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

Question 1. Select the appropriate alternative answer for the following questions.
(i) If 6 : 5 = y : 20, then what will be the value of y?
(A) 15
(B) 24
(C) 18
(D) 22.5
Answer: (B) 24
In simple words: To find 'y', set up the proportion \( \frac{6}{5} = \frac{y}{20} \), then cross-multiply and solve for y. \( 6 \times 20 = 5 \times y \), so \( 120 = 5y \), which gives \( y = 24 \).

🎯 Exam Tip: For ratio problems, ensure units are consistent or converted to the same unit before calculation. Setting up the proportion correctly is crucial for accuracy.

Question 1. (ii) What is the ratio of 1 mm to 1 cm ?
(A) 1: 100
(B) 10: 1
(C) 1:10
(D) 100: 1
Answer: (C) 1:10
In simple words: Since 1 cm equals 10 mm, the ratio of 1 mm to 1 cm is 1 mm to 10 mm, which simplifies to 1:10.

🎯 Exam Tip: Always convert quantities to the same unit before calculating their ratio to avoid errors. Remember the common unit conversions like 1 cm = 10 mm.

Question 1. (iii) The ages of Jatin, Nitin and Mohasin are 16, 24 and 36 years respectively. What is the ratio of Nitin's age to Mohasin's age ?
(A) 3:2
(B) 2:3
(C) 4:3
(D) 3:4
Answer: (B) 2:3
In simple words: Nitin's age is 24 and Mohasin's age is 36. The ratio \( \frac{24}{36} \) simplifies to \( \frac{2}{3} \) by dividing both numbers by their greatest common divisor, 12.

🎯 Exam Tip: When finding a ratio, ensure you put the quantities in the correct order as specified (e.g., Nitin's age to Mohasin's age, not the other way around) and simplify the ratio to its lowest terms.

Question 1. (iv) 24 bananas were distributed between Shubham and Anil in the ratio 3 : 5, then how many bananas did Shubham get?
(A) 8
(B) 15
(C) 12<
(D) 9
Answer: (D) 9
In simple words: The total ratio parts are \( 3 + 5 = 8 \). Shubham gets 3 parts out of 8, so multiply the total bananas (24) by \( \frac{3}{8} \) to find his share. \( 24 \times \frac{3}{8} = 9 \).

🎯 Exam Tip: When distributing a quantity in a given ratio, first find the sum of the ratio parts. Then, multiply the total quantity by the fraction representing each individual's share.

Question 1. (v) What is the mean proportional of 4 and 25?
(A) 6
(B) 8
(C) 10
(D) 12
Answer: (C) 10
In simple words: The mean proportional of two numbers is the square root of their product. So, for 4 and 25, it's \( \sqrt{4 \times 25} = \sqrt{100} = 10 \).

🎯 Exam Tip: Remember the formula for mean proportional (also known as geometric mean): for two numbers 'a' and 'b', the mean proportional 'x' is given by \( x = \sqrt{ab} \).

Question 2. For the following numbers write the ratio of first number to second number in the reduced form. [1 Mark each]
(i) 21,48
(ii) 36,90
(iii) 65,117
(iv) 138,161
(v) 114,133
Answer:
Solution:
(i) 21,48
Ratio \( = \frac{21}{48} = \frac{3 \times 7}{3 \times 16} = \frac{7}{16} = 7:16 \)
(ii) 36,90
Ratio \( = \frac{36}{90} = \frac{18 \times 2}{18 \times 5} = \frac{2}{5} = 2:5 \)
(iii) 65,117
Ratio \( = \frac{65}{117} = \frac{13 \times 5}{13 \times 9} = \frac{5}{9} = 5:9 \)
(iv) 138,161
Ratio \( = \frac{138}{161} = \frac{23 \times 6}{23 \times 7} = \frac{6}{7} = 6:7 \)
(v) 114,133
Ratio \( = \frac{114}{133} = \frac{19 \times 6}{19 \times 7} = \frac{6}{7} = 6:7 \)
In simple words: To find the ratio in reduced form, express the numbers as a fraction and divide both the numerator and denominator by their greatest common divisor until no common factors remain.

🎯 Exam Tip: Always simplify ratios to their lowest possible form by identifying and dividing by common factors. This demonstrates a complete understanding of ratio reduction.

Question 3. Write the following ratios in the reduced form.
(i) Radius to the diameter of a circle.
(ii) The ratio of diagonal to the length of a rectangle, having length 4 cm and breadth 3 cm.
(iii) The ratio of numbers denoting perimeter to area of a square, having side 4 cm.
Answer:
Solution:
(i) Radius to the diameter of a circle.
Let radius of the circle be r
then, diameter = 2 x radius = 2r
Ratio of radius to diameter of circle
\( = \frac{\text{Radius of the circle}}{\text{Diameter of the circle}} \)
\( = \frac{r}{2r} \)
\( = \frac{1}{2} \)
\( = 1:2 \)
\( \therefore \) Ratio of radius to diameter of circle is 1 : 2.

ℹ️ चित्र व्याख्या (Diagram Explanation): एक आयत ABCD दिखाया गया है, जहाँ AB इसकी लंबाई है और BC इसकी चौड़ाई है। AC आयत का विकर्ण (diagonal) है। AB की लंबाई 4 cm और BC की चौड़ाई 3 cm है।

(ii) The ratio of diagonal to the length of a rectangle, having length 4 cm and breadth 3 cm.
Let ABCD be a rectangle.
In \( \triangle \)ABC, \( \angle \)B = 90°
\( \text{AC}^2 = \text{AB}^2 + \text{BC}^2 \) ... [Pythagoras theorem]
\( = 4^2 + 3^2 = 16 + 9 \)
\( \therefore \text{AC}^2 = 25 \)
AC = 5 ... [Taking square root on both side]
The ratio of diagonal to the length of a rectangle \( = \frac{\text{AC}}{\text{AB}} \)
\( = \frac{5}{4} \)
\( = 5:4 \)
\( \therefore \) The ratio of diagonal to the length of a rectangle is 5 : 4.
(iii) The ratio of numbers denoting perimeter to area of a square, having side 4 cm.
side of square = 4cm
Perimeter of square = 4 x side = 4 x 4 = 16 cm
Area of square = (side)\( ^2 \) = (4)\( ^2 \) = 16 cm\( ^2 \)
Ratio of numbers denoting perimeter to area of square
\( = \frac{\text{Perimeter of the square}}{\text{Area of the square}} \)
\( = \frac{16}{16} = \frac{1}{1} \)
\( = 1:1 \)
\( \therefore \) The ratio of numbers denoting perimeter to area of a square is 1 : 1.
In simple words: For ratios involving geometric figures, first identify the relevant formulas (e.g., diameter = 2r, Pythagoras theorem, perimeter/area of square). Calculate the quantities using the given dimensions, then form and simplify the ratio.

🎯 Exam Tip: Clearly state the formulas used for geometric calculations. For ratios, ensure consistent units (e.g., cm) and simplify the final ratio to its simplest form.

Question 4. Check whether the following numbers are in continued proportion.
(i) 2, 4, 8
(ii) 1, 2, 3
(iii) 9, 12, 16
(iv) 3, 5, 8
Answer:
Solution:
If a, b, c are in continued proportion then b\( ^2 \) = ac.
(i) 2, 4, 8
Let, a = 2, b = 4 and c = 8
Here, b\( ^2 \) = 4\( ^2 \) = 16
ac = 2 x 8 = 16
\( \therefore \) b\( ^2 \) = ac
\( \therefore \) 2, 4,8 are in continued proportion.
(ii) 1, 2, 3
Let, a = 1, b = 2 and c = 3
Here, b\( ^2 \) = 2\( ^2 \) = 4
ac = 1 x 3 = 3
\( \therefore \) b\( ^2 \neq \) ac
\( \therefore \) 1, 2,3 are not in continued proportion.
(iii) 9, 12, 16
Let, a = 9, b = 12 and c = 16
Here, b\( ^2 \) = 12\( ^2 \) = 144
ac = 9 x 16 = 144
\( \therefore \) b\( ^2 \) = ac
\( \therefore \) 9, 12, 16 are in continued proportion.
(iv) 3, 5, 8
Let, a = 3, b = 5 and c = 8
Here, b\( ^2 \) = 5\( ^2 \) = 25
ac = 3 x 8 = 24
\( \therefore \) b\( ^2 \neq \) ac
\( \therefore \) 3, 5, 8 are not in continued proportion.
In simple words: Three numbers a, b, c are in continued proportion if the square of the middle term (b\( ^2 \)) is equal to the product of the first and third terms (ac).

🎯 Exam Tip: To check for continued proportion, simply calculate b\( ^2 \) and ac. If they are equal, the numbers are in continued proportion; otherwise, they are not. Show all calculation steps clearly.

Question 5. a, b, c are in continued proportion. If a = 3 and c = 27, then find b.
Answer:
Solution:
a, b, c are in continued proportion. ...[Given]
\( \therefore \) b\( ^2 \) = ac
\( \therefore \) b\( ^2 \) = 3 x 27 ...[: a = 3 and c = 27]
\( \therefore \) b\( ^2 \) = 81
\( \therefore \) b = 9 ...[Taking square root of both sides]
In simple words: For numbers in continued proportion, the square of the middle term equals the product of the other two. Given a=3 and c=27, b\( ^2 \) = 3 * 27 = 81, so b = \( \sqrt{81} \) = 9.

🎯 Exam Tip: When solving for an unknown in a continued proportion, always write down the fundamental relation \( b^2 = ac \) first. Remember to take the square root to find 'b'.

Question 6. Convert the following ratios into percentages.
(i) 37:500
(ii) \( \frac{5}{8} \)
(iii) \( \frac{22}{30} \)
(iv) \( \frac{5}{16} \)
(v) \( \frac{144}{1200} \)
Answer:
Solution:
(i) Let 37: 500 = x%
\( \therefore \frac{37}{500} = \frac{x}{100} \)
\( \therefore x = \frac{37}{500} \times 100 = \frac{37}{5} \)
\( \therefore x = 7.4\% \)
(ii) Let \( \frac{5}{8} \) = x%
\( \therefore \frac{5}{8} = \frac{x}{100} \)
\( \therefore x = \frac{5}{8} \times 100 = \frac{5}{2} \times 25 \)
\( = 62.5\% \)
(iii) Let \( \frac{22}{30} \) = x%
\( \therefore \frac{22}{30} = \frac{x}{100} \)
\( \therefore x = \frac{22}{30} \times 100 = \frac{22}{3} \times 10 \)
\( = 73.33\% \)
(iv) Let \( \frac{5}{16} \) = x%
\( \therefore \frac{5}{16} = \frac{x}{100} \)
\( \therefore x = \frac{5}{16} \times 100 = \frac{5}{4} \times 25 \)
\( = 31.25\% \)
(v) Let \( \frac{144}{1200} \) = x%
\( \therefore \frac{144}{1200} = \frac{x}{100} \)
\( \therefore x = \frac{144}{1200} \times 100 = \frac{144}{12} \)
\( = 12\% \)
In simple words: To convert a ratio or fraction to a percentage, multiply the value by 100. For example, a ratio a:b is \( \frac{a}{b} \times 100\% \).

🎯 Exam Tip: Ensure precise calculations when converting to percentages, especially when dealing with fractions that result in repeating decimals. Rounding to an appropriate number of decimal places might be required based on instructions.

Question 7. Write the ratio of first quantity to second quantity in the reduced form.
(i) 1024 MB, 1.2 GB [(1024 MB = 1GB)]
(ii) Rs. 17, Rs. 25 and 60 paise
(iii) 5 dozen, 120 units
(iv) 4 sq.m, 800 sq.cm
(v) 1.5 kg, 2500 gm
Answer:
Solution:
(i) 1024 MB, 1.2 GB
1024 MB = 1 GB
Ratio \( = \frac{1024 \text{ MB}}{1.2 \text{ GB}} = \frac{1 \text{ GB}}{1.2 \text{ GB}} = \frac{1}{1.2} \)
\( = \frac{10}{12} = \frac{2 \times 5}{2 \times 6} \)
\( = \frac{5}{6} = 5:6 \)
(ii) Rs. 17, Rs. 25 and 60 paise
Rs. 17 = 17 x 100 paise = 1700 paise
Rs. 25 and 60 paise = (25x 100) paise + 60 paise
= (2500 + 60) paise
= 2560 paise
Ratio \( = \frac{\text{Rs. } 17}{\text{Rs. } 25 \text{ and } 60 \text{ paise}} \)
\( = \frac{1700 \text{ paise}}{2560 \text{ paise}} \)
\( = \frac{1700}{2560} \)
\( = \frac{20 \times 85}{20 \times 128} \)
\( = \frac{85}{128} \)
\( = 85: 128 \)
(iii) 5 dozen, 120 units
5 dozen = 5 x 12 units = 60 units
Ratio \( = \frac{5 \text{ dozen}}{120 \text{ units}} = \frac{60 \text{ units}}{120 \text{ units}} \)
\( = \frac{60}{120} \)
\( = \frac{60 \times 1}{60 \times 2} = \frac{1}{2} \)
\( = 1:2 \)
(iv) 4 sq.m, 800 sq.cm
4 sq.m = 4 x 10000 sq.cm = 40000 sq.cm
Ratio \( = \frac{4 \text{sq.m}}{800 \text{sq.cm}} = \frac{40000 \text{ sq.cm}}{800 \text{ sq.cm}} \)
\( = \frac{40000}{800} \)
\( = \frac{400}{8} \)
\( = \frac{8 \times 50}{8 \times 1} = \frac{50}{1} \)
\( = 50:1 \)
(v) 1.5 kg, 2500 gm
1.5 kg = 1.5 x 1000 gm = 1500gm
Ratio \( = \frac{1.5 \text{kg}}{2500 \text{ gm}} = \frac{1500 \text{gm}}{2500 \text{gm}} \)
\( = \frac{1500}{2500} \)
\( = \frac{15}{25} \)
\( = \frac{5 \times 3}{5 \times 5} = \frac{3}{5} \)
\( = 3:5 \)
In simple words: To find the ratio between two quantities, first convert them to the same units, then express them as a fraction and simplify to the lowest terms.

🎯 Exam Tip: Unit conversion is a common pitfall; double-check conversions (e.g., kg to gm, m\( ^2 \) to cm\( ^2 \), Rs. to paise). Always simplify ratios to their lowest whole number form for full marks.

Question 8. If \( \frac{a}{b} = \frac{2}{3} \), then find the values of the following expressions.
(i) \( \frac{4a+3b}{3b} \)
(ii) \( \frac{5a^2+2b^2}{5a^2-2b^2} \)
(iii) \( \frac{a^3+b^3}{b^3} \)
(iv) \( \frac{7b-4a}{7b+4a} \)
Answer:
Solution:
(i) \( \frac{a}{b} = \frac{2}{3} \) ...[Given]
Multiplying both sides by \( \frac{4}{3} \)
\( \frac{4a}{3b} = \frac{2}{3} \times \frac{4}{3} \)
\( \therefore \frac{4a}{3b} = \frac{8}{9} \)
By componendo
\( \frac{4a+3b}{3b} = \frac{8+9}{9} \)
\( \frac{4a+3b}{3b} = \frac{17}{9} \)
(ii) \( \frac{a}{b} = \frac{2}{3} \) ...[Given]
Squaring both sides
\( \left(\frac{a}{b}\right)^2 = \left(\frac{2}{3}\right)^2 \)
\( \therefore \frac{a^2}{b^2} = \frac{4}{9} \)
Multiplying both sides by \( \frac{5}{2} \)
\( \frac{a^2}{b^2} \times \frac{5}{2} = \frac{4}{9} \times \frac{5}{2} \)
\( \therefore \frac{5a^2}{2b^2} = \frac{10}{9} \)
By componendo - dividendo
\( \frac{5a^2+2b^2}{5a^2-2b^2} = \frac{10+9}{10-9} \)
\( = \frac{19}{1} \)
\( \frac{5a^2+2b^2}{5a^2-2b^2} = 19 \)
(iii) \( \frac{a}{b} = \frac{2}{3} \) ...[Given]
Cubing both sides
\( \left(\frac{a}{b}\right)^3 = \left(\frac{2}{3}\right)^3 \)
\( \therefore \frac{a^3}{b^3} = \frac{8}{27} \)
By componendo
\( \frac{a^3+b^3}{b^3} = \frac{8+27}{27} \)
\( \therefore \frac{a^3+b^3}{b^3} = \frac{35}{27} \)
(iv) \( \frac{a}{b} = \frac{2}{3} \) ...[Given]
By invertendo
\( \frac{b}{a} = \frac{3}{2} \)
Multiplying both sides by \( \frac{7}{4} \)
\( \frac{b}{a} \times \frac{7}{4} = \frac{3}{2} \times \frac{7}{4} \)
\( \therefore \frac{7b}{4a} = \frac{21}{8} \)
By componendo - dividendo
\( \frac{7b+4a}{7b-4a} = \frac{21+8}{21-8} \)
\( \therefore \frac{7b+4a}{7b-4a} = \frac{29}{13} \)
By invertendo
\( \therefore \frac{7b-4a}{7b+4a} = \frac{13}{29} \)
In simple words: To evaluate expressions involving ratios like \( \frac{a}{b} = \frac{2}{3} \), use properties of ratio and proportion such as componendo, dividendo, and invertendo, often after squaring, cubing, or multiplying by constants to match the expression's form.

🎯 Exam Tip: Master the properties of ratio and proportion (invertendo, alternando, componendo, dividendo, componendo-dividendo) as they are essential for simplifying complex expressions without substituting direct values, especially for multiple-choice questions or proofs.

Question 9. If a, b, c, d are in proportion, then prove that
(i) \( \frac{11a^2+9ac}{11b^2+9bd} = \frac{a^2+3ac}{b^2+3bd} \)
(ii) \( \frac{\sqrt{a^2+5c^2}}{\sqrt{b^2+5d^2}} = \frac{a}{b} \)
(iii) \( \frac{a^2+ab+b^2}{a^2-ab+b^2} = \frac{c^2+cd+d^2}{c^2-cd+d^2} \)
Answer:
Solution:
a, b, c, d are in proportion. ...[Given]
\( \therefore \frac{a}{b} = \frac{c}{d} \)
Let \( \frac{a}{b} = \frac{c}{d} = k \)
\( \therefore a = bk \) and \( c = dk \) ...(i)
(i) L.H.S. \( = \frac{11a^2+9ac}{11b^2+9bd} \)
\( = \frac{11(bk)^2+9(bk)(dk)}{11b^2+9bd} \) ...[From (i)]
\( = \frac{11b^2k^2+9bdk^2}{11b^2+9bd} \)
\( = \frac{k^2(11b^2+9bd)}{11b^2+9bd} \)
\( = k^2 \)
R.H.S. \( = \frac{a^2+3ac}{b^2+3bd} \)
\( = \frac{(bk)^2+3(bk)(dk)}{b^2+3bd} \) ...[From (i)]
\( = \frac{b^2k^2+3bdk^2}{b^2+3bd} \)
\( = \frac{k^2(b^2+3bd)}{b^2+3bd} \)
\( = k^2 \)
\( \therefore \) L.H.S. = R.H.S.
\( \therefore \frac{11a^2+9ac}{11b^2+9bd} = \frac{a^2+3ac}{b^2+3bd} \)

(ii) L.H.S. \( = \sqrt{\frac{a^2+5c^2}{b^2+5d^2}} \)
\( = \sqrt{\frac{(bk)^2+5(dk)^2}{b^2+5d^2}} \) ...[From (i)]
\( = \sqrt{\frac{b^2k^2+5d^2k^2}{b^2+5d^2}} \)
\( = \sqrt{\frac{k^2(b^2+5d^2)}{b^2+5d^2}} \)
\( = \sqrt{k^2} \)
\( = k \)
R.H.S. \( = \frac{a}{b} \)
\( = \frac{bk}{b} \) ...[From (i)]
\( = k \)
\( \therefore \) L.H.S. = R.H.S.
\( \therefore \sqrt{\frac{a^2+5c^2}{b^2+5d^2}} = \frac{a}{b} \)

(iii) L.H.S. \( = \frac{a^2+ab+b^2}{a^2-ab+b^2} \)
\( = \frac{(bk)^2+(bk)b+b^2}{(bk)^2-(bk)b+b^2} \) ...[From (i)]
\( = \frac{b^2k^2+b^2k+b^2}{b^2k^2-b^2k+b^2} \)
\( = \frac{b^2(k^2+k+1)}{b^2(k^2-k+1)} \)
\( = \frac{k^2+k+1}{k^2-k+1} \)
R.H.S. \( = \frac{c^2+cd+d^2}{c^2-cd+d^2} \)
\( = \frac{(dk)^2+(dk)d+d^2}{(dk)^2-(dk)d+d^2} \) ...[From (i)]
\( = \frac{d^2k^2+d^2k+d^2}{d^2k^2-d^2k+d^2} \)
\( = \frac{d^2(k^2+k+1)}{d^2(k^2-k+1)} \)
\( = \frac{k^2+k+1}{k^2-k+1} \)
\( \therefore \) L.H.S. = R.H.S.
\( \therefore \frac{a^2+ab+b^2}{a^2-ab+b^2} = \frac{c^2+cd+d^2}{c^2-cd+d^2} \)
In simple words: To prove the given proportionalities, use the property that if a, b, c, d are in proportion, then \( \frac{a}{b} = \frac{c}{d} = k \). Substitute \( a=bk \) and \( c=dk \) into both sides of each equation and show that L.H.S. equals R.H.S.

🎯 Exam Tip: For proof-based questions involving proportions, the method of `k`-substitution (i.e., \( \frac{a}{b} = \frac{c}{d} = k \)) is highly effective. Always show step-by-step substitution and simplification for both L.H.S. and R.H.S.

Question 10. If a, b, c are in continued proportion, then prove that
(i) \( \frac{a}{a+2b} = \frac{a-2b}{a-4c} \)
(ii) \( \frac{b}{b+c} = \frac{a-b}{a-c} \)
Answer:
Solution:
a, b, c are in continued proportion. ...[Given]
\( \therefore \frac{a}{b} = \frac{b}{c} \)
Let \( \frac{a}{b} = \frac{b}{c} = k \)
\( \therefore b = ck \)
\( \therefore a = bk \)
\( = (ck)k \) ...[From (j)]
\( a = ck^2 \) ...(ii)
(i) L.H.S. \( = \frac{a}{a+2b} \)
\( = \frac{ck^2}{ck^2+2ck} \) ...[From (i) and (ii)]
\( = \frac{ck^2}{ck(k+2)} \)
\( = \frac{k}{k+2} \)
R.H.S. \( = \frac{a-2b}{a-4c} \)
\( = \frac{ck^2-2ck}{ck^2-4c} \) ...[From (i) and (ii)]
\( = \frac{ck(k-2)}{c(k^2-4)} \)
\( = \frac{k(k-2)}{(k+2)(k-2)} \)
...[ \( a^2-b^2 = (a+b)(a-b) \) ]
\( = \frac{k}{k+2} \)
\( \therefore \) L.H.S. = R.H.S.
\( \therefore \frac{a}{a+2b} = \frac{a-2b}{a-4c} \)
(ii) L.H.S. \( = \frac{b}{b+c} \)
\( = \frac{ck}{ck+c} \) ...[From (i) and (ii)]
\( = \frac{ck}{c(k+1)} \)
\( = \frac{k}{k+1} \)
R.H.S. \( = \frac{a-b}{a-c} \)
\( = \frac{ck^2-ck}{ck^2-c} \) ...[From (i) and (ii)]
\( = \frac{ck(k-1)}{c(k^2-1)} \)
\( = \frac{k(k-1)}{(k+1)(k-1)} \)
...[ \( a^2-b^2 = (a+b)(a-b) \) ]
\( = \frac{k}{k+1} \)
\( \therefore \) L.H.S. = R.H.S.
\( \therefore \frac{b}{b+c} = \frac{a-b}{a-c} \)
In simple words: For numbers in continued proportion (a, b, c), it means \( \frac{a}{b} = \frac{b}{c} \). Let this common ratio be 'k', so \( b=ck \) and \( a=bk=ck^2 \). Substitute these values into the expressions for L.H.S. and R.H.S. and simplify them to show equality.

🎯 Exam Tip: When dealing with continued proportion, establishing the `k`-values correctly (\( b=ck \) and \( a=ck^2 \)) is the first critical step. Simplify expressions by factoring common terms, especially `c` and `k`, and identify common algebraic identities like \( k^2-1 = (k-1)(k+1) \).

Question 10. i. If a, b, c are in continued proportion, then prove that

\( \frac{a}{a+2b} = \frac{a-2b}{a-4c} \) Solution:
a, b, c are in continued proportion. ... [Given]
\( \frac{a}{b} = \frac{b}{c} = k \)
Let \( \frac{a}{b} = k \)
\( \implies a = bk \)
Also, Let \( \frac{b}{c} = k \)
\( \implies b = ck \)
Substitute \( b = ck \) into \( a = bk \):
\( a = (ck)k \)
\( \implies a = ck^2 \) ...(ii)
L.H.S. = \( \frac{a}{a+2b} \)
Substitute \( a=ck^2 \) and \( b=ck \):
\( = \frac{ck^2}{ck^2+2ck} \)
\( = \frac{ck^2}{ck(k+2)} \)
\( = \frac{k}{k+2} \) ...[From (i) and (ii)]
R.H.S. = \( \frac{a-2b}{a-4c} \)
Substitute \( a=ck^2 \) and \( b=ck \):
\( = \frac{ck^2-2ck}{ck^2-4c} \)
\( = \frac{ck(k-2)}{c(k^2-4)} \)
\( = \frac{k(k-2)}{(k+2)(k-2)} \) ...[ \( a^2-b^2 = (a+b)(a-b) \) ]
\( = \frac{k}{k+2} \) ...[From (i) and (ii)]
L.H.S. = R.H.S.
In simple words: This proof demonstrates how to use the property of continued proportion (\( a/b = b/c = k \)) to substitute values and simplify both sides of the given equation, showing they are equal to the same ratio \( k/(k+2) \).

🎯 Exam Tip: When proving identities involving continued proportion, always start by setting up \( a=bk \) and \( b=ck \) (or \( a=ck^2 \)) from the common ratio \( k \). This substitution simplifies complex expressions efficiently.

Question 10. ii. If a, b, c are in continued proportion, then prove that

\( \frac{b}{b+c} = \frac{a-b}{a-c} \) Solution:
a, b, c are in continued proportion. ... [Given]
\( \frac{a}{b} = \frac{b}{c} = k \)
Let \( \frac{a}{b} = k \)
\( \implies a = bk \)
Also, Let \( \frac{b}{c} = k \)
\( \implies b = ck \)
Substitute \( b = ck \) into \( a = bk \):
\( a = (ck)k \)
\( \implies a = ck^2 \) ...(ii)
L.H.S. = \( \frac{b}{b+c} \)
Substitute \( b=ck \):
\( = \frac{ck}{ck+c} \)
\( = \frac{ck}{c(k+1)} \)
\( = \frac{k}{k+1} \) ...[From (i) and (ii)]
R.H.S. = \( \frac{a-b}{a-c} \)
Substitute \( a=ck^2 \) and \( b=ck \):
\( = \frac{ck^2-ck}{ck^2-c} \)
\( = \frac{ck(k-1)}{c(k^2-1)} \)
\( = \frac{k(k-1)}{(k+1)(k-1)} \) ...[ \( a^2-b^2 = (a+b)(a-b) \) ]
\( = \frac{k}{k+1} \) ...[From (i) and (ii)]
L.H.S. = R.H.S.
In simple words: This proof uses the definition of continued proportion to express 'a' and 'b' in terms of 'c' and 'k'. By substituting these into the given equation, both the left-hand side and right-hand side simplify to \( k/(k+1) \), thus proving the equality.

🎯 Exam Tip: Remember the algebraic identity for difference of squares \( k^2-1 = (k-1)(k+1) \) as it's often useful in simplifying ratio expressions during proofs.

Question 11. Solve:

\( \frac{12x^2+18x+42}{18x^2+12x+58} = \frac{2x+3}{3x+2} \) Solution:
If \( x = 0 \) then
L.H.S \( = \frac{12(0)^2+18(0)+42}{18(0)^2+12(0)+58} = \frac{42}{58} = \frac{21}{29} \)
R.H.S \( = \frac{2(0)+3}{3(0)+2} = \frac{3}{2} \)
\( \frac{21}{29} = \frac{3}{2} \), which is a contradiction
Therefore, \( x \ne 0 \)
Let \( \frac{12x^2+18x+42}{18x^2+12x+58} = \frac{2x+3}{3x+2} = k \) ...(i)
We multiply the numerator and denominator of the second ratio by \( 6x \), as \( x \ne 0 \):
\( k = \frac{6x(2x+3)}{6x(3x+2)} = \frac{12x^2+18x}{18x^2+12x} \)
Using the Theorem on Equal Ratios, \( \frac{a}{b} = \frac{c}{d} = \frac{a-c}{b-d} \):
\( k = \frac{(12x^2+18x+42)-(12x^2+18x)}{(18x^2+12x+58)-(18x^2+12x)} \)
\( k = \frac{12x^2+18x+42-12x^2-18x}{18x^2+12x+58-18x^2-12x} \)
\( k = \frac{42}{58} = \frac{2 \times 21}{2 \times 29} = \frac{21}{29} \)
From (i), we have \( \frac{2x+3}{3x+2} = k \)
So, \( \frac{2x+3}{3x+2} = \frac{21}{29} \)
\( 29(2x+3) = 21(3x+2) \)
\( 58x + 87 = 63x + 42 \)
\( 87 - 42 = 63x - 58x \)
\( 45 = 5x \)
\( x = 9 \)
Therefore, \( x = 9 \) is the solution of the given equation.
In simple words: To solve this complex ratio equation, we first check for \( x=0 \) to establish a non-zero solution. Then, we manipulate one of the ratios by multiplying by \( 6x \) to create common terms, which allows us to apply the theorem on equal ratios. This simplifies the equation to a simple linear form, which is then solved for \( x \).

🎯 Exam Tip: The theorem on equal ratios is a powerful tool for solving equations involving multiple ratios. Remember to identify terms that can be manipulated to cancel out or simplify when applying the theorem.

Question 12. If \( \frac{2x-3y}{3z+y} = \frac{z-y}{z-x} = \frac{x+3z}{2y-3x} \), then prove that every ratio = \( \frac{x}{y} \).
Answer: Let \( k = \frac{2x-3y}{3z+y} = \frac{z-y}{z-x} = \frac{x+3z}{2y-3x} \)
Using the Theorem on Equal Ratios, if \( \frac{a}{b} = \frac{c}{d} = \frac{e}{f} \), then \( \frac{a+mc+ne}{b+md+nf} \) is also equal to k.
\( k = \frac{(2x-3y) - 3(z-y) + (x+3z)}{(3z+y) - 3(z-x) + (2y-3x)} \)
\( k = \frac{2x-3y-3z+3y+x+3z}{3z+y-3z+3x+2y-3x} \)
\( k = \frac{3x}{3y} \)
\( k = \frac{x}{y} \)
Therefore, every ratio \( = \frac{x}{y} \).
In simple words: This proof uses a specialized application of the theorem on equal ratios. By selecting appropriate multipliers for the numerators and denominators (here, -3 for the second ratio and +1 for the first and third), the terms involving 'z' and 'y' (for numerator) and 'z' and 'x' (for denominator) cleverly cancel out, leaving only 'x' in the numerator and 'y' in the denominator, proving the desired equality.

🎯 Exam Tip: For proofs involving multiple equal ratios, identifying the correct coefficients (like -3 here) to multiply the numerators and denominators before summing them is crucial to achieve the desired simplification.

Question 13. If \( \frac{by+cz}{b^2+c^2} = \frac{cz+ax}{c^2+a^2} = \frac{ax+by}{a^2+b^2} \), then prove \( \frac{x}{a} = \frac{y}{b} = \frac{z}{c} \).
Answer: Let \( k = \frac{by+cz}{b^2+c^2} = \frac{cz+ax}{c^2+a^2} = \frac{ax+by}{a^2+b^2} \) ...(i)
Using the Theorem on Equal Ratios:
\( k = \frac{(by+cz)-(cz+ax)+(ax+by)}{(b^2+c^2)-(c^2+a^2)+(a^2+b^2)} \)
\( k = \frac{by+cz-cz-ax+ax+by}{b^2+c^2-c^2-a^2+a^2+b^2} \)
\( k = \frac{2by}{2b^2} \)
\( k = \frac{y}{b} \) ...(ii)
Again, using the Theorem on Equal Ratios:
\( k = \frac{-(by+cz)+(cz+ax)+(ax+by)}{-(b^2+c^2)+(c^2+a^2)+(a^2+b^2)} \)
\( k = \frac{-by-cz+cz+ax+ax+by}{-b^2-c^2+c^2+a^2+a^2+b^2} \)
\( k = \frac{2ax}{2a^2} \)
\( k = \frac{x}{a} \) ...(iii)
Again, using the Theorem on Equal Ratios:
\( k = \frac{(by+cz)+(cz+ax)-(ax+by)}{(b^2+c^2)+(c^2+a^2)-(a^2+b^2)} \)
\( k = \frac{by+cz+cz+ax-ax-by}{b^2+c^2+c^2+a^2-a^2-b^2} \)
\( k = \frac{2cz}{2c^2} \)
\( k = \frac{z}{c} \) ...(iv)
From (ii), (iii) and (iv):
\( \frac{x}{a} = \frac{y}{b} = \frac{z}{c} \)
Hence Proved.
In simple words: This proof involves applying the theorem on equal ratios multiple times. By strategically adding and subtracting the numerators and their corresponding denominators, we isolate the components \( y/b \), \( x/a \), and \( z/c \) sequentially. Since all these simplified expressions are equal to the same constant \( k \), it establishes their equality to each other.

🎯 Exam Tip: When proving complex ratio equalities, look for patterns that allow you to use the theorem on equal ratios to isolate individual variables. Careful manipulation of signs and grouping of terms is essential.

Question 1. Take 5 pieces of card paper. Write the following statements, one on each paper.
a, b, c, d are positive numbers and \( \frac{a}{b} = \frac{c}{d} \) is given. Which of the above statements are true or false, write at the back of each card, if false explain why. (Textbook pg. no. 70)
Answer: i. \( \frac{a+b}{b} = \frac{c+d}{d} \)
True
ii. \( \frac{a}{a-b} = \frac{c}{c-d} \)
True
iii. \( \frac{ac}{bd} = \frac{a^2}{b^2} \)
False
Here, numerator and denominator are multiplied by two different numbers a and b.
iv. \( \frac{c-a}{d-b} = \frac{c}{d} \)
False
Here, different numbers a and b are subtracted from numerator and denominator.
v. \( \frac{a^2c}{b^2d} = \frac{a^3}{b^3} \)
True
In simple words: This question tests understanding of basic properties of ratios (Componendo, Invertendo, etc.) and how they apply. Each statement needs to be evaluated based on the initial condition \( a/b = c/d \), and if false, an explanation is required to justify why the equality doesn't hold generally.

🎯 Exam Tip: For true/false questions on ratio properties, always refer back to the fundamental definition \( a/b = c/d \). Use \( a=bk \) and \( c=dk \) to test the validity of each statement rigorously.

Question 2. In the following activity, the values of a and b can be changed. That is by changing a : b we can create many examples. Teachers should give lot of practice to the students and encourage them to construct their own examples. (Textbook pg. no. 70)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक गतिविधि आरेख है जो विभिन्न अनुपात और उनके मानों को दर्शाता है। इसमें \( \frac{5a^2+2b^2}{5a^2-2b^2} = \frac{77}{13} \), \( \frac{3a}{4b} = \frac{9}{16} \), \( \frac{2a-b}{2a+b} = \frac{1}{5} \), \( \frac{a^2+b^2}{b^2} = \frac{25}{16} \), \( \frac{a}{b} = \frac{3}{4} \), और \( \frac{a}{2b} = \frac{3}{8} \) जैसे समीकरणों को बक्सों में दिखाया गया है, जो 'a' और 'b' के बीच के संबंध को दर्शाता है।
Answer: This is an activity designed for students to explore ratios by changing the values of 'a' and 'b' and observing the resulting ratios. Students should actively participate in creating and solving their own examples based on the provided framework.
In simple words: This is a hands-on exercise where students learn about ratios by experimenting with different numbers for 'a' and 'b' and seeing how various related ratios change, encouraging practical understanding.

🎯 Exam Tip: Activity-based questions are designed to build conceptual understanding. Focus on the relationships between the variables and how ratio properties apply, rather than just memorizing formulas.

Question 3. Observe the political map of India from a Geography text book. Study the scale of this map. From the given scale find the straight line distances between various cities like i. New Delhi to Bengaluru ii. Mumbai to Kolkata iii. Jaipur to Bhuvaneshvar. (Textbook pg. no. 77) [Students should attempt the above activity on their own.]
Answer: This is a practical activity for students. They need to refer to a political map of India, understand its scale, and then measure the straight-line distances between the specified cities using a ruler and convert them to actual distances based on the map's scale. Students are encouraged to complete this activity independently.
In simple words: This activity teaches students to use a map's scale to calculate real-world distances between cities, combining geography and ratio skills.

🎯 Exam Tip: When dealing with map scale problems, ensure you understand the ratio of map distance to actual distance. Pay attention to units (e.g., cm on map to km in reality) and convert them correctly for accurate calculations.