Maharashtra Board Class 9 Maths Part 1 Algebra Chapter 4 Ratio and Proportion Set 4.4 Solutions

Question 1. Fill in the blanks of the following. i. \( \frac{x}{7} = \frac{y}{3} = \frac{3x+5y}{.......} = \frac{7x-9y}{.......} \) ii. \( \frac{a}{3} = \frac{b}{4} = \frac{c}{7} = \frac{a-2b+3c}{.......} = \frac{.......}{6-8+14} \)
Answer: i. \( \frac{x}{7} = \frac{y}{3} = \frac{3x+5y}{21+15} = \frac{7x-9y}{49-27} \)
Solution: \( \frac{x}{7} = \frac{y}{3} \)
\( \implies \frac{3x}{3 \times 7} = \frac{5y}{5 \times 3} \) (Multiplying numerator and denominator by 3 and 5 respectively)
\( \implies \frac{3x}{21} = \frac{5y}{15} \)
From theorem on equal ratios, \( \frac{x}{7} = \frac{y}{3} = \frac{3x+5y}{21+15} = \frac{3x+5y}{36} \)
Similarly, \( \frac{x}{7} = \frac{y}{3} \)
\( \implies \frac{7x}{7 \times 7} = \frac{-9y}{-9 \times 3} \) (Multiplying numerator and denominator by 7 and -9 respectively)
\( \implies \frac{7x}{49} = \frac{-9y}{-27} \)
From theorem on equal ratios, \( \frac{x}{7} = \frac{y}{3} = \frac{7x-9y}{49-27} = \frac{7x-9y}{22} \)
Thus, \( \frac{x}{7} = \frac{y}{3} = \frac{3x+5y}{36} = \frac{7x-9y}{22} \) ii. \( \frac{a}{3} = \frac{b}{4} = \frac{c}{7} = \frac{a-2b+3c}{3-8+21} = \frac{2a-2b+2c}{6-8+14} \)
Solution: \( \frac{a}{3} = \frac{b}{4} = \frac{c}{7} \)
\( \implies \frac{a}{3} = \frac{-2b}{-2 \times 4} = \frac{3c}{3 \times 7} \)
\( \implies \frac{a}{3} = \frac{-2b}{-8} = \frac{3c}{21} \)
From theorem on equal ratios, \( \frac{a}{3} = \frac{b}{4} = \frac{c}{7} = \frac{a-2b+3c}{3-8+21} = \frac{a-2b+3c}{16} \)
Similarly, \( \frac{a}{3} = \frac{b}{4} = \frac{c}{7} \)
\( \implies \frac{2a}{2 \times 3} = \frac{-2b}{-2 \times 4} = \frac{2c}{2 \times 7} \)
\( \implies \frac{2a}{6} = \frac{-2b}{-8} = \frac{2c}{14} \)
From theorem on equal ratios, \( \frac{a}{3} = \frac{b}{4} = \frac{c}{7} = \frac{2a-2b+2c}{6-8+14} = \frac{2a-2b+2c}{12} \)
So, the complete equality is \( \frac{a}{3} = \frac{b}{4} = \frac{c}{7} = \frac{a-2b+3c}{16} = \frac{2a-2b+2c}{12} \)
In simple words: This question applies the theorem on equal ratios, where if several ratios are equal, then each ratio is equal to the ratio formed by adding/subtracting any corresponding multiples of numerators and denominators.

🎯 Exam Tip: Remember to apply the theorem on equal ratios correctly by multiplying both the numerator and denominator by the same non-zero number before adding or subtracting.

Question 2. If \( 5m - n = 3m + 4n \), then find the values of the following expressions. i. \( \frac{m^2 + n^2}{m^2-n^2} \) ii. \( \frac{3m + 4n}{3m-4n} \)
Answer: Solution: Given: \( 5m - n = 3m + 4n \)
\( \implies 5m - 3m = 4n + n \)
\( \implies 2m = 5n \)
\( \implies \frac{m}{n} = \frac{5}{2} \) i. To find \( \frac{m^2 + n^2}{m^2-n^2} \)
From \( \frac{m}{n} = \frac{5}{2} \)
Squaring both sides: \( \frac{m^2}{n^2} = \frac{5^2}{2^2} = \frac{25}{4} \)
By componendo-dividendo: \( \frac{m^2 + n^2}{m^2-n^2} = \frac{25+4}{25-4} = \frac{29}{21} \)
So, \( \frac{m^2 + n^2}{m^2-n^2} = 29:21 \) ii. To find \( \frac{3m + 4n}{3m-4n} \)
From \( \frac{m}{n} = \frac{5}{2} \)
Multiply both sides by \( \frac{3}{4} \): \( \frac{m}{n} \times \frac{3}{4} = \frac{5}{2} \times \frac{3}{4} \)
\( \implies \frac{3m}{4n} = \frac{15}{8} \)
By componendo-dividendo: \( \frac{3m + 4n}{3m-4n} = \frac{15+8}{15-8} = \frac{23}{7} \)
So, \( \frac{3m + 4n}{3m-4n} = 23:7 \)
In simple words: First, simplify the given equation to find the ratio \( \frac{m}{n} \). Then, use properties of ratios like squaring and componendo-dividendo to find the values of the required expressions.

🎯 Exam Tip: Componendo-dividendo is a powerful tool for simplifying ratio expressions. Ensure you apply it correctly: if \( \frac{a}{b} = \frac{c}{d} \), then \( \frac{a+b}{a-b} = \frac{c+d}{c-d} \).

Question 3. Solve: i. If \( a(y + z) = b(z + x) = c(x + y) \) and out of a, b, c no two of them are equal, then show that, \( \frac{y-z}{a(b-c)} = \frac{z-x}{b(c-a)} = \frac{x-y}{c(a-b)} \)
Answer: Solution: Here, no two of a, b and c are equal. Therefore, values of \( (b - c) \), \( (c - a) \) and \( (a - b) \) are not zero. Given: \( a(y + z) = b(z + x) = c(x + y) \)
Divide each term by abc: \( \frac{a(y+z)}{abc} = \frac{b(z+x)}{abc} = \frac{c(x+y)}{abc} \)
\( \implies \frac{y+z}{bc} = \frac{z+x}{ac} = \frac{x+y}{ab} \)
Let \( \frac{y+z}{bc} = \frac{z+x}{ac} = \frac{x+y}{ab} = k \) ...(i)
From (i), \( k = \frac{x+y}{ab} = \frac{z+x}{ac} \)
By theorem on equal ratios, \( k = \frac{(x+y)-(z+x)}{ab-ac} = \frac{x+y-z-x}{a(b-c)} = \frac{y-z}{a(b-c)} \) ...(ii)
From (i), \( k = \frac{y+z}{bc} = \frac{x+y}{ab} \)
By theorem on equal ratios, \( k = \frac{(y+z)-(x+y)}{bc-ab} = \frac{y+z-x-y}{b(c-a)} = \frac{z-x}{b(c-a)} \) ...(iii)
From (i), \( k = \frac{z+x}{ac} = \frac{y+z}{bc} \)
By theorem on equal ratios, \( k = \frac{(z+x)-(y+z)}{ac-bc} = \frac{z+x-y-z}{c(a-b)} = \frac{x-y}{c(a-b)} \) ...(iv)
From (ii), (iii) and (iv), we get: \( \frac{y-z}{a(b-c)} = \frac{z-x}{b(c-a)} = \frac{x-y}{c(a-b)} \)
Hence Proved.
In simple words: To prove the equality, first simplify the given equation by dividing by 'abc' to get equal ratios. Then, apply the theorem on equal ratios multiple times by subtracting different numerators and their corresponding denominators to derive the required expressions.

🎯 Exam Tip: When given multiple equal ratios, dividing by a common term (like 'abc' here) can often simplify the expression and reveal a common ratio 'k', which is useful for applying further theorems.

ii. If \( \frac{x}{3x-y-z} = \frac{y}{3y-z-x} = \frac{z}{3z-x-y} \) and \( x + y + z \neq 0 \), then show that the value of each ratio is equal to 1.
Answer: Solution: Let \( \frac{x}{3x-y-z} = \frac{y}{3y-z-x} = \frac{z}{3z-x-y} = k \)
By theorem on equal ratios (adding numerators and denominators): \( k = \frac{x+y+z}{(3x-y-z)+(3y-z-x)+(3z-x-y)} \)
\( k = \frac{x+y+z}{3x-y-z+3y-z-x+3z-x-y} \)
\( k = \frac{x+y+z}{(3x-x-x)+(3y-y-y)+(3z-z-z)} \)
\( k = \frac{x+y+z}{x+y+z} \)
Since \( x + y + z \neq 0 \), we can cancel out \( (x+y+z) \) from numerator and denominator.
\( k = 1 \)
Therefore, each ratio is equal to 1.
In simple words: By using the theorem of equal ratios and summing up all numerators and denominators, the expression simplifies to \( \frac{x+y+z}{x+y+z} \). Since the sum is not zero, the ratio simplifies to 1.

🎯 Exam Tip: For problems involving three or more equal ratios, always consider adding all numerators and all denominators. This often leads to significant simplification if a common factor exists.

iii. If \( \frac{x}{3x-y-z} = \frac{y}{3y-z-x} = \frac{z}{3z-x-y} \) and \( x + y + z \neq 0 \), then show that \( \frac{a+b}{2} \). (This seems to be a mixed-up question from a different context. The previous question iii was completed. Let's assume this is a new sub-part for question 3, but the question statement seems incomplete, and it's asking to show something that's not an equality of ratios.) The OCR provided an 'ab/2' which is 'a+b/2'. Given the next part starts with "Let \( \frac{ax + by}{x+y} \)", this is clearly a separate, unrelated problem or a typo in the original text. I will process the given text for what it seems to be aiming at, which is to show that a combination of these ratios also equals \( \frac{a+b}{2} \). I will interpret "then show that \( \frac{a+b}{2} \)" as "then show that the value of each ratio is \( \frac{a+b}{2} \)". This aligns with the solution provided which ends with \( k = \frac{a+b}{2} \).

Question 3. iii. If \( \frac{ax + by}{x+y} = \frac{bx + az}{x+z} = \frac{ay + bz}{y+z} \) and \( x + y + z \neq 0 \), then show that each ratio is equal to \( \frac{a+b}{2} \).
Answer: Solution: Let \( \frac{ax + by}{x+y} = \frac{bx + az}{x+z} = \frac{ay + bz}{y+z} = k \)
By theorem on equal ratios (adding all numerators and denominators): \( k = \frac{(ax + by) + (bx + az) + (ay + bz)}{(x+y) + (x+z) + (y+z)} \)
\( k = \frac{ax+by+bx+az+ay+bz}{2x+2y+2z} \)
\( k = \frac{(ax+bx) + (ay+by) + (az+bz)}{2(x+y+z)} \)
\( k = \frac{x(a+b) + y(a+b) + z(a+b)}{2(x+y+z)} \)
\( k = \frac{(a+b)(x+y+z)}{2(x+y+z)} \)
Since \( x+y+z \neq 0 \), we can cancel out \( (x+y+z) \).
\( k = \frac{a+b}{2} \)
Therefore, each ratio is equal to \( \frac{a+b}{2} \).
In simple words: By applying the theorem on equal ratios and summing all numerators and denominators, the common factor \( (x+y+z) \) appears, allowing the expression to simplify to \( \frac{a+b}{2} \).

🎯 Exam Tip: When the sum of numerators and denominators contains a common factor, this method is very efficient. Always look for opportunities to factorize and simplify.

iv. If \( \frac{y+z}{a} = \frac{z+x}{b} = \frac{x+y}{c} \), then show that \( \frac{x}{b+c-a} = \frac{y}{c+a-b} = \frac{z}{a+b-c} \).
Answer: Solution: Let \( \frac{y+z}{a} = \frac{z+x}{b} = \frac{x+y}{c} = k \) ...(i)
Consider the expression for x:
From (i), \( k = \frac{(z+x)+(x+y)-(y+z)}{b+c-a} \)
\( k = \frac{z+x+x+y-y-z}{b+c-a} \)
\( k = \frac{2x}{b+c-a} \) ...(ii)
Consider the expression for y:
From (i), \( k = \frac{(x+y)+(y+z)-(z+x)}{c+a-b} \)
\( k = \frac{x+y+y+z-z-x}{c+a-b} \)
\( k = \frac{2y}{c+a-b} \) ...(iii)
Consider the expression for z:
From (i), \( k = \frac{(y+z)+(z+x)-(x+y)}{a+b-c} \)
\( k = \frac{y+z+z+x-x-y}{a+b-c} \)
\( k = \frac{2z}{a+b-c} \) ...(iv)
From (ii), (iii) and (iv), we have: \( k = \frac{2x}{b+c-a} = \frac{2y}{c+a-b} = \frac{2z}{a+b-c} \)
Divide each ratio by 2: \( \implies \frac{x}{b+c-a} = \frac{y}{c+a-b} = \frac{z}{a+b-c} \)
Hence Proved.
In simple words: Assume the given ratios equal 'k'. Then, by skillfully applying the theorem on equal ratios (adding two numerators and subtracting the third, and doing the same for denominators), we can derive expressions for \( \frac{2x}{...} \), \( \frac{2y}{...} \), and \( \frac{2z}{...} \). Dividing by 2 then gives the desired result.

🎯 Exam Tip: The 'addition-subtraction' application of the theorem on equal ratios is crucial here. Practice identifying which terms to add and which to subtract to isolate variables like x, y, and z.

v. If \( \frac{3x-5y}{5z+3y} = \frac{x+5z}{y-5x} = \frac{y-z}{x-z} \), then show that every ratio is equal to \( \frac{x}{y} \).
Answer: Solution: Let \( \frac{3x-5y}{5z+3y} = \frac{x+5z}{y-5x} = \frac{y-z}{x-z} = k \)
Multiply numerator and denominator of the third ratio by 5:
\( k = \frac{3x-5y}{5z+3y} = \frac{x+5z}{y-5x} = \frac{5(y-z)}{5(x-z)} = \frac{5y-5z}{5x-5z} \)
By theorem on equal ratios, \( k = \frac{(3x-5y)+(x+5z)+(5y-5z)}{(5z+3y)+(y-5x)+(5x-5z)} \)
\( k = \frac{3x-5y+x+5z+5y-5z}{5z+3y+y-5x+5x-5z} \)
\( k = \frac{(3x+x) + (-5y+5y) + (5z-5z)}{(5z-5z) + (3y+y) + (-5x+5x)} \)
\( k = \frac{4x}{4y} \)
\( k = \frac{x}{y} \)
Therefore, each ratio is equal to \( \frac{x}{y} \).
In simple words: Introduce a factor of 5 to the third ratio to create common terms in both numerator and denominator when applying the theorem on equal ratios. This allows for cancellation and simplification to \( \frac{x}{y} \).

🎯 Exam Tip: Sometimes, you need to multiply a ratio's numerator and denominator by a constant to align terms for cancellation using the theorem on equal ratios. Look for coefficients that can be made common.

Question 4. Solve: i. \( \frac{16x^2-20x+9}{8x^2+12x+21} = \frac{4x-5}{2x+3} \) ii. \( \frac{5y^2+40y-12}{5y+10y^2-4} = \frac{y+8}{1+2y} \)
Answer: i. \( \frac{16x^2-20x+9}{8x^2+12x+21} = \frac{4x-5}{2x+3} \)
Solution: Case 1: If \( x = 0 \)
L.H.S. \( = \frac{16(0)^2-20(0)+9}{8(0)^2+12(0)+21} = \frac{9}{21} = \frac{3}{7} \)
R.H.S. \( = \frac{4(0)-5}{2(0)+3} = \frac{-5}{3} \)
Since \( \frac{3}{7} \neq \frac{-5}{3} \), \( x=0 \) is not a solution. Case 2: If \( x \neq 0 \)
Let \( \frac{16x^2-20x+9}{8x^2+12x+21} = \frac{4x-5}{2x+3} = k \) ...(i)
Multiply numerator and denominator of the second ratio by \( 4x \):
\( \frac{4x-5}{2x+3} = \frac{4x(4x-5)}{4x(2x+3)} = \frac{16x^2-20x}{8x^2+12x} \)
So, \( k = \frac{16x^2-20x+9}{8x^2+12x+21} = \frac{16x^2-20x}{8x^2+12x} \)
By theorem on equal ratios (subtracting the second ratio from the first):
\( k = \frac{(16x^2-20x+9)-(16x^2-20x)}{(8x^2+12x+21)-(8x^2+12x)} \)
\( k = \frac{16x^2-20x+9-16x^2+20x}{8x^2+12x+21-8x^2-12x} \)
\( k = \frac{9}{21} = \frac{3}{7} \)
Now, substitute \( k = \frac{3}{7} \) back into (i):
\( \frac{4x-5}{2x+3} = \frac{3}{7} \)
Cross-multiply:
\( 7(4x-5) = 3(2x+3) \)
\( 28x - 35 = 6x + 9 \)
\( 28x - 6x = 9 + 35 \)
\( 22x = 44 \)
\( x = \frac{44}{22} \)
\( x = 2 \)
Thus, \( x = 2 \) is the solution of the given equation.
In simple words: First, check if \( x=0 \) is a solution. If not, set the given ratios equal to a constant 'k'. Modify one ratio (by multiplying by \( 4x \)) to match parts of the other, then use the theorem on equal ratios to simplify for 'k'. Finally, substitute 'k' back to solve for x.

🎯 Exam Tip: When dealing with algebraic ratios, always consider a potential solution \( x=0 \) separately if it makes the denominators undefined or leads to a contradiction. Multiplying by a variable like \( 4x \) assumes \( x \neq 0 \).

ii. \( \frac{5y^2+40y-12}{5y+10y^2-4} = \frac{y+8}{1+2y} \)
Answer: Solution: Case 1: If \( y = 0 \)
L.H.S. \( = \frac{5(0)^2+40(0)-12}{5(0)+10(0)^2-4} = \frac{-12}{-4} = 3 \)
R.H.S. \( = \frac{0+8}{1+2(0)} = \frac{8}{1} = 8 \)
Since \( 3 \neq 8 \), \( y=0 \) is a contradiction, so \( y=0 \) is not a solution. Case 2: If \( y \neq 0 \)
Let \( \frac{5y^2+40y-12}{5y+10y^2-4} = \frac{y+8}{1+2y} = k \) ...(i)
Multiply numerator and denominator of the second ratio by \( 5y \):
\( \frac{y+8}{1+2y} = \frac{5y(y+8)}{5y(1+2y)} = \frac{5y^2+40y}{5y+10y^2} \)
So, \( k = \frac{5y^2+40y-12}{5y+10y^2-4} = \frac{5y^2+40y}{5y+10y^2} \)
By theorem on equal ratios (subtracting the second ratio from the first):
\( k = \frac{(5y^2+40y-12)-(5y^2+40y)}{(5y+10y^2-4)-(5y+10y^2)} \)
\( k = \frac{5y^2+40y-12-5y^2-40y}{5y+10y^2-4-5y-10y^2} \)
\( k = \frac{-12}{-4} = 3 \)
Now, substitute \( k = 3 \) back into (i):
\( \frac{y+8}{1+2y} = 3 \)
Cross-multiply:
\( y+8 = 3(1+2y) \)
\( y+8 = 3+6y \)
\( 8-3 = 6y-y \)
\( 5 = 5y \)
\( y = 1 \)
Thus, \( y = 1 \) is the solution of the given equation.
In simple words: First, check for \( y=0 \). Then, equate the given ratios to 'k'. Manipulate the second ratio by multiplying its numerator and denominator by \( 5y \) to align terms with the first ratio. Apply the theorem on equal ratios to find 'k', then solve for 'y'.

🎯 Exam Tip: The strategy of multiplying by a variable term (\( 5y \) in this case) to create common expressions in the numerators and denominators is key to simplifying these types of ratio equations. Always verify initial assumptions like \( y \neq 0 \).