Selina Concise Solutions for ICSE Class 10 Mathematics Chapter 7 Ratio And Proportion Including Properties Uses

Question 1. If a : b = 5 : 3, find: \( \frac{5a - 3b}{5a + 3b} \)
Answer:
Given: a : b = 5 : 3

Step 1: From the ratio, we get \( \frac{a}{b} = \frac{5}{3} \)

Step 2: Substitute in the expression.
\( \frac{5a - 3b}{5a + 3b} = \frac{5(\frac{a}{b}) - 3}{5(\frac{a}{b}) + 3} \) (Dividing each term by b)

Step 3: Put \( \frac{a}{b} = \frac{5}{3} \)
\( = \frac{5(\frac{5}{3}) - 3}{5(\frac{5}{3}) + 3} \)

Step 4: Simplify the numerator and denominator.
\( = \frac{\frac{25}{3} - 3}{\frac{25}{3} + 3} = \frac{\frac{25 - 9}{3}}{\frac{25 + 9}{3}} = \frac{16}{34} = \frac{8}{17} \)

In simple words: We used the given ratio to find the value of a/b. Then we put this value in the given fraction and calculated step by step.

📝 Teacher's Note: Show students how dividing each term by 'b' helps us use the ratio directly. This is a common trick in ratio problems.

🎯 Exam Tip: Always write the given ratio first. Then show each step clearly. The final answer should be in simplest form.

Question 2. If x : y = 4 : 7, find the value of (3x + 2y) : (5x + y)
Answer:
Given: x : y = 4 : 7

Step 1: From the ratio, we get \( \frac{x}{y} = \frac{4}{7} \)

Step 2: Find the ratio of the given expressions.
\( \frac{3x + 2y}{5x + y} = \frac{3(\frac{x}{y}) + 2}{5(\frac{x}{y}) + 1} \) (Dividing each term by y)

Step 3: Put \( \frac{x}{y} = \frac{4}{7} \)
\( = \frac{3(\frac{4}{7}) + 2}{5(\frac{4}{7}) + 1} = \frac{\frac{12}{7} + 2}{\frac{20}{7} + 1} \)

Step 4: Simplify.
\( = \frac{\frac{12 + 14}{7}}{\frac{20 + 7}{7}} = \frac{26}{27} \)

In simple words: We found x/y from the given ratio. Then we used this value to find the ratio of the two given expressions.

📝 Teacher's Note: Remind students that when we divide by y, we get x/y terms which we can replace using the given ratio.

🎯 Exam Tip: Write the final answer as a ratio. So (3x + 2y) : (5x + y) = 26 : 27.

Question 3. If a : b = 3 : 8, find the value of \( \frac{4a + 3b}{6a - b} \)
Answer:
Given: a : b = 3 : 8

Step 1: From the ratio, we get \( \frac{a}{b} = \frac{3}{8} \)

Step 2: Substitute in the expression.
\( \frac{4a + 3b}{6a - b} = \frac{4(\frac{a}{b}) + 3}{6(\frac{a}{b}) - 1} \) (Dividing each term by b)

Step 3: Put \( \frac{a}{b} = \frac{3}{8} \)
\( = \frac{4(\frac{3}{8}) + 3}{6(\frac{3}{8}) - 1} = \frac{\frac{3}{2} + 3}{\frac{9}{4} - 1} = \frac{\frac{9}{2}}{\frac{5}{4}} = \frac{18}{5} \)

In simple words: We used the given ratio to replace a/b in our fraction. Then we did the calculations to get the final answer.

📝 Teacher's Note: Show students how to add fractions with different denominators. This helps them understand each step clearly.

🎯 Exam Tip: Check your fraction arithmetic carefully. Many marks are lost due to calculation errors.

Question 4. If (a - b) : (a + b) = 1 : 11, find the ratio (5a + 4b + 15) : (5a - 4b + 3)
Answer:
Given: (a - b) : (a + b) = 1 : 11

Step 1: From the given ratio:
\( \frac{a - b}{a + b} = \frac{1}{11} \)

Step 2: Cross multiply to solve.
11(a - b) = (a + b)
11a - 11b = a + b
10a = 12b
\( \frac{a}{b} = \frac{12}{10} = \frac{6}{5} \)

Step 3: Let a = 6k and b = 5k for some value k.

Step 4: Find the required ratio.
\( \frac{5a + 4b + 15}{5a - 4b + 3} = \frac{5(6k) + 4(5k) + 15}{5(6k) - 4(5k) + 3} \)
\( = \frac{30k + 20k + 15}{30k - 20k + 3} = \frac{50k + 15}{10k + 3} = \frac{5(10k + 3)}{10k + 3} = 5 \)

Hence, (5a + 4b + 15) : (5a - 4b + 3) = 5 : 1

In simple words: We used the given ratio to find the relationship between a and b. Then we used this to calculate the new ratio.

📝 Teacher's Note: Show students how to use cross multiplication to solve ratios. This makes the algebra much easier.

🎯 Exam Tip: Always substitute back to check if your answer makes sense. This helps catch calculation mistakes.

Question 5. Find the number which bears the same ratio to \( \frac{7}{33} \) that \( \frac{8}{21} \) does to \( \frac{4}{9} \)
Answer:
Let the required number be \( \frac{x}{y} \)

Step 1: Set up the proportion.
Ratio of \( \frac{8}{21} \) to \( \frac{4}{9} = \frac{8/21}{4/9} = \frac{8}{21} \times \frac{9}{4} = \frac{6}{7} \)

Step 2: The required number has the same ratio to \( \frac{7}{33} \)
\( \frac{x/y}{7/33} = \frac{6}{7} \)

Step 3: Solve for x/y.
\( \frac{x}{y} = \frac{6}{7} \times \frac{7}{33} = \frac{6 \times 7}{7 \times 33} = \frac{2}{11} \)

Hence, the required number is \( \frac{2}{11} \)

In simple words: We found the ratio between the given fractions. Then we used this same ratio to find the unknown number.

📝 Teacher's Note: Explain that "same ratio" means equal proportions. Use simple examples like speed ratios to make this clear.

🎯 Exam Tip: Always write "Let the required number be x/y" at the start. This shows the examiner your method clearly.

Question 6. If \( \frac{m + n}{m + 3n} = \frac{2}{3} \), find: \( \frac{2n^2}{3m^2 + mn} \)
Answer:
Given: \( \frac{m + n}{m + 3n} = \frac{2}{3} \)

Step 1: Cross multiply to find the relationship.
3(m + n) = 2(m + 3n)
3m + 3n = 2m + 6n
m = 3n

Step 2: Find \( \frac{m}{n} = 3 \)

Step 3: Substitute in the required expression.
\( \frac{2n^2}{3m^2 + mn} = \frac{2}{3(\frac{m}{n})^2 + (\frac{m}{n})} \) (Dividing by \( n^2 \))

Step 4: Put \( \frac{m}{n} = 3 \)
\( = \frac{2}{3(3)^2 + (3)} = \frac{2}{27 + 3} = \frac{2}{30} = \frac{1}{15} \)

In simple words: We found the relationship between m and n. Then we used this to simplify the given expression step by step.

📝 Teacher's Note: Show students how dividing by n² helps convert the expression into ratios. This is a useful technique.

🎯 Exam Tip: Always simplify your final fraction. Write 1/15, not 2/30. Examiners prefer simplified answers.

Question 7. Find \( \frac{x}{y} \), when \( x^2 + 6y^2 = 5xy \)
Answer:
Given: \( x^2 + 6y^2 = 5xy \)

Step 1: Divide both sides by \( y^2 \)
\( \frac{x^2}{y^2} + 6 = \frac{5xy}{y^2} \)
\( (\frac{x}{y})^2 + 6 = 5(\frac{x}{y}) \)

Step 2: Let \( \frac{x}{y} = a \)
\( a^2 + 6 = 5a \)
\( a^2 - 5a + 6 = 0 \)

Step 3: Factorize the quadratic equation.
\( (a - 2)(a - 3) = 0 \)
\( a = 2 \) or \( a = 3 \)

Hence, \( \frac{x}{y} = 2 \) or \( 3 \)

In simple words: We changed the equation into a simple quadratic by substituting x/y with a variable. Then we solved it like a normal quadratic equation.

📝 Teacher's Note: Remind students that when we divide by y², we must assume y ≠ 0. This is important in ratio problems.

🎯 Exam Tip: Always write both solutions when you get a quadratic equation. Don't forget to write "or" between the answers.

Question 8. If the ratio between 8 and 11 is the same as the ratio of 2x - y to x + 2y, find the value of \( \frac{7x}{9y} \).
Answer:
Step 1: Set up the proportion.
\( \frac{2x - y}{x + 2y} = \frac{8}{11} \)

Step 2: Cross multiply.
22x - 11y = 8x + 16y

Step 3: Simplify and solve for x in terms of y.
22x - 8x = 16y + 11y
14x = 27y
\( \implies \frac{x}{y} = \frac{27}{14} \)

Step 4: Find the required value.
\( \frac{7x}{9y} = \frac{7 \times 27}{9 \times 14} = \frac{189}{126} = \frac{3}{2} \)

In simple words: We set up equal ratios and cross multiply to find the relationship between x and y. Then we substitute this relationship to find the answer.

📝 Teacher's Note: When two ratios are equal, we can cross multiply. Teach students to always write the ratios first, then cross multiply step by step.

🎯 Exam Tip: Always simplify fractions at the end. Show each step clearly. Write the final answer as a simplified fraction.

Question 9. Divide Rs 1,290 into A, B and C such that A is \( \frac{2}{5} \) of B and B : C = 4 : 3.
Answer:
Step 1: Express the given ratios.
Given B : C = 4 : 3
\( \implies \frac{B}{C} = \frac{4}{3} \)

Step 2: Find ratio of A to B.
A = \( \frac{2}{5} \)B
\( \implies \frac{A}{B} = \frac{2}{5} \)

Step 3: Express all parts in terms of common ratio.
\( \frac{A}{B} = \frac{2}{5} = \frac{2 \times 4}{5 \times 4} = \frac{8}{20} \)
\( \frac{B}{C} = \frac{4}{3} = \frac{4 \times 5}{3 \times 5} = \frac{20}{15} \)

Step 4: Find A : B : C.
A : B : C = 8 : 20 : 15

Step 5: Calculate actual amounts.
Total parts = 8 + 20 + 15 = 43
A's share = 8x = 8 × 30 = Rs. 240
B's share = 20x = 20 × 30 = Rs. 600
C's share = 15x = 15 × 30 = Rs. 450

(where x = 1290 ÷ 43 = 30)

In simple words: We found how A, B and C are related to each other, then divided the total money in that same ratio.

📝 Teacher's Note: When finding combined ratios, make sure to use a common multiplier so all three parts can be compared easily. Check that the final amounts add up to the total.

🎯 Exam Tip: Always verify your answer by adding all parts to check if they equal the total given amount. Show the value of x clearly.

Question 10. A school has 630 students. The ratio of the number of boys to the number of girls is 3 : 2. This ratio changes to 7 : 5 after the admission of 90 new students. Find the number of newly admitted boys.
Answer:
Step 1: Find initial numbers.
Let the number of boys be 3x and girls be 2x
3x + 2x = 630
5x = 630
x = 126

Number of boys = 3x = 3 × 126 = 378
Number of girls = 2x = 2 × 126 = 252

Step 2: Find numbers after admission.
After admission of 90 new students, total = 630 + 90 = 720
New ratio of boys : girls = 7 : 5
Let new number of boys be 7y and girls be 5y
7y + 5y = 720
12y = 720
y = 60

New number of boys = 7y = 7 × 60 = 420
New number of girls = 5y = 5 × 60 = 300

Step 3: Find newly admitted boys.
Number of newly admitted boys = 420 - 378 = 42

In simple words: First we found how many boys and girls were there initially. Then we found how many are there after new admissions. The difference gives us newly admitted boys.

📝 Teacher's Note: Break this into two parts - before and after admission. Students often get confused trying to do everything at once. Solve step by step.

🎯 Exam Tip: Always check that initial boys + initial girls = 630 and final boys + final girls = 720. This prevents calculation errors.

Question 11. What quantity must be subtracted from each term of the ratio 9: 17 to make it equal to 1: 3?
Answer:
Step 1: Let x be the quantity to be subtracted.
After subtraction, the ratio becomes: (9 - x) : (17 - x)

Step 2: Set up the equation.
\( \frac{9 - x}{17 - x} = \frac{1}{3} \)

Step 3: Cross multiply and solve.
3(9 - x) = 1(17 - x)
27 - 3x = 17 - x
27 - 17 = 3x - x
10 = 2x
x = 5

Step 4: Verify the answer.
New ratio = (9 - 5) : (17 - 5) = 4 : 12 = 1 : 3 ✓

Therefore, the required number which should be subtracted is 5.

In simple words: We subtracted the same number from both parts of the ratio and made it equal to 1:3. Then we solved to find that number.

📝 Teacher's Note: Always verify the final answer by substituting back. Students should check that (9-5):(17-5) really equals 1:3.

🎯 Exam Tip: Write "Let x be subtracted from each term" clearly at the start. Always verify your answer at the end.

Question 12. The monthly pocket money of Ravi and Sanjeev are in the ratio 5 : 7. Their expenditures are in the ratio 3 : 5. If each saves Rs. 80 every month, find their monthly pocket money.
Answer:
Step 1: Set up the variables.
Let pocket money of Ravi = 5k and Sanjeev = 7k
Let expenditure of Ravi = 3m and Sanjeev = 5m

Step 2: Use the savings information.
Savings = Pocket money - Expenditure
For Ravi: 5k - 3m = 80 ... (1)
For Sanjeev: 7k - 5m = 80 ... (2)

Step 3: Solve the equations.
From equation (1): 5k - 3m = 80
From equation (2): 7k - 5m = 80

Solving these equations: k = 40, m = 40

Step 4: Find the pocket money.
Ravi's monthly pocket money = 5k = 5 × 40 = Rs. 200
Sanjeev's monthly pocket money = 7k = 7 × 40 = Rs. 280

In simple words: We used two variables for pocket money and expenditure ratios. Since both save Rs. 80, we made equations and solved them.

📝 Teacher's Note: This is a two-variable problem. Teach students to use different letters (k and m) for the two different ratios. Then use the savings condition to make equations.

🎯 Exam Tip: Check your answer: Ravi gets 200, spends 120, saves 80. Sanjeev gets 280, spends 200, saves 80. Both save 80 as required.

Question 13. The work done by (x – 2) men in (4x + 1) days and the work done by (4x + 1) men in (2x – 3) days are in the ratio 3: 8. Find the value of x.
Answer:
Step 1: Set up work calculations.
Assuming each man does 1 unit of work per day:
Work done by (x - 2) men in (4x + 1) days = (x - 2)(4x + 1) units
Work done by (4x + 1) men in (2x - 3) days = (4x + 1)(2x - 3) units

Step 2: Set up the ratio equation.
\( \frac{(x - 2)(4x + 1)}{(4x + 1)(2x - 3)} = \frac{3}{8} \)

Step 3: Simplify by canceling common factor.
\( \frac{x - 2}{2x - 3} = \frac{3}{8} \)

Step 4: Cross multiply and solve.
8(x - 2) = 3(2x - 3)
8x - 16 = 6x - 9
8x - 6x = 16 - 9
2x = 7
x = 3.5

In simple words: Total work = number of men × number of days. We set up the ratio of total work and solved for x.

📝 Teacher's Note: Work problems use the formula: Work = Men × Days. Help students see that (4x + 1) cancels out from both numerator and denominator.

🎯 Exam Tip: Always assume each man does 1 unit of work per day unless stated otherwise. Cancel common factors before cross multiplying to make calculations easier.

Question 14. The bus fare between two cities is increased in the ratio 7: 9. Find the increase in the fare, if: (i) the original fare is Rs 245; (ii) the increased fare is Rs 207.
Answer:
(i) Given original fare = Rs 245:
Increased fare = \( \frac{9}{7} \times 245 = Rs. 315 \)
Increase in fare = Rs 315 - Rs 245 = Rs 70

(ii) Given increased fare = Rs 207:
Rs 207 = \( \frac{9}{7} \) × original fare
Original fare = \( Rs. 207 \times \frac{7}{9} = Rs. 161 \)
Increase in fare = Rs 207 - Rs 161 = Rs 46

In simple words: The ratio 7:9 means new fare is 9/7 times the old fare. We use this to find either the new fare or old fare, then subtract to get the increase.

📝 Teacher's Note: Ratio 7:9 means if old fare was 7 units, new fare is 9 units. So new fare = (9/7) × old fare. Practice both types - finding new from old and old from new.

🎯 Exam Tip: In ratio problems, clearly identify which value you know and which you need to find. Use the ratio fraction correctly - 9/7 for increase, 7/9 for decrease.

Question 15. By increasing the cost of entry ticket to a fair in the ratio 10: 13, the number of visitors to the fair has decreased in the ratio 6: 5. In what ratio has the total collection increased or decreased?
Answer:
Step 1: Set up the variables.
Let initial ticket cost = 10x and present cost = 13x
Let initial visitors = 6y and present visitors = 5y

Step 2: Calculate total collections.
Initial collection = 10x × 6y = 60xy
Present collection = 13x × 5y = 65xy

Step 3: Find the ratio.
Ratio of total collection = 60xy : 65xy = 12 : 13

Therefore, the total collection has increased in the ratio 12: 13.

In simple words: Collection = ticket price × number of visitors. Even though visitors decreased, the price increase was bigger, so total collection increased.

📝 Teacher's Note: Total collection depends on both price and number of visitors. Use the formula: Collection = Price × Visitors. Teach students to set up both ratios clearly.

🎯 Exam Tip: Always write Collection = Price × Number of visitors at the start. Check if the ratio shows increase (second number bigger) or decrease (second number smaller).

Question 16. In a basket, the ratio between the number of oranges and the number of apples is 7: 13. If 8 oranges and 11 apples are eaten, the ratio between the number of oranges and the number of apples becomes 1: 2. Find the original number of oranges and the original number of apples in the basket.
Answer:
Step 1: Set up initial quantities.
Let original oranges = 7x and original apples = 13x

Step 2: Set up equation after eating some fruits.
After eating: oranges = 7x - 8, apples = 13x - 11
New ratio = 1 : 2
\( \frac{7x - 8}{13x - 11} = \frac{1}{2} \)

Step 3: Cross multiply and solve.
2(7x - 8) = 1(13x - 11)
14x - 16 = 13x - 11
14x - 13x = 16 - 11
x = 5

Step 4: Find original quantities.
Original oranges = 7x = 7 × 5 = 35
Original apples = 13x = 13 × 5 = 65

In simple words: We started with oranges and apples in ratio 7:13. After eating some, the ratio changed to 1:2. We used this to find the original numbers.

📝 Teacher's Note: When quantities are removed from a ratio, set up the equation with the new quantities. Always verify: (35-8):(65-11) = 27:54 = 1:2 ✓

🎯 Exam Tip: Write the original ratio first, then the new ratio after removal. Set up the equation carefully and verify your answer at the end.

Question 17. In a mixture of 126 kg of milk and water, milk and water are in ratio 5 : 2. How much water must be added to the mixture to make this ratio 3 : 2?
Answer:
Step 1: Find initial quantities.
Milk : Water = 5 : 2
Quantity of milk = \( 126 \times \frac{5}{7} = 90 \text{ kg} \)
Quantity of water = 126 - 90 = 36 kg

Step 2: Set up equation for new ratio.
Let x kg of water be added
New ratio = 3 : 2
After adding water: milk = 90 kg, water = (36 + x) kg
\( \frac{90}{36 + x} = \frac{3}{2} \)

Step 3: Cross multiply and solve.
2 × 90 = 3 × (36 + x)
180 = 108 + 3x
180 - 108 = 3x
72 = 3x
x = 24

Therefore, 24 kg of water must be added.

In simple words: We found how much milk and water were there initially. Then we calculated how much water to add to get the new ratio 3:2.

📝 Teacher's Note: In mixture problems, first find the actual quantities from the given ratio. Remember that only water is added, milk quantity stays the same.

🎯 Exam Tip: Always find actual quantities first, not just ratios. Verify: 90:(36+24) = 90:60 = 3:2 ✓. Show that milk quantity doesn't change.

Question 18.
(A) If A: B = 3: 4 and B: C = 6: 7, find:
(i) A: B: C
(ii) A: C
(B) If A : B = 2 : 5 and A : C = 3 : 4, find
(i) A : B : C
Answer:
(A)
(i)
\( \frac{A}{B} = \frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12} \)
\( \frac{B}{C} = \frac{6}{7} = \frac{6 \times 2}{7 \times 2} = \frac{12}{14} \)
A : B : C = 9 : 12 : 14

(ii)
\( \frac{A}{B} = \frac{3}{4} \)
\( \frac{B}{C} = \frac{6}{7} \)
\( \frac{A}{C} = \frac{A}{B} \times \frac{B}{C} = \frac{3}{4} \times \frac{6}{7} = \frac{9}{14} \)
∴ A : C = 9 : 14

(B)
(i)
To compare 3 ratios, the consequent of the first ratio and the antecedent of the 2nd ratio must be made equal.
Given that A:B = 2:5 and A:C= 3:4
Interchanging the first ratio, we have,
B:A= 5:2 and A:C= 3:4
L.C.M. of 2 and 3 is 6.
⇒ B : A= 5 × 3 : 2 × 3 and A : C=3 × 2 : 4 × 2
⇒ B : A=15 : 6 and A : C=6 : 8
⇒ B : A : C = 15 : 6 : 8
⇒ A : B : C = 6 : 15 : 8
In simple words: To find three-way ratios, we make the middle term equal in both ratios. Then we write them in order.

📝 Teacher's Note: Show students how to use LCM to make common terms. Draw boxes to show A, B, C values clearly. This helps students see the pattern.

🎯 Exam Tip: Always write the LCM step clearly. Check your final answer by verifying the original ratios still work.

Question 19(i).
If 3A = 4B = 6C; find A: B: C.
Answer:
3A = 4B = 6C
3A = 4B ⇒ \( \frac{A}{B} = \frac{4}{3} \)
4B = 6C ⇒ \( \frac{B}{C} = \frac{6}{4} = \frac{3}{2} \)
Hence, A: B: C = 4: 3: 2
In simple words: When three things are equal, we can find how they compare to each other. We take two at a time and solve.

📝 Teacher's Note: Tell students to pick any two parts from the equal expression. Solve them separately, then combine. It's like solving two simple puzzles.

🎯 Exam Tip: Write each step clearly. Show 3A = 4B first, then 4B = 6C. Don't try to do everything in one step.

Question 19(ii).
If 2a = 3b and 4b = 5c, find: a : c.
Answer:
We have,
2a = 3b ⇒ \( \frac{a}{b} = \frac{3}{2} \)
And 4b = 5c ⇒ \( \frac{b}{c} = \frac{5}{4} \)
Now, \( \frac{a}{b} = \frac{3}{2} = \frac{3 \times 5}{2 \times 5} = \frac{15}{10} \) and \( \frac{b}{c} = \frac{5}{4} = \frac{5 \times 2}{4 \times 2} = \frac{10}{8} \)
⇒ a : b : c = 15 : 10 : 8
⇒ a : c = 15 : 8
In simple words: We find how a compares to b, then how b compares to c. Finally, we can find how a compares to c directly.

📝 Teacher's Note: Show students that we need a common value for 'b' in both ratios. Use simple numbers to demonstrate this chain method.

🎯 Exam Tip: Make the middle term (b) equal in both ratios. Then you can directly compare a and c. Write all working clearly.

Question 20.
Find the compound ratio of:
(i) 2: 3, 9: 14 and 14: 27
(ii) 2a: 3b, mn: x² and x: n.
(iii) \( \sqrt{2} \) : 1, 3 : \( \sqrt{5} \) and \( \sqrt{20} \) : 9.
Answer:
(i) Required compound ratio = 3 × 8: 5 × 15
= \( \frac{3 \times 8}{5 \times 15} \)
= \( \frac{8}{25} \) = 8 : 25

(ii) Required compound ratio = 2 × 9 × 14: 3 × 14 × 27
= \( \frac{2 \times 9 \times 14}{3 \times 14 \times 27} \)
= \( \frac{2}{9} \) = 2 : 9

(iii) Required compound ratio = 2a × mn × x: 3b × x² × n
= \( \frac{2a \times mn \times x}{3b \times x^2 \times n} \)
= \( \frac{2am}{3bx} \) = 2am : 3bx

(iv) Required compound ratio = \( \sqrt{2} \times 3 \times \sqrt{20} \) : \( 1 \times \sqrt{5} \times 9 \)
= \( \frac{\sqrt{2} \times 3 \times \sqrt{20}}{1 \times \sqrt{5} \times 9} \)
= \( \frac{\sqrt{2} \times \sqrt{4}}{3} \)
= \( \frac{2\sqrt{2}}{3} \) = \( 2\sqrt{2} \) : 3
In simple words: Compound ratio means we multiply the first parts together and the second parts together. Like combining multiple fractions into one.

📝 Teacher's Note: Show students to multiply all first terms, then all second terms. Use simple examples like 2:3 and 4:5 gives 8:15 first.

🎯 Exam Tip: Always simplify your final answer. Cancel common factors. Write the compound ratio in its simplest form for full marks.

Question 21.
Find duplicate ratio of:
(i) 3: 4 (ii) 3√3 : 2√5
Answer:
(i) Duplicate ratio of 3 : 4 = 3² : 4² = 9 : 16
(ii) Duplicate ratio of 3√3 : 2√5 = (3√3)² : (2√5)² = 27 : 20
In simple words: Duplicate ratio means we square both parts of the ratio. If we have a:b, duplicate ratio is a²:b².

📝 Teacher's Note: Tell students "duplicate" means "double the power". So if original has power 1, duplicate has power 2. Show with simple examples like 2:3 becomes 4:9.

🎯 Exam Tip: Square each term separately. Be careful with expressions like (3√3)². Show your working step by step.

Question 22.
Find the triplicate ratio of:
(i) 1: 3 (ii) \( \frac{m}{2} \) : \( \frac{n}{3} \)
Answer:
(i) Triplicate ratio of 1: 3 = 1³: 3³ = 1: 27
(ii) Triplicate ratio of \( \frac{m}{2} \) : \( \frac{n}{3} \)
= \( \left(\frac{m}{2}\right)^3 \) : \( \left(\frac{n}{3}\right)^3 \) = \( \frac{m^3}{8} \) : \( \frac{n^3}{27} \) = \( \frac{m^3}{8} \div \frac{n^3}{27} \) = 27m³ : 8n³
In simple words: Triplicate ratio means we cube both parts (raise to power 3). If we have a:b, triplicate ratio is a³:b³.

📝 Teacher's Note: Explain "triplicate" means "power of 3". Like duplicate (power 2), but now power 3. Use 2:3 becoming 8:27 as example.

🎯 Exam Tip: Cube each term carefully. For fractions, cube the numerator and denominator separately. Simplify your final answer.

Question 23.
Find sub-duplicate ratio of:
(i) 9: 16 (ii) (x - y)⁴: (x + y)⁶
Answer:
(i) Sub-duplicate ratio of 9: 16 = √9 : √16 = 3 : 4
(ii) Sub-duplicate ratio of (x - y)⁴: (x + y)⁶
= \( \sqrt{(x - y)^4} \) : \( \sqrt{(x + y)^6} \) = (x - y)² : (x + y)³
In simple words: Sub-duplicate ratio means we take square root of both parts. If we have a:b, sub-duplicate ratio is √a:√b.

📝 Teacher's Note: "Sub-duplicate" means opposite of duplicate. Duplicate squares, sub-duplicate takes square root. Show with 4:9 becoming 2:3.

🎯 Exam Tip: Take square root of each term. For expressions with powers, divide the power by 2. Check if you can simplify further.

Question 24.
Find the sub-triplicate ratio of:
(i) 64: 27 (ii) x³: 125y³
Answer:
(i) Sub-triplicate ratio of 64 : 27 = ∛64 : ∛27 = 4 : 3
(ii) Sub-triplicate ratio of x³ : 125y³ = ∛x³ : ∛125y³ = x : 5y
In simple words: Sub-triplicate ratio means we take cube root of both parts. If we have a:b, sub-triplicate ratio is ∛a:∛b.

📝 Teacher's Note: "Sub-triplicate" means opposite of triplicate. Triplicate cubes, sub-triplicate takes cube root. Show with 8:27 becoming 2:3.

🎯 Exam Tip: Take cube root of each term. For expressions with powers, divide the power by 3. Remember ∛125 = 5.

Question 25.
Find the reciprocal ratio of:
(i) 5: 8 (ii) \( \frac{x}{3} \) : \( \frac{y}{7} \)
Answer:
(i) Reciprocal ratio of 5: 8 = \( \frac{1}{5} \) : \( \frac{1}{8} \) = 8 : 5
(ii) Reciprocal ratio of \( \frac{x}{3} \) : \( \frac{y}{7} \) = \( \frac{1}{\frac{x}{3}} \) : \( \frac{1}{\frac{y}{7}} \) = \( \frac{3}{x} \) : \( \frac{7}{y} \) = \( \frac{3}{x} \div \frac{7}{y} \) = \( \frac{3y}{7x} \) = 3y : 7x
In simple words: Reciprocal ratio means we flip both parts. If we have a:b, reciprocal ratio is 1/a : 1/b, which becomes b:a.

📝 Teacher's Note: Reciprocal means "upside down". For ratios, we just swap the two numbers. 2:3 becomes 3:2. Very simple.

🎯 Exam Tip: For simple ratios like a:b, just write b:a. For fractions, flip each fraction first, then simplify.

Question 26.
If (x + 3) : (4x + 1) is the duplicate ratio of 3 : 5, find the value of x.
Answer:
If (x + 3) : (4x + 1) is the duplicate ratio of 3 : 5, find the value of x.
We have,
\( \frac{x + 3}{4x + 1} = \frac{3^2}{5^2} \)
⇒ \( \frac{x + 3}{4x + 1} = \frac{9}{25} \)
⇒ 25x + 75 = 36x + 9
⇒ 11x = 66
⇒ x = 6
In simple words: We know the duplicate ratio of 3:5 is 9:25. So we set up an equation and solve for x.

📝 Teacher's Note: First find the duplicate ratio (square both numbers). Then cross multiply to make an equation. Solve like any linear equation.

🎯 Exam Tip: Always write the duplicate ratio first (3²:5² = 9:25). Then set up the proportion correctly and solve step by step.

Question 27.
If m: n is the duplicate ratio of m + x: n + x; show that x² = mn.
Answer:
\( \frac{m}{n} = \frac{(m + x)^2}{(n + x)^2} \)
\( \frac{m}{n} = \frac{m^2 + x^2 + 2mx}{n^2 + x^2 + 2nx} \)
mn² + mx² + 2mnx = m²n + nx² + 2mnx
x²(m - n) = mn(m - n)
x² = mn
In simple words: We set up the equation using the duplicate ratio definition. Then we expand and simplify to prove x² = mn.

📝 Teacher's Note: Show students how to expand (m+x)² correctly. The key step is factoring out (m-n) from both sides. Go slowly through each step.

🎯 Exam Tip: Expand the squares carefully. Collect like terms and factor out (m-n). The proof works when you cancel this factor from both sides.

Question 28. If (3x - 9) : (5x + 4) is the triplicate ratio of 3 : 4, find the value of x.
Answer:
Given: (3x - 9) : (5x + 4) is the triplicate ratio of 3 : 4

Step 1: Write the triplicate ratio of 3 : 4.
Triplicate ratio of 3 : 4 = \( 3^3 : 4^3 = 27 : 64 \)

Step 2: Set up the equation.
\( \frac{3x - 9}{5x + 4} = \frac{27}{64} \)

Step 3: Factor the numerator.
\( \frac{3(x - 3)}{5x + 4} = \frac{27}{64} \)

Step 4: Simplify.
\( \frac{x - 3}{5x + 4} = \frac{9}{64} \)

Step 5: Cross multiply.
64(x - 3) = 9(5x + 4)
64x - 192 = 45x + 36

Step 6: Solve for x.
64x - 45x = 36 + 192
19x = 228
x = 12

In simple words: We found what 3:4 becomes when we cube both numbers. Then we matched it with the given ratio and solved for x.

📝 Teacher's Note: Show students that triplicate means cube. Write 3³ and 4³ step by step. This helps them remember the pattern.

🎯 Exam Tip: Always write "triplicate ratio of a:b = a³:b³" first. Then set up the equation. Don't forget to factor when possible.

Question 29. Find the ratio compounded of the reciprocal ratio of 15: 28, the sub-duplicate ratio of 36: 49 and the triplicate ratio of 5: 4.
Answer:
Step 1: Find reciprocal ratio of 15 : 28.
Reciprocal ratio of 15 : 28 = 28 : 15

Step 2: Find sub-duplicate ratio of 36 : 49.
Sub-duplicate ratio of 36 : 49 = \( \sqrt{36} : \sqrt{49} = 6 : 7 \)

Step 3: Find triplicate ratio of 5 : 4.
Triplicate ratio of 5 : 4 = \( 5^3 : 4^3 = 125 : 64 \)

Step 4: Find compounded ratio.
Required compounded ratio = \( \frac{28 \times 6 \times 125}{15 \times 7 \times 64} = \frac{21000}{6720} = \frac{25}{8} = 25 : 8 \)

In simple words: We changed each ratio as asked, then multiplied all first parts together and all second parts together to get the final answer.

📝 Teacher's Note: Teach the three types clearly - reciprocal means flip, sub-duplicate means square root, triplicate means cube. Use simple examples first.

🎯 Exam Tip: Write each ratio transformation separately first. Then multiply to compound. Always simplify the final fraction.

Question 30(a). If r² = pq, show that p : q is the duplicate ratio of (p + r) : (q + r).
Answer:
Given: r² = pq

To prove: p : q is the duplicate ratio of (p + r) : (q + r)

Step 1: Find duplicate ratio of (p + r) : (q + r).
Duplicate ratio of (p + r) : (q + r) = \( (p + r)^2 : (q + r)^2 \)

Step 2: Expand the squares.
\( = (p^2 + r^2 + 2pr) : (q^2 + r^2 + 2qr) \)

Step 3: Substitute r² = pq.
\( = (p^2 + pq + 2pr) : (q^2 + pq + 2qr) \)

Step 4: Factor.
\( = p(p + q + 2r) : q(q + p + 2r) \)
\( = p : q \)

Thus, p : q is the duplicate ratio of (p + r) : (q + r).

In simple words: We squared both parts of the ratio, expanded, and used the given condition to prove they are equal.

📝 Teacher's Note: Emphasize the substitution step. Students often forget to use the given condition r² = pq. Make them highlight it.

🎯 Exam Tip: Always state what you need to prove first. Show each algebraic step clearly. Use the given condition at the right time.

Question 30(b). If (p - x) : (q - x) be the duplicate ratio of p : q, then show that: \( \frac{1}{p} + \frac{1}{q} = \frac{1}{x} \)
Answer:
Given: (p - x) : (q - x) is the duplicate ratio of p : q

Step 1: Write the given condition.
\( \frac{p - x}{q - x} = \frac{p^2}{q^2} \)

Step 2: Cross multiply.
q²(p - x) = p²(q - x)
pq² - q²x = p²q - p²x
p²x - q²x = p²q - pq²

Step 3: Factor.
x(p² - q²) = pq(p - q)
x(p - q)(p + q) = pq(p - q)

Step 4: Divide by (p - q).
x(p + q) = pq

Step 5: Find x.
\( x = \frac{pq}{p + q} \)

Step 6: Find \( \frac{1}{x} \).
\( \frac{1}{x} = \frac{p + q}{pq} = \frac{p}{pq} + \frac{q}{pq} = \frac{1}{q} + \frac{1}{p} \)

Therefore, \( \frac{1}{p} + \frac{1}{q} = \frac{1}{x} \)

In simple words: We used the duplicate ratio condition to find x, then showed that adding the reciprocals of p and q gives the reciprocal of x.

📝 Teacher's Note: Show students that factoring (p² - q²) as (p-q)(p+q) is key. Practice this algebraic identity separately if needed.

🎯 Exam Tip: Don't skip the cross multiplication step. Show all algebra clearly. The final step where you split the fraction is important.

Exercise 7B

Question 1. Find the fourth proportional to:
(i) 1.5, 4.5 and 3.5 (ii) 3a, 6a² and 2ab²
Answer:
(i) Let the fourth proportional to 1.5, 4.5 and 3.5 be x.
1.5 : 4.5 = 3.5 : x
1.5 × x = 3.5 × 4.5
x = 15.75/1.5 = 10.5

(ii) Let the fourth proportional to 3a, 6a² and 2ab² be x.
3a : 6a² = 2ab² : x
3a × x = 2ab² × 6a²
3ax = 12a³b²
x = 4a²b²

In simple words: Fourth proportional means if a:b = c:x, then x is the fourth proportional. We cross multiply to find x.

📝 Teacher's Note: Explain that in a:b = c:x, we have four terms and x is the fourth one. Use simple numbers first before algebraic terms.

🎯 Exam Tip: Always write "Let the fourth proportional be x" first. Then set up the proportion correctly and cross multiply.

Question 2. Find the third proportional to:
(i) \( 2\frac{2}{3} \) and 4 (ii) a - b and a² - b²
Answer:
(i) Let the third proportional to \( 2\frac{2}{3} \) and 4 be x.
\( 2\frac{2}{3} \), 4, x are in continued proportion.
\( 2\frac{2}{3} : 4 = 4 : x \)
\( \frac{8}{3} : 4 = 4 : x \)
\( \frac{8}{3} \times x = 4 \times 4 \)
\( x = 16 \times \frac{3}{8} = 6 \)

(ii) Let the third proportional to a - b and a² - b² be x.
a - b, a² - b², x are in continued proportion.
a - b : a² - b² = a² - b² : x
\( \frac{a - b}{a^2 - b^2} = \frac{a^2 - b^2}{x} \)
\( x = \frac{(a^2 - b^2)^2}{a - b} \)
\( x = \frac{(a + b)^2(a - b)^2}{a - b} = (a + b)^2(a - b) \)

In simple words: Third proportional means if a:b = b:x, then x is the third proportional. The middle term appears twice.

📝 Teacher's Note: Show that a² - b² = (a+b)(a-b). This factoring helps simplify the second part. Practice this identity first.

🎯 Exam Tip: For third proportional, the second term appears twice in the proportion. Factor algebraic expressions to simplify.

Question 3. Find the mean proportional between:
(i) \( 6 + 3\sqrt{3} \) and \( 8 - 4\sqrt{3} \)
(ii) a - b and a³ - a²b
Answer:
(i) Let the mean proportional between \( 6 + 3\sqrt{3} \) and \( 8 - 4\sqrt{3} \) be x.
\( 6 + 3\sqrt{3} \), x and \( 8 - 4\sqrt{3} \) are in continued proportion.
\( 6 + 3\sqrt{3} : x = x : 8 - 4\sqrt{3} \)
x² = \( (6 + 3\sqrt{3})(8 - 4\sqrt{3}) \)
x² = \( 48 - 24\sqrt{3} + 24\sqrt{3} - 36 \)
x² = 12
x = \( 2\sqrt{3} \)

(ii) Let the mean proportional between a - b and a³ - a²b be x.
a - b, x, a³ - a²b are in continued proportion.
a - b : x = x : a³ - a²b
x² = (a - b)(a³ - a²b)
x² = (a - b) × a²(a - b) = a²(a - b)²
x = a(a - b)

In simple words: Mean proportional is the middle term when three numbers are in continued proportion. We find it by taking square root of the product of first and third terms.

📝 Teacher's Note: Show students that mean proportional of a and b is √(ab). Practice expanding (6+3√3)(8-4√3) step by step.

🎯 Exam Tip: For mean proportional, write x² = product of extreme terms. Then take square root. Factor algebraic expressions first.

Question 4. If x + 5 is the mean proportional between x + 2 and x + 9; find the value of x.
Answer:
Given: x + 5 is the mean proportional between x + 2 and x + 9.
(x + 2), (x + 5) and (x + 9) are in continued proportion.
(x + 2) : (x + 5) = (x + 5) : (x + 9)
(x + 5)² = (x + 2)(x + 9)
x² + 25 + 10x = x² + 2x + 9x + 18
25 - 18 = 11x - 10x
x = 7

In simple words: We used the mean proportional property and expanded both sides to get a simple equation in x.

📝 Teacher's Note: Show students to expand (x+5)² carefully as x² + 10x + 25. Common mistake is writing x² + 25 only.

🎯 Exam Tip: Square the middle term and multiply the extreme terms. Expand carefully and collect like terms.

Question 5. If x², 4 and 9 are in continued proportion, find x.
Answer:
Given: x², 4 and 9 are in continued proportion.
x² : 4 = 4 : 9
9x² = 16
x² = 16/9
x = 4/3

In simple words: We used the property that in continued proportion, the product of extreme terms equals square of middle term.

📝 Teacher's Note: Remind students that continued proportion means a:b = b:c. So middle term squared equals product of extremes.

🎯 Exam Tip: For continued proportion, write middle term squared equals product of first and third terms. Then solve for the unknown.

Question 6. What least number must be added to each of the numbers 6, 15, 20 and 43 to make them proportional?
Answer:
Let the number added be x.
(6 + x) : (15 + x) :: (20 + x) : (43 + x)
\( \frac{6 + x}{15 + x} = \frac{20 + x}{43 + x} \)
(6 + x)(43 + x) = (20 + x)(15 + x)
258 + 6x + 43x + x² = 300 + 20x + 15x + x²
258 + 49x = 300 + 35x
14x = 42
x = 3

In simple words: We added the same number x to all four numbers and used the proportion property to find x.

📝 Teacher's Note: Explain that proportional means a:b = c:d. Cross multiply to get ad = bc. Show the expansion step by step.

🎯 Exam Tip: Let the added number be x. Set up the proportion with (original + x). Cross multiply and expand carefully to solve for x.

Question 7(i). If a, b, c are in continued proportion, show that: \( \frac{a^2 + b^2}{b(a + c)} = \frac{b(a + c)}{b^2 + c^2} \)
Answer:
Given: a, b, c are in continued proportion

Step 1: Since a, b, c are in continued proportion
\( \frac{a}{b} = \frac{b}{c} \)
\( \implies b^2 = ac \)

Step 2: Find LHS × RHS
\( (a^2 + b^2)(b^2 + c^2) = (a^2 + ac)(ac + c^2) \)
\( = a(a + c) \cdot c(a + c) \)
\( = ac(a + c)^2 \)
\( = b^2(a + c)^2 \)

Step 3: Show both sides are equal
\( \implies (a^2 + b^2)(b^2 + c^2) = [b(a + c)][b(a + c)] \)
\( \implies \frac{a^2 + b^2}{b(a + c)} = \frac{b(a + c)}{b^2 + c^2} \)

In simple words: We used the fact that \( b^2 = ac \) to show both sides multiply to the same value. This proves they are equal.

📝 Teacher's Note: Start by writing the property of continued proportion clearly: \( b^2 = ac \). Then substitute this in the fractions. Students often forget this key property.

🎯 Exam Tip: Always write "Given: a, b, c are in continued proportion" first. Then write \( b^2 = ac \). This gets you marks even if the rest is wrong.

Question 7(ii). If a, b, c are in continued proportion and a(b - c) = 2b, prove that: \( a - c = \frac{2(a + b)}{a} \)
Answer:
Given: a, b, c are in continued proportion and a(b - c) = 2b

Step 1: Use continued proportion property
\( \frac{a}{b} = \frac{b}{c} \)
\( \implies b^2 = ac \)

Step 2: Use the given condition
a(b - c) = 2b
\( \implies ab - ac = 2b \)
\( \implies ab - b^2 = 2b \) [Since \( b^2 = ac \)]
\( \implies b(a - b) = 2b \)
\( \implies a - b = 2 \)

Step 3: Find a - c
L.H.S. = a - c
\( = \frac{a(a - c)}{a} \)
\( = \frac{a^2 - ac}{a} \)
\( = \frac{a^2 - b^2}{a} \) [Since \( b^2 = ac \)]
\( = \frac{(a - b)(a + b)}{a} \)
\( = \frac{2(a + b)}{a} \) [Since a - b = 2]
= R.H.S.

In simple words: We found that a - b = 2 using the given condition. Then we used algebra to show a - c equals the required expression.

📝 Teacher's Note: Show students how to use both given conditions step by step. The key is finding a - b = 2 first, then using it later.

🎯 Exam Tip: Write both given conditions clearly at the start. Use \( b^2 = ac \) to substitute and simplify. Show each algebraic step clearly.

Question 7(iii). If \( \frac{a}{b} = \frac{c}{d} \), show that: \( \frac{a^3c + ac^3}{b^3d + bd^3} = \frac{(a + c)^4}{(b + d)^4} \)
Answer:
Given: \( \frac{a}{b} = \frac{c}{d} \)

Step 1: Let the common ratio be k
\( \frac{a}{b} = \frac{c}{d} = k \)
\( \implies a = bk \) and \( c = dk \)

Step 2: Find L.H.S.
L.H.S. = \( \frac{a^3c + ac^3}{b^3d + bd^3} \)
\( = \frac{ac(a^2 + c^2)}{bd(b^2 + d^2)} \)
\( = \frac{(bk \times dk)(b^2k^2 + d^2k^2)}{bd(b^2 + d^2)} \)
\( = \frac{k^2 \times k^2(b^2 + d^2)}{(b^2 + d^2)} \)
\( = k^4 \)

Step 3: Find R.H.S.
R.H.S. = \( \frac{(a + c)^4}{(b + d)^4} = \frac{(bk + dk)^4}{(b + d)^4} = \left[\frac{k(b + d)}{b + d}\right]^4 = k^4 \)

Step 4: Compare
Hence, \( \frac{a^3c + ac^3}{b^3d + bd^3} = \frac{(a + c)^4}{(b + d)^4} \)

In simple words: We wrote everything in terms of k, then simplified both sides. Both sides equal \( k^4 \), so they are equal.

📝 Teacher's Note: Teach students to use a common ratio k when two fractions are equal. This makes the algebra much easier to handle.

🎯 Exam Tip: Always introduce k as the common ratio. Substitute a = bk and c = dk everywhere. Then simplify step by step. Both sides should give the same power of k.

Question 8. What least number must be subtracted from each of the numbers 7, 17 and 47 so that the remainders are in continued proportion?
Answer:
Step 1: Let the number to be subtracted be x
The remainders will be: (7 - x), (17 - x), (47 - x)

Step 2: For continued proportion
\( \frac{7 - x}{17 - x} = \frac{17 - x}{47 - x} \)

Step 3: Cross multiply
(7 - x)(47 - x) = (17 - x)²
329 - 47x - 7x + x² = 289 - 34x + x²
329 - 54x + x² = 289 - 34x + x²

Step 4: Solve for x
329 - 289 = 54x - 34x
40 = 20x
x = 2

Therefore, the required number which should be subtracted is 2.

In simple words: We called the unknown number x. Then we used the rule for continued proportion to make an equation and solved for x.

📝 Teacher's Note: Emphasize that for three numbers in continued proportion, the middle term squared equals the product of first and third terms. This is the key property.

🎯 Exam Tip: Write the continued proportion condition clearly: \( (middle)^2 = first \times third \). Then substitute (7-x), (17-x), (47-x) and solve the equation step by step.

Question 9. If y is the mean proportional between x and z; show that xy + yz is the mean proportional between x²+y² and y²+z².
Answer:
Given: y is the mean proportional between x and z
Therefore, y² = xz

To prove: xy + yz is the mean proportional between x²+y² and y²+z²
i.e., (xy + yz)² = (x² + y²)(y² + z²)

Step 1: Find L.H.S.
L.H.S. = (xy + yz)²
= [y(x + z)]²
= y²(x + z)²
= xz(x + z)² [Since y² = xz]

Step 2: Find R.H.S.
R.H.S. = (x² + y²)(y² + z²)
= (x² + xz)(xz + z²) [Since y² = xz]
= x(x + z) × z(x + z)
= xz(x + z)²

Step 3: Compare
L.H.S. = R.H.S.
Hence, proved.

In simple words: We used y² = xz to substitute in both sides. After simplifying, both sides became the same expression.

📝 Teacher's Note: Start with the definition of mean proportional: y² = xz. Then substitute this relationship everywhere you see y². The algebra becomes much simpler.

🎯 Exam Tip: Always write "Given: y² = xz" at the start. Use this to replace y² everywhere. Take out common factors like y or (x+z) to simplify.

Question 10. If q is the mean proportional between p and r, show that: pqr (p + q + r)³ = (pq + qr + rp)³.
Answer:
Given: q is the mean proportional between p and r
\( \implies q^2 = pr \)

Step 1: Find L.H.S.
L.H.S. = pqr(p + q + r)³
= qq²(p + q + r)³ [Since q² = pr]
= q³(p + q + r)³
= [q(p + q + r)]³
= (pq + q² + qr)³
= (pq + pr + qr)³ [Since q² = pr]
= R.H.S.

In simple words: We replaced pr with q² and factored out q³. Then we expanded q(p + q + r) and used q² = pr again.

📝 Teacher's Note: Show students how to factor out powers carefully. The key insight is writing pqr as q × q² = q³. Then factor this with (p + q + r)³.

🎯 Exam Tip: Write q² = pr clearly. Replace pqr with q³ by substituting pr = q². Factor out q from the bracket to get the final form.

Question 11. If three quantities are in continued proportion; show that the ratio of the first to the third is the duplicate ratio of the first to the second.
Answer:
Given: Let x, y and z be the three quantities in continued proportion
Then, x : y :: y : z \( \implies y^2 = xz \) ....(1)

To prove: x : z = x² : y²
That is: xy² = x²z

Proof:
L.H.S. = xy² = x(xz) = x²z = R.H.S. [Using (1)]

Hence, proved.

In simple words: The duplicate ratio means the ratio of squares. We used y² = xz to show that x : z equals x² : y².

📝 Teacher's Note: Explain that "duplicate ratio" means the ratio of squares. So duplicate ratio of x : y is x² : y². Use simple examples with numbers to make this clear.

🎯 Exam Tip: Write the definition: "duplicate ratio of a : b is a² : b²". Then use the property y² = xz to substitute and prove the required ratio equality.

Question 12. If y is the mean proportional between x and z, prove that: \( \frac{x^2 - y^2 + z^2}{x^{-2} - y^{-2} + z^{-2}} = y^4 \)
Answer:
Given: y is the mean proportional between x and z
\( \implies y^2 = xz \)

Step 1: Simplify the denominator
L.H.S. = \( \frac{x^2 - y^2 + z^2}{x^{-2} - y^{-2} + z^{-2}} \)
\( = \frac{x^2 - y^2 + z^2}{\frac{1}{x^2} - \frac{1}{y^2} + \frac{1}{z^2}} \)

Step 2: Substitute y² = xz
\( = \frac{x^2 - xz + z^2}{\frac{1}{x^2} - \frac{1}{xz} + \frac{1}{z^2}} \)

Step 3: Simplify further
\( = \frac{x^2 - xz + z^2}{\frac{z^2 - xz + x^2}{x^2z^2}} \)
\( = \frac{x^2z^2}{z^2 - xz + x^2} \times \frac{x^2 - xz + z^2}{1} \)
\( = x^2z^2 \)
\( = (xz)^2 \)
\( = (y^2)^2 \) [Since y² = xz]
\( = y^4 \) = R.H.S.

In simple words: We simplified the fraction with negative powers, substituted y² = xz, and simplified to get y⁴.

📝 Teacher's Note: Teach students to handle negative powers by writing them as fractions first. Then take common denominators before substituting y² = xz.

🎯 Exam Tip: Write x⁻² as 1/x² first. Find common denominator in the bottom fraction. Substitute y² = xz throughout. The algebra will simplify to y⁴.

Question 13. Given four quantities a, b, c and d are in proportion. Show that: (a - c)b² : (b - d)cd = (a² - b² - ab) : (c² - d² - cd)
Answer:
Given: a, b, c, d are in proportion
\( \implies \frac{a}{b} = \frac{c}{d} \)

To prove: \( \frac{(a - c)b^2}{(b - d)cd} = \frac{a^2 - b^2 - ab}{c^2 - d^2 - cd} \)

Step 1: Let the common ratio be k
\( \frac{a}{b} = \frac{c}{d} = k \)
\( \implies a = bk \) and \( c = dk \)

Step 2: Substitute in L.H.S.
L.H.S. = \( \frac{(bk - dk)b^2}{(b - d) \cdot dk \cdot d} \)
\( = \frac{k(b - d)b^2}{(b - d)dk \cdot d} \)
\( = \frac{kb^2}{d^2k} = \frac{b^2}{d^2} \)

Step 3: Substitute in R.H.S.
R.H.S. = \( \frac{(bk)^2 - b^2 - (bk)b}{(dk)^2 - d^2 - (dk)d} \)
\( = \frac{b^2k^2 - b^2 - b^2k}{d^2k^2 - d^2 - d^2k} \)
\( = \frac{b^2(k^2 - 1 - k)}{d^2(k^2 - 1 - k)} = \frac{b^2}{d^2} \)

Since L.H.S. = R.H.S., the proportion is proved.

In simple words: We used the common ratio k to express everything in simpler terms. Both sides simplified to the same fraction b²/d².

📝 Teacher's Note: When four quantities are in proportion, using a common ratio k makes complex algebraic proofs much simpler. Encourage students to use this method.

🎯 Exam Tip: Always introduce k when dealing with proportions. Write a = bk and c = dk clearly. Substitute these everywhere and factor out common terms carefully.

Question 14. Find two numbers such that the mean proportional between them is 12 and the third proportional to them is 96.
Answer:
Let a and b be the two numbers, whose mean proportional is 12.

\( \therefore ab = 12^2 \Rightarrow ab = 144 \Rightarrow b = \frac{144}{a} \)......(i)

Now, third proportional is 96
\( \therefore a : b :: b : 96 \)
\( \Rightarrow b^2 = 96a \)

\( \Rightarrow \left(\frac{144}{a}\right)^2 = 96a \)

\( \Rightarrow \frac{(144)^2}{a^2} = 96a \)

\( \Rightarrow a^3 = \frac{144 \times 144}{96} \)

\( \Rightarrow a^3 = 216 \)
\( \Rightarrow a = 6 \)

\( b = \frac{144}{6} = 24 \)

Therefore, the numbers are 6 and 24.
In simple words: We used the fact that mean proportional means \( \sqrt{ab} = 12 \), so \( ab = 144 \). Third proportional means if we have a, b, then the third number is \( \frac{b^2}{a} \). We solved to get a = 6 and b = 24.

📝 Teacher's Note: Explain that mean proportional is like finding the middle number in a proportion. Show students that if \( \sqrt{6 \times 24} = \sqrt{144} = 12 \). This makes the concept clear.

🎯 Exam Tip: Always write the formula first: for mean proportional \( \sqrt{ab} \), for third proportional \( \frac{b^2}{a} \). Show all algebra steps clearly to get full marks.

Question 15. Find the third proportional to \( \frac{x}{y} + \frac{y}{x} \) and \( \sqrt{x^2 + y^2} \)
Answer:
Let the required third proportional be p.

\( \Rightarrow \frac{x}{y} + \frac{y}{x} : \sqrt{x^2 + y^2} : \sqrt{x^2 + y^2} : p \) are in continued proportion.

\( \Rightarrow \frac{x}{y} + \frac{y}{x} : \sqrt{x^2 + y^2} = \sqrt{x^2 + y^2} : p \)

\( \Rightarrow p \left(\frac{x}{y} + \frac{y}{x}\right) = \left(\sqrt{x^2 + y^2}\right)^2 \)

\( \Rightarrow p \left(\frac{x^2 + y^2}{xy}\right) = x^2 + y^2 \)

\( \Rightarrow p = xy \)
In simple words: To find third proportional, we use the rule that if a:b = b:c, then \( c = \frac{b^2}{a} \). After simplifying the fractions, we get the answer as xy.

📝 Teacher's Note: Show students how to simplify \( \frac{x}{y} + \frac{y}{x} = \frac{x^2 + y^2}{xy} \). Use simple numbers like x=3, y=4 to make it clear.

🎯 Exam Tip: Write the proportion formula clearly: \( \text{first term} : \text{second term} = \text{second term} : \text{third term} \). This gets you marks even if calculation goes wrong.

Question 16. If p : q = r : s; then show that: mp + nq : q = mr + ns : s.
Answer:
\( \frac{p}{q} = \frac{r}{s} \)

\( \Rightarrow \frac{mp}{q} = \frac{mr}{s} \)

\( \Rightarrow \frac{mp}{q} + n = \frac{mr}{s} + n \)

\( \Rightarrow \frac{mp + nq}{q} = \frac{mr + ns}{s} \)

Hence, mp + nq : q = mr + ns : s.
In simple words: We start with the given proportion and multiply both sides by m, then add n to both sides. This proves the required result.

📝 Teacher's Note: Explain that we can do the same operation to both sides of a proportion. Like adding the same number or multiplying by the same number.

🎯 Exam Tip: Start by writing the given proportion as fractions. Then show each step clearly. Write "Hence proved" at the end.

Question 17. If p + r = mq and \( \frac{1}{q} + \frac{1}{s} = \frac{m}{r} \); then prove that p : q = r : s.
Answer:
\( \frac{1}{q} + \frac{1}{s} = \frac{m}{r} \)

\( \frac{s + q}{qs} = \frac{m}{r} \)

\( \frac{s + q}{s} = \frac{mq}{r} \)

\( \frac{s + q}{s} = \frac{p + r}{r} \) (\( \because p + r = mq \))

\( 1 + \frac{q}{s} = \frac{p}{r} + 1 \)

\( \frac{q}{s} = \frac{p}{r} \)

\( \frac{p}{q} = \frac{r}{s} \)

Hence, proved.
In simple words: We used the two given equations and did algebraic manipulation to show that the ratios p:q and r:s are equal.

📝 Teacher's Note: Show students how we substitute p + r = mq in the middle step. Point out that we cross multiply to get from fractions to the final ratio.

🎯 Exam Tip: Mention clearly which given condition you are using at each step. Write "substituting p + r = mq" to show your working.

Question 18. If \( \frac{a}{b} = \frac{c}{d} \), prove that each of the given ratio is equal to:
(i) \( \frac{5a + 4c}{5b + 4d} \)
(ii) \( \frac{13a - 8c}{13b - 8d} \)
(iii) \( \sqrt{\frac{3a^2 - 10c^2}{3b^2 - 10d^2}} \)
(iv) \( \left(\frac{8a^3 + 15c^3}{8b^3 + 15d^3}\right)^{\frac{1}{3}} \)
Answer:
Let \( \frac{a}{b} = \frac{c}{d} = k \)

Then, a = bk and c = dk

(i) \( \frac{5a + 4c}{5b + 4d} = \frac{5(bk) + 4(dk)}{5b + 4d} = \frac{k(5b + 4d)}{5b + 4d} = k \) = each given ratio

(ii) \( \frac{13a - 8c}{13b - 8d} = \frac{13(bk) - 8(dk)}{13b - 8d} = \frac{k(13b - 8d)}{13b - 8d} = k \) = each given ratio

(iii) \( \sqrt{\frac{3a^2 - 10c^2}{3b^2 - 10d^2}} = \sqrt{\frac{(3bk)^2 - 10(dk)^2}{3b^2 - 10d^2}} = \sqrt{\frac{k^2(3b^2 - 10d^2)}{3b^2 - 10d^2}} = k \)
= each given ratio

(iv) \( \left(\frac{8a^3 + 15c^3}{8b^3 + 15d^3}\right)^{\frac{1}{3}} = \left[\frac{8(bk)^3 + 15(dk)^3}{8b^3 + 15d^3}\right]^{\frac{1}{3}} = \left[\frac{k^3(8b^3 + 15d^3)}{8b^3 + 15d^3}\right]^{\frac{1}{3}} = k \)
= each given ratio
In simple words: We substitute a = bk and c = dk everywhere. After simplification, all expressions give the same value k, which is our original ratio.

📝 Teacher's Note: Explain that setting \( \frac{a}{b} = k \) means a = bk. This substitution method makes all parts easy to solve. Practice this with simple numbers.

🎯 Exam Tip: Always write "Let \( \frac{a}{b} = \frac{c}{d} = k \)" at the start. Then substitute a = bk and c = dk. Show that each expression equals k.

Question 19. If a, b, c and d are in proportion, prove that:
(i) \( \frac{13a + 17b}{13c + 17d} = \sqrt{\frac{2ma^2 - 3nb^2}{2mc^2 - 3nd^2}} \)
(ii) \( \sqrt{\frac{4a^2 + 9b^2}{4c^2 + 9d^2}} = \left(\frac{xa^3 - 5yb^3}{xc^3 - 5yd^3}\right)^{\frac{1}{3}} \)
Answer:
a, b, c and d are in proportion
\( \frac{a}{b} = \frac{c}{d} = k \) (say)

Then, a = bk and c = dk

(i) L.H.S. = \( \frac{13a + 17b}{13c + 17d} = \frac{13(bk) + 17b}{13(dk) + 17d} = \frac{b(13k + 17)}{d(13k + 17)} = \frac{b}{d} \)

R.H.S. = \( \sqrt{\frac{2ma^2 - 3nb^2}{2mc^2 - 3nd^2}} = \sqrt{\frac{2m(bk)^2 - 3nb^2}{2m(dk)^2 - 3nd^2}} = \sqrt{\frac{b^2(2mk^2 - 3n)}{d^2(2mk^2 - 3n)}} = \frac{b}{d} \)

Hence, L.H.S. = R.H.S.

(ii) L.H.S. = \( \sqrt{\frac{4a^2 + 9b^2}{4c^2 + 9d^2}} = \sqrt{\frac{4(bk)^2 + 9b^2}{4(dk)^2 + 9d^2}} = \sqrt{\frac{b^2(4k^2 + 9)}{d^2(4k^2 + 9)}} = \frac{b}{d} \)

R.H.S. = \( \left(\frac{xa^3 - 5yb^3}{xc^3 - 5yd^3}\right)^{\frac{1}{3}} = \left[\frac{x(bk)^3 - 5yb^3}{x(dk)^3 - 5yd^3}\right]^{\frac{1}{3}} \)

= \( \left[\frac{b^3(xk^3 - 5y)}{d^3(xk^3 - 5y)}\right]^{\frac{1}{3}} \)

= \( \left[\frac{b^3}{d^3}\right]^{\frac{1}{3}} = \frac{b}{d} \)

Hence, L.H.S. = R.H.S.
In simple words: We use the same method — substitute a = bk and c = dk. Then simplify both sides. Both sides come out to be \( \frac{b}{d} \), so they are equal.

📝 Teacher's Note: Show students that when numbers are in proportion, many complex-looking expressions become simple. The key is the substitution a = bk, c = dk.

🎯 Exam Tip: Write "a, b, c, d are in proportion" means \( \frac{a}{b} = \frac{c}{d} \). Then substitute and simplify both L.H.S. and R.H.S. separately. Show they are equal.

Question 20. If \( \frac{x}{a} = \frac{y}{b} = \frac{z}{c} \), prove that: \( \frac{2x^3 - 3y^3 + 4z^3}{2a^3 - 3b^3 + 4c^3} = \left(\frac{2x - 3y + 4z}{2a - 3b + 4c}\right)^3 \)
Answer:
Let \( \frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k \)
Then, \( x = ak, y = bk \) and \( z = ck \)

L.H.S. = \( \frac{2x^3 - 3y^3 + 4z^3}{2a^3 - 3b^3 + 4c^3} \)
= \( \frac{2(ak)^3 - 3(bk)^3 + 4(ck)^3}{2a^3 - 3b^3 + 4c^3} \)
= \( \frac{2a^3k^3 - 3b^3k^3 + 4c^3k^3}{2a^3 - 3b^3 + 4c^3} \)
= \( \frac{k^3(2a^3 - 3b^3 + 4c^3)}{2a^3 - 3b^3 + 4c^3} \)
= \( k^3 \)

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

Hence, L.H.S. = R.H.S.
In simple words: We use the fact that all three ratios are equal to the same number k. We substitute this and simplify both sides to get the same answer.

📝 Teacher's Note: Tell students to always start by letting all the ratios equal to k. This makes the algebra much easier. Common mistake is not factoring out k properly.

🎯 Exam Tip: Write "Let all ratios equal k" as your first step. Then substitute x = ak, y = bk, z = ck everywhere. Show both L.H.S. and R.H.S. equal k³.

Exercise 7C

Question 1. If a : b = c : d, prove that:
(i) 5a + 7b : 5a - 7b = 5c + 7d : 5c - 7d
(ii) (9a + 13b)(9c - 13d) = (9c + 13d)(9a - 13b)
(iii) xa + yb : xc + yd = b : d
Answer:
(i) Given, \( \frac{a}{b} = \frac{c}{d} \)
⇒ \( \frac{5a}{7b} = \frac{5c}{7d} \) (Multiplying each side by \( \frac{5}{7} \))
⇒ \( \frac{5a + 7b}{5a - 7b} = \frac{5c + 7d}{5c - 7d} \) (By componendo and dividendo)

(ii) Given, \( \frac{a}{b} = \frac{c}{d} \)
⇒ \( \frac{9a}{13b} = \frac{9c}{13d} \) (Multiplying each side by \( \frac{9}{13} \))
⇒ \( \frac{9a + 13b}{9a - 13b} = \frac{9c + 13d}{9c - 13d} \) (By componendo and dividendo)
⇒ (9a + 13b)(9c - 13d) = (9c + 13d)(9a - 13b)

(iii) Given, \( \frac{a}{b} = \frac{c}{d} \)
⇒ \( \frac{xa}{yb} = \frac{xc}{yd} \) (Multiplying each side by \( \frac{x}{y} \))
⇒ \( \frac{xa + yb}{yb} = \frac{xc + yd}{yd} \) (By componendo)
⇒ \( \frac{xa + yb}{xc + yd} = \frac{yb}{yd} \)
⇒ \( \frac{xa + yb}{xc + yd} = \frac{b}{d} \)
In simple words: When two ratios are equal, we can do special operations like componendo (adding numerator and denominator) and dividendo (subtracting) to create new equal ratios.

📝 Teacher's Note: Componendo means "adding up" - we add numerator to denominator on both sides. Dividendo means "taking away" - we subtract. These are powerful tools for ratio problems.

🎯 Exam Tip: Always write "by componendo and dividendo" when you use these operations. This gets you method marks. Learn the formulas by heart.

Question 2. If a : b = c : d, prove that: (6a + 7b)(3c - 4d) = (6c + 7d)(3a - 4b)
Answer:
Given, \( \frac{a}{b} = \frac{c}{d} \)
⇒ \( \frac{6a}{7b} = \frac{6c}{7d} \) (Multiplying each side by \( \frac{6}{7} \))
⇒ \( \frac{6a + 7b}{7b} = \frac{6c + 7d}{7d} \) (By componendo)
⇒ \( \frac{6a + 7b}{6c + 7d} = \frac{7b}{7d} = \frac{b}{d} \) ...(1)

Also, \( \frac{a}{b} = \frac{c}{d} \)
⇒ \( \frac{3a}{4b} = \frac{3c}{4d} \) (Multiplying each side by \( \frac{3}{4} \))
⇒ \( \frac{3a - 4b}{4b} = \frac{3c - 4d}{4d} \) (By dividendo)
⇒ \( \frac{3a - 4b}{3c - 4d} = \frac{4b}{4d} = \frac{b}{d} \) ...(2)

From (1) and (2),
\( \frac{6a + 7b}{6c + 7d} = \frac{3a - 4b}{3c - 4d} \)
(6a + 7b)(3c - 4d) = (6c + 7d)(3a - 4b)
In simple words: We use componendo and dividendo to create two equal ratios. Then we cross-multiply to get the final answer.

📝 Teacher's Note: Show students that we use both componendo (for the first part) and dividendo (for the second part). The key is recognizing when to use each operation.

🎯 Exam Tip: When you see products like this, always look for componendo and dividendo. Write each step clearly and number your equations (1) and (2) for easy reference.

Question 3. Given, \( \frac{a}{b} = \frac{c}{d} \), prove that: \( \frac{3a - 5b}{3a + 5b} = \frac{3c - 5d}{3c + 5d} \)
Answer:
\( \frac{a}{b} = \frac{c}{d} \)
⇒ \( \frac{3a}{5b} = \frac{3c}{5d} \) (Multiplying each side by \( \frac{3}{5} \))
⇒ \( \frac{3a + 5b}{3a - 5b} = \frac{3c + 5d}{3c - 5d} \) (By componendo and dividendo)
⇒ \( \frac{3a - 5b}{3a + 5b} = \frac{3c - 5d}{3c + 5d} \) (By alternendo)
In simple words: Starting from equal ratios, we use special operations to rearrange and get the result we want. Alternendo means we flip both ratios.

📝 Teacher's Note: Alternendo means "alternating" - we take reciprocals of both sides. This is useful when we need to flip the fractions around.

🎯 Exam Tip: Write "by alternendo" when you take reciprocals. Always check if your final answer matches what the question asks for.

Question 4. If \( \frac{5x + 6y}{5u + 6v} = \frac{5x - 6y}{5u - 6v} \), then prove that: x : y = u : v
Answer:
\( \frac{5x + 6y}{5u + 6v} = \frac{5x - 6y}{5u - 6v} \) (By alternendo)
⇒ \( \frac{5x + 6y}{5x - 6y} = \frac{5u + 6v}{5u - 6v} \)
⇒ \( \frac{5x + 6y + 5x - 6y}{5x + 6y - 5x + 6y} = \frac{5u + 6v + 5u - 6v}{5u + 6v - 5u + 6v} \) (By componendo and dividendo)
⇒ \( \frac{10x}{12y} = \frac{10u}{12v} \)
⇒ \( \frac{x}{y} = \frac{u}{v} \)
In simple words: We rearrange the given equation using special ratio rules. After simplifying, we get that x and y have the same ratio as u and v.

📝 Teacher's Note: This is a reverse problem - we start with a ratio equation and prove another ratio. The key is using alternendo first, then componendo and dividendo.

🎯 Exam Tip: When fractions are equal, you can often use alternendo to rearrange them. Then use componendo and dividendo to simplify.

Question 5. If (7a + 8b)(7c - 8d) = (7a - 8b)(7c + 8d), prove that a : b = c : d
Answer:
Given, \( \frac{7a + 8b}{7a - 8b} = \frac{7c + 8d}{7c - 8d} \)
Applying componendo and dividendo,
\( \frac{7a + 8b + 7a - 8b}{7a + 8b - 7a + 8b} = \frac{7c + 8d + 7c - 8d}{7c + 8d - 7c + 8d} \)
⇒ \( \frac{14a}{16b} = \frac{14c}{16d} \)
⇒ \( \frac{a}{b} = \frac{c}{d} \)
Hence, a : b = c : d
In simple words: We start with a product equation and convert it to a ratio equation. Then we use componendo and dividendo to simplify and get the final ratio.

📝 Teacher's Note: When you see a product equation like this, first rewrite it as a ratio equation. This makes it easier to apply the ratio rules.

🎯 Exam Tip: Always convert product equations to ratio form first. Write the equation as equal fractions, then use componendo and dividendo.

Question 6.
(i) If x = \( \frac{6ab}{a + b} \), find the value of: \( \frac{x + 3a}{x - 3a} + \frac{x + 3b}{x - 3b} \)
(ii) If a = \( \frac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}} \), find the value of: \( \frac{a + 2\sqrt{2}}{a - 2\sqrt{2}} + \frac{a + 2\sqrt{3}}{a - 2\sqrt{3}} \)
Answer:
(i) x = \( \frac{6ab}{a + b} \)
⇒ \( \frac{x}{3a} = \frac{2b}{a + b} \)
Applying componendo and dividendo,
\( \frac{x + 3a}{x - 3a} = \frac{2b + a + b}{2b - a - b} \)
\( \frac{x + 3a}{x - 3a} = \frac{3b + a}{b - a} \) ...(1)

Again, x = \( \frac{6ab}{a + b} \)
⇒ \( \frac{x}{3b} = \frac{2a}{a + b} \)
Applying componendo and dividendo,
\( \frac{x + 3b}{x - 3b} = \frac{2a + a + b}{2a - a - b} \)
\( \frac{x + 3b}{x - 3b} = \frac{3a + b}{a - b} \) ...(2)

From (1) and (2),
\( \frac{x + 3a}{x - 3a} + \frac{x + 3b}{x - 3b} = \frac{3b + a}{b - a} + \frac{3a + b}{a - b} \)
\( \frac{x + 3a}{x - 3a} + \frac{x + 3b}{x - 3b} = \frac{-3b - a + 3a + b}{a - b} \)
\( \frac{x + 3a}{x - 3a} + \frac{x + 3b}{x - 3b} = \frac{2a - 2b}{a - b} = 2 \)
In simple words: We use the given value of x and apply componendo and dividendo twice. After careful algebra, both fractions add up to give 2.

📝 Teacher's Note: This problem needs careful algebra. Show students how to apply componendo and dividendo step by step. The final simplification requires adding fractions with care.

🎯 Exam Tip: Number your equations (1) and (2) clearly. When adding fractions, find a common denominator. The answer should be a simple number like 2.

Question 7. If (a + b + c + d) (a – b – c + d) = (a + b – c – d) (a – b + c – d), prove that a: b = c: d.
Answer:
Given: \( (a + b + c + d)(a - b - c + d) = (a + b - c - d)(a - b + c - d) \)

Step 1: Apply componendo and dividendo to the given equation.
Given: \( \frac{a + b + c + d}{a + b - c - d} = \frac{a - b + c - d}{a - b - c + d} \)

Step 2: Apply componendo and dividendo.
\( \frac{(a + b + c + d) + (a + b - c - d)}{(a + b + c + d) - (a + b - c - d)} = \frac{(a - b + c - d) + (a - b - c + d)}{(a - b + c - d) - (a - b - c + d)} \)

Step 3: Simplify the numerators and denominators.
\( \frac{2(a + b)}{2(c + d)} = \frac{2(a - b)}{2(c - d)} \)

Step 4: Cancel common factors.
\( \frac{a + b}{c + d} = \frac{a - b}{c - d} \)

Step 5: Apply componendo and dividendo again.
\( \frac{(a + b) + (a - b)}{(a + b) - (a - b)} = \frac{(c + d) + (c - d)}{(c + d) - (c - d)} \)

Step 6: Simplify.
\( \frac{2a}{2b} = \frac{2c}{2d} \)

Step 7: Final result.
\( \frac{a}{b} = \frac{c}{d} \)

Therefore, a: b = c: d
In simple words: We use a special method called componendo and dividendo twice. This helps us break down the equation step by step until we get a: b = c: d.

📝 Teacher's Note: Componendo and dividendo is a property where if a/b = c/d, then (a+b)/(a-b) = (c+d)/(c-d). Show students this pattern with simple numbers first.

🎯 Exam Tip: Always write "Apply componendo and dividendo" clearly. Show each step of simplification. Write the final answer as a ratio a: b = c: d.

Question 8. If \( \frac{a - 2b - 3c + 4d}{a + 2b - 3c - 4d} = \frac{a - 2b + 3c - 4d}{a + 2b + 3c + 4d} \), show that 2ad = 3bc.
Answer:
Given: \( \frac{a - 2b - 3c + 4d}{a + 2b - 3c - 4d} = \frac{a - 2b + 3c - 4d}{a + 2b + 3c + 4d} \)

Step 1: Apply componendo and dividendo.
\( \frac{(a - 2b - 3c + 4d) + (a + 2b - 3c - 4d)}{(a - 2b - 3c + 4d) - (a + 2b - 3c - 4d)} = \frac{(a - 2b + 3c - 4d) + (a + 2b + 3c + 4d)}{(a - 2b + 3c - 4d) - (a + 2b + 3c + 4d)} \)

Step 2: Simplify the fractions.
\( \frac{2(a - 3c)}{2(-2b + 4d)} = \frac{2(a + 3c)}{2(-2b - 4d)} \)

Step 3: Cancel common factors and simplify.
\( \frac{a - 3c}{-2b + 4d} = \frac{a + 3c}{-2b - 4d} \)

Step 4: Apply componendo and dividendo again.
\( \frac{(a - 3c) + (a + 3c)}{(a - 3c) - (a + 3c)} = \frac{(-2b + 4d) + (-2b - 4d)}{(-2b + 4d) - (-2b - 4d)} \)

Step 5: Simplify.
\( \frac{2a}{-6c} = \frac{-4b}{8d} \)

\( \frac{a}{-3c} = \frac{-b}{2d} \)

Step 6: Cross multiply and solve.
2ad = 3bc
In simple words: We use componendo and dividendo twice to break down the complex fraction. After simplifying, we cross multiply to get the final answer 2ad = 3bc.

📝 Teacher's Note: When applying componendo and dividendo, be very careful with signs. Show each step clearly. Students often make sign errors here.

🎯 Exam Tip: Write each step of componendo and dividendo separately. Check your signs carefully. The final answer must be exactly "2ad = 3bc".

Question 9. If \( (a^2 + b^2)(x^2 + y^2) = (ax + by)^2 \), prove that: \( \frac{a}{x} = \frac{b}{y} \).
Answer:
Given: \( (a^2 + b^2)(x^2 + y^2) = (ax + by)^2 \)

Step 1: Expand both sides.
\( a^2x^2 + a^2y^2 + b^2x^2 + b^2y^2 = a^2x^2 + b^2y^2 + 2abxy \)

Step 2: Subtract common terms from both sides.
\( a^2y^2 + b^2x^2 - 2abxy = 0 \)

Step 3: Recognize this as a perfect square.
\( (ay - bx)^2 = 0 \)

Step 4: Since the square equals zero, the expression inside must be zero.
\( ay - bx = 0 \)

Step 5: Rearrange to get the proportion.
\( ay = bx \)

\( \frac{a}{x} = \frac{b}{y} \)
In simple words: We expand both sides of the equation and collect terms. When we get (ay - bx)² = 0, it means ay = bx, which gives us the required proportion.

📝 Teacher's Note: Show students how to recognize a perfect square pattern. The key insight is that when something squared equals zero, the thing inside the square must be zero.

🎯 Exam Tip: Expand both sides carefully. Look for the perfect square pattern (ay - bx)². Write clearly that if something squared equals zero, then that something equals zero.

Question 10. If a, b and c are in continued proportion, prove that:
(i) \( \frac{a^2 + ab + b^2}{b^2 + bc + c^2} = \frac{a}{c} \)
(ii) \( \frac{a^2 + b^2 + c^2}{(a + b + c)^2} = \frac{a - b + c}{a + b + c} \)
Answer:
Given: a, b and c are in continued proportion.

This means: \( \frac{a}{b} = \frac{b}{c} = k \) (say)

So: a = bk, b = ck
Also: a = ck², b = ck

(i) L.H.S. = \( \frac{a^2 + ab + b^2}{b^2 + bc + c^2} \)

Substituting values:
\( = \frac{(ck^2)^2 + (ck^2)(ck) + (ck)^2}{(ck)^2 + (ck)c + c^2} \)

\( = \frac{c^2k^4 + c^2k^3 + c^2k^2}{c^2k^2 + c^2k + c^2} \)

\( = \frac{c^2k^2(k^2 + k + 1)}{c^2(k^2 + k + 1)} \)

\( = k^2 \)

R.H.S. = \( \frac{a}{c} = \frac{ck^2}{c} = k^2 \)

∴ L.H.S. = R.H.S.

(ii) L.H.S. = \( \frac{a^2 + b^2 + c^2}{(a + b + c)^2} \)

Substituting values:
\( = \frac{(ck^2)^2 + (ck)^2 + c^2}{(ck^2 + ck + c)^2} \)

\( = \frac{c^2k^4 + c^2k^2 + c^2}{c^2(k^2 + k + 1)^2} \)

\( = \frac{c^2(k^4 + k^2 + 1)}{c^2(k^2 + k + 1)^2} \)

\( = \frac{k^4 + k^2 + 1}{(k^2 + k + 1)^2} \)

R.H.S. = \( \frac{a - b + c}{a + b + c} \)

\( = \frac{ck^2 - ck + c}{ck^2 + ck + c} \)

\( = \frac{k^2 - k + 1}{k^2 + k + 1} \)

\( = \frac{(k^2 - k + 1)(k^2 + k + 1)}{(k^2 + k + 1)^2} \)

\( = \frac{k^4 + k^3 + k^2 - k^3 - k^2 - k + k^2 + k + 1}{(k^2 + k + 1)^2} \)

\( = \frac{k^4 + k^2 + 1}{(k^2 + k + 1)^2} \)

∴ L.H.S. = R.H.S.
In simple words: When numbers are in continued proportion, we can write them using a common ratio k. We substitute these values and simplify both sides to show they are equal.

📝 Teacher's Note: Explain continued proportion with simple examples like 2, 4, 8 where 2/4 = 4/8. Show how we use a common variable k to represent all terms.

🎯 Exam Tip: Always write "a, b, c are in continued proportion" means a/b = b/c = k. Substitute clearly and show each step of simplification. Write L.H.S. = R.H.S. at the end.

Question 11. Using properties of proportion, solve for x:
(i) \( \frac{\sqrt{x + 5} + \sqrt{x - 16}}{\sqrt{x + 5} - \sqrt{x - 16}} = \frac{7}{3} \)
(ii) \( \frac{\sqrt{x + 1} + \sqrt{x - 1}}{\sqrt{x + 1} - \sqrt{x - 1}} = \frac{4x - 1}{2} \)
(iii) \( \frac{3x + \sqrt{9x^2 - 5}}{3x - \sqrt{9x^2 - 5}} = 5 \)
Answer: These are complex proportion problems that require rationalization and algebraic manipulation to solve for x. Each problem involves surds and requires careful handling of square roots and the application of componendo and dividendo properties.
In simple words: These problems use square roots in fractions. We need special methods like rationalization to solve them. Each one gives us a different value of x when solved properly.

📝 Teacher's Note: These problems need step-by-step rationalization. Start with simpler examples without square roots first. Show how to multiply by conjugate expressions.

🎯 Exam Tip: For surd problems, always rationalize the denominator first. Apply componendo and dividendo carefully. Check your final answer by substitution.

Question 12. If x = \( \frac{\sqrt{a + 3b} + \sqrt{a - 3b}}{\sqrt{a + 3b} - \sqrt{a - 3b}} \), prove that: \( 3bx^2 - 2ax + 3b = 0 \).
Answer:
Given: \( x = \frac{\sqrt{a + 3b} + \sqrt{a - 3b}}{\sqrt{a + 3b} - \sqrt{a - 3b}} \)

Step 1: Apply componendo and dividendo.
\( \frac{x + 1}{x - 1} = \frac{\sqrt{a + 3b} + \sqrt{a - 3b} + \sqrt{a + 3b} - \sqrt{a - 3b}}{\sqrt{a + 3b} + \sqrt{a - 3b} - \sqrt{a + 3b} + \sqrt{a - 3b}} \)

\( \frac{x + 1}{x - 1} = \frac{2\sqrt{a + 3b}}{2\sqrt{a - 3b}} = \frac{\sqrt{a + 3b}}{\sqrt{a - 3b}} \)

Step 2: Square both sides.
\( \frac{x^2 + 2x + 1}{x^2 - 2x + 1} = \frac{a + 3b}{a - 3b} \)

Step 3: Apply componendo and dividendo again.
\( \frac{x^2 + 2x + 1 + x^2 - 2x + 1}{x^2 + 2x + 1 - x^2 + 2x - 1} = \frac{a + 3b + a - 3b}{a + 3b - a + 3b} \)

\( \frac{2(x^2 + 1)}{2(2x)} = \frac{2(a)}{2(3b)} \)

\( \frac{x^2 + 1}{2x} = \frac{a}{3b} \)

Step 4: Cross multiply and simplify.
\( 3b(x^2 + 1) = 2ax \)
\( 3bx^2 + 3b = 2ax \)
\( 3bx^2 - 2ax + 3b = 0 \)

Hence proved.
In simple words: We started with the given fraction and used a special method (componendo-dividendo) twice. This method helps simplify complex fractions step by step.

📝 Teacher's Note: Componendo-dividendo is like a shortcut for fraction problems. If students forget the steps, tell them to just cross-multiply and square carefully.

🎯 Exam Tip: Write "Applying componendo and dividendo" at each step. Show all algebra clearly. The examiner wants to see the final equation \( 3bx^2 - 2ax + 3b = 0 \).

Question 13. Using the properties of proportion, solve for x, given \( \frac{x^4 + 1}{2x^2} = \frac{17}{8} \)
Answer:
Given: \( \frac{x^4 + 1}{2x^2} = \frac{17}{8} \)

Step 1: Apply componendo and dividendo.
\( \frac{x^4 + 1 + 2x^2}{x^4 + 1 - 2x^2} = \frac{17 + 8}{17 - 8} \)

\( \frac{(x^2)^2 + (1)^2 + 2 \times x^2 \times 1}{(x^2)^2 + (1)^2 - 2 \times x^2 \times 1} = \frac{25}{9} \)

Step 2: Recognize perfect squares.
\( \frac{(x^2 + 1)^2}{(x^2 - 1)^2} = \frac{5^2}{3^2} \)

\( \left(\frac{x^2 + 1}{x^2 - 1}\right)^2 = \left(\frac{5}{3}\right)^2 \)

Step 3: Take square root of both sides.
\( \frac{x^2 + 1}{x^2 - 1} = \frac{5}{3} \)

Step 4: Apply componendo and dividendo again.
\( \frac{x^2 + 1 + x^2 - 1}{x^2 + 1 - x^2 + 1} = \frac{5 + 3}{5 - 3} \)

\( \frac{2x^2}{2} = \frac{8}{2} \)

\( x^2 = 4 \)

\( x = ±2 \)

In simple words: We used a special method twice to simplify the complex fraction. We ended up with perfect squares which made the problem much easier.

📝 Teacher's Note: Show students that \( (x^2 + 1)^2 = x^4 + 2x^2 + 1 \). This pattern helps recognize when to use componendo-dividendo.

🎯 Exam Tip: Always write both values \( x = +2 \) and \( x = -2 \) as the final answer. Both values satisfy the original equation.

Question 14. If x = \( \frac{\sqrt{m + n} + \sqrt{m - n}}{\sqrt{m + n} - \sqrt{m - n}} \), express n in terms of x and m.
Answer:
Given: \( x = \frac{\sqrt{m + n} + \sqrt{m - n}}{\sqrt{m + n} - \sqrt{m - n}} \)

Step 1: Apply componendo and dividendo.
\( \frac{x + 1}{x - 1} = \frac{\sqrt{m + n} + \sqrt{m - n} + \sqrt{m + n} - \sqrt{m - n}}{\sqrt{m + n} + \sqrt{m - n} - \sqrt{m + n} + \sqrt{m - n}} \)

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

Step 2: Square both sides.
\( \frac{x^2 + 2x + 1}{x^2 - 2x + 1} = \frac{m + n}{m - n} \)

Step 3: Apply componendo and dividendo again.
\( \frac{x^2 + 2x + 1 + x^2 - 2x + 1}{x^2 + 2x + 1 - x^2 + 2x - 1} = \frac{m + n + m - n}{m + n - m + n} \)

\( \frac{2x^2 + 2}{4x} = \frac{2m}{2n} \)

\( \frac{x^2 + 1}{2x} = \frac{m}{n} \)

Step 4: Solve for n.
\( \frac{x^2 + 1}{2mx} = \frac{1}{n} \)

\( n = \frac{2mx}{x^2 + 1} \)

In simple words: We used the same method as the previous problems. We arranged the equation step by step until we got n alone on one side.

📝 Teacher's Note: This is the same pattern as Question 12. Students should see that the method stays the same even when letters change from a, b to m, n.

🎯 Exam Tip: Write the final answer clearly: \( n = \frac{2mx}{x^2 + 1} \). This exact form gets full marks.

Question 15. If \( \frac{x^3 + 3xy^2}{3x^2y + y^3} = \frac{m^3 + 3mn^2}{3m^2n + n^3} \), show that: nx = my.
Answer:
Given: \( \frac{x^3 + 3xy^2}{3x^2y + y^3} = \frac{m^3 + 3mn^2}{3m^2n + n^3} \)

Step 1: Factor the numerators and denominators.
Numerator: \( x^3 + 3xy^2 = x(x^2 + 3y^2) \)
Denominator: \( 3x^2y + y^3 = y(3x^2 + y^2) \)

Similarly: \( m^3 + 3mn^2 = m(m^2 + 3n^2) \)
\( 3m^2n + n^3 = n(3m^2 + n^2) \)

Step 2: Rewrite the equation.
\( \frac{x(x^2 + 3y^2)}{y(3x^2 + y^2)} = \frac{m(m^2 + 3n^2)}{n(3m^2 + n^2)} \)

\( \frac{x}{y} \cdot \frac{x^2 + 3y^2}{3x^2 + y^2} = \frac{m}{n} \cdot \frac{m^2 + 3n^2}{3m^2 + n^2} \)

Step 3: For this equation to hold, we need:
\( \frac{x}{y} = \frac{m}{n} \) and \( \frac{x^2 + 3y^2}{3x^2 + y^2} = \frac{m^2 + 3n^2}{3m^2 + n^2} \)

The second condition is automatically satisfied if the first condition is true.

Step 4: From \( \frac{x}{y} = \frac{m}{n} \):
Cross multiply: \( xn = my \)
Therefore: \( nx = my \)

Hence proved.
In simple words: We factored the complex fractions into simpler parts. When two ratios are equal, their parts must be equal too.

📝 Teacher's Note: Show students how to factor \( x^3 + 3xy^2 = x(x^2 + 3y^2) \). This factoring skill is very important for many algebra problems.

🎯 Exam Tip: Always show the factoring step clearly. Write "Hence proved" or "Hence nx = my" at the end to show you finished the proof.

Exercise 7D

Question 1. If a: b = 3: 5, find: (10a + 3b): (5a + 2b)
Answer:
Given: \( \frac{a}{b} = \frac{3}{5} \)

Step 1: Express the required ratio
\( \frac{10a + 3b}{5a + 2b} \)

Step 2: Substitute \( \frac{a}{b} = \frac{3}{5} \)
\( = \frac{10\left(\frac{3}{5}\right) + 3}{5\left(\frac{3}{5}\right) + 2} \)

Step 3: Simplify
\( = \frac{10 \times \frac{3}{5} + 3}{5 \times \frac{3}{5} + 2} \)
\( = \frac{6 + 3}{3 + 2} \)
\( = \frac{9}{5} \)

Therefore: (10a + 3b): (5a + 2b) = 9: 5
In simple words: When we know the ratio of two numbers, we can find the ratio of any expressions made from them. We just substitute the given ratio and calculate.

📝 Teacher's Note: Show students to substitute the given ratio value first. Then do simple arithmetic. Common mistake is not converting fractions properly.

🎯 Exam Tip: Always write "Given" first and show each substitution step clearly. Write the final answer as a ratio like 9:5.

Question 2. If 5x + 6y: 8x + 5y = 8: 9, find x: y.
Answer:
Given: \( \frac{5x + 6y}{8x + 5y} = \frac{8}{9} \)

Step 1: Cross multiply
\( 9(5x + 6y) = 8(8x + 5y) \)

Step 2: Expand both sides
\( 45x + 54y = 64x + 40y \)

Step 3: Collect like terms
\( 64x - 45x = 54y - 40y \)
\( 19x = 14y \)

Step 4: Find the ratio
\( \frac{x}{y} = \frac{14}{19} \)

Therefore: x: y = 14: 19
In simple words: We cross multiply to remove fractions. Then we collect all x terms on one side and all y terms on the other side to find the ratio.

📝 Teacher's Note: Teach cross multiplication clearly first. Show that when a/b = c/d, then ad = bc. Students often make sign errors when moving terms.

🎯 Exam Tip: Write each step of cross multiplication and expansion clearly. Check your arithmetic by substituting back into the original equation.

Question 3. If (3x - 4y): (2x - 3y) = (5x - 6y): (4x - 5y), find x: y.
Answer:
Given: \( \frac{3x - 4y}{2x - 3y} = \frac{5x - 6y}{4x - 5y} \)

Step 1: Cross multiply
\( (3x - 4y)(4x - 5y) = (5x - 6y)(2x - 3y) \)

Step 2: Apply componendo and dividendo
\( \frac{3x - 4y + 2x - 3y}{3x - 4y - 2x + 3y} = \frac{5x - 6y + 4x - 5y}{5x - 6y - 4x + 5y} \)

Step 3: Simplify
\( \frac{5x - 7y}{x - y} = \frac{9x - 11y}{x - y} \)

Step 4: Since denominators are equal
\( 5x - 7y = 9x - 11y \)
\( 11y - 7y = 9x - 5x \)
\( 4y = 4x \)

Step 5: Find the ratio
\( \frac{x}{y} = \frac{1}{1} \)

Therefore: x: y = 1: 1
In simple words: We use a special method called componendo and dividendo to solve this type of equation. It helps us find that x and y are equal.

📝 Teacher's Note: Componendo and dividendo is a useful tool for proportion problems. If a/b = c/d, then (a+b)/(a-b) = (c+d)/(c-d). Practice this method well.

🎯 Exam Tip: Write "Applying componendo and dividendo" clearly. This method saves time in proportion problems. Always check that x:y = 1:1 means x = y.

Question 4. Find the:
(i) duplicate ratio of \( 2\sqrt{2} : 3\sqrt{5} \)
(ii) triplicate ratio of 2a: 3b
(iii) sub-duplicate ratio of \( 9x^2a^4 : 25y^6b^2 \)
(iv) sub-triplicate ratio of 216: 343
(v) reciprocal ratio of 3: 5
(vi) ratio compounded of the duplicate ratio of 5: 6, the reciprocal ratio of 25: 42 and the sub-duplicate ratio of 36: 49.
Answer:

(i) Duplicate ratio of \( 2\sqrt{2} : 3\sqrt{5} \)
Duplicate ratio = \( (2\sqrt{2})^2 : (3\sqrt{5})^2 = 8 : 45 \)

(ii) Triplicate ratio of 2a: 3b
Triplicate ratio = \( (2a)^3 : (3b)^3 = 8a^3 : 27b^3 \)

(iii) Sub-duplicate ratio of \( 9x^2a^4 : 25y^6b^2 \)
Sub-duplicate ratio = \( \sqrt{9x^2a^4} : \sqrt{25y^6b^2} = 3xa^2 : 5y^3b \)

(iv) Sub-triplicate ratio of 216: 343
Sub-triplicate ratio = \( \sqrt[3]{216} : \sqrt[3]{343} = 6 : 7 \)

(v) Reciprocal ratio of 3: 5
Reciprocal ratio = 5: 3

(vi) Required compound ratio
Duplicate ratio of 5: 6 = 25: 36
Reciprocal ratio of 25: 42 = 42: 25
Sub-duplicate ratio of 36: 49 = 6: 7
Required compound ratio = \( \frac{25 \times 42 \times 6}{36 \times 25 \times 7} = 1 : 1 \)

In simple words: Duplicate means square both numbers. Triplicate means cube both numbers. Sub-duplicate means take square root. Sub-triplicate means take cube root. Reciprocal means flip the ratio.

📝 Teacher's Note: Use simple examples first. Show that duplicate ratio of 2:3 is 4:9 (square both). Make sure students know that sub-duplicate means square root of both terms.

🎯 Exam Tip: Remember the definitions clearly. Duplicate = square, Triplicate = cube, Sub-duplicate = square root, Sub-triplicate = cube root. For compound ratios, multiply all numerators and denominators.

Question 5. Find the value of x, if:
(i) (2x + 3): (5x - 38) is the duplicate ratio of \( \sqrt{5} : \sqrt{6} \)
(ii) (2x + 1): (3x + 13) is the sub-duplicate ratio of 9: 25.
(iii) (3x - 7): (4x + 3) is the sub-triplicate ratio of 8: 27.
Answer:

(i) Given: (2x + 3): (5x - 38) is the duplicate ratio of \( \sqrt{5} : \sqrt{6} \)
Duplicate ratio of \( \sqrt{5} : \sqrt{6} \) = 5: 6
\( \frac{2x + 3}{5x - 38} = \frac{5}{6} \)
Cross multiplying: \( 6(2x + 3) = 5(5x - 38) \)
\( 12x + 18 = 25x - 190 \)
\( 25x - 12x = 190 + 18 \)
\( 13x = 208 \)
\( x = 16 \)

(ii) Given: (2x + 1): (3x + 13) is the sub-duplicate ratio of 9: 25
Sub-duplicate ratio of 9: 25 = 3: 5
\( \frac{2x + 1}{3x + 13} = \frac{3}{5} \)
Cross multiplying: \( 5(2x + 1) = 3(3x + 13) \)
\( 10x + 5 = 9x + 39 \)
\( 10x - 9x = 39 - 5 \)
\( x = 34 \)

(iii) Given: (3x - 7): (4x + 3) is the sub-triplicate ratio of 8: 27
Sub-triplicate ratio of 8: 27 = 2: 3
\( \frac{3x - 7}{4x + 3} = \frac{2}{3} \)
Cross multiplying: \( 3(3x - 7) = 2(4x + 3) \)
\( 9x - 21 = 8x + 6 \)
\( 9x - 8x = 6 + 21 \)
\( x = 27 \)

In simple words: First find what the duplicate, sub-duplicate, or sub-triplicate ratio is. Then set up an equation and solve for x using cross multiplication.

📝 Teacher's Note: Teach students to first calculate the required ratio type before setting up the equation. Common mistake is forgetting to find the correct ratio first.

🎯 Exam Tip: Always write what type of ratio you need to find first. Show the calculation of that ratio clearly. Then set up and solve the equation step by step.

Question 6. What quantity must be added to each term of the ratio x: y so that it may become equal to c: d?
Answer:
Let the required quantity be p.

Then we have: \( \frac{x + p}{y + p} = \frac{c}{d} \)

Cross multiplying: \( d(x + p) = c(y + p) \)
\( dx + pd = cy + cp \)
\( pd - cp = cy - dx \)
\( p(d - c) = cy - dx \)

Therefore: \( p = \frac{cy - dx}{d - c} \)

In simple words: We add the same number p to both parts of the ratio. Then we set it equal to the target ratio and solve for p.

📝 Teacher's Note: Show students that we add the same quantity to both numerator and denominator. This is different from multiplying both by the same number.

🎯 Exam Tip: Let the quantity be p. Set up the equation (x+p):(y+p) = c:d. Cross multiply and solve for p. Write the answer as a fraction.

Question 7. A woman reduces her weight in the ratio 7: 5. What does her weight become if originally it was 84 kg?
Answer:
Let the reduced weight be x.
Original weight = 84 kg

Given: 84: x = 7: 5
\( \frac{84}{x} = \frac{7}{5} \)

Cross multiplying: \( 84 \times 5 = 7 \times x \)
\( x = \frac{84 \times 5}{7} = \frac{420}{7} = 60 \)

Therefore, her reduced weight is 60 kg.
In simple words: The ratio 7:5 means for every 7 kg original weight, the new weight is 5 kg. So we find what 5 kg is when 7 kg is 84 kg.

📝 Teacher's Note: Help students understand that reducing weight in ratio 7:5 means new weight is 5/7 of the original weight. This is a simple proportion problem.

🎯 Exam Tip: Set up the proportion correctly. Original weight : New weight = 7 : 5. Always check that your answer is smaller than the original weight since it's a reduction.

Question 8. If 15(2x² - y²) = 7xy, find x: y; if x and y both are positive.
Answer:
Given: \( 15(2x^2 - y^2) = 7xy \)

Step 1: Expand
\( 30x^2 - 15y^2 = 7xy \)

Step 2: Rearrange
\( 30x^2 - 7xy - 15y^2 = 0 \)

Step 3: Divide by \( y^2 \)
\( 30\left(\frac{x}{y}\right)^2 - 7\left(\frac{x}{y}\right) - 15 = 0 \)

Step 4: Let \( \frac{x}{y} = a \)
\( 30a^2 - 7a - 15 = 0 \)

Step 5: Factor the quadratic
\( (5a + 3)(6a - 5) = 0 \)

Step 6: Solve for a
\( a = -\frac{3}{5} \) or \( a = \frac{5}{6} \)

Since x and y are both positive, \( \frac{x}{y} = \frac{5}{6} \)

Therefore: x: y = 5: 6
In simple words: We change the equation into a form where we can find x/y. Since both x and y are positive, we take the positive solution.

📝 Teacher's Note: Show students how dividing by y² helps convert this into a quadratic in x/y. This is a common technique for ratio problems with quadratic equations.

🎯 Exam Tip: When both variables are positive, reject any negative ratio solutions. Always check your answer by substituting back into the original equation.

Question 9. Find the:
(i) fourth proportional to 2xy, x² and y².
(ii) third proportional to a² – b² and a + b.
(iii) mean proportional to (x – y) and (x³ – x²y).
Answer:
(i) Fourth proportional to 2xy, x², y²:
Let the fourth proportional to 2xy, x² and y² be n.
\( \Rightarrow 2xy : x² = y² : n \)
\( \Rightarrow 2xy \times n = x² \times y² \)
\( \Rightarrow n = \frac{x²y²}{2xy} = \frac{xy}{2} \)

(ii) Third proportional to a² – b², a + b:
Let the third proportional to a² – b² and a + b be n.
\( \Rightarrow a² – b², a + b \) and n are in continued proportion.
\( \Rightarrow a² – b² : a + b = a + b : n \)
\( \Rightarrow n = \frac{(a + b)²}{a² – b²} = \frac{(a + b)²}{(a + b)(a – b)} = \frac{a + b}{a – b} \)

(iii) Mean proportional to (x – y), (x³ – x²y):
Let the mean proportional to (x – y) and (x³ – x²y) be n.
\( \Rightarrow (x – y), n, (x³ – x²y) \) are in continued proportion
\( \Rightarrow (x – y) : n = n : (x³ – x²y) \)
\( \Rightarrow n² = (x – y)(x³ – x²y) \)
\( \Rightarrow n² = x²(x – y)(x – y) \)
\( \Rightarrow n² = x²(x – y)² \)
\( \Rightarrow n = x(x – y) \)
In simple words: We use the proportional formulas. Fourth proportional means a:b = c:d where d is the fourth. Third proportional means a:b = b:c where c is the third. Mean proportional means a:b = b:c where b is the mean.

📝 Teacher's Note: Draw a simple diagram showing how proportionals work. Use numbers like 2:4 = 4:8 to show the pattern. Students understand better with concrete examples.

🎯 Exam Tip: Always write the proportion formula first. Then substitute values carefully. Show each step clearly to get full marks.

Question 10. Find two numbers such that the mean proportional between them is 14 and third proportional to them is 112.
Answer:
Let the required numbers be a and b.
Given, 14 is the mean proportional between a and b.
\( \Rightarrow a : 14 = 14 : b \)
\( \Rightarrow ab = 196 \)
\( \Rightarrow a = \frac{196}{b} \) ...(1)

Also, given, third proportional to a and b is 112.
\( \Rightarrow a : b = b : 112 \)
\( \Rightarrow b² = 112a \) ...(2)

Using (1), we have:
\( b² = 112 \times \frac{196}{b} \)
\( b³ = (14)²(2)³ \)
b = 28
From (1),
\( a = \frac{196}{28} = 7 \)

Thus, the two numbers are 7 and 28.
In simple words: We use two conditions given. Mean proportional means the middle number. Third proportional means the last number in the ratio. We solve two equations to find both numbers.

📝 Teacher's Note: Show students how to set up two equations from the given conditions. Use simple examples like mean proportional between 4 and 9 is 6 because 4:6 = 6:9.

🎯 Exam Tip: Write both conditions as separate equations. Label them (1) and (2). Show substitution steps clearly. Check your answer by substituting back.

Question 11. If x and y be unequal and x: y is the duplicate ratio of x + z and y + z, prove that z is mean proportional between x and y.
Answer:
Given, \( \frac{x}{y} = \frac{(x + z)²}{(y + z)²} \)
x(y² + z² + 2yz) = y(x² + z² + 2xz)
xy² + xz² + 2xyz = x²y + yz² + 2xyz
xy² + xz² = x²y + yz²
xy² – x²y = yz² – xz²
xy(y – x) = z²(y – x)
xy = z²
Hence, z is mean proportional between x and y.
In simple words: Duplicate ratio means square of the ratio. We expand both sides and simplify. The result shows that z² = xy, which means z is the mean proportional.

📝 Teacher's Note: Explain that duplicate ratio means squaring the ratio. Show students how to expand (a+b)² step by step. Common mistake is forgetting the middle terms.

🎯 Exam Tip: Write "duplicate ratio" means square. Expand brackets carefully. Show each algebra step. End with "Hence z is mean proportional" to get full marks.

Question 12. If x = \( \frac{2ab}{a + b} \), find the value of \( \frac{x + a}{x – a} + \frac{x + b}{x – b} \).
Answer:
\( x = \frac{2ab}{a + b} \)
\( \frac{x}{a} = \frac{2b}{a + b} \)
Applying componendo and dividendo,
\( \frac{x + a}{x – a} = \frac{2b + a + b}{2b – a – b} \)
\( \frac{x + a}{x – a} = \frac{3b + a}{b – a} \) ...(1)

Also, \( x = \frac{2ab}{a + b} \)
\( \frac{x}{b} = \frac{2a}{a + b} \)
Applying componendo and dividendo,
\( \frac{x + b}{x – b} = \frac{2a + a + b}{2a – a – b} \)
\( \frac{x + b}{x – b} = \frac{3a + b}{a – b} \) ...(2)

From (1) and (2),
\( \frac{x + a}{x – a} + \frac{x + b}{x – b} = \frac{3b + a}{b – a} + \frac{3a + b}{a – b} \)
\( \frac{x + a}{x – a} + \frac{x + b}{x – b} = \frac{3b + a}{b – a} + \frac{-(3a + b)}{b – a} \)
\( \frac{x + a}{x – a} + \frac{x + b}{x – b} = \frac{3b + a – 3a – b}{b – a} \)
\( \frac{x + a}{x – a} + \frac{x + b}{x – b} = \frac{2b – 2a}{b – a} = \frac{2(b – a)}{b – a} = 2 \)
In simple words: We use componendo and dividendo rule. This rule helps us change fractions into a useful form. Then we add the two parts and simplify to get 2.

📝 Teacher's Note: Teach componendo and dividendo as: if a/b = c/d, then (a+b)/(a-b) = (c+d)/(c-d). Practice with simple numbers first before using algebra.

🎯 Exam Tip: Write "Applying componendo and dividendo" clearly. Show the rule. Be careful with signs when adding fractions. The final answer should be a simple number.

Question 13. If (4a + 9b) (4c – 9d) = (4a – 9b) (4c + 9d), prove that: a: b = c: d.
Answer:
Given, \( \frac{4a + 9b}{4a – 9b} = \frac{4c + 9d}{4c – 9d} \)
Applying componendo and dividendo,
\( \frac{4a + 9b + 4a – 9b}{4a + 9b – 4a + 9b} = \frac{4c + 9d + 4c – 9d}{4c + 9d – 4c + 9d} \)
\( \frac{8a}{18b} = \frac{8c}{18d} \)
\( \frac{a}{b} = \frac{c}{d} \)
In simple words: We rearrange the given equation as a ratio. Then we use componendo and dividendo to simplify both sides. The algebra cancels out to give us a:b = c:d.

📝 Teacher's Note: Show students how to rearrange the product equation into ratio form first. Then apply componendo and dividendo carefully. The terms cancel out nicely.

🎯 Exam Tip: Start by writing the equation as a ratio. Apply componendo and dividendo correctly. Show each step of simplification. End with "Hence a:b = c:d".

Question 14. If \( \frac{a}{b} = \frac{c}{d} \), show that: (a + b): (c + d) = \( \sqrt{a² + b²} : \sqrt{c² + d²} \)
Answer:
Let \( \frac{a}{b} = \frac{c}{d} = k \) (say)
\( \Rightarrow a = bk, c = dk \)
L.H.S. = \( \frac{a + b}{c + d} \)
= \( \frac{bk + b}{dk + d} \)
= \( \frac{b(k + 1)}{d(k + 1)} \)
= \( \frac{b}{d} \)

R.H.S. = \( \frac{\sqrt{a² + b²}}{\sqrt{c² + d²}} \)
= \( \frac{\sqrt{(bk)² + b²}}{\sqrt{(dk)² + d²}} \)
= \( \frac{\sqrt{b²(k² + 1)}}{\sqrt{d²(k² + 1)}} \)
= \( \frac{\sqrt{b²}}{\sqrt{d²}} \)
= \( \frac{b}{d} \)

∴ L.H.S. = R.H.S
In simple words: We assume the ratio equals some number k. Then we substitute this in both sides. After simplifying, both sides become b/d, so they are equal.

📝 Teacher's Note: Explain that letting a ratio equal k is a standard trick. Show how to factor out common terms from square roots. Students often forget to simplify √(b²) = b.

🎯 Exam Tip: Always write "Let a/b = c/d = k". Show L.H.S. and R.H.S. separately. Simplify step by step. End with "Hence L.H.S. = R.H.S." or "Hence proved."

Question 15. There are 36 members in a student council in a school and the ratio of the number of boys to the number of girls is 3: 1. How many more girls should be added to the council so that the ratio of the number of boys to the number of girls may be 9: 5?
Answer:
Ratio of number of boys to the number of girls = 3: 1
Let the number of boys be 3x and number of girls be x.
3x + x = 36
4x = 36
x = 9

So, number of boys = 3 × 9 = 27
Number of girls = 9

Let y girls be added to make the ratio 9: 5
\( \frac{27}{9 + y} = \frac{9}{5} \)
27 × 5 = 9 × (9 + y)
135 = 81 + 9y
54 = 9y
y = 6

Therefore, 6 more girls should be added to the council.
In simple words: First we find how many boys and girls are there now. Boys = 27, Girls = 9. Then we find how many girls to add so the new ratio becomes 9:5. Answer is 6 more girls.

📝 Teacher's Note: Draw a simple table showing before and after. Show students how the number of boys stays the same but girls increases. Use cross multiplication for the final step.

🎯 Exam Tip: First find the actual numbers of boys and girls. Then set up the equation with unknown y. Use cross multiplication. Always check your answer by substituting back.

Question 15.
Answer:
∴ Number of boys = 27
Number of girls = 9
Let n number of girls be added to the council.
From given information, we have:

\( \frac{27}{9+n} = \frac{9}{5} \)

Cross multiplying:
135 = 81 + 9n
9n = 54
n = 6

Thus, 6 girls are added to the council.
In simple words: We need to make the ratio of boys to girls equal to 9:5. So we add some girls and solve the equation to find how many.

📝 Teacher's Note: Show students that when we add girls, only the denominator changes. The number of boys stays the same. This makes the problem easier.

🎯 Exam Tip: Always write "Let n = number of girls to be added" first. Then set up the ratio equation carefully. Cross multiply to solve.

Question 16. If 7x - 15y = 4x + y, find the value of x: y. Hence, use componendo and dividendo to find the values of:
(i) \( \frac{9x + 5y}{9x - 5y} \)
(ii) \( \frac{3x^2 + 2y^2}{3x^2 - 2y^2} \)
Answer:

Step 1: Find x:y from the given equation
7x - 15y = 4x + y
7x - 4x = y + 15y
3x = 16y
\( \frac{x}{y} = \frac{16}{3} \)

Step 2: Solve part (i)
(i) \( \frac{x}{y} = \frac{16}{3} \)

\( \Rightarrow \frac{9x}{5y} = \frac{144}{15} \) (Multiplying both sides by \( \frac{9}{5} \))

Applying componendo and dividendo:
\( \Rightarrow \frac{9x + 5y}{9x - 5y} = \frac{144 + 15}{144 - 15} = \frac{159}{129} = \frac{53}{43} \)

Step 3: Solve part (ii)
(ii) \( \frac{x}{y} = \frac{16}{3} \)

\( \Rightarrow \frac{x^2}{y^2} = \frac{256}{9} \)

\( \Rightarrow \frac{3x^2}{2y^2} = \frac{768}{18} = \frac{128}{3} \) (Multiplying both sides by \( \frac{3}{2} \))

Applying componendo and dividendo:
\( \Rightarrow \frac{3x^2 + 2y^2}{3x^2 - 2y^2} = \frac{128 + 3}{128 - 3} = \frac{131}{125} \)

In simple words: First we find the ratio x:y from the given equation. Then we use this ratio with componendo-dividendo rule to find the required expressions.

📝 Teacher's Note: Componendo-dividendo is a shortcut method. If a/b = c/d, then (a+b)/(a-b) = (c+d)/(c-d). Practice this rule with simple numbers first.

🎯 Exam Tip: Always find x:y first. Then apply componendo-dividendo carefully. Write each step clearly to avoid calculation mistakes.

Question 17. If \( \frac{4m + 3n}{4m - 3n} = \frac{7}{4} \), use properties of proportion to find:
(i) m:n
(ii) \( \frac{2m^2 - 11n^2}{2m^2 + 11n^2} \)
Answer:

Step 1: Find m:n
(i) Given, \( \frac{4m + 3n}{4m - 3n} = \frac{7}{4} \)

Applying componendo and dividendo:
\( \frac{4m + 3n + 4m - 3n}{4m + 3n - 4m + 3n} = \frac{7 + 4}{7 - 4} \)
\( \frac{8m}{6n} = \frac{11}{3} \)
\( \frac{m}{n} = \frac{11}{4} \)

Step 2: Solve part (ii)
(ii) \( \frac{m}{n} = \frac{11}{4} \)
\( \frac{m^2}{n^2} = \frac{121}{16} \)

\( \frac{2m^2}{11n^2} = \frac{2 \times 121}{11 \times 16} = \frac{11}{8} \) (Multiplying both sides by \( \frac{2}{11} \))

Applying componendo and dividendo:
\( \frac{2m^2 + 11n^2}{2m^2 - 11n^2} = \frac{11 + 8}{11 - 8} = \frac{19}{3} \)

Applying invertendo:
\( \frac{2m^2 - 11n^2}{2m^2 + 11n^2} = \frac{3}{19} \)

In simple words: We use componendo-dividendo on the given ratio to find m:n. Then we use this to find the second expression by applying the same rule again.

📝 Teacher's Note: Invertendo means flipping the fraction. If a/b = c/d, then b/a = d/c. This is useful when the question asks for the reverse ratio.

🎯 Exam Tip: Apply componendo-dividendo step by step. Don't try to do multiple steps at once. Check if you need invertendo at the end.

Question 18. If x, y, z are in continued proportion, prove that \( \frac{(x + y)^2}{(y + z)^2} = \frac{x}{z} \).
Answer:

∵ x, y, z are in continued proportion.
∴ \( \frac{x}{y} = \frac{y}{z} \Rightarrow y^2 = zx \)....(1)

Therefore,
\( \frac{x + y}{y} = \frac{y + z}{z} \) (By componendo)

\( \Rightarrow \frac{x + y}{y + z} = \frac{y}{z} \) (By alternendo)

\( \Rightarrow \frac{(x + y)^2}{(y + z)^2} = \frac{y^2}{z^2} \) (Squaring both sides)

\( \Rightarrow \frac{(x + y)^2}{(y + z)^2} = \frac{zx}{z^2} \) [From (1)]

\( \Rightarrow \frac{(x + y)^2}{(y + z)^2} = \frac{x}{z} \)

Hence Proved.

In simple words: When three numbers are in continued proportion, the middle term squared equals the product of first and third terms. We use this property and ratio rules to prove the given statement.

📝 Teacher's Note: Continued proportion means a:b = b:c. This gives us b² = ac. Use this key property in all continued proportion problems.

🎯 Exam Tip: Always write "y² = zx" first when dealing with continued proportion. This is the most important relationship. Then apply componendo and alternendo.

Question 19. Given x = \( \frac{\sqrt{a^2 + b^2} + \sqrt{a^2 - b^2}}{\sqrt{a^2 + b^2} - \sqrt{a^2 - b^2}} \), use componendo and dividendo to prove that \( b^2 = \frac{2a^2x}{x^2 + 1} \).
Answer:

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

By componendo and dividendo,
\( \frac{x + 1}{x - 1} = \frac{\sqrt{a^2 + b^2} + \sqrt{a^2 - b^2} + \sqrt{a^2 + b^2} - \sqrt{a^2 - b^2}}{\sqrt{a^2 + b^2} + \sqrt{a^2 - b^2} - \sqrt{a^2 + b^2} + \sqrt{a^2 - b^2}} \)

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

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

By componendo and dividendo,
\( \frac{(x^2 + 2x + 1) + (x^2 - 2x + 1)}{(x^2 + 2x + 1) - (x^2 - 2x + 1)} = \frac{(a^2 + b^2) + (a^2 - b^2)}{(a^2 + b^2) - (a^2 - b^2)} \)

\( \Rightarrow \frac{2(x^2 + 1)}{4x} = \frac{2a^2}{2b^2} \)

\( \Rightarrow \frac{x^2 + 1}{2x} = \frac{a^2}{b^2} \)

\( \Rightarrow b^2 = \frac{2a^2x}{x^2 + 1} \)

Hence Proved.

In simple words: We apply componendo-dividendo twice. First time to simplify the expression, second time after squaring both sides to get the final result.

📝 Teacher's Note: This problem uses componendo-dividendo twice. Show students that when we add/subtract the numerator and denominator, terms cancel out nicely.

🎯 Exam Tip: Apply componendo-dividendo step by step. Don't skip the squaring step. Write all algebraic simplifications clearly to avoid errors.

Question 20. If \( \frac{x^2 + y^2}{x^2 - y^2} = 2\frac{1}{8} \), find:
(i) \( \frac{x}{y} \)
(ii) \( \frac{x^3 + y^3}{x^3 - y^3} \)
Answer:

Step 1: Convert mixed number and find x/y
(i) Given, \( \frac{x^2 + y^2}{x^2 - y^2} = 2\frac{1}{8} = \frac{17}{8} \)

Applying componendo and dividendo:
\( \frac{x^2 + y^2 + x^2 - y^2}{x^2 + y^2 - x^2 + y^2} = \frac{17 + 8}{17 - 8} \)
\( \frac{2x^2}{2y^2} = \frac{25}{9} \)
\( \frac{x^2}{y^2} = \frac{25}{9} \)
\( \frac{x}{y} = \frac{5}{3} = 1\frac{2}{3} \)

Step 2: Find the required expression
(ii) \( \frac{x^3 + y^3}{x^3 - y^3} \)

\( = \frac{(\frac{x}{y})^3 + 1}{(\frac{x}{y})^3 - 1} \)

\( = \frac{(\frac{5}{3})^3 + 1}{(\frac{5}{3})^3 - 1} \)

\( = \frac{\frac{125}{27} + 1}{\frac{125}{27} - 1} \)

\( = \frac{\frac{125 + 27}{27}}{\frac{125 - 27}{27}} = \frac{152}{98} \)

In simple words: First we find x:y using componendo-dividendo. Then we substitute this ratio into the second expression and simplify.

📝 Teacher's Note: When we have x³ + y³, we can factor it as (x + y)(x² - xy + y²). But here it's easier to divide by y³ and use the x/y ratio.

🎯 Exam Tip: Convert mixed numbers to improper fractions first. Always simplify the final answer to its lowest terms if possible.

Question 21. Using componendo and dividendo find the value of x: \( \frac{\sqrt{3x + 4} + \sqrt{3x - 5}}{\sqrt{3x + 4} - \sqrt{3x - 5}} = 9 \)
Answer:
Given: \( \frac{\sqrt{3x + 4} + \sqrt{3x - 5}}{\sqrt{3x + 4} - \sqrt{3x - 5}} = \frac{9}{1} \)

Step 1: Apply componendo and dividendo.
\( \frac{\sqrt{3x + 4} + \sqrt{3x - 5} + \sqrt{3x + 4} - \sqrt{3x - 5}}{\sqrt{3x + 4} + \sqrt{3x - 5} - \sqrt{3x + 4} + \sqrt{3x - 5}} = \frac{9 + 1}{9 - 1} \)

Step 2: Simplify numerator and denominator.
\( \frac{2\sqrt{3x + 4}}{2\sqrt{3x - 5}} = \frac{10}{8} \)

Step 3: Further simplification.
\( \frac{\sqrt{3x + 4}}{\sqrt{3x - 5}} = \frac{5}{4} \)

Step 4: Square both sides.
\( \frac{3x + 4}{3x - 5} = \frac{25}{16} \)

Step 5: Cross multiply.
\( 16(3x + 4) = 25(3x - 5) \)
\( 48x + 64 = 75x - 125 \)
\( 75x - 48x = 64 + 125 \)
\( 27x = 189 \)

Step 6: Solve for x.
\( x = \frac{189}{27} = 7 \)

Therefore, x = 7
In simple words: We used componendo and dividendo rule to split the fraction into two parts. Then we simplified step by step to find x equals 7.

📝 Teacher's Note: Make students remember the componendo-dividendo formula: if a/b = c/d, then (a+b)/(a-b) = (c+d)/(c-d). Show them how to apply it carefully step by step.

🎯 Exam Tip: Always write "Applying componendo and dividendo" as your first step. Check your final answer by putting it back in the original equation.

Question 22. If \( x = \frac{\sqrt{a+1} + \sqrt{a-1}}{\sqrt{a+1} - \sqrt{a-1}} \), using properties of proportion, show that: \( x^2 - 2ax + 1 = 0 \)
Answer:
Given: \( x = \frac{\sqrt{a+1} + \sqrt{a-1}}{\sqrt{a+1} - \sqrt{a-1}} \)

Step 1: Apply Componendo-Dividendo.
\( \frac{x+1}{x-1} = \frac{(\sqrt{a+1} + \sqrt{a-1}) + (\sqrt{a+1} - \sqrt{a-1})}{(\sqrt{a+1} + \sqrt{a-1}) - (\sqrt{a+1} - \sqrt{a-1})} \)

Step 2: Simplify numerator and denominator.
\( \frac{x+1}{x-1} = \frac{2\sqrt{a+1}}{2\sqrt{a-1}} = \frac{\sqrt{a+1}}{\sqrt{a-1}} \)

Step 3: Square both sides of the equation.
\( \left(\frac{x+1}{x-1}\right)^2 = \frac{a+1}{a-1} \)

Step 4: Expand and simplify.
\( (x+1)^2(a-1) = (x-1)^2(a+1) \)
\( (x^2+2x+1)(a-1) = (x^2-2x+1)(a+1) \)
\( a(x^2+2x+1) - (x^2+2x+1) = a(x^2-2x+1) + (x^2-2x+1) \)

Step 5: Collect like terms.
\( 4ax = 2x^2 + 2 \)
\( 2ax = x^2 + 1 \)
\( x^2 - 2ax + 1 = 0 \)

Hence proved: \( x^2 - 2ax + 1 = 0 \)
In simple words: We used the componendo-dividendo rule to change the given fraction into a simpler form. Then we did algebra to prove the required equation.

📝 Teacher's Note: This is a standard proof question. Students should first apply componendo-dividendo, then square both sides. Show them how to expand brackets carefully.

🎯 Exam Tip: Write "Hence proved" at the end. Show every algebraic step clearly. Don't skip any steps in the expansion of brackets.

Question 23. Given \( \frac{x^3 + 12x}{6x^2 + 8} = \frac{y^3 + 27y}{9y^2 + 27} \). Using componendo and dividendo, find x : y.
Answer:
Given: \( \frac{x^3 + 12x}{6x^2 + 8} = \frac{y^3 + 27y}{9y^2 + 27} \)

Step 1: Apply componendo and dividendo.
\( \frac{x^3 + 12x + 6x^2 + 8}{x^3 + 12x - 6x^2 - 8} = \frac{y^3 + 27y + 9y^2 + 27}{y^3 + 27y - 9y^2 - 27} \)

Step 2: Factor the expressions.
\( \frac{x^3 + 3(1)(4)x + 3(1)(2)x^2 + 2^3}{x^3 + 3(1)(4)x - 3(1)(2)x^2 - 2^3} = \frac{y^3 + 3(1)(9)y + 3(1)(3)y^2 + 3^3}{y^3 + 3(1)(9)y - 3(1)(3)y^2 - 3^3} \)

Step 3: Recognize perfect cube patterns.
\( \frac{(x + 2)^3}{(x - 2)^3} = \frac{(y + 3)^3}{(y - 3)^3} \)

Step 4: Take cube root of both sides.
\( \frac{x + 2}{x - 2} = \frac{y + 3}{y - 3} \)

Step 5: Apply componendo and dividendo again.
\( \frac{x + 2 + x - 2}{x + 2 - x + 2} = \frac{y + 3 + y - 3}{y + 3 - y + 3} \)
\( \frac{2x}{4} = \frac{2y}{6} \)

Step 6: Simplify.
\( \frac{x}{2} = \frac{y}{3} \)

Step 7: Apply alternendo property.
\( \frac{x}{y} = \frac{2}{3} \)

Therefore, x : y = 2 : 3
In simple words: We used componendo-dividendo twice to simplify the complex fractions. Finally we found that x and y are in the ratio 2:3.

📝 Teacher's Note: This question needs students to recognize perfect cube patterns like \( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \). Practice factoring cubes with students.

🎯 Exam Tip: Look for perfect cube patterns when you see expressions with x³ terms. Always write the final ratio in simplest form like 2:3.

Question 24. Let \( \frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k \)
Answer:
Given: \( \frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k \)

This means: x = ak, y = bk, z = ck

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

Step 1: Substitute the values.
\( = \frac{(ak)^3}{a^3} + \frac{(bk)^3}{b^3} + \frac{(ck)^3}{c^3} \)

Step 2: Simplify each term.
\( = \frac{a^3k^3}{a^3} + \frac{b^3k^3}{b^3} + \frac{c^3k^3}{c^3} \)

\( = k^3 + k^3 + k^3 = 3k^3 \)

R.H.S. = \( \frac{3xyz}{abc} \)

Step 3: Substitute x = ak, y = bk, z = ck.
\( = \frac{3(ak)(bk)(ck)}{abc} = \frac{3abck^3}{abc} = 3k^3 \)

Step 4: Compare L.H.S. and R.H.S.
L.H.S. = R.H.S. = \( 3k^3 \)

Hence proved: \( \frac{x^3}{a^3} + \frac{y^3}{b^3} + \frac{z^3}{c^3} = \frac{3xyz}{abc} \)
In simple words: Since all three fractions equal k, we can write x, y, z in terms of k. When we substitute and simplify, both sides become 3k³.

📝 Teacher's Note: This is a standard proportional identity. Make students understand that when ratios are equal, we can introduce a common factor k. Show them how to substitute carefully.

🎯 Exam Tip: Always introduce the common ratio k when you see equal ratios. Write "Hence proved" at the end with the original statement.

Question 25. Given that b is the mean proportion between a and c. \( \frac{a}{b} = \frac{b}{c} = k \)
Answer:
Given: b is the mean proportion between a and c.
This means: \( \frac{a}{b} = \frac{b}{c} = k \)

From this: b = ck, a = bk = (ck)k = ck²

L.H.S. = \( \frac{a^4 - a^2b^2 - b^4}{b^4 - b^2c^2 - c^4} \)

Step 1: Substitute a = ck² and b = ck.
\( = \frac{(ck^2)^4 - (ck^2)^2(ck)^2 - (ck)^4}{(ck)^4 - (ck)^2c^2 - c^4} \)

Step 2: Expand the powers.
\( = \frac{c^4k^8 - c^4k^6 - c^4k^4}{c^4k^4 - c^4k^2 - c^4} \)

Step 3: Factor out c⁴.
\( = \frac{c^4k^4(k^4 - k^2 - 1)}{c^4(k^4 - k^2 - 1)} \)

Step 4: Cancel common factors.
\( = k^4 \) ... (i)

R.H.S. = \( \frac{a^2}{c^2} = \frac{(ck^2)^2}{c^2} = \frac{c^2k^4}{c^2} = k^4 \) ... (ii)

Step 5: Compare results.
From (i) and (ii): L.H.S. = R.H.S. = k⁴

Hence proved: \( \frac{a^4 - a^2b^2 - b^4}{b^4 - b^2c^2 - c^4} = \frac{a^2}{c^2} \)
In simple words: Since b is the mean proportion between a and c, we can write a and c in terms of k. After substituting and simplifying, both sides equal k⁴.

📝 Teacher's Note: Mean proportion means b² = ac, so b/c = a/b. Students often forget this relationship. Practice more examples of mean proportions.

🎯 Exam Tip: When b is mean proportion between a and c, always write b² = ac first. Then introduce the common ratio k for easier calculation.

Question 26. \( \frac{7m - 2n}{7m - 2n} = \frac{5}{3} \)
Answer:
Given: \( \frac{7m - 2n}{7m - 2n} = \frac{5}{3} \)

Step 1: Apply componendo and dividendo.
\( \frac{7m - 2n - 7m - 2n}{7m - 2n - 7m - 2n} = \frac{5 - 3}{5 - 3} \)
\( \frac{14m}{4n} = \frac{8}{2} \)

Step 2: Simplify.
\( \frac{m}{n} = \frac{8 \times 4}{2 \times 14} = \frac{8}{7} \)

Therefore: m : n = 8 : 7

ii. From (i), \( \frac{m}{n} = \frac{8}{7} \)

\( \frac{m^2}{n^2} = \frac{64}{49} \)

Step 3: Apply componendo and dividendo.
\( \frac{m^2 - n^2}{m^2 - n^2} = \frac{64 - 49}{64 - 49} = \frac{64 - 49}{64 - 49} \)
\( \frac{m^2 - n^2}{m^2 - n^2} = \frac{113}{15} = 7\frac{8}{15} \)

In simple words: We used componendo-dividendo to find the ratio of m to n first. Then we used it again to find the ratio of (m²-n²) to (m²-n²).

📝 Teacher's Note: There seems to be an error in the original question - the numerator and denominator are the same in the given fraction. Check if this should be two different expressions.

🎯 Exam Tip: Always check if the question makes mathematical sense. If numerator and denominator are identical, the fraction always equals 1, not 5/3.

Question 27.

i. (2x2 - 5y2): xy = 1: 3
Answer:
Given: \( (2x^2 - 5y^2) : xy = 1 : 3 \)

Step 1: Write the proportion as a fraction.
\( \frac{2x^2 - 5y^2}{xy} = \frac{1}{3} \)

Step 2: Separate the fraction.
\( \frac{2x}{y} - \frac{5y}{x} = \frac{1}{3} \)

Step 3: Put \( \frac{x}{y} = a \), we get
\( 2a - 5 \cdot \frac{1}{a} = \frac{1}{3} \)

Step 4: Multiply through by 3a.
\( 3[2a^2 - 5] = a \)
\( 6a^2 - a - 15 = 0 \)
\( 6a^2 - 9a - 10a - 15 = 0 \)
\( 3a(2a - 3) - 5(2a - 3) = 0 \)
\( (2a - 3)(3a - 5) = 0 \)

Step 5: Solve for a.
\( (2a - 3) = 0 \) or \( (3a - 5) = 0 \)
\( a = \frac{3}{2} \) or \( a = \frac{5}{3} \)

Since \( a = -\frac{3}{2} \) is not acceptable, as x and y both are positive.
\( a = \frac{5}{3} \Rightarrow \frac{x}{y} = \frac{5}{3} \)
\( \Rightarrow x : y = 5 : 3 \)

In simple words: We changed the ratio into a fraction. Then we substituted one variable to make it easier. We got a quadratic equation and solved it to find the ratio of x to y.

📝 Teacher's Note: When dealing with ratios involving squares, substitution helps a lot. Put one ratio as 'a' and solve step by step. Students often forget to check if negative values make sense.

🎯 Exam Tip: Always write "x : y =" at the end. Show all steps clearly. Check if your answer makes sense - negative ratios may not be valid in some problems.

ii. \( 16 \left(\frac{a - x}{a - x}\right)^3 = \frac{a - x}{a - x} \)
Answer:
Step 1: Simplify the given equation.
\( 16 = \left(\frac{a - x}{a - x}\right)^4 \)

Step 2: Take the fourth root on both sides.
\( (2)^4 = \left(\frac{a - x}{a - x}\right)^4 \)

Step 3: This gives us.
\( \frac{a - x}{a - x} = ±2 \)

In simple words: We moved all terms to one side and took the fourth root. This gave us two possible values: +2 and -2.

📝 Teacher's Note: When you have even powers like 4, remember there are positive and negative solutions. Draw this on the board: \( x^4 = 16 \) gives \( x = ±2 \).

🎯 Exam Tip: Always write ± when taking even roots. Don't forget the negative solution. Write both answers clearly.

Question 28.
If \( \frac{x}{a} = \frac{y}{b} = \frac{z}{c} \), prove that: \( \frac{ax - by}{(a + b)(x - y)} + \frac{by - cz}{(b + c)(y - z)} + \frac{cz - ax}{(c + a)(z - x)} = 3 \)
Answer:
Solution:

Step 1: Let \( \frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k \) (say)
\( \Rightarrow x = ak, y = bk, z = ck \)

Step 2: Substitute in L.H.S.
\( = \frac{ax - by}{(a + b)(x - y)} + \frac{by - cz}{(b + c)(y - z)} + \frac{cz - ax}{(c + a)(z - x)} \)

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

Step 3: Simplify each fraction.
\( = \frac{k(a^2 - b^2)}{k(a + b)(a - b)} + \frac{k(b^2 - c^2)}{k(b + c)(b - c)} + \frac{k(c^2 - a^2)}{k(c + a)(c - a)} \)

Step 4: Cancel k and factorize.
\( = \frac{k(a^2 - b^2)}{k(a^2 - b^2)} + \frac{k(b^2 - c^2)}{k(b^2 - c^2)} + \frac{k(c^2 - a^2)}{k(c^2 - a^2)} \)

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

In simple words: We set all three ratios equal to k. Then we substituted these values in the big expression. After simplifying, each fraction became 1, so we got 1 + 1 + 1 = 3.

📝 Teacher's Note: This is a standard technique - set all ratios equal to k. Then substitute everywhere. The algebra becomes much easier. Practice this method with simpler examples first.

🎯 Exam Tip: Write "Let the common ratio = k" at the start. Show each substitution step clearly. The examiner wants to see the method, not just the final answer.

Question 29.
If q is the mean proportional between p and r, prove that: \( \frac{p^3 + q^3 + r^3}{p^2q^2r^2} = \frac{1}{p^3} + \frac{1}{q^3} + \frac{1}{r^3} \)
Answer:
Solution:

Since, q is the mean proportional between p and r,
\( q^2 = pr \)

Step 1: Start with L.H.S.
\( L.H.S. = \frac{p^3 + q^3 + r^3}{p^2q^2r^2} \)

Step 2: Substitute \( q^2 = pr \).
\( = \frac{p^3 + (pr)q + r^3}{p^2(pr)^2} \)

\( = \frac{p^3 + pqr + r^3}{p^3r^3} \)

Step 3: Separate into fractions.
\( = \frac{1}{r^3} + \frac{q}{p^2r^2} + \frac{1}{p^3} \)

\( = \frac{1}{r^3} + \frac{q}{(q^2)^2} + \frac{1}{p^3} \)

\( = \frac{1}{r^3} + \frac{1}{q^3} + \frac{1}{p^3} \)

\( = R.H.S. \)

In simple words: Since q is the mean proportional, we have q² = pr. We used this relationship to substitute in the fraction and simplified it step by step to get the right side.

📝 Teacher's Note: Mean proportional means the middle term when three numbers are in proportion. If a, b, c are in proportion, then b² = ac. This is the key relationship to remember.

🎯 Exam Tip: Write "q² = pr" first thing when you see mean proportional. Use this to substitute wherever you see q². Show each substitution step clearly.

Question 30.
If a, b and c are in continued proportion, prove that: a: c = (a² + b²) : (b² + c²)
Answer:
Solution:

Given, a, b and c are in continued proportion.
\( \Rightarrow a: b = b: c \)

Step 1: Set up the common ratio.
Let \( \frac{a}{b} = \frac{b}{c} = k \) (say)
\( \Rightarrow a = bk, b = ck \)
\( \Rightarrow a = ck^2, b = ck \)

Step 2: Find L.H.S.
Now, \( L.H.S. = \frac{a}{c} = \frac{ck^2}{c} = k^2 \)

Step 3: Find R.H.S.
\( R.H.S. = \frac{a^2 + b^2}{b^2 + c^2} \)

\( = \frac{(ck^2)^2 + (ck)^2}{(ck)^2 + c^2} \)

\( = \frac{c^2k^4 + c^2k^2}{c^2k^2 + c^2} \)

\( = \frac{c^2k^2(k^2 + 1)}{c^2(k^2 + 1)} \)

\( = k^2 \)

\( \therefore L.H.S. = R.H.S. \)

In simple words: Three numbers in continued proportion means the first ratio equals the second ratio. We used substitution to show that both sides of the equation give the same value k².

📝 Teacher's Note: Continued proportion means a:b = b:c, so b is the mean proportional. Use substitution method - set the common ratio as k and express everything in terms of k.

🎯 Exam Tip: Write "Let the common ratio = k" and express all terms using k. Show that L.H.S. = R.H.S. = k². This method works for all continued proportion problems.