Frank Brothers Solutions for ICSE Class 10 Mathematics Chapter 9 Ratio And Proportion

Exercise 9.1

Question 1. Find the compounded ratio of 560:240, 216:60, 540:18.
Answer:
Step 1: Simplify each ratio to its lowest terms.
\( \frac{560}{240} = \frac{56}{24} = \frac{28}{12} = \frac{14}{6} = \frac{7}{3} \)

\( \frac{216}{60} = \frac{108}{30} = \frac{54}{15} = \frac{18}{5} \)

\( \frac{540}{18} = \frac{54}{18} \times \frac{10}{1} = \frac{54}{18} = 3 \)
But wait, \( \frac{540}{18} = 30 \) and \( \frac{30}{1} = \frac{30}{1} \)
Actually: \( \frac{540}{18} = 30 = \frac{30}{1} \)

Step 2: Multiply all the ratios together.
Compounded ratio = \( \frac{7}{3} \times \frac{18}{5} \times \frac{30}{1} \)

Step 3: Simplify the multiplication.
\( = \frac{7 \times 18 \times 30}{3 \times 5 \times 1} = \frac{3780}{15} = 252 \)

Therefore, the compounded ratio = 252:1
In simple words: We first made each ratio simpler. Then we multiplied all the first parts together and all the second parts together. This gave us one big ratio.

📝 Teacher's Note: Show students that finding compound ratio means multiplying fractions. First simplify each ratio, then multiply like normal fractions. Cancel common factors to make calculation easy.

🎯 Exam Tip: Always simplify each ratio first. Then multiply carefully. Write your final answer as "a:b" format. Check your arithmetic twice.

Question 3. If (5m+n):(3m+n) = 3:8, find m:n.
Answer:
Step 1: Set up the proportion.
\( \frac{5m+n}{3m+n} = \frac{3}{8} \)

Step 2: Cross multiply.
\( 8(5m+n) = 3(3m+n) \)
\( 40m + 8n = 9m + 3n \)

Step 3: Collect like terms.
\( 40m - 9m = 3n - 8n \)
\( 31m = -5n \)

Step 4: Find the ratio m:n.
\( \frac{m}{n} = \frac{-5}{31} \)

Therefore, m:n = -5:31
In simple words: We wrote the given proportion as a fraction equation. Then we cross multiplied and solved for the relationship between m and n.

📝 Teacher's Note: Teach cross multiplication clearly. When we have a/b = c/d, then a×d = b×c. Students often forget this step.

🎯 Exam Tip: Write "Cross multiply" as your step. Show all algebra clearly. Don't worry if the answer has negative numbers - ratios can be negative.

Question 4. (i) Annual expenses to annual income.
Answer:
Given: Monthly income = Rs 5,000, Monthly savings = Rs 800

Step 1: Find annual income.
Annual income = Monthly income × 12 = Rs 5,000 × 12 = Rs 60,000

Step 2: Find monthly expenses.
Monthly expenses = Rs 5,000 - 800 = Rs 4,200

Step 3: Find annual expenses.
Annual expenses = Monthly expenses × 12 = Rs 4,200 × 12 = Rs 50,400

Step 4: Find the ratio.
\( \frac{\text{Annual expenses}}{\text{Annual income}} = \frac{50,400}{60,000} = \frac{504}{600} = \frac{21}{25} \)

Therefore, Annual expenses : Annual income = 21:25

(ii) Monthly savings to monthly expenses.
Monthly expenses = Rs 5,000 - 800 = Rs 4,200

\( \frac{\text{Monthly savings}}{\text{Monthly expenses}} = \frac{800}{4,200} = \frac{8}{42} = \frac{4}{21} \)

Therefore, Monthly savings : Monthly expenses = 4:21
In simple words: We found how much money is spent and earned in a year. Then we compared them as a ratio. Same for monthly savings and expenses.

📝 Teacher's Note: Help students understand that expenses = income - savings. Make them write this formula first. Use simple numbers to practice.

🎯 Exam Tip: Always write "Given" and list what you know. Show each calculation step. Simplify your final ratio to lowest terms.

Question 5. If a+b : a-b = 11:8, find a:b.
Answer:
Step 1: Set up the proportion.
\( \frac{a+b}{a-b} = \frac{11}{8} \)

Step 2: Cross multiply.
\( 8(a+b) = 11(a-b) \)
\( 8a + 8b = 11a - 11b \)

Step 3: Collect like terms.
\( 8b + 11b = 11a - 8a \)
\( 19b = 3a \)

Step 4: Find the ratio.
\( \frac{a}{b} = \frac{19}{3} \)

Therefore, a:b = 19:3
In simple words: We used cross multiplication to solve the equation. Then we rearranged to find how a and b are related.

📝 Teacher's Note: Show students how to collect terms on different sides. Move all 'a' terms to one side and all 'b' terms to the other side.

🎯 Exam Tip: Write each step clearly. When collecting terms, be careful with plus and minus signs. Double check your final answer.

Question 6. If p:q = 2:5 and q:r = 4:3, find p:r.
Answer:
Step 1: Write the given ratios as fractions.
\( \frac{p}{q} = \frac{2}{5} \) and \( \frac{q}{r} = \frac{4}{3} \)

Step 2: Multiply the ratios to find p:r.
\( \frac{p}{q} \times \frac{q}{r} = \frac{p}{r} \)
\( \frac{2}{5} \times \frac{4}{3} = \frac{8}{15} \)

Therefore, p:r = 8:15
In simple words: When we multiply two ratios, the middle term (q) cancels out. We get the ratio of the first to the last.

📝 Teacher's Note: Show students that q appears in both ratios. When we multiply the fractions, q cancels out. This is like a chain - p connects to q, q connects to r.

🎯 Exam Tip: Write the ratios as fractions first. Multiply them directly. The common term will cancel out automatically.

Question 7. If a:e = 5:12, e:i = 8:3 and i:u = 9:16, find a:u.
Answer:
Step 1: Write the ratios as fractions.
\( \frac{a}{e} = \frac{5}{12} \), \( \frac{e}{i} = \frac{8}{3} \), \( \frac{i}{u} = \frac{9}{16} \)

Step 2: Multiply all three ratios.
\( \frac{a}{e} \times \frac{e}{i} \times \frac{i}{u} = \frac{a}{u} \)
\( \frac{5}{12} \times \frac{8}{3} \times \frac{9}{16} = \frac{5 \times 8 \times 9}{12 \times 3 \times 16} \)

Step 3: Calculate.
\( = \frac{360}{576} = \frac{10}{16} = \frac{5}{8} \)

Therefore, a:u = 5:8
In simple words: We multiplied all three ratios together. The middle terms (e and i) cancelled out, leaving us with a:u.

📝 Teacher's Note: This is like a chain: a connects to e, e connects to i, i connects to u. When we multiply, the middle parts cancel out.

🎯 Exam Tip: Write all ratios as fractions. Multiply them in one step. Cancel common factors to simplify your work.

Question 8. Find the compounded ratio of 15:16 and 8:5.
Answer:
Compounded ratio = \( \frac{15}{16} \times \frac{8}{5} = \frac{15 \times 8}{16 \times 5} = \frac{120}{80} = \frac{3}{2} \)

Therefore, compounded ratio = 3:2

Also find the compounded ratio of (a²-b²):(a²+b²) and (a⁴-b⁴):(a+b)⁴
Compounded ratio = \( \frac{a^2-b^2}{a^2+b^2} \times \frac{a^4-b^4}{(a+b)^4} \)

= \( \frac{(a^2-b^2)(a^4-b^4)}{(a^2+b^2)(a+b)^4} \)

= \( \frac{(a-b)(a+b)(a^2+b^2)(a^2-b^2)}{(a^2+b^2)(a+b)^4} \)

= \( \frac{(a-b)(a+b)(a-b)(a+b)}{(a+b)^4} \)

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

Therefore, compounded ratio = (a-b)²:(a+b)²

Also find the compounded ratio of 3:5, 7:9 and 15:28
Compounded ratio = \( \frac{3}{5} \times \frac{7}{9} \times \frac{15}{28} = \frac{3 \times 7 \times 15}{5 \times 9 \times 28} = \frac{315}{1260} = \frac{1}{4} \)

Therefore, compounded ratio = 1:4
In simple words: To find compounded ratio, we multiply all the ratios together like fractions.

📝 Teacher's Note: For algebraic expressions, teach students to factorize first. For numerical ratios, multiply directly and then simplify.

🎯 Exam Tip: Always simplify your final answer. For algebraic expressions, factorize completely before canceling common factors.

Question 15. Two numbers are in the ratio 7:10. If 8 is added to each number, the ratio becomes 3:4. Find the two numbers.
Answer:
Step 1: Let the two numbers be 7x and 10x (since ratio is 7:10).

Step 2: Set up equation for the new ratio.
\( \frac{7x + 8}{10x + 8} = \frac{3}{4} \)

Step 3: Cross multiply.
\( 4(7x + 8) = 3(10x + 8) \)
\( 28x + 32 = 30x + 24 \)

Step 4: Solve for x.
\( 32 - 24 = 30x - 28x \)
\( 8 = 2x \)
\( x = 4 \)

Step 5: Find the numbers.
First number = 7x = 7 × 4 = 28
Second number = 10x = 10 × 4 = 40

Therefore, the two numbers are 28 and 40.
In simple words: We used a variable x to represent both numbers. Then we made an equation using the new ratio after adding 8 to both numbers.

📝 Teacher's Note: Teach students to use the same variable for both numbers when they are in a given ratio. This makes the algebra much easier.

🎯 Exam Tip: Always let the numbers be 7x and 10x (not just 7 and 10). Set up your equation carefully for the new condition. Check your answer by substituting back.

Answer 16.
Answer:
Let the two numbers be \( 3x \) and \( 4x \), since the ratio between them is 3:4.

Now,
\( (3x)^2 + (4x)^2 = 1225 \)
\( \implies 9x^2 + 16x^2 = 1225 \)
\( \implies 25x^2 = 1225 \)
\( \implies x^2 = 49 \)
\( \implies x = 7 \)

\( \therefore 3x = 21 \)
\( 4x = 28 \)

Therefore, the two numbers are 21 and 28.
In simple words: We called the numbers 3x and 4x because their ratio is 3:4. Then we used the fact that sum of their squares is 1225 to find x.

📝 Teacher's Note: Always start ratio problems by calling the numbers like 3x and 4x. This makes the math much easier. Students often forget to take square root at the end.

🎯 Exam Tip: Write "Let the numbers be 3x and 4x" first. Then substitute in the given condition. Always check your answer by putting back the values.

Answer 17.
Answer:
Let the two numbers be \( 5x \) and \( 7x \), since the ratio between them is 5:7.

Now,
\( (7x)^2 - (5x)^2 = 600 \)
\( \implies 49x^2 - 25x^2 = 600 \)
\( \implies 24x^2 = 600 \)
\( \implies x^2 = 25 \)
\( \implies x = 5 \)

\( \therefore 5x = 25 \)
\( 7x = 35 \)

Therefore, the two numbers are 25 and 35.
In simple words: We used the same method as before. This time we used difference of squares instead of sum of squares.

📝 Teacher's Note: Point out that this uses the formula \( a^2 - b^2 = (a+b)(a-b) \). But the direct method shown here is simpler for students.

🎯 Exam Tip: When you see "difference of squares", expand it as shown. Don't try to factorize unless asked. The direct method gets marks faster.

Answer 18.
Answer:
Let \( x \) be subtracted from each term such that

\( \frac{39 - x}{89 - x} = \frac{2}{5} \)
\( \implies 195 - 5x = 178 - 2x \)
\( \implies 3x = 17 \)
\( \implies x = \frac{17}{3} \)

\( \frac{17}{3} \) should be subtracted from each term.
In simple words: We made the two fractions equal and solved for x. Cross multiplication helps us get rid of fractions.

📝 Teacher's Note: Teach cross multiplication clearly. Many students make errors in this step. Show them: if a/b = c/d, then ad = bc.

🎯 Exam Tip: Always cross multiply when you have equal fractions. Write the cross multiplication step clearly. Check your answer by substituting back.

Answer 19.
Answer:
Let \( x \) be added to each term such that

\( \frac{19 + x}{51 + x} = \frac{3}{7} \)
\( \implies 133 + 7x = 153 + 3x \)
\( \implies 4x = 20 \)
\( \implies x = 5 \)

5 should be added to each term.
In simple words: Same method as before, but this time we add x instead of subtract x. Cross multiplication gives us the answer.

📝 Teacher's Note: Compare this with the previous question. Show students that the method is exactly the same - only + and - change.

🎯 Exam Tip: Read the question carefully. Is it "add" or "subtract"? This changes the sign in your equation. Always double-check this.

Answer 20.
Answer:
Let \( x \) be added to each term such that

\( \frac{(p + q) + x}{(p - q) + x} = \frac{(p + q)^2}{(p - q)^2} \)
\( \implies (p + q + x)(p^2 - 2pq + q^2) = (p^2 + 2pq + q^2)(p - q + x) \)
\( \implies p^3 + p^2q + p^2x + q^2p + q^3 + q^2x - 2p^2q - 2pq^2 - 2pqx \)
\( = p^3 - p^2q + p^2x + q^2p - q^3 + q^2x + 2p^2q - 2pq^2 + 2pqx \)
\( \implies p^2q + q^3 - 2p^2q - 2pqx = -p^2q - q^3 + 2p^2q + 2pqx \)
\( \implies 2q^3 - p^2q = 4pqx \)
\( \implies 2q(q^2 - p^2) = 4pqx \)
\( \implies x = \frac{(q^2 - p^2)}{2p} \)

\( \frac{(q^2 - p^2)}{2p} \) should be added to each term.
In simple words: This is a harder version of the same type. We cross multiply and then solve the algebra carefully step by step.

📝 Teacher's Note: This is quite advanced for weaker students. Focus on the method - cross multiply, expand, collect like terms. Don't worry if they can't do this independently.

🎯 Exam Tip: Take your time with algebra steps. Write each step clearly. If you make a mistake early, your whole answer will be wrong.

Answer 21.
Answer:
\( \frac{3x - 4}{2x + 5} = \frac{(3)^2}{(4)^2} \)
\( \implies \frac{3x - 4}{2x + 5} = \frac{9}{16} \)
\( \implies 48x - 64 = 18x + 45 \)
\( \implies 30x = 109 \)
\( \implies x = \frac{109}{30} \)

\( x = \frac{109}{30} \)
In simple words: We squared 3 and 4 to get 9 and 16. Then we cross multiplied and solved for x.

📝 Teacher's Note: Make sure students see that (3)² means 3×3 = 9, and (4)² means 4×4 = 16. Some students forget this basic step.

🎯 Exam Tip: Square the numbers first, then cross multiply. Don't try to do both steps together - you might make mistakes.

Answer 22.
Answer:
\( \frac{5x + 3}{3x + 1} = \frac{(4)^3}{(3)^3} \)
\( \implies \frac{5x + 3}{3x + 1} = \frac{64}{27} \)
\( \implies 135x + 81 = 192x + 64 \)
\( \implies 57x = 17 \)
\( \implies x = \frac{17}{57} \)

\( x = \frac{17}{57} \)
In simple words: This time we cubed 4 and 3 to get 64 and 27. Then we followed the same cross multiplication method.

📝 Teacher's Note: Show students that 4³ = 4×4×4 = 64 and 3³ = 3×3×3 = 27. Practice these small calculations first.

🎯 Exam Tip: Be careful with cubes. 4³ is not 12, it's 64. Calculate this step separately to avoid errors.

Answer 23.
Answer:
\( \frac{p}{q} = \frac{(p + r)^2}{(q + r)^2} \)
\( \implies \frac{p}{q} = \frac{p^2 + r^2 + 2pr}{q^2 + r^2 + 2qr} \)
\( \implies \frac{p}{q} = \frac{p^2 + pq + 2pr}{q^2 + pq + 2qr} \)
\( \implies pq^2 + p^2q + 2pqr = p^2q + pq^2 + 2pqr \)

Since LHS = RHS, therefore p:q is the duplicate ratio of (p+r) : (q+r)
In simple words: We expanded both sides and showed they are equal. This proves that the ratios are the same - they are duplicate ratios.

📝 Teacher's Note: "Duplicate ratio" means the square of the ratio. If a:b = 2:3, then the duplicate ratio is 4:9. This question proves they are the same.

🎯 Exam Tip: When you prove two things are equal, expand both sides and show LHS = RHS. Write "Since LHS = RHS" clearly for marks.

Answer 24.
Answer:
\( \frac{2}{3} = \frac{x}{7,47,300} \)
\( \implies 14,94,600 = 3x \)
\( \implies x = 4,98,200 \)

Sum distributed = Rs (7,47,300+4,98,200) = Rs 12,45,500
In simple words: We used proportion to find x. The total amount distributed is the sum of both parts.

📝 Teacher's Note: In ratio problems, always find all parts first, then add them to get the total. Students often forget this last step.

🎯 Exam Tip: Cross multiply carefully with big numbers. Double-check your arithmetic. Write the final answer clearly with "Rs" and commas.

Answer 25.
Answer:
\( \frac{5}{7} = \frac{60}{x} \)
\( \implies 5x = 420 \)
\( \implies x = 84 \)

Number = 84+60 = 144
In simple words: The smaller part is 60, which is 5/7 of the larger part x. We solve for x, then add both parts.

📝 Teacher's Note: Make sure students understand that 5:7 means one part is 5/7 of the other part. This is different from 5/12 of the total.

🎯 Exam Tip: In ratio 5:7, the first part is 5/7 times the second part. Don't confuse this with 5/12 of the whole.

Answer 26.
Answer:
Let the number of Re 1, Rs 2 and Rs 5 coins be 3x, 7x and 11x

The sum of their value =
\( 3x + (2 \times 7x) + (5 \times 11x) = Rs1800 \)
\( 3x + 14x + 55x = 1800 \)
\( 72x = 1800 \)
\( x = 25 \)

\( 3x = 3 \times 25 = 75 \)
\( 7x = 7 \times 25 = 175 \)
\( 11x = 11 \times 25 = 275 \)
total = 75 + 175 + 275 = 525

Total number of coins = 525
In simple words: We used the ratio 3:7:11 to call the numbers 3x, 7x, 11x. Then we calculated their total value and solved for x.

📝 Teacher's Note: In coin problems, multiply each type of coin by its value. Re 1 coin = Re 1, Rs 2 coin = Rs 2, etc. Then add up the total value.

🎯 Exam Tip: Don't mix up number of coins and value of coins. Re 1 coins contribute 1x each, Rs 2 coins contribute 2x each to the total value.

Answer 27.
Answer:
Let present ages be 4x and 3x.

\( \frac{4x + 9}{3x + 9} = \frac{23}{18} \)
\( \implies 72x + 162 = 69x + 207 \)
\( \implies 3x = 45 \)
\( \implies x = 15 \)

\( \therefore 3x = 45 \)
\( 4x = 60 \)

Hence, their present ages are 60 years and 45 years.
In simple words: We used the ratio 4:3 for present ages. After 9 years, both ages increase by 9. We used the future ratio to find x.

📝 Teacher's Note: In age problems, always be clear about present age vs future age. Add the same number of years to both people's ages.

🎯 Exam Tip: Write "present ages" and "after 9 years" clearly. Don't confuse which ratio applies to which time period.

Answer 28.
Answer:
Let money saved by the boy and her sister be 5x and 4x and let the sister need to save Rs y more

\( \frac{5x + 100}{4x + y} = \frac{5}{4} \)
\( \implies 20x + 400 = 20x + 5y \)
\( \implies 5y = 400 \)
\( \implies y = 80 \)

The sister needs to save Rs 80 more.
In simple words: The boy gets Rs 100 extra. For the ratio to stay 5:4, we find how much extra the sister needs.

📝 Teacher's Note: This is a tricky problem. The ratio changes when only one person gets extra money. Set up the equation carefully.

🎯 Exam Tip: Read carefully - only one person gets the extra amount. Don't add the same amount to both people.

Answer 29.
Answer:
Quantity of milk = \( 60 \times \frac{11}{15} = 44 \) litres

Quantity of water in it = 60-44 = 16 litres

New ratio = 11:6

Let quantity of water to be added further be x litres.

Then, milk : water = \( \frac{44}{16 + x} \)

Now,
\( \frac{44}{16 + x} = \frac{11}{6} \)
\( \implies 264 = 176 + 11x \)
\( \implies 11x = 88 \)
\( \implies x = 8 \)

Therefore, 8 litres of water need to be added further.
In simple words: First we found how much milk and water is already there. Then we found how much extra water to add to get the new ratio.

📝 Teacher's Note: In mixture problems, first find the actual quantities of each component. Then set up equations for the new ratio.

🎯 Exam Tip: Always find actual quantities first (not just ratios). When adding something, only that component increases - the other stays the same.

Exercise 9.2

Answer 30.
Answer:
Let successful candidates be \( 7x \) and unsuccessful candidates be \( 5x \).

Now,
\[ \frac{7x + 10}{5x + 20} = \frac{4}{3} \]
\[ \Rightarrow 21x + 30 = 20x + 80 \]
\[ \Rightarrow x = 50 \]

\( 7x = 7 \times 50 = 350 \)
\( 5x = 5 \times 50 = 250 \)
\( \therefore 7x + 5x = 350 + 250 = 600 \)

Therefore, there were 600 candidates originally.
In simple words: We used algebra to find the number of candidates. The ratio changed when some passed, so we made equations and solved step by step.

📝 Teacher's Note: Show students how ratios work with real numbers. Use examples like marbles in bags to make ratios clear. Common mistake is not setting up the equation correctly.

🎯 Exam Tip: Always define variables clearly (let successful = 7x). Show each step of cross multiplication. Write the final answer in a complete sentence.

Answer 5.
Answer:
Since x, 12 and 16 are in continued proportion
\[ \Rightarrow x : 12 :: 12 : 16 \]
\[ \Rightarrow 16x = 12 \times 12 \]
\[ \Rightarrow 16x = 144 \]
\[ \Rightarrow x = 9 \]

\( x = 9 \)
In simple words: When three numbers are in continued proportion, the middle number squared equals the product of first and last numbers.

📝 Teacher's Note: Teach students the formula for continued proportion: if a, b, c are in continued proportion, then b² = ac. This makes solving much faster.

🎯 Exam Tip: Write "continued proportion" clearly and use the formula b² = ac. Show cross multiplication steps to get full marks.

Answer 6.
Answer:
Since \( \frac{1}{12} \), x and \( \frac{1}{75} \) are in continued proportion
\[ \Rightarrow \frac{1}{12} : x :: x : \frac{1}{75} \]
\[ \Rightarrow x^2 = \frac{1}{12} \times \frac{1}{75} \]
\[ \Rightarrow x = \sqrt{\frac{1}{900}} \]
\[ \Rightarrow x = \frac{1}{30} \]

\( x = \frac{1}{30} \)
In simple words: We found the middle number by taking the square root of the product of the first and last fractions.

📝 Teacher's Note: When dealing with fractions in continued proportion, multiply the fractions first, then take square root. Practice with simple fractions before complex ones.

🎯 Exam Tip: Show the square root step clearly. Write the final answer as a simplified fraction. Check your work by substituting back.

Answer 7.
Answer:
Since y is the mean proportion between x and z
\[ y^2 = xz \]
LHS
\[ xyz(x + y + z)^3 \]
\[ = yy^2(x + y + z)^3 \]
\[ = y^3(x + y + z)^3 \]
\[ = [y(x + y + z)]^3 \]
\[ = (xy + y^2 + yz)^3 \]
\[ = (xy + yz + xz)^3 = RHS \]

LHS = RHS
In simple words: We used the property that y² = xz to prove both sides are equal. We expanded and simplified step by step.

📝 Teacher's Note: Emphasize that mean proportion means y² = xz. Show students how to substitute this relationship to simplify complex expressions.

🎯 Exam Tip: Start by writing the given condition (y² = xz). Show LHS and RHS separately. Use substitution clearly to get full marks.

Answer 8.
Answer:
Since y is the mean proportion between x and z
Therefore, \( y^2 = xz \)

Now, we have to prove that xy+yz is the mean proportional between \( x^2+y^2 \) and \( y^2+z^2 \).

\[ (xy + yz)^2 = (x^2 + y^2)(y^2 + z^2) \]
LHS = \( (xy + yz)^2 \)
\[ = [y(x + z)]^2 \]
\[ = y^2(x + z)^2 \]
\[ = xz(x + z)^2 \]
RHS = \( (x^2 + y^2)(y^2 + z^2) \)
\[ = (x^2 + xz)(xz + z^2) \]
\[ = x(x + z)z(x + z) \]
\[ = xz(x + z)^2 \]

LHS=RHS
In simple words: We proved that the square of the middle term equals the product of the outer terms, which is the definition of mean proportion.

📝 Teacher's Note: Break down complex algebraic proofs into small steps. Show students how to factor expressions to reveal patterns. Practice substitution techniques.

🎯 Exam Tip: Write what you need to prove at the start. Use the given condition (y² = xz) throughout. Show LHS = RHS clearly.

Answer 9.
Answer:
Let x, y and z are the three quantities which are in continued proportion.
Then, \( x:y::y:z \Rightarrow y^2 = xz \)

Now, we have to prove that
\[ x:z = x^2:y^2 \]
\[ \Rightarrow xy^2 = x^2z \]
LHS
\[ = xy^2 = x(xz) = x^2z = RHS \]
LHS = RHS
In simple words: We used the continued proportion property (y² = xz) to show that the given ratio is true by substitution.

📝 Teacher's Note: Remind students that in continued proportion, the middle term squared equals the product of the other two terms. This is the key to solving such problems.

🎯 Exam Tip: Write the continued proportion condition first. Use substitution method clearly. Show that LHS equals RHS step by step.

Answer 11.
Answer:
\[ \frac{a}{b} = \frac{c}{d} \Rightarrow \frac{ax}{b} = \frac{cx}{d} \]

Adding y to both sides:
\[ \Rightarrow \frac{ax}{b} + y = \frac{cx}{d} + y \]
\[ \Rightarrow \frac{ax + by}{b} = \frac{cx + dy}{d} \]
\[ \Rightarrow (ax + by) : b = (cx + dy) : d \]

Proved.
In simple words: If two fractions are equal, they stay equal when we add the same number to both. This is a basic property of equal ratios.

📝 Teacher's Note: Show students that adding the same value to numerator keeps the ratio property intact. Use simple numbers first before algebraic expressions.

🎯 Exam Tip: Start with the given condition clearly. Show each algebraic step. Write "Proved" at the end to show completion.

Answer 12.
Answer:
\[ \frac{a + c}{b} = \frac{5}{1} \]
\[ \Rightarrow a + c = 5b \]

\[ \frac{bc + cd}{bd} = \frac{5}{1} \]
\[ \Rightarrow bc + cd = 5bd \]
\[ \Rightarrow bc + cd = (a + c)d \]
\[ \Rightarrow bc + cd = ad + cd \]
\[ \Rightarrow bc = ad \]
\[ \Rightarrow \frac{a}{b} = \frac{c}{d} \]

Hence a:b = c:d
In simple words: We used algebra to show that if certain conditions are met, then the two ratios must be equal.

📝 Teacher's Note: This is a reverse proof - we start with a condition and prove the ratios are equal. Show students how to work backwards from given information.

🎯 Exam Tip: Use substitution carefully. Show how the algebra leads to the final ratio. Write the conclusion in ratio form clearly.

Answer 13.
Answer:
Let x be subtracted from each number so that 20-x, 29-x, 84-x and 129-x are in proportion.

\[ \therefore \frac{20 - x}{29 - x} = \frac{84 - x}{129 - x} \]
\[ \Rightarrow (20 - x)(129 - x) = (29 - x)(84 - x) \]
\[ \Rightarrow 2580 - 129x - 20x + x^2 = 2436 - 84x - 29x + x^2 \]
\[ \Rightarrow 2580 - 149x = 2436 - 113x \]
\[ \Rightarrow 36x = 144 \]
\[ \Rightarrow x = 4 \]

Hence, 4 is to be subtracted from 20, 29, 84 and 129 for them to be in proportion.
In simple words: We found what number to subtract so that all four numbers form a proportion (equal ratios).

📝 Teacher's Note: Explain that in proportion, the product of extremes equals product of means. Students often forget to expand brackets correctly.

🎯 Exam Tip: Set up the proportion equation first. Cross multiply carefully. Check your answer by substituting back into the original ratios.

Answer 14.
Answer:
Let a and b be the two numbers, whose mean proportional is 12.
\[ \therefore ab = 12^2 \Rightarrow ab = 144 \Rightarrow b = \frac{144}{a} \ldots\ldots(i) \]

Now, third proportional is 324
\[ \therefore a : b :: b : 324 \]
\[ \Rightarrow b^2 = 324a \]
\[ \Rightarrow \left(\frac{144}{a}\right)^2 = 324a \]
\[ \Rightarrow \frac{(144)^2}{a^2} = 324a \]
\[ \Rightarrow a^3 = \frac{144 \times 144}{324} \]
\[ \Rightarrow a^3 = 64 \]
\[ \Rightarrow a = 4 \]

\[ b = \frac{144}{a} = \frac{144}{4} = 36 \]

Therefore, numbers are 4 and 36
In simple words: We used the mean proportional and third proportional properties to find the two unknown numbers by solving equations.

📝 Teacher's Note: Teach students that mean proportional of a,b is √(ab) and third proportional follows the pattern a:b::b:c. Practice with smaller numbers first.

🎯 Exam Tip: Define variables clearly. Use the mean proportional formula (√ab = 12). Show substitution steps clearly for full marks.

Answer 15.
Answer:
Let a and b be the two numbers, whose mean proportional is 18.
\[ \therefore ab = 18^2 \Rightarrow ab = 324 \Rightarrow b = \frac{324}{a} \ldots\ldots(i) \]

Now, third proportional is 486
\[ \therefore a : b :: b : 486 \]
\[ \Rightarrow b^2 = 486a \]
\[ \Rightarrow \left(\frac{324}{a}\right)^2 = 486a \]
\[ \Rightarrow \frac{(324)^2}{a^2} = 486a \]
\[ \Rightarrow a^3 = \frac{324 \times 324}{486} \]
\[ \Rightarrow a^3 = 216 \]
\[ \Rightarrow a = 6 \]

\[ b = \frac{324}{a} = \frac{324}{6} = 54 \]

Therefore, numbers are 6 and 54
In simple words: Similar to the previous problem, we found two numbers using their mean proportional and third proportional values.

📝 Teacher's Note: This follows the same pattern as the previous problem. Help students see the pattern so they can solve similar problems faster.

🎯 Exam Tip: Follow the same method as Answer 14. Be careful with arithmetic when finding cube roots. Double-check by substituting back.