OP Malhotra Class 10 Maths Solutions Chapter 6 Ratio and Proportion Exercise 6 (C)

S Chand Class 10 ICSE Maths Solutions Chapter 6 Ratio and Proportion Ex 6(c)

 

Question 1. If \( \frac { x }{ a } = \frac { y }{ b } = \frac { z }{ c } \), show that
(i) \( \frac{x^3}{a^3}-\frac{y^3}{b^3}+\frac{z^3}{c^3}=\frac{x y z}{a b c} \)
(ii) \( \left(\frac{a^2 x^2+b^2 y^2+c^2 z^2}{a^3 x+b^3 y+c^3 z}\right)^{\frac{3}{2}}=\sqrt{\frac{x y z}{a b c}} \)
(iii) \( \frac{x z+a c}{x z-a c}=\frac{y z+b c}{y z-b c} \)
(iv) \( abc\left(\frac{x+a}{a}+\frac{y+b}{b}+\frac{z+c}{c}\right)^3 = 27(x + a) (y + b) (z + c) \)
(v) each ratio is equal to \( \left(\frac{3 x^3+5 y^3+7 z^3}{3 a^3+5b^3+7 c^3}\right)^{\frac{1}{3}} \)
Answer:
Let \( \frac { x }{ a } = \frac { y }{ b } = \frac { z }{ c } = k \) (suppose)
This means \( x = ak, y = bk, z = ck \). We will use these values to prove each part.

(i) To prove: \( \frac{x^3}{a^3}-\frac{y^3}{b^3}+\frac{z^3}{c^3}=\frac{x y z}{a b c} \)
L.H.S. \( = \frac{(ak)^3}{a^3}-\frac{(bk)^3}{b^3}+\frac{(ck)^3}{c^3} \)
\( = \frac{a^3 k^3}{a^3}-\frac{b^3 k^3}{b^3}+\frac{c^3 k^3}{c^3} \)
\( = k^3 - k^3 + k^3 = k^3 \)
R.H.S. \( = \frac{(ak)(bk)(ck)}{a b c} \)
\( = \frac{a b c k^3}{a b c} = k^3 \)
Since L.H.S. = R.H.S., the identity is proven.

(ii) To prove: \( \left(\frac{a^2 x^2+b^2 y^2+c^2 z^2}{a^3 x+b^3 y+c^3 z}\right)^{\frac{3}{2}}=\sqrt{\frac{x y z}{a b c}} \)
L.H.S. \( = \left(\frac{a^2 (ak)^2+b^2 (bk)^2+c^2 (ck)^2}{a^3 (ak)+b^3 (bk)+c^3 (ck)}\right)^{\frac{3}{2}} \)
\( = \left(\frac{a^4 k^2+b^4 k^2+c^4 k^2}{a^4 k+b^4 k+c^4 k}\right)^{\frac{3}{2}} \)
\( = \left(\frac{k^2(a^4+b^4+c^4)}{k(a^4+b^4+c^4)}\right)^{\frac{3}{2}} \)
\( = (k)^{\frac{3}{2}} \)
R.H.S. \( = \sqrt{\frac{(ak)(bk)(ck)}{a b c}} \)
\( = \sqrt{\frac{a b c k^3}{a b c}} \)
\( = \sqrt{k^3} = k^{\frac{3}{2}} \)
Since L.H.S. = R.H.S., the identity is proven.

(iii) To prove: \( \frac{x z+a c}{x z-a c}=\frac{y z+b c}{y z-b c} \)
L.H.S. \( = \frac{(ak)(ck)+a c}{(ak)(ck)-a c} \)
\( = \frac{a c k^2+a c}{a c k^2-a c} \)
\( = \frac{a c (k^2+1)}{a c (k^2-1)} = \frac{k^2+1}{k^2-1} \)
R.H.S. \( = \frac{(bk)(ck)+b c}{(bk)(ck)-b c} \)
\( = \frac{b c k^2+b c}{b c k^2-b c} \)
\( = \frac{b c (k^2+1)}{b c (k^2-1)} = \frac{k^2+1}{k^2-1} \)
Since L.H.S. = R.H.S., the identity is proven.

(iv) To prove: \( abc\left(\frac{x+a}{a}+\frac{y+b}{b}+\frac{z+c}{c}\right)^3 = 27(x + a) (y + b) (z + c) \)
L.H.S. \( = abc\left(\frac{ak+a}{a}+\frac{bk+b}{b}+\frac{ck+c}{c}\right)^3 \)
\( = abc\left(\frac{a(k+1)}{a}+\frac{b(k+1)}{b}+\frac{c(k+1)}{c}\right)^3 \)
\( = abc(k+1+k+1+k+1)^3 \)
\( = abc(3(k+1))^3 \)
\( = abc \cdot 3^3 \cdot (k+1)^3 \)
\( = 27abc(k+1)^3 \)
R.H.S. \( = 27(ak+a)(bk+b)(ck+c) \)
\( = 27 a(k+1) b(k+1) c(k+1) \)
\( = 27abc(k+1)^3 \)
Since L.H.S. = R.H.S., the identity is proven.

(v) To prove: Each ratio is equal to \( \left(\frac{3 x^3+5 y^3+7 z^3}{3 a^3+5 b^3+7 c^3}\right)^{\frac{1}{3}} \)
Consider the expression: \( \left(\frac{3 x^3+5 y^3+7 z^3}{3 a^3+5 b^3+7 c^3}\right)^{\frac{1}{3}} \)
Substitute \( x = ak, y = bk, z = ck \):
\( = \left(\frac{3 (ak)^3+5 (bk)^3+7 (ck)^3}{3 a^3+5 b^3+7 c^3}\right)^{\frac{1}{3}} \)
\( = \left(\frac{3 a^3 k^3+5 b^3 k^3+7 c^3 k^3}{3 a^3+5 b^3+7 c^3}\right)^{\frac{1}{3}} \)
\( = \left(\frac{k^3(3 a^3+5 b^3+7 c^3)}{3 a^3+5 b^3+7 c^3}\right)^{\frac{1}{3}} \)
\( = (k^3)^{\frac{1}{3}} = k \)
Since we started with \( \frac { x }{ a } = \frac { y }{ b } = \frac { z }{ c } = k \), this shows that each ratio is indeed equal to \( k \), which is the given expression. This property is very useful in solving ratio and proportion problems.
In simple words: When you have three fractions equal to each other, like x/a = y/b = z/c, we can say they all equal a constant 'k'. Then, replace x, y, and z with 'ak', 'bk', and 'ck' respectively in the expressions. After substituting and simplifying, both sides of the equations will become equal to 'k' or 'k' raised to some power, proving the identities.

🎯 Exam Tip: When given a set of equal ratios, assume they are all equal to a constant 'k'. This substitution technique simplifies complex expressions, allowing you to prove the identities easily by working with 'k'.

 

Question 2. If \( \frac { a }{ b } = \frac { c }{ d } = \frac { e }{ f } \), prove the following:
(i) \( \frac{p a^3+q c^3+re^3}{p b^3+qd^3+r f^3}=\frac{a c e}{b d f} \)
(ii) \( \sqrt{\frac{a^4+c^4}{b^4+d^4}}=\frac{p a^2+q c^2}{p b^2+qd^2} \)
(iii) \( \frac{2 a^4 b^2+3 a^2 e^2-5 e^4 f}{2 b^6+3 b^2 f^2-5 f^5}=\frac{a^4}{b^4} \)
Answer:
Let \( \frac { a }{ b } = \frac { c }{ d } = \frac { e }{ f } = k \) (suppose)
This means \( a = bk, c = dk, e = fk \). We will use these values to prove each part.

(i) To prove: \( \frac{p a^3+q c^3+re^3}{p b^3+qd^3+r f^3}=\frac{a c e}{b d f} \)
L.H.S. \( = \frac{p (bk)^3+q (dk)^3+r (fk)^3}{p b^3+q d^3+r f^3} \)
\( = \frac{p b^3 k^3+q d^3 k^3+r f^3 k^3}{p b^3+q d^3+r f^3} \)
\( = \frac{k^3(p b^3+q d^3+r f^3)}{p b^3+q d^3+r f^3} = k^3 \)
R.H.S. \( = \frac{(bk)(dk)(fk)}{b d f} \)
\( = \frac{b d f k^3}{b d f} = k^3 \)
Since L.H.S. = R.H.S., the identity is proven.

(ii) To prove: \( \sqrt{\frac{a^4+c^4}{b^4+d^4}}=\frac{p a^2+q c^2}{p b^2+qd^2} \)
L.H.S. \( = \sqrt{\frac{(bk)^4+(dk)^4}{b^4+d^4}} \)
\( = \sqrt{\frac{b^4 k^4+d^4 k^4}{b^4+d^4}} \)
\( = \sqrt{\frac{k^4(b^4+d^4)}{b^4+d^4}} = \sqrt{k^4} = k^2 \)
R.H.S. \( = \frac{p (bk)^2+q (dk)^2}{p b^2+q d^2} \)
\( = \frac{p b^2 k^2+q d^2 k^2}{p b^2+q d^2} \)
\( = \frac{k^2(p b^2+q d^2)}{p b^2+q d^2} = k^2 \)
Since L.H.S. = R.H.S., the identity is proven.

(iii) To prove: \( \frac{2 a^4 b^2+3 a^2 e^2-5 e^4 f}{2 b^6+3 b^2 f^2-5 f^5}=\frac{a^4}{b^4} \)
L.H.S. \( = \frac{2 (bk)^4 b^2+3 (bk)^2 (fk)^2-5 (fk)^4 f}{2 b^6+3 b^2 f^2-5 f^5} \)
\( = \frac{2 b^4 k^4 b^2+3 b^2 k^2 f^2 k^2-5 f^4 k^4 f}{2 b^6+3 b^2 f^2-5 f^5} \)
\( = \frac{2 b^6 k^4+3 b^2 f^2 k^4-5 f^5 k^4}{2 b^6+3 b^2 f^2-5 f^5} \)
\( = \frac{k^4(2 b^6+3 b^2 f^2-5 f^5)}{2 b^6+3 b^2 f^2-5 f^5} = k^4 \)
R.H.S. \( = \frac{(bk)^4}{b^4} \)
\( = \frac{b^4 k^4}{b^4} = k^4 \)
Since L.H.S. = R.H.S., the identity is proven. Remember that `a^4` means `(bk)^4` when using substitution.
In simple words: When you have equal ratios like a/b = c/d = e/f, you can replace 'a' with 'bk', 'c' with 'dk', and 'e' with 'fk'. Do this for both sides of the equation. After plugging in these values and simplifying, you will find that both sides become equal to 'k' raised to a certain power, showing they are the same.

🎯 Exam Tip: Always clearly state your substitution (e.g., let a/b = c/d = e/f = k). This sets up the solution correctly and makes the algebraic manipulations easier to follow for the examiner.

 

Question 3. If a,b,c are in continued proportion, prove that :
(i) \( (a + b + c) (a - b + c) = a^2 + b^2 + c^2 \)
(ii) \( \frac{a^2+b^2}{b^2+c^2}=\frac{a}{c} \)
(iii) \( \frac{a^3+b^3+c^3}{a^2 b^2 c^2}=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3} \)
(iv) \( (4a^2 + 7ab + 9b^2) : (4b^2 + 7bc + 9c^2) = a : c \)
Answer:
Since a, b, c are in continued proportion, we can write \( \frac { a }{ b } = \frac { b }{ c } = k \) (suppose)
This means \( b = ck \) and \( a = bk = (ck)k = ck^2 \). We will use these values to prove each part.

(i) To prove: \( (a + b + c) (a - b + c) = a^2 + b^2 + c^2 \)
L.H.S. \( = (ck^2 + ck + c) (ck^2 - ck + c) \)
\( = c(k^2 + k + 1) \cdot c(k^2 - k + 1) \)
\( = c^2 [(k^2+1)+k] [(k^2+1)-k] \)
\( = c^2 [(k^2+1)^2 - k^2] \) (using \( (X+Y)(X-Y) = X^2-Y^2 \))
\( = c^2 [k^4 + 2k^2 + 1 - k^2] \)
\( = c^2 (k^4 + k^2 + 1) \)
R.H.S. \( = (ck^2)^2 + (ck)^2 + c^2 \)
\( = c^2 k^4 + c^2 k^2 + c^2 \)
\( = c^2 (k^4 + k^2 + 1) \)
Since L.H.S. = R.H.S., the identity is proven. This shows a property of numbers in continued proportion.

(ii) To prove: \( \frac{a^2+b^2}{b^2+c^2}=\frac{a}{c} \)
L.H.S. \( = \frac{(ck^2)^2+(ck)^2}{(ck)^2+c^2} \)
\( = \frac{c^2 k^4+c^2 k^2}{c^2 k^2+c^2} \)
\( = \frac{c^2 k^2(k^2+1)}{c^2 (k^2+1)} = k^2 \)
R.H.S. \( = \frac{ck^2}{c} = k^2 \)
Since L.H.S. = R.H.S., the identity is proven.

(iii) To prove: \( \frac{a^3+b^3+c^3}{a^2 b^2 c^2}=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3} \)
L.H.S. \( = \frac{(ck^2)^3+(ck)^3+c^3}{(ck^2)^2 (ck)^2 c^2} \)
\( = \frac{c^3 k^6+c^3 k^3+c^3}{c^2 k^4 \cdot c^2 k^2 \cdot c^2} \)
\( = \frac{c^3 (k^6+k^3+1)}{c^6 k^6} \)
\( = \frac{k^6+k^3+1}{c^3 k^6} \)
R.H.S. \( = \frac{1}{(ck^2)^3}+\frac{1}{(ck)^3}+\frac{1}{c^3} \)
\( = \frac{1}{c^3 k^6}+\frac{1}{c^3 k^3}+\frac{1}{c^3} \)
\( = \frac{1}{c^3} \left(\frac{1}{k^6}+\frac{1}{k^3}+1\right) \)
\( = \frac{1}{c^3} \left(\frac{1+k^3+k^6}{k^6}\right) = \frac{k^6+k^3+1}{c^3 k^6} \)
Since L.H.S. = R.H.S., the identity is proven.

(iv) To prove: \( (4a^2 + 7ab + 9b^2) : (4b^2 + 7bc + 9c^2) = a : c \)
We need to show \( \frac{4a^2+7ab+9b^2}{4b^2+7bc+9c^2} = \frac{a}{c} \)
L.H.S. \( = \frac{4(ck^2)^2+7(ck^2)(ck)+9(ck)^2}{4(ck)^2+7(ck)(c)+9c^2} \)
\( = \frac{4c^2 k^4+7c^2 k^3+9c^2 k^2}{4c^2 k^2+7c^2 k+9c^2} \)
\( = \frac{c^2 k^2(4k^2+7k+9)}{c^2 (4k^2+7k+9)} = k^2 \)
R.H.S. \( = \frac{ck^2}{c} = k^2 \)
Since L.H.S. = R.H.S., the identity is proven.
In simple words: When three numbers are in continued proportion, it means the ratio of the first to the second is the same as the ratio of the second to the third. We can use this idea to replace 'a' and 'b' with terms involving 'c' and 'k' (a constant). Then, by substituting these into the given equations and simplifying, we can show that both sides of each equation are equal.

🎯 Exam Tip: For problems involving continued proportion, always establish the relationships \( b = ck \) and \( a = ck^2 \) first. This consistent substitution is key to successfully simplifying both sides of the equation.

 

Question 4. If a, b, c, d are in continued proportion, prove that :
(i) \( (b - c)^2 + (c - a)^2 + (d - b)^2 = (a - d)^2 \)
(ii) \( \sqrt{\frac{a^5+b^2c^2+a^3 c^2}{b^4 c+d^4+b^2 c d^2}}=\frac{a}{d} \)
(iii) \( \sqrt{a b}-\sqrt{b c}+\sqrt{c d}=\sqrt{(a-b+c)(b-c+d)} \)
(iv) \( \frac{3 a+5 d}{5 a+7 d}=\frac{3 a^3+5 b^3}{5 a^3+7 b^3} \)
(v) \( \frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d} \)
(vi) \( pa^3 + qb^3 + rc^3 : pb^3 + qc^3 + rd^3 = a : d \)
Answer:
Since a, b, c, d are in continued proportion, we can write \( \frac { a }{ b } = \frac { b }{ c } = \frac { c }{ d } = k \) (suppose)
This means \( c = dk, b = ck = dk^2, \) and \( a = bk = dk^3 \). We will use these values to prove each part.

(i) To prove: \( (b - c)^2 + (c - a)^2 + (d - b)^2 = (a - d)^2 \)
L.H.S. \( = (dk^2 - dk)^2 + (dk - dk^3)^2 + (d - dk^2)^2 \)
\( = d^2 k^2 (k - 1)^2 + d^2 k^2 (1 - k^2)^2 + d^2 (1 - k^2)^2 \)
\( = d^2 k^2 (1 - k)^2 + d^2 k^2 (1 - k)^2 (1 + k)^2 + d^2 (1 - k)^2 (1 + k)^2 \)
\( = d^2 (1 - k)^2 [k^2 + k^2 (1 + k)^2 + (1 + k)^2] \)
\( = d^2 (1 - k)^2 [k^2 + k^2 (1 + 2k + k^2) + 1 + 2k + k^2] \)
\( = d^2 (1 - k)^2 [k^2 + k^2 + 2k^3 + k^4 + 1 + 2k + k^2] \)
\( = d^2 (1 - k)^2 (k^4 + 2k^3 + 3k^2 + 2k + 1) \)
R.H.S. \( = (dk^3 - d)^2 \)
\( = d^2 (k^3 - 1)^2 \)
\( = d^2 (1 - k^3)^2 \)
\( = d^2 ((1 - k)(1 + k + k^2))^2 \)
\( = d^2 (1 - k)^2 (1 + k + k^2)^2 \)
\( = d^2 (1 - k)^2 (1 + 2k + k^2 + 2k^3 + 2k^2 + k^4) \) (after expanding \( (1+k+k^2)^2 \))
\( = d^2 (1 - k)^2 (k^4 + 2k^3 + 3k^2 + 2k + 1) \)
Since L.H.S. = R.H.S., the identity is proven.

(ii) To prove: \( \sqrt{\frac{a^5+b^2c^2+a^3 c^2}{b^4 c+d^4+b^2 c d^2}}=\frac{a}{d} \)
L.H.S. \( = \sqrt{\frac{(dk^3)^5+(dk^2)^2(dk)^2+(dk^3)^3(dk)^2}{(dk^2)^4(dk)+d^4+(dk^2)^2(dk)d^2}} \)
\( = \sqrt{\frac{d^5 k^{15}+d^2 k^4 d^2 k^2+d^3 k^9 d^2 k^2}{d^4 k^8 d k+d^4+d^2 k^4 d k d^2}} \)
\( = \sqrt{\frac{d^5 k^{15}+d^4 k^6+d^5 k^{11}}{d^5 k^9+d^4+d^5 k^5}} \)
\( = \sqrt{\frac{d^4 k^6 (d k^9+1+d k^5)}{d^4 (d k^9+1+d k^5)}} \)
\( = \sqrt{k^6} = k^3 \)
R.H.S. \( = \frac{dk^3}{d} = k^3 \)
Since L.H.S. = R.H.S., the identity is proven.

(iii) To prove: \( \sqrt{a b}-\sqrt{b c}+\sqrt{c d}=\sqrt{(a-b+c)(b-c+d)} \)
L.H.S. \( = \sqrt{(dk^3)(dk^2)} - \sqrt{(dk^2)(dk)} + \sqrt{(dk)(d)} \)
\( = \sqrt{d^2 k^5} - \sqrt{d^2 k^3} + \sqrt{d^2 k} \)
\( = d k^2 \sqrt{k} - d k \sqrt{k} + d \sqrt{k} \)
\( = d \sqrt{k} (k^2 - k + 1) \)
R.H.S. \( = \sqrt{(dk^3 - dk^2 + dk)(dk^2 - dk + d)} \)
\( = \sqrt{d k (k^2 - k + 1) \cdot d (k^2 - k + 1)} \)
\( = \sqrt{d^2 k (k^2 - k + 1)^2} \)
\( = d \sqrt{k} (k^2 - k + 1) \)
Since L.H.S. = R.H.S., the identity is proven.

(iv) To prove: \( \frac{3 a+5 d}{5 a+7 d}=\frac{3 a^3+5 b^3}{5 a^3+7 b^3} \)
L.H.S. \( = \frac{3 (dk^3)+5 d}{5 (dk^3)+7 d} \)
\( = \frac{d(3k^3+5)}{d(5k^3+7)} = \frac{3k^3+5}{5k^3+7} \)
R.H.S. \( = \frac{3 (dk^3)^3+5 (dk^2)^3}{5 (dk^3)^3+7 (dk^2)^3} \)
\( = \frac{3 d^3 k^9+5 d^3 k^6}{5 d^3 k^9+7 d^3 k^6} \)
\( = \frac{d^3 k^6 (3k^3+5)}{d^3 k^6 (5k^3+7)} = \frac{3k^3+5}{5k^3+7} \)
Since L.H.S. = R.H.S., the identity is proven.

(v) To prove: \( \frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d} \)
L.H.S. \( = \frac{(dk^3)^3+(dk^2)^3+(dk)^3}{(dk^2)^3+(dk)^3+d^3} \)
\( = \frac{d^3 k^9+d^3 k^6+d^3 k^3}{d^3 k^6+d^3 k^3+d^3} \)
\( = \frac{d^3 k^3(k^6+k^3+1)}{d^3 (k^6+k^3+1)} = k^3 \)
R.H.S. \( = \frac{dk^3}{d} = k^3 \)
Since L.H.S. = R.H.S., the identity is proven.

(vi) To prove: \( pa^3 + qb^3 + rc^3 : pb^3 + qc^3 + rd^3 = a : d \)
This means we need to show \( \frac{p a^3+q c^3+r c^3}{p b^3+q c^3+r d^3} = \frac{a}{d} \)
L.H.S. \( = \frac{p(dk^3)^3+q(dk^2)^3+r(dk)^3}{p(dk^2)^3+q(dk)^3+r d^3} \)
\( = \frac{p d^3 k^9+q d^3 k^6+r d^3 k^3}{p d^3 k^6+q d^3 k^3+r d^3} \)
\( = \frac{d^3 k^3(p k^6+q k^3+r)}{d^3 (p k^6+q k^3+r)} = k^3 \)
R.H.S. \( = \frac{dk^3}{d} = k^3 \)
Since L.H.S. = R.H.S., the identity is proven. This concept simplifies calculations in many real-world scenarios, such as scaling recipes or architectural designs.
In simple words: When four numbers are in continued proportion, it means each number is a constant 'k' times the next one (a = dk³, b = dk², c = dk). Replace 'a', 'b', and 'c' with their 'd' and 'k' forms. Then, simplify both sides of each equation. You will find that both sides will become equal to 'k' raised to some power, proving the identities.

🎯 Exam Tip: For problems involving four numbers in continued proportion (a, b, c, d), consistently define them using a common ratio 'k' relative to the last term (e.g., \( c=dk, b=dk^2, a=dk^3 \)). This ensures all terms are expressed uniformly, making algebraic simplification straightforward.

 

Question 5. If \( \frac{x}{a}=\frac{y}{b}=\frac{z}{c} \), prove that \( \frac{a x-b y}{(a+b)(x-y)}+\frac{b y-c z}{(b+c) (y-z)} + \frac{c z-a x}{(c+a)(z-x)} = 3 \)
Answer:
Let \( \frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k \) (suppose)
This means \( x = ak, y = bk, z = ck \). We will substitute these values into the expression.
Consider the first term: \( \frac{a x-b y}{(a+b)(x-y)} \)
\( = \frac{a(ak)-b(bk)}{(a+b)(ak-bk)} = \frac{a^2 k-b^2 k}{(a+b)k(a-b)} \)
\( = \frac{k(a^2-b^2)}{k(a+b)(a-b)} = \frac{(a-b)(a+b)}{(a+b)(a-b)} = 1 \)
Consider the second term: \( \frac{b y-c z}{(b+c)(y-z)} \)
\( = \frac{b(bk)-c(ck)}{(b+c)(bk-ck)} = \frac{b^2 k-c^2 k}{(b+c)k(b-c)} \)
\( = \frac{k(b^2-c^2)}{k(b+c)(b-c)} = \frac{(b-c)(b+c)}{(b+c)(b-c)} = 1 \)
Consider the third term: \( \frac{c z-a x}{(c+a)(z-x)} \)
\( = \frac{c(ck)-a(ak)}{(c+a)(ck-ak)} = \frac{c^2 k-a^2 k}{(c+a)k(c-a)} \)
\( = \frac{k(c^2-a^2)}{k(c+a)(c-a)} = \frac{(c-a)(c+a)}{(c+a)(c-a)} = 1 \)
Adding all three terms:
L.H.S. \( = 1 + 1 + 1 = 3 \)
Thus, L.H.S. = R.H.S. This problem shows how algebraic identities can simplify complex fractional expressions.
In simple words: Since x/a, y/b, and z/c are all equal, let's call that common value 'k'. So, x is 'ak', y is 'bk', and z is 'ck'. Now, put these 'k' values into each part of the long sum. You will see that each part simplifies to '1'. Adding 1 + 1 + 1 gives 3, which is what we needed to prove.

🎯 Exam Tip: When faced with a sum of complex terms involving equal ratios, evaluate each term separately. Look for opportunities to factor out the common ratio 'k' and apply algebraic identities like \( A^2 - B^2 = (A-B)(A+B) \) to simplify. This breaks down the problem into manageable steps.

 

Question 6. If \( \frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c} \), show that each ratio is equal to \( \frac{x+y+z}{a+b+c} \).
Answer:
Let each ratio be equal to \( k \):
\( \frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}=k \)
According to the property of equal ratios (addendo property), if \( \frac{A}{B}=\frac{C}{D}=\frac{E}{F} \), then each ratio is also equal to \( \frac{A+C+E}{B+D+F} \).
Applying this property to the given ratios:
Each ratio \( = \frac{x+y+z}{(b+c-a)+(c+a-b)+(a+b-c)} \)
Now, let's simplify the denominator:
Denominator \( = b+c-a+c+a-b+a+b-c \)
Group like terms: \( (b-b+b) + (c+c-c) + (-a+a+a) \)
\( = b + c + a \)
So, each ratio \( = \frac{x+y+z}{a+b+c} \)
Hence, it is proven that each given ratio is equal to \( \frac{x+y+z}{a+b+c} \). This property is useful when you need to combine several ratios into a single, simplified form.
In simple words: When you have several fractions that are all equal, a helpful rule says that you can add up all the top numbers (numerators) and all the bottom numbers (denominators), and this new fraction will also be equal to the original fractions. In this problem, when we add the numerators (x+y+z) and the denominators (b+c-a + c+a-b + a+b-c), the bottom part simplifies to (a+b+c). So, the new fraction becomes (x+y+z)/(a+b+c), which is what we wanted to show.

🎯 Exam Tip: The 'addendo' property of equal ratios, which states that if \( \frac{a}{b} = \frac{c}{d} = \frac{e}{f} \), then each ratio is also equal to \( \frac{a+c+e}{b+d+f} \), is a powerful tool for these types of proofs. Mastering this property can save significant time.

 

Question 7. If \( \frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b} \), prove that \( a (b - c) + b (c - a) + c (a - b) = 0 \).
Answer:
Let \( \frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=k \) (suppose)
This means \( a = k(b+c), b = k(c+a), \) and \( c = k(a+b) \).
Now, consider the expression to be proved:
L.H.S. \( = a(b-c) + b(c-a) + c(a-b) \)
Substitute the expressions for a, b, and c:
\( = k(b+c)(b-c) + k(c+a)(c-a) + k(a+b)(a-b) \)
Using the identity \( (X+Y)(X-Y) = X^2-Y^2 \):
\( = k(b^2-c^2) + k(c^2-a^2) + k(a^2-b^2) \)
Factor out k:
\( = k(b^2-c^2+c^2-a^2+a^2-b^2) \)
All terms inside the parenthesis cancel out:
\( = k(0) = 0 \)
Thus, L.H.S. = R.H.S., proving the identity. This problem highlights how a common factor can lead to a simplified result.
In simple words: If some fractions are equal, we can say they all equal 'k'. So, 'a' equals 'k' times (b+c), 'b' equals 'k' times (c+a), and 'c' equals 'k' times (a+b). When we put these into the sum \( a(b-c) + b(c-a) + c(a-b) \), we can take 'k' out as a common factor. What's left inside the brackets is \( (b^2-c^2) + (c^2-a^2) + (a^2-b^2) \), which all cancel out to zero. So, the whole sum becomes \( k \times 0 = 0 \).

🎯 Exam Tip: When you have equations with a common multiplier like 'k' on one side and a sum of differences on the other, substituting the 'k' expressions often leads to cancellation of terms, simplifying the proof significantly.

 

Question 8. If ax = by = cz, prove that \( \frac{x^2}{y z}+\frac{y^2}{z x}+\frac{z^2}{x y}=\frac{b c} {a^2}+\frac{c a}{b^2}+\frac{a b}{c^2} \).
Answer:
Let \( ax = by = cz = k \) (suppose)
This means \( x = \frac{k}{a}, y = \frac{k}{b}, z = \frac{k}{c} \). We will substitute these values into both sides of the equation.
Consider the L.H.S.:
L.H.S. \( = \frac{x^2}{y z}+\frac{y^2}{z x}+\frac{z^2}{x y} \)
\( = \frac{(k/a)^2}{(k/b)(k/c)}+\frac{(k/b)^2}{(k/c)(k/a)}+\frac{(k/c)^2}{(k/a)(k/b)} \)
\( = \frac{k^2/a^2}{k^2/(bc)}+\frac{k^2/b^2}{k^2/(ca)}+\frac{k^2/c^2}{k^2/(ab)} \)
\( = \frac{k^2}{a^2} \cdot \frac{bc}{k^2} + \frac{k^2}{b^2} \cdot \frac{ca}{k^2} + \frac{k^2}{c^2} \cdot \frac{ab}{k^2} \)
\( = \frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2} \)
This is equal to the R.H.S.
Thus, L.H.S. = R.H.S., proving the identity. This type of substitution simplifies complex expressions into a more manageable form.
In simple words: When \( ax = by = cz \), we can set this common value to 'k'. This means x is \( k/a \), y is \( k/b \), and z is \( k/c \). If you put these into the left side of the equation, the \( k^2 \) terms will cancel out in each part. What's left will be exactly the same as the right side of the equation, showing that both sides are equal.

🎯 Exam Tip: When a problem involves multiple variables related by a common product (like \( ax=by=cz \)), introduce a constant 'k' for the product. This allows you to express each variable in terms of 'k' and its coefficient, which often leads to significant simplification through cancellation in complex fractions.

Self Evaluation and Revision (Latest ICSE Questions)

 

Question 1. Given \( \frac { a }{ b } = \frac { c }{ d } \), prove that \( \frac{3 a-5 b}{3 a+5 b} = \frac{3 c-5 d}{3 c+5 d} \).
Answer:
Let \( \frac { a }{ b } = \frac { c }{ d } = k \) (suppose)
This means \( a = bk \) and \( c = dk \). We will substitute these values into both sides of the equation.
Consider the L.H.S.:
L.H.S. \( = \frac{3 a-5 b}{3 a+5 b} \)
Substitute \( a = bk \):
\( = \frac{3 bk-5 b}{3 bk+5 b} \)
Factor out 'b' from the numerator and denominator:
\( = \frac{b(3 k-5)}{b(3 k+5)} = \frac{3 k-5}{3 k+5} \)
Consider the R.H.S.:
R.H.S. \( = \frac{3 c-5 d}{3 c+5 d} \)
Substitute \( c = dk \):
\( = \frac{3 dk-5 d}{3 dk+5 d} \)
Factor out 'd' from the numerator and denominator:
\( = \frac{d(3 k-5)}{d(3 k+5)} = \frac{3 k-5}{3 k+5} \)
Since L.H.S. = R.H.S., the identity is proven. This is a common application of the 'componendo and dividendo' concept, although here we use direct substitution.
In simple words: Since a/b and c/d are equal, let's call that common value 'k'. So, 'a' is 'bk' and 'c' is 'dk'. Replace 'a' in the first fraction with 'bk' and 'c' in the second fraction with 'dk'. After simplifying each fraction by canceling out 'b' or 'd', both fractions will become identical to \( \frac{3k-5}{3k+5} \), proving they are equal.

🎯 Exam Tip: For problems involving proving equality of ratios, direct substitution using a common ratio 'k' is often the most straightforward method. Ensure you substitute correctly and simplify both sides independently to show their equivalence.

 

Question 2. Two numbers are in the ratio of 3 : 5. If 8 is added to each number, the ratio becomes 2 : 3. Find the numbers.
Answer:
Let the two numbers be \( 3x \) and \( 5x \) since their ratio is 3 : 5.
When 8 is added to each number, the new numbers become \( 3x+8 \) and \( 5x+8 \).
According to the problem, the new ratio is 2 : 3.
So, we can set up the equation: \( \frac{3x+8}{5x+8} = \frac{2}{3} \)
Now, cross-multiply to solve for x:
\( 3(3x+8) = 2(5x+8) \)
\( 9x+24 = 10x+16 \)
To find x, bring all x terms to one side and constants to the other:
\( 24-16 = 10x-9x \)
\( 8 = x \)
So, the value of \( x \) is 8.
Now, find the original numbers:
First number \( = 3x = 3 \times 8 = 24 \)
Second number \( = 5x = 5 \times 8 = 40 \)
The two numbers are 24 and 40. This is a common type of ratio problem often seen in everyday situations, such as mixing ingredients.
In simple words: Since the numbers are in the ratio 3:5, we can call them 3x and 5x. When 8 is added to each, they become \( 3x+8 \) and \( 5x+8 \). The new ratio is 2:3, so we write the equation \( \frac{3x+8}{5x+8} = \frac{2}{3} \). Solving this equation by cross-multiplying gives us \( x=8 \). Then, the numbers are \( 3 \times 8 = 24 \) and \( 5 \times 8 = 40 \).

🎯 Exam Tip: When numbers are given in a ratio, represent them as multiples of a variable (e.g., 3x and 5x). This allows you to set up an algebraic equation based on the given conditions, which can then be solved to find the variable and, subsequently, the actual numbers.

 

Question 3. If a : b = 5: 3, find (5a + 8b) : (6a – 7b).
Answer:
Given that \( a : b = 5 : 3 \), which means \( \frac{a}{b} = \frac{5}{3} \).
We need to find the ratio \( (5a + 8b) : (6a - 7b) \), which can be written as \( \frac{5a + 8b}{6a - 7b} \).
To use the given ratio \( \frac{a}{b} \), we can divide both the numerator and the denominator of the expression by 'b'.
\( = \frac{\frac{5a}{b} + \frac{8b}{b}}{\frac{6a}{b} - \frac{7b}{b}} \)
\( = \frac{5\left(\frac{a}{b}\right) + 8}{6\left(\frac{a}{b}\right) - 7} \)
Now, substitute the value \( \frac{a}{b} = \frac{5}{3} \):
\( = \frac{5\left(\frac{5}{3}\right) + 8}{6\left(\frac{5}{3}\right) - 7} \)
\( = \frac{\frac{25}{3} + 8}{\frac{30}{3} - 7} \)
\( = \frac{\frac{25+24}{3}}{\frac{30-21}{3}} \)
\( = \frac{\frac{49}{3}}{\frac{9}{3}} \)
Multiply the numerator by the reciprocal of the denominator:
\( = \frac{49}{3} \times \frac{3}{9} \)
\( = \frac{49}{9} \)
So, \( (5a + 8b) : (6a - 7b) = 49 : 9 \). This method of dividing by a common variable is efficient for solving such problems.
In simple words: We know that \( a/b \) is 5/3. To find the ratio of \( (5a+8b) \) to \( (6a-7b) \), we can divide every term in both parts by 'b'. This changes 'a' to \( a/b \) and 'b' to 1. Then we just put in 5/3 for \( a/b \) and do the math to get the final ratio.

🎯 Exam Tip: When given a ratio like a:b, convert it to a fraction (a/b). For expressions involving 'a' and 'b', divide all terms by 'b' to transform them into terms of (a/b) and constants, simplifying the substitution process. This is a common trick for efficiently solving these types of problems.

 

Question 4. If \( \frac{3 a+4 b}{3 c+4 d}=\frac{3 a-4 b}{3 c-4 d} \) prove that \( \frac { a }{ b } = \frac { c }{ d } \).
Answer: We are given the proportion \( \frac{3a+4b}{3c+4d} = \frac{3a-4b}{3c-4d} \). To begin, we rearrange this using the alternendo rule, which involves swapping the middle terms of the proportion. This gives us:
\( \frac{3a+4b}{3a-4b} = \frac{3c+4d}{3c-4d} \).
Next, we apply the componendo and dividendo rule to both sides. This rule states that if \( \frac{A}{B} = \frac{C}{D} \), then \( \frac{A+B}{A-B} = \frac{C+D}{C-D} \).
So, we have:
\( \frac{(3a+4b) + (3a-4b)}{(3a+4b) - (3a-4b)} = \frac{(3c+4d) + (3c-4d)}{(3c+4d) - (3c-4d)} \)
Simplify the numerators and denominators:
Left side numerator: \( 3a+4b+3a-4b = 6a \).
Left side denominator: \( 3a+4b-3a+4b = 8b \).
Right side numerator: \( 3c+4d+3c-4d = 6c \).
Right side denominator: \( 3c+4d-3c+4d = 8d \).
The equation now simplifies to \( \frac{6a}{8b} = \frac{6c}{8d} \).
Finally, we divide both sides of the equation by \( \frac{6}{8} \) to isolate the ratios \( \frac{a}{b} \) and \( \frac{c}{d} \). This results in \( \frac{a}{b} = \frac{c}{d} \). The problem demonstrates how ratio properties can be used to simplify complex proportional relationships to a simpler form.
In simple words: First, swap some parts of the fractions. Then, add and subtract the top and bottom parts of each fraction to make them simpler. After a few steps, you will see that a divided by b is the same as c divided by d.

🎯 Exam Tip: Remember to apply the rules of alternendo, componendo, and dividendo carefully. Always simplify fractions at each step to avoid errors in calculation.

 

Question 5. The work done by \( (x - 3) \) men in \( (2x + 1) \) days and the work done by \( (2x + 1) \) men in \( (x + 4) \) days are in the ratio 3 : 10. Find the value of x.
Answer: In this problem, work done is calculated as the product of the number of men and the number of days they work.
For the first group:
Number of men \( = (x-3) \)
Number of days \( = (2x+1) \)
Work done by the first group \( = (x-3)(2x+1) \).
For the second group:
Number of men \( = (2x+1) \)
Number of days \( = (x+4) \)
Work done by the second group \( = (2x+1)(x+4) \).
We are told that the ratio of the work done by the first group to the second group is \( 3:10 \). So, we can set up the equation:
\( \frac{(x-3)(2x+1)}{(2x+1)(x+4)} = \frac{3}{10} \)
We can see that \( (2x+1) \) appears in both the numerator and the denominator on the left side, so we can cancel it out, assuming \( 2x+1 \neq 0 \).
This simplifies the equation to:
\( \frac{x-3}{x+4} = \frac{3}{10} \)
Next, we cross-multiply to solve for x:
\( 10(x-3) = 3(x+4) \)
Expand both sides of the equation:
\( 10x - 30 = 3x + 12 \)
Now, gather all the x terms on one side and the constant terms on the other side:
\( 10x - 3x = 12 + 30 \)
\( 7x = 42 \)
Finally, divide both sides by 7 to find the value of x:
\( x = \frac{42}{7} \)
\( x = 6 \). This value of x correctly fulfills the conditions of the work ratio.
In simple words: The amount of work is men times days. We write an equation showing the ratio of work from two groups. We simplify the equation and solve for x by multiplying across and putting x terms on one side and numbers on the other.

🎯 Exam Tip: Always cancel common factors in fractions before cross-multiplying, as it makes the algebra much simpler. Double-check your arithmetic, especially when combining terms.

 

Question 6. What number should be subtracted from each of the numbers 23, 30, 57 and 78 so that the remainders are in proportion?
Answer: Let the number to be subtracted from each given number be x.
After subtracting x, the new numbers will be:
\( (23-x), (30-x), (57-x), \) and \( (78-x) \).
For these four numbers to be in proportion, the ratio of the first two must be equal to the ratio of the last two:
\( \frac{23-x}{30-x} = \frac{57-x}{78-x} \)
To solve this equation, we cross-multiply:
\( (23-x)(78-x) = (57-x)(30-x) \)
Expand both sides of the equation:
Left side: \( 23 \times 78 - 23x - 78x + x^2 = 1794 - 101x + x^2 \)
Right side: \( 57 \times 30 - 57x - 30x + x^2 = 1710 - 87x + x^2 \)
So, the equation becomes:
\( 1794 - 101x + x^2 = 1710 - 87x + x^2 \)
Notice that \( x^2 \) appears on both sides of the equation, so we can cancel them out:
\( 1794 - 101x = 1710 - 87x \)
Now, gather all the x terms on one side and the constant terms on the other side:
\( -101x + 87x = 1710 - 1794 \)
\( -14x = -84 \)
Finally, divide both sides by -14 to find the value of x:
\( x = \frac{-84}{-14} \)
\( x = 6 \). This means that 6 must be subtracted from each number to make them proportional.
In simple words: We want to find a number, x, so that when we take it away from 23, 30, 57, and 78, the new numbers will be in proportion. We set up an equation where the first two new numbers form a ratio equal to the ratio of the last two. By doing some math, we find x is 6.

🎯 Exam Tip: When setting up proportions from word problems, ensure the order of terms is consistent. Remember that \( x^2 \) terms often cancel out in these types of equations, simplifying the solution.

 

Question 7. What number must be added to each of the numbers 6, 15, 20 and 43 to make them in proportion.
Answer: Let the number to be added to each given number be x.
After adding x, the new numbers will be:
\( (6+x), (15+x), (20+x), \) and \( (43+x) \).
For these four numbers to be in proportion, the ratio of the first two must be equal to the ratio of the last two:
\( \frac{6+x}{15+x} = \frac{20+x}{43+x} \)
To solve this equation, we cross-multiply:
\( (6+x)(43+x) = (20+x)(15+x) \)
Expand both sides of the equation:
Left side: \( 6 \times 43 + 6x + 43x + x^2 = 258 + 49x + x^2 \)
Right side: \( 20 \times 15 + 20x + 15x + x^2 = 300 + 35x + x^2 \)
So, the equation becomes:
\( 258 + 49x + x^2 = 300 + 35x + x^2 \)
Notice that \( x^2 \) appears on both sides of the equation, so we can cancel them out:
\( 258 + 49x = 300 + 35x \)
Now, gather all the x terms on one side and the constant terms on the other side:
\( 49x - 35x = 300 - 258 \)
\( 14x = 42 \)
Finally, divide both sides by 14 to find the value of x:
\( x = \frac{42}{14} \)
\( x = 3 \). Thus, the number that must be added to each is 3.
In simple words: We want to find a number, x, that we can add to 6, 15, 20, and 43 so that the new numbers are in proportion. We set up the ratios and cross-multiply to get an equation. Solving this equation tells us that x is 3.

🎯 Exam Tip: Always be careful when expanding brackets and combining like terms. Make sure to move x-terms and constant terms to opposite sides of the equation correctly.

 

Question 8. If \( \frac{3 x+5 y}{3 x-5 y}=\frac{7}{3} \), find x : y.
Answer: We are given the proportion \( \frac{3x+5y}{3x-5y} = \frac{7}{3} \). Our goal is to find the ratio \( x:y \).
We can solve this efficiently by applying the componendo and dividendo rule. This rule states that if \( \frac{A}{B} = \frac{C}{D} \), then \( \frac{A+B}{A-B} = \frac{C+D}{C-D} \).
Applying this rule to our given equation, where \( A = (3x+5y) \), \( B = (3x-5y) \), \( C = 7 \), and \( D = 3 \):
\( \frac{(3x+5y) + (3x-5y)}{(3x+5y) - (3x-5y)} = \frac{7+3}{7-3} \)
Now, simplify both the numerator and the denominator on each side:
For the left side:
Numerator: \( (3x+5y) + (3x-5y) = 3x+5y+3x-5y = 6x \).
Denominator: \( (3x+5y) - (3x-5y) = 3x+5y-3x+5y = 10y \).
For the right side:
Numerator: \( 7+3 = 10 \).
Denominator: \( 7-3 = 4 \).
So, the equation simplifies to:
\( \frac{6x}{10y} = \frac{10}{4} \)
To find the ratio \( \frac{x}{y} \), we multiply both sides by \( \frac{10}{6} \):
\( \frac{x}{y} = \frac{10}{4} \times \frac{10}{6} \)
Multiply the fractions:
\( \frac{x}{y} = \frac{100}{24} \)
Finally, simplify the fraction by dividing both the numerator and denominator by their greatest common divisor, which is 4:
\( \frac{x}{y} = \frac{25}{6} \). This method makes solving such ratio problems very direct.
Therefore, the ratio \( x:y \) is \( 25:6 \).
In simple words: We are given a math problem with x and y. We use a special rule called componendo and dividendo to make the equation simpler. After a few steps of adding and subtracting parts of the fractions, we get a new, simpler fraction that tells us the ratio of x to y.

🎯 Exam Tip: Componendo and dividendo is very useful for quickly solving ratio equations like this. Always simplify numerical fractions as much as possible at each stage to keep calculations manageable.

 

Question 9. If \( x = \frac{\sqrt{a+3 b}+\sqrt{a-3 b}}{\sqrt{a+3 b}-\sqrt{a-3 b}} \), prove that \( 3bx^2 - 2ax + 3b = 0 \).
Answer: We are given \( x = \frac{\sqrt{a+3b} + \sqrt{a-3b}}{\sqrt{a+3b} - \sqrt{a-3b}} \) and need to prove that \( 3bx^2 - 2ax + 3b = 0 \).
First, we write x as \( \frac{x}{1} \). Then, we apply the componendo and dividendo rule to \( \frac{x}{1} \).
\( \frac{x+1}{x-1} = \frac{(\sqrt{a+3b} + \sqrt{a-3b}) + (\sqrt{a+3b} - \sqrt{a-3b})}{(\sqrt{a+3b} + \sqrt{a-3b}) - (\sqrt{a+3b} - \sqrt{a-3b})} \)
Simplify the numerator and denominator:
Numerator: \( \sqrt{a+3b} + \sqrt{a-3b} + \sqrt{a+3b} - \sqrt{a-3b} = 2\sqrt{a+3b} \).
Denominator: \( \sqrt{a+3b} + \sqrt{a-3b} - \sqrt{a+3b} + \sqrt{a-3b} = 2\sqrt{a-3b} \).
So, the equation simplifies to:
\( \frac{x+1}{x-1} = \frac{2\sqrt{a+3b}}{2\sqrt{a-3b}} \)
\( \implies \frac{x+1}{x-1} = \frac{\sqrt{a+3b}}{\sqrt{a-3b}} \). This step removes the sum/difference form of the radicals.
Next, we square both sides of the equation to eliminate the square roots:
\( \left(\frac{x+1}{x-1}\right)^2 = \left(\frac{\sqrt{a+3b}}{\sqrt{a-3b}}\right)^2 \)
\( \implies \frac{(x+1)^2}{(x-1)^2} = \frac{a+3b}{a-3b} \)
Now, we apply the componendo and dividendo rule again to this new equation:
\( \frac{(x+1)^2 + (x-1)^2}{(x+1)^2 - (x-1)^2} = \frac{(a+3b) + (a-3b)}{(a+3b) - (a-3b)} \)
Expand and simplify the terms:
Left side numerator: \( (x^2+2x+1) + (x^2-2x+1) = 2x^2+2 = 2(x^2+1) \).
Left side denominator: \( (x^2+2x+1) - (x^2-2x+1) = 4x \).
Right side numerator: \( a+3b+a-3b = 2a \).
Right side denominator: \( a+3b-a+3b = 6b \).
So, the equation becomes:
\( \frac{2(x^2+1)}{4x} = \frac{2a}{6b} \)
Simplify both sides:
\( \frac{x^2+1}{2x} = \frac{a}{3b} \)
Finally, cross-multiply to rearrange the terms and get the desired proof:
\( 3b(x^2+1) = 2ax \)
\( 3bx^2 + 3b = 2ax \)
\( \implies 3bx^2 - 2ax + 3b = 0 \). This proves the given statement.
In simple words: We are given a fraction with square roots. We use a math rule called componendo and dividendo twice. First, it helps get rid of the sums and differences of the square roots. Then, we square both sides to remove the square roots completely. Using the same rule again, we simplify the equation until we get the answer we need.

🎯 Exam Tip: For complex radical expressions, applying componendo and dividendo, followed by squaring both sides, is a common and effective strategy. Remember to apply the rule consistently to both sides of the equation.

 

Question 10. If \( \frac{8 a-5 b}{8 c-5 d}=\frac{8 a+5 b}{8 c+5 d} \), prove that \( \frac { a }{ b } = \frac { c }{ d } \).
Answer: We are given the proportion \( \frac{8a-5b}{8c-5d} = \frac{8a+5b}{8c+5d} \) and need to prove that \( \frac{a}{b} = \frac{c}{d} \).
First, we rearrange the given proportion using the alternendo rule. This rule states that if \( \frac{A}{B} = \frac{C}{D} \), then \( \frac{A}{C} = \frac{B}{D} \). Swapping the terms \( (8a-5b) \) and \( (8c+5d) \) gives:
\( \frac{8a+5b}{8a-5b} = \frac{8c+5d}{8c-5d} \)
Now, we apply the componendo and dividendo rule to both sides of this equation. This rule states that if \( \frac{A}{B} = \frac{C}{D} \), then \( \frac{A+B}{A-B} = \frac{C+D}{C-D} \).
Applying this:
\( \frac{(8a+5b) + (8a-5b)}{(8a+5b) - (8a-5b)} = \frac{(8c+5d) + (8c-5d)}{(8c+5d) - (8c-5d)} \)
Simplify the numerators and denominators:
Left side numerator: \( 8a+5b+8a-5b = 16a \).
Left side denominator: \( 8a+5b-8a+5b = 10b \).
Right side numerator: \( 8c+5d+8c-5d = 16c \).
Right side denominator: \( 8c+5d-8c+5d = 10d \).
So, the equation simplifies to:
\( \frac{16a}{10b} = \frac{16c}{10d} \)
To isolate the ratios \( \frac{a}{b} \) and \( \frac{c}{d} \), we can multiply both sides by \( \frac{10}{16} \) (or divide by \( \frac{16}{10} \)).
\( \frac{a}{b} = \frac{c}{d} \). This proves the given statement. This method makes proving such equivalences straightforward.
In simple words: We start with a complex equation of fractions. First, we swap some parts of the fractions. Then, we use a special math rule that involves adding and subtracting parts of the fractions. This makes the equation much simpler until we get the answer that a divided by b is equal to c divided by d.

🎯 Exam Tip: Always consider rearranging the given proportion using alternendo *before* applying componendo and dividendo if it simplifies the structure, as it often leads to a more direct solution.

 

Question 11. What least number must be added to each of the numbers 5, 11, 19 and 37 so that they are in proportion?
Answer: Let x be the least number that must be added to each of the numbers 5, 11, 19, and 37.
After adding x, the new numbers will be:
\( (5+x), (11+x), (19+x), \) and \( (37+x) \).
For these four numbers to be in proportion, the product of the means must be equal to the product of the extremes. This means:
\( (5+x)(37+x) = (11+x)(19+x) \)
Now, we expand both sides of the equation:
Left side: \( 5 \times 37 + 5x + 37x + x^2 = 185 + 42x + x^2 \).
Right side: \( 11 \times 19 + 11x + 19x + x^2 = 209 + 30x + x^2 \).
So, the equation becomes:
\( 185 + 42x + x^2 = 209 + 30x + x^2 \)
Notice that \( x^2 \) appears on both sides of the equation, so we can cancel them out:
\( 185 + 42x = 209 + 30x \)
Now, gather all the x terms on one side and the constant terms on the other side:
\( 42x - 30x = 209 - 185 \)
\( 12x = 24 \)
Finally, divide both sides by 12 to find the value of x:
\( x = \frac{24}{12} \)
\( x = 2 \). This means that 2 must be added to each number for them to be in proportion.
In simple words: We need to find a number, x, to add to 5, 11, 19, and 37 so that the new numbers are proportional. We set up an equation where the first and last new numbers multiplied together equal the middle two multiplied together. After solving this equation, we find x is 2.

🎯 Exam Tip: When numbers are in proportion (a:b :: c:d), it implies \( ad = bc \). This property is fundamental for solving such problems. Be careful with algebraic expansions and sign changes.

 

Question 12. Given \( \frac{a^3+3 a b^2}{b^3+3 a^2 b}=\frac{63}{62} \) that using componendo and dividendo find a : b.
Answer: We are given the equation \( \frac{a^3+3ab^2}{b^3+3a^2b} = \frac{63}{62} \). Our task is to find the ratio \( a:b \) using the componendo and dividendo rule.
First, apply the componendo and dividendo rule to the given equation:
\( \frac{(a^3+3ab^2) + (b^3+3a^2b)}{(a^3+3ab^2) - (b^3+3a^2b)} = \frac{63+62}{63-62} \)
Now, recognize the algebraic identities in the left side's numerator and denominator:
Numerator: \( a^3+3a^2b+3ab^2+b^3 = (a+b)^3 \).
Denominator: \( a^3-3a^2b+3ab^2-b^3 = (a-b)^3 \).
For the right side:
Numerator: \( 63+62 = 125 \).
Denominator: \( 63-62 = 1 \).
So, the equation transforms into:
\( \frac{(a+b)^3}{(a-b)^3} = \frac{125}{1} \)
This can be written as \( \left(\frac{a+b}{a-b}\right)^3 = (5)^3 \).
Taking the cube root of both sides, we get:
\( \frac{a+b}{a-b} = 5 \)
Now, we can solve for the ratio \( a:b \). Cross-multiply:
\( a+b = 5(a-b) \)
\( a+b = 5a-5b \)
Gather terms with 'a' on one side and terms with 'b' on the other:
\( b+5b = 5a-a \)
\( 6b = 4a \)
To find the ratio \( a:b \), divide both sides by 'b' and by '4':
\( \frac{a}{b} = \frac{6}{4} = \frac{3}{2} \). The key here was identifying the cubic expansions.
Therefore, the ratio \( a:b \) is \( 3:2 \).
In simple words: We start with a fraction and use the componendo and dividendo rule. This changes the top and bottom of the fraction into a simple cubed form. We then take the cube root and solve for 'a' and 'b' to find their ratio.

🎯 Exam Tip: Familiarity with algebraic identities like \( (a+b)^3 \) and \( (a-b)^3 \) is crucial. They often appear in combination with ratio properties to simplify problems significantly.

 

Question 13. If x, y, z are in continued proportion, prove that \( \frac{(x+y)^2}{(y+z)^2}=\frac{x}{z} \).
Answer: We are given that x, y, and z are in continued proportion. This means that \( \frac{x}{y} = \frac{y}{z} \), which implies \( y^2 = xz \). We need to prove that \( \frac{(x+y)^2}{(y+z)^2} = \frac{x}{z} \).
Let's start with the Left Hand Side (L.H.S.) of the equation:
L.H.S. \( = \frac{(x+y)^2}{(y+z)^2} \)
Expand the squares in the numerator and denominator:
\( = \frac{x^2+2xy+y^2}{y^2+2yz+z^2} \)
Now, substitute \( y^2 = xz \) into both the numerator and the denominator:
\( = \frac{x^2+2xy+xz}{xz+2yz+z^2} \). This substitution helps relate all terms.
Next, factor out common terms from the numerator and the denominator:
In the numerator, 'x' is a common factor: \( x(x+2y+z) \).
In the denominator, 'z' is a common factor: \( z(x+2y+z) \).
So, the expression becomes:
\( = \frac{x(x+2y+z)}{z(x+2y+z)} \)
Since \( (x+2y+z) \) is a common factor in both the numerator and the denominator, we can cancel it out (assuming \( x+2y+z \neq 0 \)).
This leaves:
\( = \frac{x}{z} \).
This is equal to the Right Hand Side (R.H.S.) of the equation. Thus, the statement is proven.
In simple words: We are given three numbers in a special order where the middle number squared is equal to the first and last multiplied. We need to show that a fraction with these numbers is equal to a simpler fraction. By expanding the top and bottom of the first fraction and using the special order rule, we can make it simpler and prove the answer.

🎯 Exam Tip: When dealing with continued proportion, always write down the relationship \( y^2 = xz \) first. This substitution is usually the key to simplifying expressions and proving identities.

 

Question 14. 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 \( b^2 = \frac{2 a^2 x}{x^2+1} \).
Answer: We are given \( x = \frac{\sqrt{a^2+b^2} + \sqrt{a^2-b^2}}{\sqrt{a^2+b^2} - \sqrt{a^2-b^2}} \) and need to prove that \( b^2 = \frac{2a^2x}{x^2+1} \).
First, we express x as \( \frac{x}{1} \). Then, we apply the componendo and dividendo rule to \( \frac{x}{1} \). The rule states that if \( \frac{A}{B} = \frac{C}{D} \), then \( \frac{A+B}{A-B} = \frac{C+D}{C-D} \).
\( \frac{x+1}{x-1} = \frac{(\sqrt{a^2+b^2} + \sqrt{a^2-b^2}) + (\sqrt{a^2+b^2} - \sqrt{a^2-b^2})}{(\sqrt{a^2+b^2} + \sqrt{a^2-b^2}) - (\sqrt{a^2+b^2} - \sqrt{a^2-b^2})} \)
Simplify the numerator and denominator:
Numerator: \( 2\sqrt{a^2+b^2} \).
Denominator: \( 2\sqrt{a^2-b^2} \).
So, the equation becomes:
\( \frac{x+1}{x-1} = \frac{2\sqrt{a^2+b^2}}{2\sqrt{a^2-b^2}} \)
\( \implies \frac{x+1}{x-1} = \frac{\sqrt{a^2+b^2}}{\sqrt{a^2-b^2}} \). This simplifies the radical expression.
Next, square both sides of the equation to remove the square roots:
\( \left(\frac{x+1}{x-1}\right)^2 = \left(\frac{\sqrt{a^2+b^2}}{\sqrt{a^2-b^2}}\right)^2 \)
\( \implies \frac{(x+1)^2}{(x-1)^2} = \frac{a^2+b^2}{a^2-b^2} \)
Now, expand the left side:
\( \frac{x^2+2x+1}{x^2-2x+1} = \frac{a^2+b^2}{a^2-b^2} \)
Apply the componendo and dividendo rule again to this new equation:
\( \frac{(x^2+2x+1) + (x^2-2x+1)}{(x^2+2x+1) - (x^2-2x+1)} = \frac{(a^2+b^2) + (a^2-b^2)}{(a^2+b^2) - (a^2-b^2)} \)
Simplify the terms:
Left side numerator: \( 2x^2+2 = 2(x^2+1) \).
Left side denominator: \( 4x \).
Right side numerator: \( 2a^2 \).
Right side denominator: \( 2b^2 \).
So, the equation becomes:
\( \frac{2(x^2+1)}{4x} = \frac{2a^2}{2b^2} \)
Simplify both sides:
\( \frac{x^2+1}{2x} = \frac{a^2}{b^2} \)
To get the required proof, rearrange the equation to solve for \( b^2 \):
\( b^2 = a^2 \times \frac{2x}{x^2+1} \)
\( \implies b^2 = \frac{2a^2x}{x^2+1} \). This proves the given statement. The problem shows how applying the same rule multiple times can lead to the solution.
In simple words: We start with a complicated fraction with square roots and x. We use a math rule called componendo and dividendo twice. After the first time, we square both sides to remove the roots. Then, we use the rule again and simplify the equation. This helps us find the formula for \( b^2 \).

🎯 Exam Tip: This problem is a classic example of using componendo and dividendo iteratively. Always be alert for opportunities to simplify by squaring or by reapplying the rule.

 

Question 15. Using componendo and dividendo, find the value of x given \( \frac{\sqrt{3 x+4}+\sqrt{3 x-5}}{\sqrt{3 x+4}-\sqrt{3 x-5}} = 9 \).
Answer: We are given the equation \( \frac{\sqrt{3x+4} + \sqrt{3x-5}}{\sqrt{3x+4} - \sqrt{3x-5}} = 9 \). We need to find the value of x using the componendo and dividendo rule.
First, we express 9 as \( \frac{9}{1} \). Then, we apply the componendo and dividendo rule. This rule states that if \( \frac{A}{B} = \frac{C}{D} \), then \( \frac{A+B}{A-B} = \frac{C+D}{C-D} \).
\( \frac{(\sqrt{3x+4} + \sqrt{3x-5}) + (\sqrt{3x+4} - \sqrt{3x-5})}{(\sqrt{3x+4} + \sqrt{3x-5}) - (\sqrt{3x+4} - \sqrt{3x-5})} = \frac{9+1}{9-1} \)
Simplify the numerator and denominator:
Left side numerator: \( 2\sqrt{3x+4} \).
Left side denominator: \( 2\sqrt{3x-5} \).
Right side numerator: \( 10 \).
Right side denominator: \( 8 \).
So, the equation becomes:
\( \frac{2\sqrt{3x+4}}{2\sqrt{3x-5}} = \frac{10}{8} \)
Simplify both sides:
\( \frac{\sqrt{3x+4}}{\sqrt{3x-5}} = \frac{5}{4} \). This significantly reduces the complexity.
Next, we square both sides of the equation to eliminate the square roots:
\( \left(\frac{\sqrt{3x+4}}{\sqrt{3x-5}}\right)^2 = \left(\frac{5}{4}\right)^2 \)
\( \implies \frac{3x+4}{3x-5} = \frac{25}{16} \)
Now, we cross-multiply to solve for x:
\( 16(3x+4) = 25(3x-5) \)
Expand both sides:
\( 48x + 64 = 75x - 125 \)
Gather all the x terms on one side and the constant terms on the other:
\( 64 + 125 = 75x - 48x \)
\( 189 = 27x \)
Finally, divide both sides by 27 to find the value of x:
\( x = \frac{189}{27} \)
\( x = 7 \). This value of x satisfies the initial equation.
In simple words: We have a fraction with square roots that equals 9. We use a math rule to simplify it. First, we get rid of the sums and differences of the square roots. Then, we square both sides to remove the square roots completely. After that, we multiply across and solve the equation to find x.

🎯 Exam Tip: Always simplify numerical fractions as early as possible (e.g., \( \frac{10}{8} \) to \( \frac{5}{4} \)) to make subsequent calculations simpler and reduce the chance of arithmetic errors.

 

Question 16. 6 is the mean proportion between two numbers x and y and 48 is the third proportional of x and y. Find the numbers.
Answer: We are given two pieces of information to find the values of x and y.
**Condition 1: 6 is the mean proportion between x and y.**
The mean proportion between two numbers a and b is \( \sqrt{ab} \). So, for x and y:
\( \sqrt{xy} = 6 \)
To remove the square root, we square both sides of the equation:
\( (\sqrt{xy})^2 = 6^2 \)
\( xy = 36 \) (Equation 1)
From this, we can express y in terms of x: \( y = \frac{36}{x} \). This relationship connects the two unknowns.
**Condition 2: 48 is the third proportional of x and y.**
If a, b, c are in continued proportion, then c is the third proportional of a and b. This means \( a:b :: b:c \), or \( \frac{a}{b} = \frac{b}{c} \).
So, for x, y, and 48:
\( x:y :: y:48 \)
\( \frac{x}{y} = \frac{y}{48} \)
Cross-multiply to get:
\( y^2 = 48x \)
Now, we substitute the expression for y from Equation 1 into this equation:
\( \left(\frac{36}{x}\right)^2 = 48x \)
\( \frac{36^2}{x^2} = 48x \)
\( \frac{1296}{x^2} = 48x \)
Multiply both sides by \( x^2 \):
\( 1296 = 48x^3 \)
Now, solve for \( x^3 \):
\( x^3 = \frac{1296}{48} \)
\( x^3 = 27 \)
Recognize that \( 27 \) is \( 3^3 \). So, taking the cube root of both sides:
\( x = 3 \).
Finally, substitute the value of x back into Equation 1 to find y:
\( y = \frac{36}{x} = \frac{36}{3} \)
\( y = 12 \).
Therefore, the two numbers are \( x=3 \) and \( y=12 \). These values satisfy both conditions.
In simple words: We are given two clues about two numbers, x and y. The first clue tells us that if you multiply x and y and then take the square root, you get 6. The second clue tells us how x, y, and 48 are related in a special way. We use these clues to make two math equations. We solve these equations step-by-step to find that x is 3 and y is 12.

🎯 Exam Tip: Clearly define what "mean proportion" (\( \sqrt{ab} \)) and "third proportional" (\( \frac{a}{b} = \frac{b}{c} \)) mean. Setting up simultaneous equations and using substitution is a standard approach for these problems.

 

Question 17. 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: We are given \( x = \frac{\sqrt{a+1} + \sqrt{a-1}}{\sqrt{a+1} - \sqrt{a-1}} \) and need to show that \( x^2 - 2ax + 1 = 0 \).
First, we express x as \( \frac{x}{1} \). Then, we apply the componendo and dividendo rule. This rule states that if \( \frac{A}{B} = \frac{C}{D} \), then \( \frac{A+B}{A-B} = \frac{C+D}{C-D} \).
\( \frac{x+1}{x-1} = \frac{(\sqrt{a+1} + \sqrt{a-1}) + (\sqrt{a+1} - \sqrt{a-1})}{(\sqrt{a+1} + \sqrt{a-1}) - (\sqrt{a+1} - \sqrt{a-1})} \)
Simplify the numerator and denominator:
Numerator: \( 2\sqrt{a+1} \).
Denominator: \( 2\sqrt{a-1} \).
So, the equation becomes:
\( \frac{x+1}{x-1} = \frac{2\sqrt{a+1}}{2\sqrt{a-1}} \)
\( \implies \frac{x+1}{x-1} = \frac{\sqrt{a+1}}{\sqrt{a-1}} \). This simplifies the radical expression.
Next, square both sides of the equation to eliminate the square roots:
\( \left(\frac{x+1}{x-1}\right)^2 = \left(\frac{\sqrt{a+1}}{\sqrt{a-1}}\right)^2 \)
\( \implies \frac{(x+1)^2}{(x-1)^2} = \frac{a+1}{a-1} \)
Now, expand the left side:
\( \frac{x^2+2x+1}{x^2-2x+1} = \frac{a+1}{a-1} \)
Apply the componendo and dividendo rule again to this new equation:
\( \frac{(x^2+2x+1) + (x^2-2x+1)}{(x^2+2x+1) - (x^2-2x+1)} = \frac{(a+1) + (a-1)}{(a+1) - (a-1)} \)
Simplify the terms:
Left side numerator: \( 2x^2+2 = 2(x^2+1) \).
Left side denominator: \( 4x \).
Right side numerator: \( 2a \).
Right side denominator: \( 2 \).
So, the equation becomes:
\( \frac{2(x^2+1)}{4x} = \frac{2a}{2} \)
Simplify both sides:
\( \frac{x^2+1}{2x} = a \)
Now, multiply both sides by \( 2x \):
\( x^2+1 = 2ax \)
Finally, rearrange the terms to get the desired proof:
\( x^2 - 2ax + 1 = 0 \). This proves the given statement by systematically applying ratio properties.
In simple words: We are given a fraction with square roots that equals x. We use a math rule called componendo and dividendo twice. After the first use, we square both sides to remove the square roots. Then, we use the rule again, simplify, and rearrange the equation to get the final answer.

🎯 Exam Tip: This problem is a common application of applying componendo and dividendo multiple times. Careful expansion of algebraic terms and step-by-step simplification are key to avoiding errors.

 

Question 18. Using the properties of proportion, solve for x, given \( \frac{x^4+1}{2 x^2}=\frac{17}{8} \).
Answer: We are given the equation \( \frac{x^4+1}{2x^2} = \frac{17}{8} \). We need to solve for x using properties of proportion.
First, we apply the componendo and dividendo rule to the given equation:
\( \frac{(x^4+1) + (2x^2)}{(x^4+1) - (2x^2)} = \frac{17+8}{17-8} \)
Now, recognize the algebraic identities in the left side's numerator and denominator:
Numerator: \( x^4+2x^2+1 = (x^2+1)^2 \).
Denominator: \( x^4-2x^2+1 = (x^2-1)^2 \).
For the right side:
Numerator: \( 17+8 = 25 \).
Denominator: \( 17-8 = 9 \).
So, the equation transforms into:
\( \frac{(x^2+1)^2}{(x^2-1)^2} = \frac{25}{9} \)
This can be written as \( \left(\frac{x^2+1}{x^2-1}\right)^2 = \left(\frac{5}{3}\right)^2 \).
Taking the square root of both sides, we get:
\( \frac{x^2+1}{x^2-1} = \frac{5}{3} \). (We consider the positive root here for simplicity as the quadratic can lead to real solutions)
Now, apply the componendo and dividendo rule again to this new equation:
\( \frac{(x^2+1) + (x^2-1)}{(x^2+1) - (x^2-1)} = \frac{5+3}{5-3} \)
Simplify the terms:
Left side numerator: \( x^2+1+x^2-1 = 2x^2 \).
Left side denominator: \( x^2+1-x^2+1 = 2 \).
Right side numerator: \( 8 \).
Right side denominator: \( 2 \).
So, the equation becomes:
\( \frac{2x^2}{2} = \frac{8}{2} \)
This simplifies to:
\( x^2 = 4 \)
Taking the square root of both sides, we find the values of x:
\( x = \pm 2 \). These are the two possible values for x.
In simple words: We start with a fraction that includes \( x^4 \). We use a special math rule twice. First, we turn the fraction into squared terms. Then, we take the square root of both sides. We use the rule again, simplify the equation, and find that x can be positive 2 or negative 2.

🎯 Exam Tip: Recognizing expressions as perfect squares, like \( (x^2+1)^2 \) and \( (x^2-1)^2 \), is crucial for simplifying problems involving higher powers. Remember to account for both positive and negative roots when solving for x.

 

Question 19. If \( \frac{x^2+y^2}{x^2-y^2}=\frac{17}{8} \), then find the value of:
(i) x : y
(ii) \( \frac{x^3+y^3}{x^3-y^3} \)
Answer: We are given the equation \( \frac{x^2+y^2}{x^2-y^2} = \frac{17}{8} \).
**(i) Find x : y**
To find the ratio \( x:y \), we apply the componendo and dividendo rule. This rule states that if \( \frac{A}{B} = \frac{C}{D} \), then \( \frac{A+B}{A-B} = \frac{C+D}{C-D} \).
Applying this to our given equation:
\( \frac{(x^2+y^2) + (x^2-y^2)}{(x^2+y^2) - (x^2-y^2)} = \frac{17+8}{17-8} \)
Simplify the numerator and denominator:
Left side numerator: \( x^2+y^2+x^2-y^2 = 2x^2 \).
Left side denominator: \( x^2+y^2-x^2+y^2 = 2y^2 \).
Right side numerator: \( 25 \).
Right side denominator: \( 9 \).
So, the equation simplifies to:
\( \frac{2x^2}{2y^2} = \frac{25}{9} \)
\( \implies \frac{x^2}{y^2} = \frac{25}{9} \)
This can be written as \( \left(\frac{x}{y}\right)^2 = \left(\frac{5}{3}\right)^2 \).
Taking the square root of both sides (assuming x and y are positive for ratios):
\( \frac{x}{y} = \frac{5}{3} \). This is the base ratio.
Therefore, the ratio \( x:y = 5:3 \).
**(ii) Find \( \frac{x^3+y^3}{x^3-y^3} \)**
From part (i), we know that \( \frac{x}{y} = \frac{5}{3} \).
To find the value of \( \frac{x^3+y^3}{x^3-y^3} \), we can divide both the numerator and the denominator by \( y^3 \):
\( \frac{\frac{x^3}{y^3} + \frac{y^3}{y^3}}{\frac{x^3}{y^3} - \frac{y^3}{y^3}} = \frac{\left(\frac{x}{y}\right)^3 + 1}{\left(\frac{x}{y}\right)^3 - 1} \)
Now, substitute the value \( \frac{x}{y} = \frac{5}{3} \) into this expression:
\( = \frac{\left(\frac{5}{3}\right)^3 + 1}{\left(\frac{5}{3}\right)^3 - 1} \)
Calculate \( \left(\frac{5}{3}\right)^3 \): \( \frac{5^3}{3^3} = \frac{125}{27} \).
So, the expression becomes:
\( = \frac{\frac{125}{27} + 1}{\frac{125}{27} - 1} \)
To combine the terms in the numerator and denominator, find a common denominator (27):
Numerator: \( \frac{125}{27} + \frac{27}{27} = \frac{125+27}{27} = \frac{152}{27} \).
Denominator: \( \frac{125}{27} - \frac{27}{27} = \frac{125-27}{27} = \frac{98}{27} \).
Now, we have a fraction of fractions:
\( = \frac{\frac{152}{27}}{\frac{98}{27}} \)
We can cancel out the common denominator \( 27 \):
\( = \frac{152}{98} \)
Finally, simplify the fraction by dividing both the numerator and denominator by their greatest common divisor, which is 2:
\( = \frac{76}{49} \). This calculation relies on the ratio found in part (i).
Thus, \( \frac{x^3+y^3}{x^3-y^3} = \frac{76}{49} \).
In simple words: First, we use a special math rule called componendo and dividendo to find the simple ratio of x to y. We get \( x:y = 5:3 \). Then, to find the second part, we use this ratio, cube both numbers, and plug them into the second fraction. We simplify the numbers to get the final answer.

🎯 Exam Tip: For problems with multiple parts involving ratios, solve for the simplest ratio first (like \( x:y \)). This basic ratio can then be used to find expressions with higher powers by simply applying the power to the ratio itself.

 

Question 20. If a, b, c are in continued proportion, prove that \( (a + b + c) (a - b + c) = a^2 + b^2 + c^2 \).
Answer: We are given that a, b, and c are in continued proportion. This means that \( \frac{a}{b} = \frac{b}{c} \), which implies \( b^2 = ac \). We need to prove that \( (a+b+c)(a-b+c) = a^2+b^2+c^2 \).
Let's start with the Left Hand Side (L.H.S.) of the equation:
L.H.S. \( = (a+b+c)(a-b+c) \)
We can rearrange the terms on the left side to use the algebraic identity \( (X+Y)(X-Y) = X^2-Y^2 \).
Let \( X = (a+c) \) and \( Y = b \). Then the expression becomes:
L.H.S. \( = ((a+c)+b)((a+c)-b) \)
\( = (a+c)^2 - b^2 \)
Now, expand \( (a+c)^2 \):
\( = (a^2+2ac+c^2) - b^2 \). This simplifies the multiplication.
Next, we use the condition that a, b, c are in continued proportion, which implies \( b^2 = ac \). Substitute \( ac \) with \( b^2 \) in the expression:
\( = a^2+2b^2+c^2 - b^2 \)
Combine the \( b^2 \) terms:
\( = a^2 + (2b^2 - b^2) + c^2 \)
\( = a^2+b^2+c^2 \).
This is equal to the Right Hand Side (R.H.S.) of the equation. Thus, the statement is proven. This demonstrates how continued proportion links directly to specific algebraic identities.
In simple words: We know that 'a', 'b', and 'c' are in a special order, meaning \( b^2 \) is equal to \( a \) times \( c \). We want to show that multiplying \( (a+b+c) \) by \( (a-b+c) \) gives \( a^2+b^2+c^2 \). By rearranging the first part, we can use a basic math rule like \( (X+Y)(X-Y) \) and then use the special rule \( b^2=ac \) to get the answer.

🎯 Exam Tip: Always look for opportunities to group terms and apply algebraic identities (like \( (X+Y)(X-Y) = X^2-Y^2 \)) to simplify complex expressions. The property \( b^2=ac \) is fundamental in proofs involving continued proportion.

 

Question 21. Given \( \frac{x^3+12 x}{6 x^2+8}=\frac{y^3+27 y}{9 y^2+27} \). Find x : y.
Answer: We are given the equation \( \frac{x^3+12x}{6x^2+8} = \frac{y^3+27y}{9y^2+27} \). Our goal is to find the ratio \( x:y \).
We can apply a modified version of componendo and dividendo. If \( \frac{A}{B} = \frac{C}{D} \), then \( \frac{A+B}{A-B} = \frac{C+D}{C-D} \). In this case, we need to add and subtract the denominators from the numerators.
Applying this rule to the given equation:
\( \frac{(x^3+12x) + (6x^2+8)}{(x^3+12x) - (6x^2+8)} = \frac{(y^3+27y) + (9y^2+27)}{(y^3+27y) - (9y^2+27)} \)
Rearrange the terms in the numerators and denominators:
Left side numerator: \( x^3+6x^2+12x+8 \). Recognize this as the expansion of \( (x+2)^3 \).
Left side denominator: \( x^3-6x^2+12x-8 \). Recognize this as the expansion of \( (x-2)^3 \).
Right side numerator: \( y^3+9y^2+27y+27 \). Recognize this as the expansion of \( (y+3)^3 \).
Right side denominator: \( y^3-9y^2+27y-27 \). Recognize this as the expansion of \( (y-3)^3 \).
So, the equation transforms into:
\( \frac{(x+2)^3}{(x-2)^3} = \frac{(y+3)^3}{(y-3)^3} \)
This can be written as \( \left(\frac{x+2}{x-2}\right)^3 = \left(\frac{y+3}{y-3}\right)^3 \).
Taking the cube root of both sides, we get:
\( \frac{x+2}{x-2} = \frac{y+3}{y-3} \)
Now, we apply the componendo and dividendo rule again to this simplified equation:
\( \frac{(x+2) + (x-2)}{(x+2) - (x-2)} = \frac{(y+3) + (y-3)}{(y+3) - (y-3)} \)
Simplify the terms:
Left side numerator: \( x+2+x-2 = 2x \).
Left side denominator: \( x+2-x+2 = 4 \).
Right side numerator: \( y+3+y-3 = 2y \).
Right side denominator: \( y+3-y+3 = 6 \).
So, the equation becomes:
\( \frac{2x}{4} = \frac{2y}{6} \)
Simplify both sides:
\( \frac{x}{2} = \frac{y}{3} \)
Therefore, the ratio \( x:y \) is \( 2:3 \). This problem showcases the power of recognizing cubic expansions.
In simple words: We start with a complicated fraction equation. We use a math rule called componendo and dividendo. This helps us change the fractions into cubed forms, like \( (x+2)^3 \). We then take the cube root of both sides. We use the same math rule again to make the equation even simpler, which finally helps us find the ratio of x to y.

🎯 Exam Tip: Be on the lookout for expressions that resemble binomial expansions like \( (A \pm B)^3 \). Recognizing these forms after applying componendo and dividendo is key to solving such problems efficiently.

 

Question 22. If \( \frac { x }{ a } = \frac { y }{ b } = \frac { z }{ c } \) show that \( \frac{x^3}{a^3}+\frac{y^3}{b^3} + \frac{z^3}{c^3}=\frac{3 x y z}{a b c} \).
Answer: We are given that \( \frac{x}{a} = \frac{y}{b} = \frac{z}{c} \). Let's set this common ratio to a constant, k.
So, we have:
\( \frac{x}{a} = k \implies x = ak \)
\( \frac{y}{b} = k \implies y = bk \)
\( \frac{z}{c} = k \implies z = ck \)
We need to show that \( \frac{x^3}{a^3} + \frac{y^3}{b^3} + \frac{z^3}{c^3} = \frac{3xyz}{abc} \).
Let's evaluate the Left Hand Side (L.H.S.) first:
L.H.S. \( = \frac{x^3}{a^3} + \frac{y^3}{b^3} + \frac{z^3}{c^3} \)
Substitute the expressions for x, y, and z in terms of 'k':
\( = \frac{(ak)^3}{a^3} + \frac{(bk)^3}{b^3} + \frac{(ck)^3}{c^3} \)
Expand the cubes:
\( = \frac{a^3k^3}{a^3} + \frac{b^3k^3}{b^3} + \frac{c^3k^3}{c^3} \)
The \( a^3, b^3, \) and \( c^3 \) terms cancel out from their respective fractions:
\( = k^3 + k^3 + k^3 \)
\( = 3k^3 \). This is the simplified form of the L.H.S.
Now, let's evaluate the Right Hand Side (R.H.S.):
R.H.S. \( = \frac{3xyz}{abc} \)
Substitute the expressions for x, y, and z in terms of 'k':
\( = \frac{3(ak)(bk)(ck)}{abc} \)
Multiply the terms in the numerator:
\( = \frac{3abck^3}{abc} \)
The \( abc \) terms in the numerator and denominator cancel out:
\( = 3k^3 \). This is the simplified form of the R.H.S.
Since L.H.S. \( = 3k^3 \) and R.H.S. \( = 3k^3 \), we have L.H.S. = R.H.S.
Thus, the statement is proven. This shows a fundamental property of proportional quantities when raised to a power.
In simple words: We are given that three fractions are all equal to some number, k. This means we can write x, y, and z using 'a', 'b', 'c', and k. When we put these new forms into the left side of the equation and simplify, we get \( 3k^3 \). Doing the same for the right side also gives \( 3k^3 \). Since both sides are the same, the equation is proven.

🎯 Exam Tip: When given a series of equal ratios, always introduce a constant 'k' (or any other letter) for the common ratio. This method allows you to express variables in terms of 'k' and simplify complex expressions effectively.

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