Maharashtra Board Class 9 Maths Part 1 Algebra Chapter 4 Ratio and Proportion Set 4.5 Solutions

Question 1.
Which number should be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion?
Answer:
Solution:
Let the number to be subtracted be \(x\).
\(\therefore (12 - x)\), \((16 - x)\) and \((21 - x)\) are in continued proportion.
\[ \frac{12-x}{16-x} = \frac{16-x}{21-x} \]
\(\therefore\) By property of equal ratios (dividendo):
\[ \frac{(12-x)-(16-x)}{16-x} = \frac{(16-x)-(21-x)}{21-x} \]
\[ \frac{12-x-16+x}{16-x} = \frac{16-x-21+x}{21-x} \]
\[ \frac{-4}{16-x} = \frac{-5}{21-x} \]
\[ \frac{4}{16-x} = \frac{5}{21-x} \]
\(\therefore 4(21-x) = 5(16-x)\)
\(\therefore 84 - 4x = 80 - 5x\)
\(\therefore 5x - 4x = 80 - 84\)
\(\therefore x = -4\)
\(\therefore -4\) should be subtracted from 12, 16 and 21 so that the resultant numbers in continued proportion.
In simple words: To find the number, we set up the continued proportion with 'x' subtracted from each term and used the dividendo property of ratios to solve for 'x'.

🎯 Exam Tip: Remember the properties of equal ratios like dividendo, componendo, and alternendo; they are crucial for solving such problems efficiently. Show each step of algebraic manipulation clearly for full marks.

Question 2.
If \((28 - x)\) is the mean proportional of \((23 - x)\) and \((19 - x)\), then find the value of \(x\).
Answer:
Solution:
\((28 - x)\) is the mean proportional of \((23 - x)\) and \((19 - x)\). ...[Given]
\(\therefore \frac{23-x}{28-x} = \frac{28-x}{19-x}\)
\(\therefore\) By property of equal ratios (dividendo):
\[ \frac{(23-x)-(28-x)}{28-x} = \frac{(28-x)-(19-x)}{19-x} \]
\[ \frac{23-x-28+x}{28-x} = \frac{28-x-19+x}{19-x} \]
\[ \frac{-5}{28-x} = \frac{9}{19-x} \]
\(\therefore -5(19 - x) = 9(28 - x)\)
\(\therefore -95 + 5x = 252 - 9x\)
\(\therefore 5x + 9x = 252 + 95\)
\(\therefore 14x = 347\)
\(\therefore x = \frac{347}{14}\)
In simple words: If a number is the mean proportional of two others, its square equals the product of the other two. We used this and the dividendo property to solve the equation for 'x'.

🎯 Exam Tip: Clearly state the given condition (mean proportional) and the property of ratios used (dividendo) to justify your steps. Double-check your arithmetic, especially with negative signs.

Question 3.
Three numbers are in continued proportion, whose mean proportional is 12 and the sum of the remaining two numbers is 26, then find these numbers.
Answer:
Solution:
Let the first number be \(x\).
\(\therefore\) Third number \( = 26 - x\)
12 is the mean proportional of \(x\) and \((26 - x)\).
\[ \frac{x}{12} = \frac{12}{26-x} \]
\(\therefore x(26 - x) = 12 \times 12\)
\(\therefore 26x - x^2 = 144\)
\(\therefore x^2 - 26x + 144 = 0\)
\(\therefore x^2 - 18x - 8x + 144 = 0\)
\(\therefore x(x - 18) - 8(x - 18) = 0\)
\(\therefore (x - 18)(x - 8) = 0\)
\(\therefore x = 18\) or \(x = 8\)
\(\therefore\) Third number \( = 26 - x = 26 - 18 = 8\) or \(26 - x = 26 - 8 = 18\)
\(\therefore\) The numbers are 18, 12, 8 or 8, 12, 18.
In simple words: We set up equations based on the definition of continued proportion and the given sum of the outer terms, resulting in a quadratic equation that yields two possible sets of numbers.

🎯 Exam Tip: When a quadratic equation gives two solutions, remember to find the corresponding 'third number' for each 'first number' to present all possible sets of numbers. Clearly state the definition of continued proportion.

Question 4.
If \((a + b + c)(a - b + c) = a^2 + b^2 + c^2\), show that \(a, b, c\) are in continued proportion.
Answer:
Solution:
\((a + b + c)(a - b + c) = a^2 + b^2 + c^2\) ...[Given]
\(\therefore a(a - b + c) + b(a - b + c) + c(a - b + c) = a^2 + b^2 + c^2\)
\(\therefore a^2 - ab + ac + ab - b^2 + bc + ac - bc + c^2 = a^2 + b^2 + c^2\)
\(\therefore a^2 + 2ac - b^2 + c^2 = a^2 + b^2 + c^2\)
\(\therefore 2ac - b^2 = b^2\)
\(\therefore 2ac = 2b^2\)
\(\therefore ac = b^2\)
\(\therefore b^2 = ac\)
\(\therefore a, b, c\) are in continued proportion.
In simple words: By expanding the given algebraic identity and simplifying, we arrive at the condition \(b^2 = ac\), which is the definition of numbers in continued proportion.

🎯 Exam Tip: Pay close attention to algebraic expansions and cancellations. The goal is to derive the condition \(b^2 = ac\), which is the hallmark of continued proportion.

Question 5.
If \(\frac{a}{b} = \frac{b}{c}\) and \(a, b, c > 0\), then show that,
(i) \((a + b + c)(b - c) = ab - c^2\)
(ii) \((a^2 + b^2)(b^2 + c^2) = (ab + bc)^2\)
(iii) \(\frac{a^2+b^2}{ab} = \frac{a+c}{b}\)
Answer:
Solution:
Let \(\frac{a}{b} = \frac{b}{c} = k\)
\(\therefore b = ck\)
\(\therefore a = bk = (ck)k\)
\(\therefore a = ck^2\) ...(ii)

i. To show: \((a + b + c)(b - c) = ab - c^2\)
L.H.S \( = (a + b + c) (b - c)\)
\( = [ck^2 + ck + c] [ck - c]\) ... [From (i) and (ii)]
\( = c(k^2 + k + 1) c (k - 1)\)
\( = c^2 (k^2 + k + 1) (k - 1)\)
R.H.S \( = ab - c^2\)
\( = (ck^2) (ck) - c^2\) ... [From (i) and (ii)]
\( = c^2k^3 - c^2\)
\( = c^2(k^3 - 1)\)
\( = c^2 (k - 1) (k^2 + k + 1)\) ... \([a^3 - b^3 = (a - b) (a^2 + ab + b^2)]\)
\(\therefore \text{L.H.S} = \text{R.H.S}\)
\(\therefore (a + b + c) (b - c) = ab - c^2\)

ii. To show: \((a^2 + b^2)(b^2 + c^2) = (ab + bc)^2\)
\(b = ck; a = ck^2\)
L.H.S \( = (a^2 + b^2) (b^2 + c^2)\)
\( = [(ck^2)^2 + (ck)^2] [(ck)^2 + c^2]\) ... [From (i) and (ii)]
\( = [c^2k^4 + c^2k^2] [c^2k^2 + c^2]\)
\( = c^2k^2 (k^2 + 1) c^2 (k^2 + 1)\)
\( = c^4k^2 (k^2 + 1)^2\)
R.H.S \( = (ab + bc)^2\)
\( = [(ck^2)(ck) + (ck)c]^2\) ... [From (i) and (ii)]
\( = [c^2k^3 + c^2k]^2\)
\( = [c^2k (k^2 + 1)]^2 = c^4(k^2 + 1)^2\)
\(\therefore \text{L.H.S} = \text{R.H.S}\)
\(\therefore (a^2 + b^2) (b^2 + c^2) = (ab + bc)^2\)

iii. To show: \(\frac{a^2+b^2}{ab} = \frac{a+c}{b}\)
\(b = ck; a = ck^2\)
L.H.S \( = \frac{a^2 + b^2}{ab}\)
\[ = \frac{(ck^2)^2 + (ck)^2}{(ck^2)(ck)} \]
\[ = \frac{c^2k^4 + c^2k^2}{c^2k^3} \]
\[ = \frac{c^2k^2 (k^2 + 1)}{c^2k^3} \]
\[ = \frac{k^2+1}{k} \]
R.H.S \( = \frac{a+c}{b}\)
\[ = \frac{ck^2 + c}{ck} \]
\[ = \frac{c(k^2 +1)}{ck} \]
\[ = \frac{k^2 + 1}{k} \]
\(\therefore \text{L.H.S} = \text{R.H.S}\)
\(\therefore \frac{a^2 + b^2}{ab} = \frac{a+c}{b}\)
In simple words: We used the property of continued proportion \(\frac{a}{b} = \frac{b}{c} = k\) to express 'a' and 'b' in terms of 'c' and 'k'. Substituting these values into both sides of each equation, we showed that L.H.S. equals R.H.S.

🎯 Exam Tip: When proving identities involving continued proportion, using the 'k' method (i.e., \(a=ck^2, b=ck\)) is often the most systematic approach. Keep track of algebraic simplifications carefully.

Question 6.
Find mean proportional of \(\frac{x+y}{x-y}\) and \(\frac{x^2-y^2}{x^2y^2}\).
Answer:
Solution:
Let \(a\) be the mean proportional of \(\frac{x+y}{x-y}\) and \(\frac{x^2-y^2}{x^2y^2}\)
\(\therefore a^2 = \frac{x+y}{x-y} \times \frac{x^2-y^2}{x^2y^2}\)
\[ = \frac{x+y}{x-y} \times \frac{(x+y)(x-y)}{x^2y^2} \]
\([\therefore a^2-b^2 = (a + b)(a - b)]\)
\[ a^2 = \frac{(x+y)^2}{x^2y^2} \]
\(\therefore a = \frac{x+y}{xy}\) ...[Taking square root of both sides]
Mean proportional of \(\frac{x+y}{x-y}\), \(\frac{x^2-y^2}{x^2y^2}\) is \(\frac{x+y}{xy}\)
In simple words: The mean proportional of two numbers is the square root of their product. We multiplied the given expressions and simplified them to find the square root, which is the mean proportional.

🎯 Exam Tip: Remember the definition of mean proportional: for two numbers \(A\) and \(B\), the mean proportional \(M\) satisfies \(M^2 = A \times B\). Factorization (like \(x^2-y^2\)) is key for simplification.