-4} + c^{-4}) = (ck^2)^2(ck)^2c^2\left((ck^2)^{-4} + (ck)^{-4} + c^{-4}\right) \)
\( = c^2k^4 \cdot c^2k^2 \cdot c^2 \cdot (c^{-4}k^{-8} + c^{-4}k^{-4} + c^{-4}) = c^6k^6 \cdot c^{-4}(k^{-8} + k^{-4} + 1) \)
\( = c^2k^6\left(\frac{1}{k^8} + \frac{1}{k^4} + 1\right) = c^2k^6 \cdot \frac{1 + k^4 + k^8}{k^8} = \frac{c^2(1 + k^4 + k^8)}{k^2} \)
R.H.S. \( = b^{-2}(a^4 + b^4 + c^4) = (ck)^{-2}((ck^2)^4 + (ck)^4 + c^4) = c^{-2}k^{-2}(c^4k^8 + c^4k^4 + c^4) \)
\( = c^{-2}k^{-2} \cdot c^4(k^8 + k^4 + 1) = c^2k^{-2}(k^8 + k^4 + 1) = \frac{c^2(1 + k^4 + k^8)}{k^2} \)
Since, L.H.S. = R.H.S., hence proved that \( a^2b^2c^2(a^{-4} + b^{-4} + c^{-4}) = b^{-2}(a^4 + b^4 + c^4) \)
(v) Since a, b, c are in continued proportion:
\( \therefore \frac{a}{b} = \frac{b}{c} = k \implies b = ck \text{ and } a = bk = ck^2 \)
L.H.S. \( = abc(a + b + c)^3 = (ck^2)(ck)c(ck^2 + ck + c)^3 = c^3k^3 \cdot c^3(k^2 + k + 1)^3 = c^6k^3(k^2 + k + 1)^3 \)
R.H.S. \( = (ab + bc + ca)^3 = ((ck^2)(ck) + (ck)c + c(ck^2))^3 = (c^2k^3 + c^2k + c^2k^2)^3 \)
\( = (c^2k)^3(k^2 + 1 + k)^3 = c^6k^3(k^2 + k + 1)^3 \)
Since, L.H.S. = R.H.S., hence proved that \( abc(a + b + c)^3 = (ab + bc + ca)^3 \)
(vi) Since a, b, c are in continued proportion:
\( \therefore \frac{a}{b} = \frac{b}{c} = k \implies b = ck \text{ and } a = bk = ck^2 \)
L.H.S. \( = (a + b + c)(a - b + c) = (ck^2 + ck + c)(ck^2 - ck + c) = c(k^2 + k + 1) \cdot c(k^2 - k + 1) \)
\( = c^2(k^2 + k + 1)(k^2 - k + 1) = c^2(k^4 - k^3 + k^2 + k^3 - k^2 + k + k^2 - k + 1) = c^2(k^4 + k^2 + 1) \)
R.H.S. \( = a^2 + b^2 + c^2 = (ck^2)^2 + (ck)^2 + c^2 = c^2k^4 + c^2k^2 + c^2 = c^2(k^4 + k^2 + 1) \)
Since, L.H.S. = R.H.S., hence proved that \( (a + b + c)(a - b + c) = a^2 + b^2 + c^2 \)
In simple words: In each part, express a and b in terms of c and k. Substitute into both L.H.S. and R.H.S., simplify by factoring out common powers of c and k, and show both sides are identical.

Exam Tip: For multi-part identity proofs, substitute the continued proportion relationships consistently in each part. Factor out common terms systematically to reduce the algebraic complexity at each step.

Question 23. If a, b, c, d are in continued proportion prove that:
(i) \( \frac{a^3 + b^3 + c^3}{b^3 + c^3 + d^3} = \frac{a}{d} \)
(ii) \( (a^2 - b^2)(c^2 - d^2) = (b^2 - c^2)^2 \)
(iii) \( (a + d)(b + c) - (a + c)(b + d) = (b - c)^2 \)
(iv) \( a : d = \text{triplicate ratio of } (a - b) : (b - c) \)
(v) \( \left(\frac{a - b}{c} + \frac{a - c}{b}\right)^2 - \left(\frac{d - b}{f} + \frac{d - c}{b}\right)^2 = (a - d)^2\left(\frac{1}{c^2} - \frac{1}{b^2}\right) \)
Answer:
(i) Since a, b, c, d are in continued proportion:
\( \therefore \frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k \)
\( \implies c = dk, b = ck = (dk)k = dk^2 \text{ and } a = bk = (ck)k = (dk)k^2 = dk^3 \)
L.H.S. \( = \frac{a^3 + b^3 + c^3}{b^3 + c^3 + d^3} = \frac{(dk^3)^3 + (dk^2)^3 + (dk)^3}{(dk^2)^3 + (dk)^3 + d^3} = \frac{d^3k^9 + d^3k^6 + d^3k^3}{d^3k^6 + d^3k^3 + d^3} \)
\( = \frac{d^3k^3(k^6 + k^3 + 1)}{d^3(k^6 + k^3 + 1)} = k^3 \)
R.H.S. \( = \frac{a}{d} = \frac{dk^3}{d} = k^3 \)
Since, L.H.S. = R.H.S., hence proved that \( \frac{a^3 + b^3 + c^3}{b^3 + c^3 + d^3} = \frac{a}{d} \)
(ii) Since a, b, c, d are in continued proportion:
\( \therefore \frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k \)
\( \implies c = dk, b = ck = (dk)k = dk^2 \text{ and } a = bk = (ck)k = (dk)k^2 = dk^3 \)
L.H.S. \( = (a^2 - b^2)(c^2 - d^2) = [(dk^3)^2 - (dk^2)^2][(dk)^2 - d^2] = (d^2k^6 - d^2k^4)(d^2k^2 - d^2) \)
\( = d^2k^4(k^2 - 1) \cdot d^2(k^2 - 1) = d^4k^4(k^2 - 1)^2 \)
R.H.S. \( = (b^2 - c^2)^2 = [(dk^2)^2 - (dk)^2]^2 = (d^2k^4 - d^2k^2)^2 = [d^2k^2(k^2 - 1)]^2 = d^4k^4(k^2 - 1)^2 \)
Since, L.H.S. = R.H.S., hence proved that \( (a^2 - b^2)(c^2 - d^2) = (b^2 - c^2)^2 \)
(iii) Since a, b, c, d are in continued proportion:
\( \therefore \frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k \)
\( \implies c = dk, b = ck = (dk)k = dk^2 \text{ and } a = bk = (ck)k = (dk)k^2 = dk^3 \)
L.H.S. \( = (a + d)(b + c) - (a + c)(b + d) = (dk^3 + d)(dk^2 + dk) - (dk^3 + dk)(dk^2 + d) \)
\( = d(k^3 + 1) \cdot dk(k + 1) - dk(k^2 + 1) \cdot d(k^2 + 1) = d^2k[(k^3 + 1)(k + 1) - (k^2 + 1)^2] \)
\( = d^2k[(k^4 + k^3 + k + 1) - (k^4 + 1 + 2k^2)] = d^2k(k^3 - 2k^2 + k) = d^2k \cdot k(k^2 - 2k + 1) \)
\( = d^2k^2(k - 1)^2 \)
R.H.S. \( = (b - c)^2 = (dk^2 - dk)^2 = [dk(k - 1)]^2 = d^2k^2(k - 1)^2 \)
Since, L.H.S. = R.H.S., hence proved that \( (a + d)(b + c) - (a + c)(b + d) = (b - c)^2 \)
(iv) Since a, b, c, d are in continued proportion:
\( \therefore \frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k \)
\( \implies c = dk, b = ck = (dk)k = dk^2 \text{ and } a = bk = (ck)k = (dk)k^2 = dk^3 \)
L.H.S. \( = a : d = \frac{a}{d} = \frac{dk^3}{d} = k^3 \)
R.H.S. \( = \text{triplicate ratio of } (a - b) : (b - c) = (a - b)^3 : (b - c)^3 = \frac{(a - b)^3}{(b - c)^3} \)
\( = \frac{(dk^3 - dk^2)^3}{(dk^2 - dk)^3} = \frac{[k(dk^2 - dk)]^3}{(dk^2 - dk)^3} = \frac{k^3(dk^2 - dk)^3}{(dk^2 - dk)^3} = k^3 \)
Since, L.H.S. = R.H.S., hence proved that \( a : d = \text{triplicate ratio of } (a - b) : (b - c) \)
(v) Since a, b, c, d are in continued proportion:
\( \therefore \frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k \)
\( \implies c = dk, b = ck = (dk)k = dk^2 \text{ and } a = bk = (ck)k = (dk)k^2 = dk^3 \)
L.H.S. \( = \left(\frac{a - b}{c} + \frac{a - c}{b}\right)^2 - \left(\frac{d - b}{c} + \frac{d - c}{b}\right)^2 \)
\( = \left(\frac{dk^3 - dk^2}{dk} + \frac{dk^3 - dk}{dk^2}\right)^2 - \left(\frac{d - dk^2}{dk} + \frac{d - dk}{dk^2}\right)^2 \)
\( = \left(\frac{dk(k^2 - k) + dk^3 - dk}{dk^2}\right)^2 - \left(\frac{d(1 - k^2) + d - dk}{dk^2}\right)^2 \)
\( = \left(\frac{dk^4 - dk}{dk^2}\right)^2 - \left(\frac{d - dk^3}{dk^2}\right)^2 = \left(\frac{dk(k^3 - 1)}{dk^2}\right)^2 - \left(\frac{d(1 - k^3)}{dk^2}\right)^2 \)
\( = \frac{d^2k^2(k^3 - 1)^2}{d^2k^4} - \frac{d^2(1 - k^3)^2}{d^2k^4} = \frac{(k^3 - 1)^2}{k^2} - \frac{(1 - k^3)^2}{k^4} \)
\( = \frac{k^6 + 1 - 2k^3}{k^2} - \frac{1 + k^6 - 2k^3}{k^4} = \frac{k^2(k^6 + 1 - 2k^3) - (1 + k^6 - 2k^3)}{k^4} \)
\( = \frac{k^8 + k^2 - 2k^5 - 1 - k^6 + 2k^3}{k^4} \)
R.H.S. \( = (a - d)^2\left(\frac{1}{c^2} - \frac{1}{b^2}\right) = (dk^3 - d)^2\left(\frac{1}{(dk)^2} - \frac{1}{(dk^2)^2}\right) \)
\( = d^2(k^3 - 1)^2\left(\frac{1}{d^2k^2} - \frac{1}{d^2k^4}\right) = d^2(k^3 - 1)^2 \cdot \frac{k^2 - 1}{d^2k^4} = \frac{(k^3 - 1)^2(k^2 - 1)}{k^4} \)
\( = \frac{(k^6 + 1 - 2k^3)(k^2 - 1)}{k^4} = \frac{k^8 + k^2 - 2k^5 - 1 - k^6 + 2k^3}{k^4} \)
Since L.H.S. = R.H.S., hence proved that \( \left(\frac{a - b}{c} + \frac{a - c}{b}\right)^2 - \left(\frac{d - b}{c} + \frac{d - c}{b}\right)^2 = (a - d)^2\left(\frac{1}{c^2} - \frac{1}{b^2}\right) \)
In simple words: Express all four variables in terms of d and k using the continued proportion relationship. Substitute systematically into each part and simplify by factoring. Match the algebra on both sides to confirm the identity holds.

Exam Tip: For longer algebraic proofs, keep careful track of exponents and factors during substitution. Factor out common terms aggressively to simplify before attempting final matching of L.H.S. and R.H.S.

Exercise 6.3

Question 1. If a : b : : c : d, prove that
(i) \( \frac{2a + 5b}{2a - 5b} = \frac{2c + 5d}{2c - 5d} \)
(ii) \( \frac{5a + 11b}{5c + 11d} = \frac{5a - 11b}{5c - 11d} \)
(iii) \( (2a + 3b)(2c - 3d) = (2a - 3b)(2c + 3d) \)
(iv) \( (la + mb) : (lc + md) :: (la - mb) : (lc - md) \)
Answer:
(i) Since \( a : b :: c : d \), we have \( \frac{a}{b} = \frac{c}{d} \).

Multiply both sides by \( \frac{2}{5} \):
\( \implies \frac{2a}{5b} = \frac{2c}{5d} \)

Using componendo and dividendo:
\( \implies \frac{2a + 5b}{2a - 5b} = \frac{2c + 5d}{2c - 5d} \)

(ii) Since \( a : b :: c : d \), we have \( \frac{a}{b} = \frac{c}{d} \).

Multiply both sides by \( \frac{11}{5} \):
\( \implies \frac{11a}{5b} = \frac{11c}{5d} \)

Using componendo and dividendo:
\( \implies \frac{5a + 11b}{5a - 11b} = \frac{5c + 11d}{5c - 11d} \)

Using alternendo:
\( \implies \frac{5a + 11b}{5c + 11d} = \frac{5a - 11b}{5c - 11d} \)

(iii) Since \( a : b :: c : d \), we have \( \frac{a}{b} = \frac{c}{d} \).

Multiply both sides by \( \frac{2}{3} \):
\( \implies \frac{2a}{3b} = \frac{2c}{3d} \)

Using componendo and dividendo:
\( \implies \frac{2a + 3b}{2a - 3b} = \frac{2c + 3d}{2c - 3d} \)

Cross-multiplying:
\( \implies (2a + 3b)(2c - 3d) = (2c + 3d)(2a - 3b) \)

(iv) Since \( a : b :: c : d \), we have \( \frac{a}{b} = \frac{c}{d} \).

Multiply both sides by \( \frac{l}{m} \):
\( \implies \frac{la}{mb} = \frac{lc}{md} \)

Using componendo and dividendo:
\( \implies \frac{la + mb}{la - mb} = \frac{lc + md}{lc - md} \)

Using alternendo:
\( \implies \frac{la + mb}{lc + md} = \frac{la - mb}{lc - md} \)
\( \implies (la + mb) : (lc + md) :: (la - mb) : (lc - md) \)
In simple words: When four numbers are in proportion, we can use the rules of componendo and dividendo (adding and subtracting numerators and denominators in matching ways) to show that other ratios formed from them are also equal.

Exam Tip: Remember that componendo and dividendo is a shortcut - always verify by cross-multiplication. Label each step clearly showing which rule you are using.

Question 2(i). If \( \frac{5u + 7v}{5x + 7y} = \frac{5u - 7v}{5x - 7y} \), show that \( \frac{x}{y} = \frac{u}{v} \).
Answer:
Given, \( \frac{5u + 7v}{5x + 7y} = \frac{5u - 7v}{5x - 7y} \)

Using alternendo (swapping the middle terms):
\( \implies \frac{5u + 7v}{5u - 7v} = \frac{5x + 7y}{5x - 7y} \)

Using componendo and dividendo:
\( \implies \frac{5u + 7v + 5u - 7v}{5u + 7v - 5u + 7v} = \frac{5x + 7y + 5x - 7y}{5x + 7y - 5x + 7y} \)

\( \implies \frac{10u}{14v} = \frac{10x}{14y} \)

Dividing both sides by \( \frac{10}{14} \):
\( \implies \frac{x}{y} = \frac{u}{v} \)
In simple words: When two fractions that are opposites are equal, we can rearrange and simplify to find that the two pairs of numbers form the same ratio.

Exam Tip: The key step is recognizing when alternendo applies - look for equal fractions where the numerators and denominators are related by a pattern.

Question 2(ii). If \( \frac{8a - 5b}{8c - 5d} = \frac{8a + 5b}{8c + 5d} \), prove that \( \frac{a}{b} = \frac{c}{d} \).
Answer:
Given, \( \frac{8a - 5b}{8c - 5d} = \frac{8a + 5b}{8c + 5d} \)

Using alternendo:
\( \implies \frac{8a - 5b}{8a + 5b} = \frac{8c - 5d}{8c + 5d} \)

Using componendo and dividendo:
\( \implies \frac{8a - 5b + 8a + 5b}{8a - 5b - 8a - 5b} = \frac{8c - 5d + 8c + 5d}{8c - 5d - 8c - 5d} \)

\( \implies \frac{16a}{-10b} = \frac{16c}{-10d} \)

Dividing the equation by \( -\frac{16}{10} \):
\( \implies \frac{a}{b} = \frac{c}{d} \)
In simple words: Start by flipping one side of the equal fractions to match the other pattern. Then use componendo and dividendo to simplify and cancel the coefficients.

Exam Tip: Pay close attention to signs when using componendo and dividendo with subtraction - negative results are expected and correct.

Question 3. If (4a + 5b)(4c - 5d) = (4a - 5b)(4c + 5d), prove that a, b, c, d are in proportion.
Answer:
Given, \( (4a + 5b)(4c - 5d) = (4a - 5b)(4c + 5d) \).

Cross-multiplying to separate the ratios:
\( \implies \frac{4a + 5b}{4a - 5b} = \frac{4c + 5d}{4c - 5d} \)

Using componendo and dividendo:
\( \implies \frac{4a + 5b + 4a - 5b}{4a + 5b - 4a + 5b} = \frac{4c + 5d + 4c - 5d}{4c + 5d - 4c + 5d} \)

\( \implies \frac{8a}{10b} = \frac{8c}{10d} \)

Dividing the equation by \( \frac{8}{10} \):
\( \implies \frac{a}{b} = \frac{c}{d} \)
\( \implies a : b :: c : d \)
In simple words: Convert the product form into a fraction form, then apply componendo and dividendo to show the four numbers are in proportion.

Exam Tip: Always state the final result using proportion notation (a : b :: c : d) when asked to prove numbers are in proportion - this makes your answer complete.

Question 4. If (pa + qb) : (pc + qd) : : (pa - qb) : (pc - qd), prove that a : b : : c : d.
Answer:
Given, \( (pa + qb) : (pc + qd) :: (pa - qb) : (pc - qd) \).

This means \( \frac{pa + qb}{pc + qd} = \frac{pa - qb}{pc - qd} \).

Using alternendo:
\( \implies \frac{pa + qb}{pa - qb} = \frac{pc + qd}{pc - qd} \)

Using componendo and dividendo:
\( \implies \frac{pa + qb + pa - qb}{pa + qb - pa + qb} = \frac{pc + qd + pc - qd}{pc + qd - pc + qd} \)

\( \implies \frac{2pa}{2qb} = \frac{2pc}{2qd} \)

Dividing the equation by \( \frac{2p}{2q} \):
\( \implies \frac{a}{b} = \frac{c}{d} \)
\( \implies a : b :: c : d \)
In simple words: When two ratios of expressions are equal in this special form, the coefficients cancel out and the original variables are shown to be in proportion.

Exam Tip: Notice that p and q are just coefficients that add structure - alternendo and componendo work the same way and eliminate them.

Question 5. If (ma + nb) : b : : (mc + nd) : d, prove that a, b, c, d are in proportion.
Answer:
Given, \( (ma + nb) : b :: (mc + nd) : d \).

This means \( \frac{ma + nb}{b} = \frac{mc + nd}{d} \).

Cross-multiplying:
\( \implies d(ma + nb) = b(mc + nd) \)
\( \implies mad + nbd = bmc + bnd \)
\( \implies mad = bmc \)

Dividing equation by m:
\( \implies ad = bc \)
\( \implies \frac{a}{b} = \frac{c}{d} \)
\( \implies a : b :: c : d \)
In simple words: Break down the proportion into an equation, expand both sides, cancel matching terms, then rearrange to show that a, b, c, and d satisfy the proportion condition.

Exam Tip: Cross-multiplication is your first tool when dealing with proportions in this form - it converts the ratio statement into an algebraic equation.

Question 6. If (11a² + 13b²)(11c² - 13d²) = (11a² - 13b²)(11c² + 13d²), prove that a : b : : c : d.
Answer:
Given, \( (11a^2 + 13b^2)(11c^2 - 13d^2) = (11a^2 - 13b^2)(11c^2 + 13d^2) \).

Cross-multiplying to form a ratio:
\( \implies \frac{11a^2 + 13b^2}{11a^2 - 13b^2} = \frac{11c^2 + 13d^2}{11c^2 - 13d^2} \)

Using componendo and dividendo:
\( \implies \frac{11a^2 + 13b^2 + 11a^2 - 13b^2}{11a^2 + 13b^2 - 11a^2 + 13b^2} = \frac{11c^2 + 13d^2 + 11c^2 - 13d^2}{11c^2 + 13d^2 - 11c^2 + 13d^2} \)

\( \implies \frac{22a^2}{26b^2} = \frac{22c^2}{26d^2} \)

Dividing the equation by \( \frac{22}{26} \):
\( \implies \frac{a^2}{b^2} = \frac{c^2}{d^2} \)

Taking square roots of both sides:
\( \implies \sqrt{\frac{a^2}{b^2}} = \sqrt{\frac{c^2}{d^2}} \)
\( \implies \frac{a}{b} = \frac{c}{d} \)
\( \implies a : b :: c : d \)
In simple words: The coefficients in front of a and c square the same way. Use componendo and dividendo to cancel them, then square root to get back to the original variables.

Exam Tip: When working with squared terms, apply componendo and dividendo first, then take square roots - doing it in this order keeps the algebra cleaner.

Question 7. If \( x = \frac{2ab}{a + b} \), find the value of \( \frac{x + a}{x - a} + \frac{x + b}{x - b} \).
Answer:
Given, \( x = \frac{2ab}{a + b} \).

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

Using componendo and dividendo on \( \frac{x}{a} = \frac{2b}{a + b} \):
\( \implies \frac{x + a}{x - a} = \frac{2b + a + b}{2b - a - b} = \frac{3b + a}{b - a} \) ... (Eq 1)

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

Using componendo and dividendo on \( \frac{x}{b} = \frac{2a}{a + b} \):
\( \implies \frac{x + b}{x - b} = \frac{2a + a + b}{2a - a - b} = \frac{3a + b}{a - b} \) ... (Eq 2)

Adding Eq 1 and Eq 2:
\( \frac{x + a}{x - a} + \frac{x + b}{x - b} = \frac{3b + a}{b - a} + \frac{3a + b}{a - b} \)
\( = \frac{3b + a}{b - a} - \frac{3a + b}{b - a} \)
\( = \frac{3b + a - 3a - b}{b - a} \)
\( = \frac{2b - 2a}{b - a} \)
\( = \frac{2(b - a)}{b - a} = 2 \)
In simple words: Find the ratio of x to each variable separately, use componendo and dividendo on each, then add the results to get the final answer.

Exam Tip: Break the given expression into smaller parts by finding x/a and x/b - this makes applying componendo and dividendo much simpler.

Question 8. If \( x = \frac{8ab}{a + b} \), find the value of \( \frac{x + 4a}{x - 4a} + \frac{x + 4b}{x - 4b} \).
Answer:
Given, \( x = \frac{8ab}{a + b} \).

\( \frac{x}{4a} = \frac{8ab/(a+b)}{4a} = \frac{2b}{a + b} \)

Using componendo and dividendo on \( \frac{x}{4a} = \frac{2b}{a + b} \):
\( \implies \frac{x + 4a}{x - 4a} = \frac{2b + a + b}{2b - a - b} = \frac{3b + a}{b - a} \) ... (Eq 1)

\( \frac{x}{4b} = \frac{8ab/(a+b)}{4b} = \frac{2a}{a + b} \)

Using componendo and dividendo on \( \frac{x}{4b} = \frac{2a}{a + b} \):
\( \implies \frac{x + 4b}{x - 4b} = \frac{2a + a + b}{2a - a - b} = \frac{3a + b}{a - b} \) ... (Eq 2)

Adding Eq 1 and Eq 2:
\( \frac{x + 4a}{x - 4a} + \frac{x + 4b}{x - 4b} = \frac{3b + a}{b - a} + \frac{3a + b}{a - b} \)
\( = \frac{3b + a}{b - a} - \frac{3a + b}{b - a} \)
\( = \frac{3b + a - 3a - b}{b - a} \)
\( = \frac{2b - 2a}{b - a} \)
\( = \frac{2(b - a)}{b - a} = 2 \)
In simple words: The method is the same as Question 7 - divide x by the coefficients of a and b, apply componendo and dividendo, then add the results together.

Exam Tip: Notice the pattern: even though the coefficient is 8 instead of 2, we divide by 4a and 4b, which gives us the same ratios as before - the answer remains 2.

Question 9. If \( x = \frac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}} \), find the value of \( \frac{x + 2\sqrt{2}}{x - 2\sqrt{2}} + \frac{x + 2\sqrt{3}}{x - 2\sqrt{3}} \).
Answer:
Given, \( x = \frac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}} \).

\( \frac{x}{2\sqrt{2}} = \frac{4\sqrt{6}/(\sqrt{2} + \sqrt{3})}{2\sqrt{2}} = \frac{2\sqrt{3}}{\sqrt{2} + \sqrt{3}} \)

Using componendo and dividendo on \( \frac{x}{2\sqrt{2}} = \frac{2\sqrt{3}}{\sqrt{2} + \sqrt{3}} \):
\( \implies \frac{x + 2\sqrt{2}}{x - 2\sqrt{2}} = \frac{2\sqrt{3} + \sqrt{2} + \sqrt{3}}{2\sqrt{3} - \sqrt{2} - \sqrt{3}} = \frac{3\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \) ... (Eq 1)

\( \frac{x}{2\sqrt{3}} = \frac{4\sqrt{6}/(\sqrt{2} + \sqrt{3})}{2\sqrt{3}} = \frac{2\sqrt{2}}{\sqrt{2} + \sqrt{3}} \)

Using componendo and dividendo on \( \frac{x}{2\sqrt{3}} = \frac{2\sqrt{2}}{\sqrt{2} + \sqrt{3}} \):
\( \implies \frac{x + 2\sqrt{3}}{x - 2\sqrt{3}} = \frac{2\sqrt{2} + \sqrt{2} + \sqrt{3}}{2\sqrt{2} - \sqrt{2} - \sqrt{3}} = \frac{3\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} \) ... (Eq 2)

Adding Eq 1 and Eq 2:
\( \frac{x + 2\sqrt{2}}{x - 2\sqrt{2}} + \frac{x + 2\sqrt{3}}{x - 2\sqrt{3}} = \frac{3\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} + \frac{3\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} \)
\( = \frac{3\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} - \frac{3\sqrt{2} + \sqrt{3}}{\sqrt{3} - \sqrt{2}} \)
\( = \frac{3\sqrt{3} + \sqrt{2} - 3\sqrt{2} - \sqrt{3}}{\sqrt{3} - \sqrt{2}} \)
\( = \frac{2\sqrt{3} - 2\sqrt{2}}{\sqrt{3} - \sqrt{2}} \)
\( = \frac{2(\sqrt{3} - \sqrt{2})}{\sqrt{3} - \sqrt{2}} = 2 \)
In simple words: Even with surds (square roots), the same method works - divide x by each coefficient, apply componendo and dividendo, then add to get 2.

Exam Tip: Surds follow the same algebraic rules as regular numbers - don't be intimidated by square roots; treat them as variables and the answer comes out cleanly.

Question 10. Using properties of proportion, find x from the following equations:
(i) \( \frac{\sqrt{2 - x} + \sqrt{2 + x}}{\sqrt{2 - x} - \sqrt{2 + x}} = 3 \)
(ii) \( \frac{\sqrt{x + 4} + \sqrt{x - 10}}{\sqrt{x + 4} - \sqrt{x - 10}} = \frac{5}{2} \)
(iii) \( \frac{\sqrt{1 + x} + \sqrt{1 - x}}{\sqrt{1 + x} - \sqrt{1 - x}} = \frac{a}{b} \)
(iv) \( \frac{\sqrt{5x} + \sqrt{2x - 6}}{\sqrt{5x} - \sqrt{2x - 6}} = 4 \)
(v) \( \frac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x} - \sqrt{a - x}} = \frac{c}{d} \)
(vi) \( \frac{a + \sqrt{a^2 - 2ax}}{a - \sqrt{a^2 - 2ax}} = b \)
Answer:
(i) Given, \( \frac{\sqrt{2 - x} + \sqrt{2 + x}}{\sqrt{2 - x} - \sqrt{2 + x}} = \frac{3}{1} \)

Applying componendo and dividendo:
\( \implies \frac{\sqrt{2 - x} + \sqrt{2 + x} + \sqrt{2 - x} - \sqrt{2 + x}}{\sqrt{2 - x} + \sqrt{2 + x} - \sqrt{2 - x} + \sqrt{2 + x}} = \frac{3 + 1}{3 - 1} \)

\( \implies \frac{2\sqrt{2 - x}}{2\sqrt{2 + x}} = \frac{4}{2} \)

\( \implies \frac{\sqrt{2 - x}}{\sqrt{2 + x}} = \frac{2}{1} \)

Squaring both sides:
\( \implies \frac{2 - x}{2 + x} = \frac{4}{1} \)
\( \implies 2 - x = 4(2 + x) \)
\( \implies 2 - x = 8 + 4x \)
\( \implies -6 = 5x \)
\( \implies x = -\frac{6}{5} \)

(ii) Given, \( \frac{\sqrt{x + 4} + \sqrt{x - 10}}{\sqrt{x + 4} - \sqrt{x - 10}} = \frac{5}{2} \)

Applying componendo and dividendo:
\( \implies \frac{\sqrt{x + 4} + \sqrt{x - 10} - \sqrt{x + 4} + \sqrt{x - 10}}{\sqrt{x + 4} + \sqrt{x - 10} + \sqrt{x + 4} - \sqrt{x - 10}} = \frac{5 - 2}{5 + 2} \)

\( \implies \frac{2\sqrt{x - 10}}{2\sqrt{x + 4}} = \frac{3}{7} \)

\( \implies \frac{\sqrt{x - 10}}{\sqrt{x + 4}} = \frac{3}{7} \)

Squaring both sides:
\( \implies \frac{x - 10}{x + 4} = \frac{9}{49} \)
\( \implies 9(x + 4) = 49(x - 10) \)
\( \implies 9x + 36 = 49x - 490 \)
\( \implies 40x = 526 \)
\( \implies x = \frac{263}{20} \)

(iii) Given, \( \frac{\sqrt{1 + x} + \sqrt{1 - x}}{\sqrt{1 + x} - \sqrt{1 - x}} = \frac{a}{b} \)

Applying componendo and dividendo:
\( \implies \frac{\sqrt{1 + x} + \sqrt{1 - x} + \sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x} - \sqrt{1 + x} + \sqrt{1 - x}} = \frac{a + b}{a - b} \)

\( \implies \frac{2\sqrt{1 + x}}{2\sqrt{1 - x}} = \frac{a + b}{a - b} \)

\( \implies \frac{\sqrt{1 + x}}{\sqrt{1 - x}} = \frac{a + b}{a - b} \)

Squaring both sides:
\( \implies \frac{1 + x}{1 - x} = \frac{(a + b)^2}{(a - b)^2} \)
\( \implies (1 + x)(a - b)^2 = (1 - x)(a + b)^2 \)

Solving for x:
\( x = \frac{(a + b)^2 - (a - b)^2}{(a + b)^2 + (a - b)^2} = \frac{4ab}{2a^2 + 2b^2} = \frac{2ab}{a^2 + b^2} \)

(iv) Given, \( \frac{\sqrt{5x} + \sqrt{2x - 6}}{\sqrt{5x} - \sqrt{2x - 6}} = 4 \)

Applying componendo and dividendo:
\( \implies \frac{\sqrt{5x} + \sqrt{2x - 6} - \sqrt{5x} + \sqrt{2x - 6}}{\sqrt{5x} + \sqrt{2x - 6} + \sqrt{5x} - \sqrt{2x - 6}} = \frac{4 - 1}{4 + 1} \)

\( \implies \frac{2\sqrt{2x - 6}}{2\sqrt{5x}} = \frac{3}{5} \)

\( \implies \frac{\sqrt{2x - 6}}{\sqrt{5x}} = \frac{3}{5} \)

Squaring both sides:
\( \implies \frac{2x - 6}{5x} = \frac{9}{25} \)
\( \implies 25(2x - 6) = 9(5x) \)
\( \implies 50x - 150 = 45x \)
\( \implies 5x = 150 \)
\( \implies x = 30 \)

(v) Given, \( \frac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x} - \sqrt{a - x}} = \frac{c}{d} \)

Applying componendo and dividendo:
\( \implies \frac{\sqrt{a + x} + \sqrt{a - x} + \sqrt{a + x} - \sqrt{a - x}}{\sqrt{a + x} + \sqrt{a - x} - \sqrt{a + x} + \sqrt{a - x}} = \frac{c + d}{c - d} \)

\( \implies \frac{2\sqrt{a + x}}{2\sqrt{a - x}} = \frac{c + d}{c - d} \)

\( \implies \frac{\sqrt{a + x}}{\sqrt{a - x}} = \frac{c + d}{c - d} \)

Squaring both sides:
\( \implies \frac{a + x}{a - x} = \frac{(c + d)^2}{(c - d)^2} \)

Solving for x:
\( x = \frac{a[(c + d)^2 - (c - d)^2]}{(c + d)^2 + (c - d)^2} = \frac{4acd}{c^2 + d^2} \)

(vi) Given, \( \frac{a + \sqrt{a^2 - 2ax}}{a - \sqrt{a^2 - 2ax}} = b \)

Applying componendo and dividendo:
\( \implies \frac{a + \sqrt{a^2 - 2ax} + a - \sqrt{a^2 - 2ax}}{a + \sqrt{a^2 - 2ax} - a + \sqrt{a^2 - 2ax}} = \frac{b + 1}{b - 1} \)

\( \implies \frac{2a}{2\sqrt{a^2 - 2ax}} = \frac{b + 1}{b - 1} \)

\( \implies \frac{a}{\sqrt{a^2 - 2ax}} = \frac{b + 1}{b - 1} \)

Squaring both sides:
\( \implies \frac{a^2}{a^2 - 2ax} = \frac{(b + 1)^2}{(b - 1)^2} \)
\( \implies a^2(b - 1)^2 = (a^2 - 2ax)(b + 1)^2 \)

Solving for x:
\( x = \frac{a[1 - (b - 1)^2/(b + 1)^2]}{2} = \frac{a[(b + 1)^2 - (b - 1)^2]}{2(b + 1)^2} = \frac{2ab}{(b + 1)^2} \)
In simple words: For each equation, apply componendo and dividendo to simplify the radicals. Then square both sides to remove the square roots and solve for x using basic algebra.

Exam Tip: Componendo and dividendo is the key tool for equations with radicals in proportions. Always apply it first before attempting to solve - it dramatically simplifies the algebra.