Ganit Prakash - Class-X - Ratio and Proportion
Let us work out 5.3 Proportions & Variables
📘 Exercise 5.3 Solutions
Proportion Laws(i) \((a^2+b^2) : (a^2-b^2) = (ac+bd) : (ac-bd)\)
(ii) \[(a^2+ab+b^2) : (a^2-ab+b^2) = (c^2+cd+d^2) : (c^2-cd+d^2)\]
(iii) \(\sqrt{a^2+c^2} : \sqrt{b^2+d^2} = (pa+qc) : (pb+qd)\)
Let \(\frac{a}{b} = \frac{c}{d} = k\) (where \(k\) is a constant).
\(\therefore a = bk\) and \(c = dk\).
(i)
L.H.S \(= \frac{a^2+b^2}{a^2-b^2} \)
\(= \frac{(bk)^2+b^2}{(bk)^2-b^2} \)
\(= \frac{b^2k^2+b^2}{b^2k^2-b^2} \)
\(= \frac{b^2(k^2+1)}{b^2(k^2-1)} \)
\(= \frac{k^2+1}{k^2-1}\)
R.H.S \(= \frac{ac+bd}{ac-bd} \)
\(= \frac{(bk)(dk)+bd}{(bk)(dk)-bd} \)
\(= \frac{bdk^2+bd}{bdk^2-bd} \)
\(= \frac{bd(k^2+1)}{bd(k^2-1)} \)
\(= \frac{k^2+1}{k^2-1}\)
\(\therefore\) L.H.S = R.H.S [Proved]
(ii)
L.H.S \(= \frac{a^2+ab+b^2}{a^2-ab+b^2} \)
\(= \frac{(bk)^2+(bk)b+b^2}{(bk)^2-(bk)b+b^2} \)
\(= \frac{b^2k^2+b^2k+b^2}{b^2k^2-b^2k+b^2} \)
\(= \frac{b^2(k^2+k+1)}{b^2(k^2-k+1)} \)
\(= \frac{k^2+k+1}{k^2-k+1}\)
R.H.S \(= \frac{c^2+cd+d^2}{c^2-cd+d^2} \)
\(= \frac{(dk)^2+(dk)d+d^2}{(dk)^2-(dk)d+d^2} \)
\(= \frac{d^2k^2+d^2k+d^2}{d^2k^2-d^2k+d^2} \)
\(= \frac{d^2(k^2+k+1)}{d^2(k^2-k+1)} \)
\(= \frac{k^2+k+1}{k^2-k+1}\)
\(\therefore\) L.H.S = R.H.S [Proved]
(iii)
L.H.S \(= \frac{\sqrt{a^2+c^2}}{\sqrt{b^2+d^2}} \)
\(= \frac{\sqrt{(bk)^2+(dk)^2}}{\sqrt{b^2+d^2}} \)
\(= \frac{\sqrt{b^2k^2+d^2k^2}}{\sqrt{b^2+d^2}} \)
\(= \frac{\sqrt{k^2(b^2+d^2)}}{\sqrt{b^2+d^2}} \)
\(= \frac{k\sqrt{b^2+d^2}}{\sqrt{b^2+d^2}} \)
\(= k\)
R.H.S \(= \frac{pa+qc}{pb+qd} \)
\(= \frac{p(bk)+q(dk)}{pb+qd} \)
\(= \frac{pbk+qdk}{pb+qd} \)
\(= \frac{k(pb+qd)}{pb+qd} \)
\(= k\)
\(\therefore\) L.H.S = R.H.S [Proved]
(i) \(\frac{x^3}{a^2} + \frac{y^3}{b^2} + \frac{z^3}{c^2} = \frac{(x+y+z)^3}{(a+b+c)^2}\)
(ii) \(\frac{x^3+y^3+z^3}{a^3+b^3+c^3} = \frac{xyz}{abc}\)
(iii) \[(a^2+b^2+c^2)(x^2+y^2+z^2) = (ax+by+cz)^2\]
Let \(\frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k\) (where \(k\) is a constant).
\(\therefore x = ak\), \(y = bk\) and \(z = ck\).
(i)
L.H.S \(= \frac{x^3}{a^2} + \frac{y^3}{b^2} + \frac{z^3}{c^2}\)
\(= \frac{(ak)^3}{a^2} + \frac{(bk)^3}{b^2} + \frac{(ck)^3}{c^2}\)
\(= \frac{a^3k^3}{a^2} + \frac{b^3k^3}{b^2} + \frac{c^3k^3}{c^2}\)
\(= ak^3 + bk^3 + ck^3\)
\(= k^3(a+b+c)\)
R.H.S \(= \frac{(x+y+z)^3}{(a+b+c)^2}\)
\(= \frac{(ak+bk+ck)^3}{(a+b+c)^2}\)
\(= \frac{[k(a+b+c)]^3}{(a+b+c)^2}\)
\(= \frac{k^3(a+b+c)^3}{(a+b+c)^2}\)
\(= k^3(a+b+c)\)
\(\therefore\) L.H.S = R.H.S [Proved]
(ii)
L.H.S \(= \frac{x^3+y^3+z^3}{a^3+b^3+c^3}\)
\(= \frac{(ak)^3+(bk)^3+(ck)^3}{a^3+b^3+c^3}\)
\(= \frac{a^3k^3+b^3k^3+c^3k^3}{a^3+b^3+c^3}\)
\(= \frac{k^3(a^3+b^3+c^3)}{a^3+b^3+c^3}\)
\(= k^3\)
R.H.S \(= \frac{xyz}{abc}\)
\(= \frac{(ak)(bk)(ck)}{abc}\)
\(= \frac{abck^3}{abc}\)
\(= k^3\)
\(\therefore\) L.H.S = R.H.S [Proved]
(iii)
L.H.S \(= (a^2+b^2+c^2)(x^2+y^2+z^2)\)
\(= (a^2+b^2+c^2)((ak)^2+(bk)^2+(ck)^2)\)
\(= (a^2+b^2+c^2)(a^2k^2+b^2k^2+c^2k^2)\)
\(= (a^2+b^2+c^2)k^2(a^2+b^2+c^2)\)
\(= k^2(a^2+b^2+c^2)^2\)
R.H.S \(= (ax+by+cz)^2 \)
\(= (a(ak)+b(bk)+c(ck))^2 \)
\(= (a^2k+b^2k+c^2k)^2 \)
\(= [k(a^2+b^2+c^2)]^2 \)
\(= k^2(a^2+b^2+c^2)^2\)
\(\therefore\) L.H.S = R.H.S [Proved]
(i) Each ratio = \(\frac{5a-7c-13e}{5b-7d-13f}\)
(ii) \[(a^2+c^2+e^2)(b^2+d^2+f^2) = (ab+cd+ef)^2\]
Let \(\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k\) (where \(k\) is a constant).
\(\therefore a = bk\), \(c = dk\) and \(e = fk\).
(i)
We need to show that \(\frac{5a-7c-13e}{5b-7d-13f}\) is equal to \(k\) (since each ratio = \(k\)).
Expression \(= \frac{5(bk)-7(dk)-13(fk)}{5b-7d-13f} \)
\(= \frac{5bk-7dk-13fk}{5b-7d-13f} \)
\(= \frac{k(5b-7d-13f)}{5b-7d-13f} = k\)
\(\therefore\) Each ratio = \(\frac{5a-7c-13e}{5b-7d-13f}\) [Proved]
(ii)
L.H.S \(= (a^2+c^2+e^2)(b^2+d^2+f^2) \)
\(= ((bk)^2+(dk)^2+(fk)^2)(b^2+d^2+f^2)\)
\(= (b^2k^2+d^2k^2+f^2k^2)(b^2+d^2+f^2) \)
\(= k^2(b^2+d^2+f^2)(b^2+d^2+f^2) \)
\(= k^2(b^2+d^2+f^2)^2\)
R.H.S \(= (ab+cd+ef)^2 \)
\(= ((bk)b+(dk)d+(fk)f)^2 \)
\(= (b^2k+d^2k+f^2k)^2 \)
\(= [k(b^2+d^2+f^2)]^2 \)
\(= k^2(b^2+d^2+f^2)^2\)
\(\therefore\) L.H.S = R.H.S [Proved]
(i) \(\left(\frac{a+b}{b+c}\right)^2 = \frac{a^2+b^2}{b^2+c^2}\)
(ii) \(a^2b^2c^2\left(\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3}\right) = a^3+b^3+c^3\)
(iii) \(\frac{abc(a+b+c)^3}{(ab+bc+ca)^3} = 1\)
Let \(\frac{a}{b} = \frac{b}{c} = k\) (where \(k\) is a constant).
\(\therefore b = ck\)
And, \(a = bk = (ck)k = ck^2\).
(i)
L.H.S \(= \left(\frac{a+b}{b+c}\right)^2\)
\(= \left(\frac{ck^2+ck}{ck+c}\right)^2\)
\(= \left(\frac{ck(k+1)}{c(k+1)}\right)^2\)
\(= (k)^2\)
\(= k^2\)
R.H.S \(= \frac{a^2+b^2}{b^2+c^2}\)
\(= \frac{(ck^2)^2+(ck)^2}{(ck)^2+c^2}\)
\(= \frac{c^2k^4+c^2k^2}{c^2k^2+c^2}\)
\(= \frac{c^2k^2(k^2+1)}{c^2(k^2+1)}\)
\(= k^2\)
\(\therefore\) L.H.S = R.H.S [Proved]
(ii)
L.H.S \(= a^2b^2c^2\left(\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3}\right)\)
\(= (ck^2)^2(ck)^2c^2 \left(\frac{1}{(ck^2)^3} + \frac{1}{(ck)^3} + \frac{1}{c^3}\right)\)
\(= (c^2k^4)(c^2k^2)(c^2) \left(\frac{1}{c^3k^6} + \frac{1}{c^3k^3} + \frac{1}{c^3}\right)\)
\(= c^6k^6 \left(\frac{1}{c^3k^6} + \frac{1}{c^3k^3} + \frac{1}{c^3}\right)\)
\(= \frac{c^6k^6}{c^3k^6} + \frac{c^6k^6}{c^3k^3} + \frac{c^6k^6}{c^3}\)
\(= c^3 + c^3k^3 + c^3k^6\)
R.H.S \(= a^3+b^3+c^3\)
\(= (ck^2)^3+(ck)^3+c^3\)
\(= c^3k^6+c^3k^3+c^3\)
\(= c^3 + c^3k^3 + c^3k^6\)
\(\therefore\) L.H.S = R.H.S [Proved]
(iii)
L.H.S \(= \frac{abc(a+b+c)^3}{(ab+bc+ca)^3}\)
\(= \frac{(ck^2)(ck)(c) \cdot (ck^2+ck+c)^3}{((ck^2)(ck)+(ck)(c)+(c)(ck^2))^3}\)
\(= \frac{c^3k^3 \cdot [c(k^2+k+1)]^3}{(c^2k^3+c^2k+c^2k^2)^3}\)
\(= \frac{c^3k^3 \cdot c^3(k^2+k+1)^3}{[c^2k(k^2+k+1)]^3}\)
\(= \frac{c^6k^3(k^2+k+1)^3}{c^6k^3(k^2+k+1)^3}\)
\(= 1\)
R.H.S \(= 1\)
\(\therefore\) L.H.S = R.H.S [Proved]
(i) \[(a^2+b^2+c^2)(b^2+c^2+d^2) = (ab+bc+cd)^2\]
(ii) \((b-c)^2+(c-a)^2+(b-d)^2 = (a-d)^2\)
Let \(\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k\) (where \(k\) is a constant).
\(\therefore c = dk\)
\(b = ck = (dk)k = dk^2\)
\(a = bk = (dk^2)k = dk^3\)
(i)
L.H.S \(= (a^2+b^2+c^2)(b^2+c^2+d^2)\)
\(= ((dk^3)^2+(dk^2)^2+(dk)^2)((dk^2)^2+(dk)^2+d^2)\)
\(= (d^2k^6+d^2k^4+d^2k^2)(d^2k^4+d^2k^2+d^2)\)
\(= d^2k^2(k^4+k^2+1) \cdot d^2(k^4+k^2+1)\)
\(= d^4k^2(k^4+k^2+1)^2\)
R.H.S \(= (ab+bc+cd)^2\)
\(= ((dk^3)(dk^2) + (dk^2)(dk) + (dk)(d))^2\)
\(= (d^2k^5+d^2k^3+d^2k)^2\)
\(= [d^2k(k^4+k^2+1)]^2\)
\(= d^4k^2(k^4+k^2+1)^2\)
\(\therefore\) L.H.S = R.H.S [Proved]
(ii)
L.H.S \(= (b-c)^2+(c-a)^2+(b-d)^2\)
\(= (b^2-2bc+c^2) + (c^2-2ca+a^2) + (b^2-2bd+d^2)\)
\(= a^2 + 2b^2 + 2c^2 + d^2 - 2bc - 2ca - 2bd\)
From the continued proportion \(\frac{a}{b} = \frac{b}{c} = \frac{c}{d}\), we have:
\(b^2 = ac\), \(c^2 = bd\), and \(bc = ad\).
Substituting \(2b^2 = 2ac\) and \(2c^2 = 2bd\) into the expression:
\(= a^2 + 2ac + 2bd + d^2 - 2bc - 2ac - 2bd\)
\(= a^2 + d^2 - 2bc\)
Now substituting \(bc = ad\):
\(= a^2 + d^2 - 2ad\)
\(= (a-d)^2\)
R.H.S \(= (a-d)^2\)
\(\therefore\) L.H.S = R.H.S [Proved]
(i) If \(\frac{m}{a} = \frac{n}{b}\), let us show that \((m^2+n^2)(a^2+b^2) = (am+bn)^2\)
(ii) If \(\frac{a}{b} = \frac{x}{y}\), let us show that, \((a+b)(a^2+b^2)x^3 = (x+y)(x^2+y^2)a^3\)
(iii) If, \(\frac{x}{\ell m-n^2} = \frac{y}{mn-\ell^2} = \frac{z}{n\ell-m^2}\), let us show that \(\ell x+my+nz = 0\)
(iv) If \(\frac{x}{b+c-a} = \frac{y}{c+a-b} = \frac{z}{a+b-c}\), let us show that \((b-c)x+(c-a)y+(a-b)z = 0\)
(v) If \(\frac{x}{y} = \frac{a+2}{a-2}\), let us show that \(\frac{x^2-y^2}{x^2+y^2} = \frac{4a}{a^2+4}\)
(vi) If \(x = \frac{8ab}{a+b}\), let us write by calculating the value of \(\left(\frac{x+4a}{x-4a} + \frac{x+4b}{x-4b}\right)\)
(i) If \(\frac{m}{a} = \frac{n}{b}\), let us show that \((m^2+n^2)(a^2+b^2) = (am+bn)^2\)
Let \(\frac{m}{a} = \frac{n}{b} = k\) (where \(k\) is a constant).
\(\therefore m = ak\) and \(n = bk\).
L.H.S \(= (m^2+n^2)(a^2+b^2)\)
\(= ((ak)^2+(bk)^2)(a^2+b^2)\)
\(= (a^2k^2+b^2k^2)(a^2+b^2)\)
\(= k^2(a^2+b^2)(a^2+b^2)\)
\(= k^2(a^2+b^2)^2\)
R.H.S \(= (am+bn)^2\)
\(= (a(ak)+b(bk))^2\)
\(= (a^2k+b^2k)^2\)
\(= [k(a^2+b^2)]^2\)
\(= k^2(a^2+b^2)^2\)
\(\therefore\) L.H.S = R.H.S [Proved]
(ii) If \(\frac{a}{b} = \frac{x}{y}\), let us show that, \((a+b)(a^2+b^2)x^3 = (x+y)(x^2+y^2)a^3\)
Given \(\frac{a}{b} = \frac{x}{y}\), this can be rearranged as \(\frac{a}{x} = \frac{b}{y}\).
Let \(\frac{a}{x} = \frac{b}{y} = k\) (where \(k\) is a constant).
\(\therefore a = kx\) and \(b = ky\).
L.H.S \(= (a+b)(a^2+b^2)x^3\)
\(= (kx+ky)((kx)^2+(ky)^2)x^3\)
\(= k(x+y)(k^2x^2+k^2y^2)x^3\)
\(= k(x+y)k^2(x^2+y^2)x^3\)
\(= k^3x^3(x+y)(x^2+y^2)\)
R.H.S \(= (x+y)(x^2+y^2)a^3\)
\(= (x+y)(x^2+y^2)(kx)^3\)
\(= (x+y)(x^2+y^2)k^3x^3\)
\(= k^3x^3(x+y)(x^2+y^2)\)
\(\therefore\) L.H.S = R.H.S [Proved]
(iii) If, \(\frac{x}{\ell m-n^2} = \frac{y}{mn-\ell^2} = \frac{z}{n\ell-m^2}\), let us show that \(\ell x+my+nz = 0\)
Let \(\frac{x}{\ell m-n^2} = \frac{y}{mn-\ell^2} = \frac{z}{n\ell-m^2} = k\) (where \(k\) is a constant).
\(\therefore x = k(\ell m-n^2)\), \(y = k(mn-\ell^2)\) and \(z = k(n\ell-m^2)\).
L.H.S \(= \ell x+my+nz\)
\(= \ell[k(\ell m-n^2)] + m[k(mn-\ell^2)] + n[k(n\ell-m^2)]\)
\(= k(\ell^2m - \ell n^2) + k(m^2n - m\ell^2) + k(n^2\ell - nm^2)\)
\(= k(\ell^2m - \ell n^2 + m^2n - m\ell^2 + n^2\ell - nm^2)\)
\(= k(0)\)
\(= 0\)
R.H.S \(= 0\)
\(\therefore\) L.H.S = R.H.S [Proved]
(iv) If \(\frac{x}{b+c-a} = \frac{y}{c+a-b} = \frac{z}{a+b-c}\), let us show that \((b-c)x+(c-a)y+(a-b)z = 0\)
Let \(\frac{x}{b+c-a} = \frac{y}{c+a-b} = \frac{z}{a+b-c} = k\) (where \(k\) is a constant).
\(\therefore x = k(b+c-a)\), \(y = k(c+a-b)\) and \(z = k(a+b-c)\).
L.H.S \(= (b-c)x+(c-a)y+(a-b)z\)
\(= (b-c)k(b+c-a) + (c-a)k(c+a-b) + (a-b)k(a+b-c)\)
\(= k [ (b-c)(b+c) - a(b-c) + (c-a)(c+a) - b(c-a) + (a-b)(a+b) - c(a-b) ]\)
\(= k [ (b^2-c^2) - ab + ac + (c^2-a^2) - bc + ab + (a^2-b^2) - ac + bc ]\)
\(= k [ b^2 - c^2 - ab + ac + c^2 - a^2 - bc + ab + a^2 - b^2 - ac + bc ]\)
\(= k(0)\)
\(= 0\)
R.H.S \(= 0\)
\(\therefore\) L.H.S = R.H.S [Proved]
(v) If \(\frac{x}{y} = \frac{a+2}{a-2}\), let us show that \(\frac{x^2-y^2}{x^2+y^2} = \frac{4a}{a^2+4}\)
Given \(\frac{x}{y} = \frac{a+2}{a-2}\)
Squaring both sides, we get:
\(\frac{x^2}{y^2} = \frac{(a+2)^2}{(a-2)^2}\)
Applying the dividendo and componendo process, we get:
\(\frac{x^2-y^2}{x^2+y^2} = \frac{(a+2)^2-(a-2)^2}{(a+2)^2+(a-2)^2}\)
Using the algebraic formulas \((A+B)^2-(A-B)^2 = 4AB\) and \((A+B)^2+(A-B)^2 = 2(A^2+B^2)\):
\(\frac{x^2-y^2}{x^2+y^2} = \frac{4 \cdot a \cdot 2}{2(a^2+2^2)}\)
\(= \frac{8a}{2(a^2+4)}\)
\(= \frac{4a}{a^2+4}\)
[Proved]
(vi) If \(x = \frac{8ab}{a+b}\), let us write by calculating the value of \(\left(\frac{x+4a}{x-4a} + \frac{x+4b}{x-4b}\right)\)
Given \(x = \frac{8ab}{a+b}\)
Dividing both sides by \(4a\), we get:
\(\frac{x}{4a} = \frac{2b}{a+b}\)
Applying componendo and dividendo:
\(\frac{x+4a}{x-4a} = \frac{2b+(a+b)}{2b-(a+b)} = \frac{a+3b}{b-a}\)
Again, dividing the given equation \(x = \frac{8ab}{a+b}\) by \(4b\), we get:
\(\frac{x}{4b} = \frac{2a}{a+b}\)
Applying componendo and dividendo:
\(\frac{x+4b}{x-4b} = \frac{2a+(a+b)}{2a-(a+b)} = \frac{3a+b}{a-b} = \frac{-(3a+b)}{b-a}\)
Now, adding the two parts together:
\(\frac{x+4a}{x-4a} + \frac{x+4b}{x-4b} = \frac{a+3b}{b-a} + \frac{-(3a+b)}{b-a}\)
\(= \frac{a+3b-3a-b}{b-a}\)
\(= \frac{2b-2a}{b-a}\)
\(= \frac{2(b-a)}{b-a}\)
\(= 2\)
Answer: 2
(i) If \(\frac{a}{3} = \frac{b}{4} = \frac{c}{7}\), let us show that \(\frac{a+b+c}{c} = 2\)
(ii) If \(\frac{a}{q-r} = \frac{b}{r-p} = \frac{c}{p-q}\), let us show that \(a+b+c = 0 = pa+qb+rc\)
(iii) If \(\frac{ax+by}{a} = \frac{bx-ay}{b}\), let us show that each ratio is equal to \(x\).
(i) If \(\frac{a}{3} = \frac{b}{4} = \frac{c}{7}\), let us show that \(\frac{a+b+c}{c} = 2\)
Let \(\frac{a}{3} = \frac{b}{4} = \frac{c}{7} = k\) (where \(k\) is a constant).
\(\therefore a = 3k\), \(b = 4k\), and \(c = 7k\).
L.H.S \(= \frac{a+b+c}{c}\)
\(= \frac{3k+4k+7k}{7k}\)
\(= \frac{14k}{7k}\)
\(= 2\)
R.H.S \(= 2\)
\(\therefore\) L.H.S = R.H.S [Proved]
(ii) If \(\frac{a}{q-r} = \frac{b}{r-p} = \frac{c}{p-q}\), let us show that \(a+b+c = 0 = pa+qb+rc\)
Let \(\frac{a}{q-r} = \frac{b}{r-p} = \frac{c}{p-q} = k\) (where \(k\) is a constant).
\(\therefore a = k(q-r)\), \(b = k(r-p)\), and \(c = k(p-q)\).
First part: \(a+b+c\)
\(= k(q-r) + k(r-p) + k(p-q)\)
\(= k(q - r + r - p + p - q)\)
\(= k(0)\)
\(= 0\)
Second part: \(pa+qb+rc\)
\(= p[k(q-r)] + q[k(r-p)] + r[k(p-q)]\)
\(= k(pq - pr + qr - pq + rp - rq)\)
\(= k(pq - pq - pr + rp + qr - rq)\)
\(= k(0)\)
\(= 0\)
\(\therefore a+b+c = 0 = pa+qb+rc\) [Proved]
(iii) If \(\frac{ax+by}{a} = \frac{bx-ay}{b}\), let us show that each ratio is equal to \(x\).
Let each ratio be equal to \(k\). Then,
\(\frac{ax+by}{a} = \frac{bx-ay}{b} = k\)
Multiplying the numerator and denominator of the first ratio by \(a\), and the second ratio by \(b\):
\(\frac{a(ax+by)}{a \cdot a} = \frac{b(bx-ay)}{b \cdot b} = k\)
\(\implies \frac{a^2x+aby}{a^2} = \frac{b^2x-aby}{b^2} = k\)
By the property of addendo (summing the numerators and denominators):
\(k = \frac{(a^2x+aby) + (b^2x-aby)}{a^2+b^2}\)
\(k = \frac{a^2x + b^2x + aby - aby}{a^2+b^2}\)
\(k = \frac{x(a^2+b^2)}{a^2+b^2}\)
\(k = x\) (Assuming \(a^2+b^2 \neq 0\))
\(\therefore\) Each ratio is equal to \(x\). [Proved]
(i) If \(\frac{a+b}{b+c} = \frac{c+d}{d+a}\), let us prove that \(c=a\) or \(a+b+c+d = 0\)
(ii) If \(\frac{x}{b+c} = \frac{y}{c+a} = \frac{z}{a+b}\), let us show that \(\frac{a}{y+z-x} = \frac{b}{z+x-y} = \frac{c}{x+y-z}\)
(iii) If \(\frac{x+y}{3a-b} = \frac{y+z}{3b-c} = \frac{z+x}{3c-a}\), let us show that \(\frac{x+y+z}{a+b+c} = \frac{ax+by+cz}{a^2+b^2+c^2}\)
(iv) If \(\frac{x}{a} = \frac{y}{b} = \frac{z}{c}\), let us show that \(\frac{x^2-yz}{a^2-bc} = \frac{y^2-zx}{b^2-ca} = \frac{z^2-xy}{c^2-ab}\)
(i) If \(\frac{a+b}{b+c} = \frac{c+d}{d+a}\), let us prove that \(c=a\) or \(a+b+c+d = 0\)
Given: \(\frac{a+b}{b+c} = \frac{c+d}{d+a}\)
By cross-multiplication, we get:
\((a+b)(d+a) = (b+c)(c+d)\)
\(\implies ad + a^2 + bd + ab = bc + bd + c^2 + cd\)
\(\implies ad + a^2 + ab = bc + c^2 + cd\)
\(\implies a^2 - c^2 + ad - cd + ab - bc = 0\)
\(\implies (a^2 - c^2) + d(a - c) + b(a - c) = 0\)
\(\implies (a-c)(a+c) + d(a-c) + b(a-c) = 0\)
Taking \((a-c)\) common:
\(\implies (a-c)(a+c+d+b) = 0\)
Therefore, either \(a - c = 0 \implies c = a\)
Or \(a + b + c + d = 0\). [Proved]
(ii) If \(\frac{x}{b+c} = \frac{y}{c+a} = \frac{z}{a+b}\), let us show that \(\frac{a}{y+z-x} = \frac{b}{z+x-y} = \frac{c}{x+y-z}\)
Let \(\frac{x}{b+c} = \frac{y}{c+a} = \frac{z}{a+b} = k\) (where \(k\) is a constant).
\(\implies x = k(b+c)\), \(y = k(c+a)\), and \(z = k(a+b)\).
Now, let us find the values of the denominators of the given expression:
\(y+z-x = k(c+a) + k(a+b) - k(b+c) = k(c+a+a+b-b-c) = k(2a) = 2ak\)
\(z+x-y = k(a+b) + k(b+c) - k(c+a) = k(a+b+b+c-c-a) = k(2b) = 2bk\)
\(x+y-z = k(b+c) + k(c+a) - k(a+b) = k(b+c+c+a-a-b) = k(2c) = 2ck\)
Now substituting these back into the ratios:
First ratio \(= \frac{a}{y+z-x} = \frac{a}{2ak} = \frac{1}{2k}\)
Second ratio \(= \frac{b}{z+x-y} = \frac{b}{2bk} = \frac{1}{2k}\)
Third ratio \(= \frac{c}{x+y-z} = \frac{c}{2ck} = \frac{1}{2k}\)
\(\therefore \frac{a}{y+z-x} = \frac{b}{z+x-y} = \frac{c}{x+y-z}\) [Proved]
(iii) If \(\frac{x+y}{3a-b} = \frac{y+z}{3b-c} = \frac{z+x}{3c-a}\), let us show that \(\frac{x+y+z}{a+b+c} = \frac{ax+by+cz}{a^2+b^2+c^2}\)
Let \(\frac{x+y}{3a-b} = \frac{y+z}{3b-c} = \frac{z+x}{3c-a} = k\) (where \(k\) is a constant).
\(\implies x+y = k(3a-b)\) --- (1)
\(\implies y+z = k(3b-c)\) --- (2)
\(\implies z+x = k(3c-a)\) --- (3)
Adding equations (1), (2), and (3):
\(2(x+y+z) = k(3a-b+3b-c+3c-a)\)
\(2(x+y+z) = k(2a+2b+2c)\)
\(2(x+y+z) = 2k(a+b+c)\)
\(\implies x+y+z = k(a+b+c)\) --- (4)
From this, L.H.S \(= \frac{x+y+z}{a+b+c} = \frac{k(a+b+c)}{a+b+c} = k\).
Now, let's find \(x, y, z\) using equation (4):
\(x = (x+y+z) - (y+z) = k(a+b+c) - k(3b-c) = k(a+b+c-3b+c) = k(a-2b+2c)\)
\(y = (x+y+z) - (z+x) = k(a+b+c) - k(3c-a) = k(a+b+c-3c+a) = k(2a+b-2c)\)
\(z = (x+y+z) - (x+y) = k(a+b+c) - k(3a-b) = k(a+b+c-3a+b) = k(-2a+2b+c)\)
Substitute these into the numerator of R.H.S:
\(ax+by+cz = a[k(a-2b+2c)] + b[k(2a+b-2c)] + c[k(-2a+2b+c)]\)
\(= k[a^2 - 2ab + 2ac + 2ab + b^2 - 2bc - 2ac + 2bc + c^2]\)
\(= k(a^2 + b^2 + c^2)\)
R.H.S \(= \frac{ax+by+cz}{a^2+b^2+c^2} = \frac{k(a^2+b^2+c^2)}{a^2+b^2+c^2} = k\).
\(\therefore\) L.H.S = R.H.S [Proved]
(iv) If \(\frac{x}{a} = \frac{y}{b} = \frac{z}{c}\), let us show that \(\frac{x^2-yz}{a^2-bc} = \frac{y^2-zx}{b^2-ca} = \frac{z^2-xy}{c^2-ab}\)
Let \(\frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k\) (where \(k\) is a constant).
\(\implies x = ak\), \(y = bk\), and \(z = ck\).
Now calculate each ratio:
First ratio \(= \frac{x^2-yz}{a^2-bc} = \frac{(ak)^2 - (bk)(ck)}{a^2-bc} = \frac{a^2k^2 - bck^2}{a^2-bc} = \frac{k^2(a^2-bc)}{a^2-bc} = k^2\)
Second ratio \(= \frac{y^2-zx}{b^2-ca} = \frac{(bk)^2 - (ck)(ak)}{b^2-ca} = \frac{b^2k^2 - cak^2}{b^2-ca} = \frac{k^2(b^2-ca)}{b^2-ca} = k^2\)
Third ratio \(= \frac{z^2-xy}{c^2-ab} = \frac{(ck)^2 - (ak)(bk)}{c^2-ab} = \frac{c^2k^2 - abk^2}{c^2-ab} = \frac{k^2(c^2-ab)}{c^2-ab} = k^2\)
Since all three expressions equal \(k^2\):
\(\therefore \frac{x^2-yz}{a^2-bc} = \frac{y^2-zx}{b^2-ca} = \frac{z^2-xy}{c^2-ab}\) [Proved]
(i) If \(\frac{3x+4y}{3u+4v} = \frac{3x-4y}{3u-4v}\), let us show that \(\frac{x}{y} = \frac{u}{v}\)
(ii) If \[(a+b+c+d) : (a+b-c-d) = (a-b+c-d) : (a-b-c+d)\], let us prove that, \(a:b = c:d\)
(i) If \(\frac{3x+4y}{3u+4v} = \frac{3x-4y}{3u-4v}\), let us show that \(\frac{x}{y} = \frac{u}{v}\)
Given: \(\frac{3x+4y}{3u+4v} = \frac{3x-4y}{3u-4v}\)
By alternendo process, interchanging the means:
\(\frac{3x+4y}{3x-4y} = \frac{3u+4v}{3u-4v}\)
Now, applying componendo and dividendo process:
\(\frac{(3x+4y) + (3x-4y)}{(3x+4y) - (3x-4y)} = \frac{(3u+4v) + (3u-4v)}{(3u+4v) - (3u-4v)}\)
\(\implies \frac{3x+4y+3x-4y}{3x+4y-3x+4y} = \frac{3u+4v+3u-4v}{3u+4v-3u+4v}\)
\(\implies \frac{6x}{8y} = \frac{6u}{8v}\)
Multiplying both sides by \(\frac{8}{6}\), we get:
\(\frac{x}{y} = \frac{u}{v}\) [Proved]
(ii) If \((a+b+c+d) : (a+b-c-d) = (a-b+c-d) : (a-b-c+d)\), let us prove that, \(a:b = c:d\)
Given: \(\frac{a+b+c+d}{a+b-c-d} = \frac{a-b+c-d}{a-b-c+d}\)
Applying componendo and dividendo process:
\(\frac{(a+b+c+d) + (a+b-c-d)}{(a+b+c+d) - (a+b-c-d)} = \frac{(a-b+c-d) + (a-b-c+d)}{(a-b+c-d) - (a-b-c+d)}\)
\(\implies \frac{2(a+b)}{2(c+d)} = \frac{2(a-b)}{2(c-d)}\)
\(\implies \frac{a+b}{c+d} = \frac{a-b}{c-d}\)
By alternendo process:
\(\frac{a+b}{a-b} = \frac{c+d}{c-d}\)
Applying componendo and dividendo process again:
\(\frac{(a+b) + (a-b)}{(a+b) - (a-b)} = \frac{(c+d) + (c-d)}{(c+d) - (c-d)}\)
\(\implies \frac{2a}{2b} = \frac{2c}{2d}\)
\(\implies \frac{a}{b} = \frac{c}{d}\)
\(\therefore a:b = c:d\) [Proved]
(i) If \(\frac{a^2}{b+c} = \frac{b^2}{c+a} = \frac{c^2}{a+b} = 1\), let us show that \(\frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c} = 1\)
(ii) If \[x^2:(by+cz) = y^2:(cz+ax) = z^2:(ax+by) = 1\], let us show that, \(\frac{a}{a+x} + \frac{b}{b+y} + \frac{c}{c+z} = 1\)
(i) If \(\frac{a^2}{b+c} = \frac{b^2}{c+a} = \frac{c^2}{a+b} = 1\), let us show that \(\frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c} = 1\)
Given: \(\frac{a^2}{b+c} = 1 \implies a^2 = b+c\)
\(\frac{b^2}{c+a} = 1 \implies b^2 = c+a\)
\(\frac{c^2}{a+b} = 1 \implies c^2 = a+b\)
L.H.S \(= \frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c}\)
Multiplying the numerator and denominator of the first term by \(a\), second term by \(b\), and third term by \(c\):
\(= \frac{a}{a(1+a)} + \frac{b}{b(1+b)} + \frac{c}{c(1+c)}\)
\(= \frac{a}{a+a^2} + \frac{b}{b+b^2} + \frac{c}{c+c^2}\)
Substituting the values of \(a^2, b^2, c^2\):
\(= \frac{a}{a+b+c} + \frac{b}{b+c+a} + \frac{c}{c+a+b}\)
\(= \frac{a}{a+b+c} + \frac{b}{a+b+c} + \frac{c}{a+b+c}\)
\(= \frac{a+b+c}{a+b+c}\)
\(= 1\)
R.H.S \(= 1\)
\(\therefore\) L.H.S = R.H.S [Proved]
(ii) If \(x^2:(by+cz) = y^2:(cz+ax) = z^2:(ax+by) = 1\), let us show that, \(\frac{a}{a+x} + \frac{b}{b+y} + \frac{c}{c+z} = 1\)
Given: \(\frac{x^2}{by+cz} = 1 \implies x^2 = by+cz\)
\(\frac{y^2}{cz+ax} = 1 \implies y^2 = cz+ax\)
\(\frac{z^2}{ax+by} = 1 \implies z^2 = ax+by\)
L.H.S \(= \frac{a}{a+x} + \frac{b}{b+y} + \frac{c}{c+z}\)
Multiplying the numerator and denominator of the first term by \(x\), second term by \(y\), and third term by \(z\):
\(= \frac{ax}{x(a+x)} + \frac{by}{y(b+y)} + \frac{cz}{z(c+z)}\)
\(= \frac{ax}{ax+x^2} + \frac{by}{by+y^2} + \frac{cz}{cz+z^2}\)
Substituting the values of \(x^2, y^2, z^2\):
\(= \frac{ax}{ax+by+cz} + \frac{by}{by+cz+ax} + \frac{cz}{cz+ax+by}\)
\(= \frac{ax}{ax+by+cz} + \frac{by}{ax+by+cz} + \frac{cz}{ax+by+cz}\)
\(= \frac{ax+by+cz}{ax+by+cz}\)
\(= 1\)
R.H.S \(= 1\)
\(\therefore\) L.H.S = R.H.S [Proved]
(i) If \(\frac{x}{xa+yb+zc} = \frac{y}{ya+zb+xc} = \frac{z}{za+xb+yc}\) and \(x+y+z \neq 0\), let us show that each ratio is equal to \(\frac{1}{a+b+c}\)
(ii) If \(\frac{x^2-yz}{a} = \frac{y^2-zx}{b} = \frac{z^2-xy}{c}\), let us prove that \((a+b+c)(x+y+z) = ax+by+cz\)
(iii) If \(\frac{a}{y+z} = \frac{b}{z+x} = \frac{c}{x+y}\), let us prove that \(\frac{a(b-c)}{y^2-z^2} = \frac{b(c-a)}{z^2-x^2} = \frac{c(a-b)}{x^2-y^2}\)
(i) If \(\frac{x}{xa+yb+zc} = \frac{y}{ya+zb+xc} = \frac{z}{za+xb+yc}\) and \(x+y+z \neq 0\), let us show that each ratio is equal to \(\frac{1}{a+b+c}\)
Let each ratio be \(k\). So,
\(k = \frac{x}{xa+yb+zc} = \frac{y}{ya+zb+xc} = \frac{z}{za+xb+yc}\)
Applying the addendo property (sum of numerators / sum of denominators):
\(k = \frac{x+y+z}{(xa+yb+zc) + (ya+zb+xc) + (za+xb+yc)}\)
\(k = \frac{x+y+z}{x(a+b+c) + y(a+b+c) + z(a+b+c)}\)
\(k = \frac{x+y+z}{(a+b+c)(x+y+z)}\)
Since \(x+y+z \neq 0\), we can cancel \((x+y+z)\) from the numerator and denominator:
\(k = \frac{1}{a+b+c}\)
\(\therefore\) Each ratio is equal to \(\frac{1}{a+b+c}\). [Proved]
(ii) If \(\frac{x^2-yz}{a} = \frac{y^2-zx}{b} = \frac{z^2-xy}{c}\), let us prove that \((a+b+c)(x+y+z) = ax+by+cz\)
Let \(\frac{x^2-yz}{a} = \frac{y^2-zx}{b} = \frac{z^2-xy}{c} = k\).
\(\implies a = \frac{x^2-yz}{k}\), \(b = \frac{y^2-zx}{k}\), and \(c = \frac{z^2-xy}{k}\).
L.H.S \(= (a+b+c)(x+y+z)\)
\(= \left( \frac{x^2-yz}{k} + \frac{y^2-zx}{k} + \frac{z^2-xy}{k} \right) (x+y+z)\)
\(= \frac{1}{k}(x^2+y^2+z^2-xy-yz-zx)(x+y+z)\)
Using the algebraic identity \((x+y+z)(x^2+y^2+z^2-xy-yz-zx) = x^3+y^3+z^3-3xyz\):
\(= \frac{1}{k}(x^3+y^3+z^3-3xyz)\)
R.H.S \(= ax+by+cz\)
\(= \left(\frac{x^2-yz}{k}\right)x + \left(\frac{y^2-zx}{k}\right)y + \left(\frac{z^2-xy}{k}\right)z\)
\(= \frac{1}{k}(x^3-xyz + y^3-xyz + z^3-xyz)\)
\(= \frac{1}{k}(x^3+y^3+z^3-3xyz)\)
\(\therefore\) L.H.S = R.H.S [Proved]
(iii) If \(\frac{a}{y+z} = \frac{b}{z+x} = \frac{c}{x+y}\), let us prove that \(\frac{a(b-c)}{y^2-z^2} = \frac{b(c-a)}{z^2-x^2} = \frac{c(a-b)}{x^2-y^2}\)
Let \(\frac{a}{y+z} = \frac{b}{z+x} = \frac{c}{x+y} = k\).
\(\implies a = k(y+z)\), \(b = k(z+x)\), and \(c = k(x+y)\).
Evaluating the first term:
\(\frac{a(b-c)}{y^2-z^2} = \frac{k(y+z) [k(z+x) - k(x+y)]}{y^2-z^2}\)
\(= \frac{k^2(y+z)(z+x-x-y)}{(y+z)(y-z)}\)
\(= \frac{k^2(y+z)(z-y)}{(y+z)(y-z)}\)
\(= \frac{-k^2(y+z)(y-z)}{(y+z)(y-z)} = -k^2\)
Evaluating the second term:
\(\frac{b(c-a)}{z^2-x^2} = \frac{k(z+x) [k(x+y) - k(y+z)]}{z^2-x^2}\)
\(= \frac{k^2(z+x)(x+y-y-z)}{(z+x)(z-x)}\)
\(= \frac{k^2(z+x)(x-z)}{(z+x)(z-x)}\)
\(= \frac{-k^2(z+x)(z-x)}{(z+x)(z-x)} = -k^2\)
Evaluating the third term:
\(\frac{c(a-b)}{x^2-y^2} = \frac{k(x+y) [k(y+z) - k(z+x)]}{x^2-y^2}\)
\(= \frac{k^2(x+y)(y+z-z-x)}{(x+y)(x-y)}\)
\(= \frac{k^2(x+y)(y-x)}{(x+y)(x-y)}\)
\(= \frac{-k^2(x+y)(x-y)}{(x+y)(x-y)} = -k^2\)
Since all three expressions evaluate to \(-k^2\):
\(\therefore \frac{a(b-c)}{y^2-z^2} = \frac{b(c-a)}{z^2-x^2} = \frac{c(a-b)}{x^2-y^2}\) [Proved]
(i) The fourth proportional of 3, 4 and 6 is
(a) 8
(b) 10
(c) 12
(d) 24
(ii) The 3rd proportional of 8 and 12 is
(a) 12
(b) 16
(c) 18
(d) 20
(iii) The mean proportional of 16 and 25 is
(a) 400
(b) 100
(c) 20
(d) 40
(iv) a is a positive number and if \(a : \frac{27}{64} = \frac{3}{4} : a\), then the value of a is
(a) \(\frac{81}{256}\)
(b) 9
(c) \(\frac{9}{16}\)
(d) \(\frac{16}{9}\)
(v) If \(2a = 3b = 4c\), then \(a : b : c\) is
(a) 3:4:6
(b) 4:3:6
(c) 3:6:4
(d) 6:4:3
(i) Let the fourth proportional be \(x\).
\(3:4 :: 6:x \implies \frac{3}{4} = \frac{6}{x}\)
\(\implies 3x = 24 \implies x = 8\)
Correct Option: (a) 8
(ii) Let the 3rd proportional be \(x\).
\(8:12 :: 12:x \implies \frac{8}{12} = \frac{12}{x}\)
\(\implies 8x = 144 \implies x = \frac{144}{8} = 18\)
Correct Option: (c) 18
(iii) Let the mean proportional be \(x\).
\(16:x :: x:25 \implies x^2 = 16 \times 25\)
\(\implies x^2 = 400 \implies x = \sqrt{400} = 20\)
Correct Option: (c) 20
(iv) Given: \(a : \frac{27}{64} = \frac{3}{4} : a\)
\(\implies \frac{a}{\frac{27}{64}} = \frac{\frac{3}{4}}{a}\)
\(\implies a \times a = \frac{27}{64} \times \frac{3}{4}\)
\(\implies a^2 = \frac{81}{256}\)
\(\implies a = \sqrt{\frac{81}{256}} = \frac{9}{16}\) (since \(a\) is a positive number)
Correct Option: (c) \(\frac{9}{16}\)
(v) Given: \(2a = 3b = 4c\)
Let \(2a = 3b = 4c = k\)
\(\therefore a = \frac{k}{2}\), \(b = \frac{k}{3}\), \(c = \frac{k}{4}\)
Now, \(a : b : c = \frac{k}{2} : \frac{k}{3} : \frac{k}{4}\)
Dividing by \(k\) and multiplying by the LCM of 2, 3, and 4 (which is 12):
\(= \left(\frac{1}{2} \times 12\right) : \left(\frac{1}{3} \times 12\right) : \left(\frac{1}{4} \times 12\right)\)
\(= 6 : 4 : 3\)
Correct Option: (d) 6:4:3
(i) The compound ratio of \(ab:c^2\), \(bc:a^2\) and \(ca:b^2\) is 1:1.
(ii) \(x^3y\), \(x^2y^2\) and \(xy^3\) are in continued proportion.
(i) The given ratios are \(ab:c^2\), \(bc:a^2\) and \(ca:b^2\).
Compound ratio = Product of antecedents : Product of consequents
\(= (ab \times bc \times ca) : (c^2 \times a^2 \times b^2)\)
\(= a^2b^2c^2 : a^2b^2c^2\)
\(= 1 : 1\)
Answer: True
(ii) The given terms are \(x^3y\), \(x^2y^2\) and \(xy^3\).
For three terms \(A, B, C\) to be in continued proportion, \(B^2\) must equal \(A \times C\).
Middle term squared = \((x^2y^2)^2 = x^4y^4\)
Product of extreme terms = \((x^3y) \times (xy^3) = x^4y^4\)
Since \((x^2y^2)^2 = (x^3y) \times (xy^3)\), they are in continued proportion.
Answer: True
(i) If the product of three positive continued proportional numbers is 64, then their mean proportional is _______ .
(ii) If \(a:2 = b:5 = c:8\), then 50% of a = 20% of b = _______ % of c.
(iii) If the mean proportional of \((x-2)\) and \((x-3)\) is \(x\), then the value of \(x\) is _______ .
(i) Let the three positive continued proportional numbers be \(a, b\), and \(c\).
Then \(\frac{a}{b} = \frac{b}{c} \implies b^2 = ac\).
Given, their product is 64. So, \(a \times b \times c = 64\).
\(\implies b \times (ac) = 64\)
\(\implies b \times b^2 = 64\)
\(\implies b^3 = 64 = 4^3\)
\(\implies b = 4\).
Answer: 4
(ii) Given \(\frac{a}{2} = \frac{b}{5} = \frac{c}{8} = k\) (where \(k\) is a constant).
\(\therefore a = 2k, b = 5k, c = 8k\).
50% of \(a\) = \(\frac{50}{100} \times 2k = k\).
20% of \(b\) = \(\frac{20}{100} \times 5k = k\).
Let \(y\%\) of \(c = k\).
\(\implies \frac{y}{100} \times 8k = k\)
\(\implies \frac{8y}{100} = 1 \implies 8y = 100 \implies y = \frac{100}{8} = \frac{25}{2} = 12\frac{1}{2}\)
Answer: \(12\frac{1}{2}\)
(iii) Given that the mean proportional of \((x-2)\) and \((x-3)\) is \(x\).
Therefore, \((x-2) : x :: x : (x-3)\)
\(\implies x^2 = (x-2)(x-3)\)
\(\implies x^2 = x^2 - 3x - 2x + 6\)
\(\implies x^2 = x^2 - 5x + 6\)
\(\implies 5x = 6 \implies x = \frac{6}{5} = 1.2\)
Answer: \(\frac{6}{5}\)
(i) If \(\frac{a}{2} = \frac{b}{3} = \frac{c}{4} = \frac{2a-3b+4c}{p}\), let us find the value of \(p\).
(ii) If \(\frac{3x - 5y}{3x + 5y} = \frac{1}{2}\), let us find the value of \(\frac{3x^2 - 5y^2}{3x^2 + 5y^2}\)
(iii) If \(a:b = 3:4\) and \(x:y = 5:7\), let us find the value of \((3ax - by) : (4by - 7ax)\)
(iv) If \(x, 12, y, 27\) are in continued proportion, let us find the positive value of \(x\) and \(y\).
(v) If \(a:b = 3:2\) and \(b:c = 3:2\), let us find the value of \((a+b):(b+c)\)
(i) Let \(\frac{a}{2} = \frac{b}{3} = \frac{c}{4} = k\) (where \(k\) is a constant).
\(\implies a = 2k\), \(b = 3k\), and \(c = 4k\).
Given \(\frac{2a-3b+4c}{p} = k\)
\(\implies \frac{2(2k) - 3(3k) + 4(4k)}{p} = k\)
\(\implies \frac{4k - 9k + 16k}{p} = k\)
\(\implies \frac{11k}{p} = k\)
\(\implies p = \frac{11k}{k} = 11\).
Answer: 11
(ii) Given \(\frac{3x - 5y}{3x + 5y} = \frac{1}{2}\)
By cross-multiplication:
\(2(3x - 5y) = 1(3x + 5y)\)
\(\implies 6x - 10y = 3x + 5y\)
\(\implies 6x - 3x = 5y + 10y\)
\(\implies 3x = 15y\)
\(\implies x = 5y \implies \frac{x}{y} = \frac{5}{1}\)
Let \(x = 5k\) and \(y = k\).
Now, \(\frac{3x^2 - 5y^2}{3x^2 + 5y^2} = \frac{3(5k)^2 - 5(k)^2}{3(5k)^2 + 5(k)^2}\)
\(= \frac{3(25k^2) - 5k^2}{3(25k^2) + 5k^2}\)
\(= \frac{75k^2 - 5k^2}{75k^2 + 5k^2}\)
\(= \frac{70k^2}{80k^2} = \frac{7}{8}\)
Answer: \(\frac{7}{8}\)
(iii) Given \(\frac{a}{b} = \frac{3}{4} \implies a = 3k\) and \(b = 4k\).
Given \(\frac{x}{y} = \frac{5}{7} \implies x = 5m\) and \(y = 7m\).
Now, substitute these into the ratio \(\frac{3ax - by}{4by - 7ax}\):
\(= \frac{3(3k)(5m) - (4k)(7m)}{4(4k)(7m) - 7(3k)(5m)}\)
\(= \frac{45km - 28km}{112km - 105km}\)
\(= \frac{17km}{7km}\)
\(= \frac{17}{7}\)
Answer: 17:7
(iv) Since \(x, 12, y, 27\) are in continued proportion, we have:
\(\frac{x}{12} = \frac{12}{y} = \frac{y}{27}\)
From the last two parts: \(\frac{12}{y} = \frac{y}{27}\)
\(\implies y^2 = 12 \times 27\)
\(\implies y^2 = 324\)
\(\implies y = 18\) (taking the positive value)
Now, from the first two parts: \(\frac{x}{12} = \frac{12}{y}\)
\(\implies \frac{x}{12} = \frac{12}{18}\)
\(\implies 18x = 144\)
\(\implies x = \frac{144}{18} = 8\)
Answer: \(x = 8, y = 18\)
(v) To find \(a:b:c\), we make \(b\) the same in both ratios.
Given \(a:b = 3:2 = (3 \times 3):(2 \times 3) = 9:6\)
Given \(b:c = 3:2 = (3 \times 2):(2 \times 2) = 6:4\)
\(\therefore a:b:c = 9:6:4\)
Let \(a = 9k\), \(b = 6k\), and \(c = 4k\).
Now, \(\frac{a+b}{b+c} = \frac{9k+6k}{6k+4k}\)
\(= \frac{15k}{10k}\)
\(= \frac{15}{10} = \frac{3}{2}\)
Answer: 3:2
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