S Chand CLASS 10 Chapter 6 Ration and proportion Exercise 6C

  Exercise 6C

Question 1

Ans:

Given,

$\begin{aligned}\frac{x}{a} &=\frac{y}{b}=\frac{z}{c} \\\text { let } \frac{x}{a}=\frac{y}{b} &=\frac{z}{c}=k .\end{aligned}$

$\therefore x=a k, y=b k$ and $z=c k$

(i) $\frac{x^{3}}{a^{3}}-\frac{y^{3}}{b^{3}}+\frac{z^{3}}{c^{3}}=\frac{x y z}{a b c}$

$\frac{(a k)^{3}}{a^{3}}$ - $\frac{(b k)^{3}}{b^{3}}$$+\frac{(c k)^{3}}{c^{3}}$

$\frac{a k\times b k\times ck}{a b c}$

$\frac{a^{3} k^{3}}{a^{2}}-\frac{b^{3} k^{3}}{ b^{3}}+\frac{c^{3} k^{3}}{c^{3}}= k^{3}$

$k^{3}-k^{2}+ k^{3}=k^{3}$

$k^{3}=k^{3}$

Hence, proved: L.H.S=R.H.S 

(ii)  $\left(\frac{a^{2} x^{2}+b^{2} y^{2}+c^{2} z^{2}}{a^{3} x+b^{3} y+c^{3} z}\right)^{\frac{3}{2}}=\sqrt{\frac{x y z}{a b c}}$

$\left(\frac{a^{2} \times a^{2} x^{2}+b^{2} \times b^{2} k^{2}+c^{2} \times c^{2} k^{2}}{a^{3} \times a k+b^{3} \times b k+c^{3}+c k}\right)^{\frac{3}{2}}=\sqrt{\frac{a k \times b k \times c k}{a b c}}$

$\left(\frac{a^{4} k^{2}+b^{4} k^{2}+c^{4} k^{2}}{a^{4} k+b^{4} k+c^{4} k}\right)^{\frac{3}{2}}=$ $\sqrt{\frac{a b c k^{3}}{a b c}}$

$\left(\frac{k^{2}\left(a^{4}+b^{4}+c^{4}\right)}{k\left(a^{4}+b^{4}+c^{4}\right)}\right)^{\frac{3}{2}}$  $=\sqrt{K^{3}}$

$(K)^{\frac{3}{2}}=\left(K^{3}\right)^{\frac{1}{2}}$

$K^{\frac{3}{2}}=K^{\frac{3}{2}} .$ 

Hence prove : L.H.S =R.H.S

(iii)

$\begin{aligned}&\frac{x z+a c}{x z-a c}=\frac{y z+b c}{y z-b c} \\&\frac{a k \times c k+a . c}{a k \times c k-a c}=\frac{b k \times c k+b c}{b k \times c k-b c} \\&\frac{a c k^{2}+a c}{a c k^{2}-a c}=\frac{b c k^{2}+b c}{b c k^{2}-b c}\end{aligned}$

$\frac{a c\left(k^{2}+1\right)}{a c\left(k^{2}-1\right)}=\frac{b c\left(k^{2}+1\right)}{b c\left(k^{2}-1\right)}$

$\frac{k^{2}+1}{k^{2}-1}=\frac{k^{2}+1}{k^{2}-1}$

L.H.S =R.H.S 

Hence proved

(iv) $a b\left(\frac{x+a}{a}+\frac{y+b}{b}+\frac{2+c}{c}\right)^{3}=27(x+a)(y+b)(2+c)$ $a b\left(\left(\frac{a k+a}{a}+\frac{b k+b}{b}+\frac{c k+c}{c}\right)^{3}=27(a k+a)(b k+b)(c k+c)\right.$

$a b c\left(\frac{a(k+1)}{a}+\frac{b(k+1)}{b}+\frac{c(k+1)}{c}\right)^{3}=$ $27 a(k+1) \cdot b(k+1)-c(k+1)$

$a b c(k+1+k+1+k+1)^{3}$ =$27 a b c(k+1)^{3}$

$a b c(3 k+3)^{3}=27 a b c(k+1)^{3}$

$a b c \cdot(3)^{3}(k+1)^{3}=27 a b c(k+1)^{3}$

$27 a b c(k+1)^{3}=27 a b c(k+1)^{3}$

L.H.S = R.H.S 

Hence proved 

(iv)$\left(\frac{3 x^{3}+5 y^{3}+7 z^{3}}{3 a^{3}+5 b^{3}+7 c^{3}}\right)^{\frac{1}{3}}$

$\left(\frac{3 \times a^{3} k^{3}+5 \times  b^{3} \times  k^{3}+7 \times c^{3} k^{3}}{3 a^{3}+5 b^{3}+7 c^{3}}\right)^{\frac{1}{3}}$

$\left(\frac{3 a^{3} k^{3}+5 b^{3} k^{3}+7 c^{3} k^{3}}{3 a^{3}+5 b^{3}+7 c^{3}}\right)^{\frac{1}{3}}$

$\left(\frac{k^{3}\left(3 a^{3}+5 b^{3}+7 b^{3}\right)}{\left(3 a^{3}+5 b^{3}+7 c^{3}\right)}\right)^{\frac{1}{3}}$

=$\left(k^{3}\right)^{\frac{1}{3}}$

=$k^{\left(3 \times \frac{1}{3}\right)}$

Hence proved 

Question 2

Ans:

Given,

$\text { let } \begin{aligned}\frac{a}{b}=\frac{c}{d} &=\frac{e}{f} \\\frac{a}{b}=\frac{c}{d} &=\frac{e}{f}=k\end{aligned}$

$\therefore a=b k, c=d k$ and $e=f k $

(i) $\frac{p a^{3}+q c^{3}+r c^{3}}{p b^{3}+q d^{3}+r f^{3}}=\frac{a c c}{b d f}$

$\frac{p b^{3} k^{3}+q d^{3} k^{3}+r j^{3} k^{3}}{p b^{3}+q d^{3}+rf^{3}}=$

$\frac{k^{3}\left(p b^{3}+q d^{3}+r f^{3}\right)}{\left(p b^{3}+q d^{3}+rf^{3}\right)}$

$k^{3}=k^{3} \text {. }$

Hence, prove.

(ii)$\sqrt{\frac{\dot{a}^{4}+c^{4}}{b^{4}+d^{4}}}=\frac{p a^{2}+q c^{2}}{p b^{2}+q d^{2}}$

$\sqrt{\frac{b^{4} k^{4}+d^{4} k^{4}}{b^{4}+d^{4}}}=\frac{p b^{2} k^{2}+q d^{2} k^{2}}{p b^{2}+q d^{2}}$

$\sqrt{\frac{k^{4}\left(b^{4}+d ^{4}\right)}{b^{4}+d^{4}}}=\frac{k^{2}\left(p b^{2}+q d^{2}\right)}{p b^{2}+q d^{2}}$

$\begin{aligned}&\sqrt{k^{4}}=k^{2} \\&\sqrt{k^{2} \times  k^{2}}=k^{2}

\\&k^{2}=k^{2}\end{aligned}$

Hence, proved

(iii) $\frac{2 a^{4} b^{2}+3 a^{2} c^{2}-5 c^{4}f}{2 b^{6}+3 b^{2} f^{2}-5f^{5}}$ $=\frac{a^{4}}{b^{4}}$

$\frac{2 b^{4} k^{4} b^{2}+3 b^{2} k^{2} f^{2} k^{2}-5 f^{4} k^{4} f}{2 b^{6}+3 b^{2} f^{2}-5 f^{5}}=$ $\frac{b^{4} k^{4}}{b^{4}}$

$\frac{2 b^{6} k^{4}+3 b^{2} f^{2} k^{4}-5 f^{5} k^{4}}{2 b^{6}+3 b^{2} f^{2}-5 f^{5}}=k^{4}$

$k^{4} \frac{\left(2 b^{6}+3 b^{2} f^{2}-5 f^{5}\right)}{2 b^{6}+3 b^{2} f^{2}-5 f^{5}}=k^{4}$

$ k^{4}=k^{4} .$

Hence, proved.

Question 3

Ans:

Given 

a, b, c are in continued proportion 

$\therefore \quad a: b:: b: c .$

$\frac{a}{b}=\frac{b}{c} .$

Let $\frac{a}{b}=\frac{b}{c}=k$

$b=c k, a=b k=c k-k=c k^{2}$

(i) $(a+b+c)(a-b+c)=a^{2}+b^{2}+c^{2} .$

$\left(c k^{2}+c k+c\right)\left(c k^{2}-ck+c)=\left(ck^{2}\right)^{2}+(c k)^{2}+c^{2}\right.$

$c\left(k^{2}+k+1\right) c\left(k^{2}-k+1\right)=c^{2} k^{4}+c^{2} k^{2}+c^{2}$

$c^{2}\left(k^{4}+k^{2}+1\right)=c^{2}\left(k^{4}+k^{2}+1\right)$

Hence proved 

(ii) $\frac{a^{2}+b^{2}}{b^{2}+c^{2}}=\frac{a}{c}$

$\frac{\left(c k^{2}\right)^{2}+(c k)^{2}}{(c k)^{2}+c^{2}}$ =$\frac{ ck^{2}}{c}$

$\frac{c^{2} k^{4}+c^{2} k^{2}}{c^{2} k^{2}+c^{2}}=\frac{c k^{2}}{c}$

$k^{2}=k^{2}$

Hence proved

(iii)  $\frac{a^{3}+b^{3}+c^{3}}{a^{2} b^{2} c^{2}}=\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}$

$\frac{\left(c k^{2}\right)^{3}+(c k)^{3}+c^{3}}{\left(c k^{2}\right)^{2} \cdot(c k)^{2} \cdot c^{2}}=\frac{b^{3} \cdot c^{3}+a^{3} \cdot c^{3}+a^{3}-b^{3}}{a^{3} b^{3} c^{3}}$

$\frac{c^{3} k^{6}+c^{3} k^{3}+c^{3}}{c^{2} k^{4} \cdot c^{2} k^{2} \cdot c^{2}}=\frac{(c k)^{3} \cdot c^{3}+\left(c k^{2}\right)^{3}-c^{3}+\left(c k^{2}\right)^{3}-(c k)^{3}}{\left(c k^{2}\right)^{3}-(c k)^{3}-c^{3}}$

$\frac{c^{3}\left(x^{6}+k^{3}+1\right)}{c^{2} k^{6} \cdot c^{2} k^{2}-c^{2}}=\frac{c^{3} k^{3} \cdot c^{3}+c^{3} k^{6} \cdot c^{3}+c^{3} k^{6} \cdot c^{3} k^{3}}{c^{3} k^{6} \cdot c^{3} k^{3} \cdot c^{3}}$

$\frac{c^{3}\left(k^{6}+k^{3}+1\right)}{c^{6}.k^{6}}=\frac{c^{6}-k^{3}+c^{6} \cdot k^{6}+c^{6} k^{9}}{c^{3} k^{6} \cdot c^{3} k^{3} \cdot c^{3}}$

$\frac{k^{6}+k^{3}+1}{c^{3} k^{6}}$ = $\frac{1+k^{3}+k^{6}}{c^{3} k^{6}}$

$\frac{1+k^{3}+k^{6}}{c^{3} k^{6}}=\frac{1+k^{3}+1k^{6}}{c^{3} k^{6}}$

Hence proved 

(iv) $\left(4 a^{2}+7 a b+9 b^{2}\right):\left(4 b^{2}+7 b c+9 c^{2}\right)=a: c .$

$\frac{4 a^{2}+7 a b+9 b^{2}}{4 b^{2}+7 b c+9 c^{2}}=\frac{a}{c} .$

$\frac{4\left(c k^{2}\right)^{2}+7 c k^{2} \cdot c k+9(c k)^{2}}{4 \cdot(c k)^{2}+7 c k \cdot c+9 \cdot c^{2}}=$ $\frac{ck^{2}}{c}$

$\frac{4 c^{2} k^{4}+7 c^{2} k^{3}+9 c^{2} k^{2}}{4 c^{2} k^{2}+7 c^{2} k+9 c^{2}}=k^{2}$

$k^{2}=k^{2}$ 

Hence proved 

Question 4

Ans: Given,

$a, b, c, d$ are in continued proportion

$\therefore \quad \frac{a}{b}=\frac{b}{c}=\frac{c}{d}$

let $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k$.

$c=d k, b=c k-d k \cdot k=d k^{2} a=b k=d k^{2} k=d k^{3}$

a, b ,c , d are in continued proportion 

$\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=k$ 

$c=d k, b=c k \Rightarrow b=d k^{2} .$

$a=d k,=d k^{2} k=d k^{3}$

$(b-c)^{2}+(c-a)^{2}+(d-b)^{2}=(a-{d})^{2}$

(image to be added)

(ii) $\sqrt{\frac{a^{5}+b^{2} c^{2}+a^{3} c^{2}}{b^{4} c+d^{4}+b^{2} c d^{2}}}$ = $\frac{a}{d}$

$\sqrt{\frac{\left(d k^{3}\right)^{5}+\left(d k^{2}\right)^{2}(d k)^{2}+\left(d k^{3}\right)^{5}-(d k)^{2}}{\left(d k^{2}\right)^{4} \cdot d k+d^{4}+\left(d k^{2}\right)^{2} \cdot d k \cdot d^{2}}}=\frac{d k^{3}}{d}$

$\sqrt{\frac{d^{5} k^{15}+d^{2} k^{4} \cdot d^{2} k^{2}+d^{3} k^{9} \cdot d^{2} k^{2}}{d^{4} k^{8} \cdot d k+d^{4}+d^{2} \cdot k^{4} \cdot d k \cdot d^{2}}}=\frac{d k^{3}}{d}$

$\sqrt{\frac{d^{5} k^{15}+d^{4} \cdot k^{6}+d^{5} k^{11}}{d^{5} k^{9}+d^{4}+d^{5} k^{5}}}=k^{3}$

$\sqrt{k^{6}}=k^{3}$

$\sqrt{k^{3} \times k^{3}}=k^{3}$

$k^{3}=k^{3}$

Hence proved 

(iii) $\sqrt{a b}-\sqrt{b c}+\sqrt{c d}=\sqrt{(a-b+c)(b-c+d)}$

$\sqrt{d k^{3} \cdot d k^{2}}-\sqrt{d k^{2} \cdot d k}+\sqrt{d k \cdot d}=\sqrt{\left(d k^{3}-d k^{2}+d k\right)\left(d k^{2}-d k+d\right)}$

$\sqrt{d^{2} k^{5}}-\sqrt{d^{2} k^{3}}+\sqrt{d^{2} k}=\sqrt{d k\left(k^{2}-k+1\right) d\left(k^{2}-k+1\right)}$

$d k^{2} \sqrt{k}-d k \sqrt{k}+d \sqrt{k}=\sqrt{d^{2} k\left(k^{2}-k+1\right)^{2}}$

$d \sqrt{k}\left(k^{2}-k+1\right)=d \sqrt{k}\left(k^{2}-k+1\right) .$

Hence proved 

(iv) $\frac{3 a+5 d}{5 a+7 d}=\frac{3 a^{3}+5 b^{3}}{5 a^{3}+7 b^{3}}$

$\frac{3 d k^{3}+5 d}{5 d k^{3}+7 d}=\frac{3\left(d k^{3}\right)^{3}+5\left(d k^{2}\right)^{3}}{5\left(d k^{3}\right)^{3}+7\left(d k^{2}\right)^{3}}$

$\frac{d\left(3 k^{3}+5\right)}{d\left(5 k^{3}+7\right)}=\frac{3 d^{3} k^{9}+5 d^{3} k^{6}}{5 d^{3} k^{9}+7 d^{3} k^{6}}$

$\frac{3 k^{3}+5}{5 k^{3}+7}=\frac{3 k^{3}+5}{5 k^{3}+7}$

Hence proved 

(v) $\frac{a^{3}+b^{3}+c^{3}}{b^{3}+c^{3}+d^{3}}=\frac{a}{d}$

$\frac{\left(d k^{3}\right)^{3}+\left(d k^{2}\right)^{3}+(d k)^{3}}{\left(d k^{2}\right)^{3}+(d k)^{3}+d^{3}}=$ $\frac{d k^{3}}{d}$

$\frac{d^{3} k^{9}+d^{3} k^{6}+d^{3} k^{3}}{d^{3} k^{6}+d^{3} k^{3}+d^{3}}=k^{3}$

$\frac{d^{3} k^{3}\left(k^{6}+k^{3}+1\right)}{d^{2}\left(k^{6}+k^{3}+1\right)}=k^{3}$

$k^{3}=k^{3}$

Hence, proved

(vi) (to be added)

Question 5

Ans: Given,

$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$

$\operatorname{let} \frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k .$

$\therefore x=a k, y=b k$ and $z=c k .$

$\frac{a x-b y}{(a+b)(x-y)}+\frac{b y-c z}{(b+c)(y-z)}+\frac{c z-a x}{c(+a)(z-x)}=3$

$\frac{a-a k-b \cdot b k}{(a+b)(a k-b k)}+\frac{b b k-c \cdot c k}{(b+c)(b k-c k)}+\frac{c(c k-a \cdot a k}{(c+a)(c k-a k)}=3$

$\frac{a^{2} k-b^{2} k}{(a+b)(a k-b k)}+\frac{b^{2} k-c^{2} k}{(b+c)(b k-c k)}+\frac{c^{2} k-a^{2} k}{(c+a)(c k-9 k)}=3$

$\frac{k^{2}\left(a^{2}-b^{2}\right)}{(a+b) k(a-b)}$+ $\frac{K\left(b^{2}-c^{2}\right)}{(b+c)) k(b-c)}$

$\frac{(a+b)(a-b)}{(a+b)(a-b)}+\left(\frac{(b+b)(b-c)}{(b+x) \cdot(b-5)}+\frac{(c+a)(c-a)}{(c+a)(-c-a)}=3\right.$

$1+1+1=3$

3=3 Hence proved

Question 6

Ans: Given,

$\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{Z}{a+b-c}$

$=\frac{x+y+z}{b+c-a+c+a-b+a+b-c}$

$=\frac{x+y+z}{a+b+c}$

Hence, proved

Question 7

Ans: Given,

$\frac{a}{b+c}=\frac{b}{c+a} = \frac{c}{a+b}$

Let $\frac{a}{b+c}=\frac{b}{c+c}=\frac{c}{a+b}=k$

$a=k(b+c), b=k(c+a), c=k(a+b)$

$\therefore a(b-c)+b(c-a)+c(a-b)=0$

$k\left(b^{2}-c^{2}\right)+k\left(c^{2}-a^{2}\right)+k\left(a^{2}-b^{2}\right)=0$

$K \times 0=0$

0=0 

Hence proved 

Question 8

Ans: Given,

$\begin{aligned} a x &=b y=c z \\ \text { Leb } a x &=b y=c z=k . \\ x &=\frac{1}{4}, y=\frac{k}{b} \text {. and } z=\frac{k}{c .} \\ \therefore \frac{x^{2}}{y z} &+\frac{y^{2}}{2 x}+\frac{z^{2}}{x y}=\frac{b c}{a^{2}}+\frac{c a}{b^{2}}+\frac{a b}{c^{2}} \end{aligned}$

$\frac{\left(\frac{k}{a}\right)^{2}}{\frac{k}{3} \times \frac{k}{c}}+\frac{\left(\frac{k}{b}\right)^{2}}{\frac{k}{c} \times \frac{k}{a}}+\frac{\left(\frac{k}{c}\right)^{2}}{\frac{k}{a} \times \frac{k}{b}}=\frac{b c}{a^{2}}+\frac{c a}{b^{2}}+\frac{a b}{c^{2}}$

$\frac{\frac{k^{2}}{a^{2}}}{\frac{k^{2}}{b c}}+\frac{\frac{k^{2}}{b^{2}}}{\frac{k^{2}}{ca^{2}}}+\frac{\frac{k^{2}}{c^{2}}}{\frac{k^{2}}{a b}}=\frac{b^{2}}{a^{2}}+\frac{c a}{b^{2}}+\frac{a b}{c^{2}}$

$\frac{b c}{a^{2}}+\frac{c a}{b^{2}}+\frac{a b}{c^{2}}=\frac{b c}{a^{2}}+\frac{c a}{b^{2}}+\frac{a b}{c^{2}}$

HENCE PROVED 

S Chand CLASS 10 Chapter 6 Ration and proportion Exercise 6B

  Exercise 6B

Question 1 

Ans: 1 

Given, 

$\frac{x}{y}=\frac{p}{q}$

According to question, $\frac{x}{y}=\frac{p}{q}$

$\frac{x}{y}=\frac{p}{q}$

Multiply both sides by $\frac{5}{7}$

$\frac{5 x}{7 y}=\frac{5 p}{7 q} .$

$\frac{5 x+7 y}{5 x-y}=\frac{5 p+7 q}{5 p-7 q}$ (by compound and divided)

hence proved 

Question 2

Ans:2 Given,

$4 a+9 b: 4 a-9 b=4 c+9 d: 4 c-9 d$.

$\frac{4 a+9 b}{4 a-9 b}=\frac{4 c+9 d}{4 c-9 d} \text {}$

$\frac{4 a+9 b+4 a-9 b}{4 a+9 b-(4 a-9 b)}=\frac{4 c+9 d+4 c-9 d}{4 c+9 d-(4 c-9 d)}$ (by compound and dividends)

$\frac{4 a+9 b+4 a-9b}{4 a+9 b-4 a+9 b}=\frac{4 c+9 d-(4 c-9 d}{4 c+9 d-4 c+9 d}$ 

$\frac{8 a}{18b}=\frac{8 c}{18 d} \text {. }$

Multiply both sides by $\frac{18}{8}$

$\frac{8a}{18b} \times \frac{18}{8}=\frac{8 c}{18 d} \times \frac{18}{8}$◘

$\frac{a}{b}=\frac{c}{d}$

$\therefore a: b:: c \mathrm{~d}$

Hence, Proved

Question 3

Ans:3 $(5 a+11 b):(5 c+11 d)::(5 a-11 b):(5 c-11 d)$

$\frac{5 a+11 b}{5 c+11 d}=\frac{5 a-11 b}{5 c-11 d}$

$\frac{5 a+11 b}{5 a-11 \cdot b}=\frac{5 c+11 d}{5 c-11 . d}$

using comp. 

$\frac{5 a+11 b+5 a-11 b}{5 a+11 b-5 a+11 b}$ = $\frac{5 c-11d+5 c-11d}{5 c+11 d-5 c+11 d}$

$\frac{10 a}{22 b}=\frac{10 c}{22 d}$ proved 

Question 4

Ans:4

 (i) Given, $m a^{2}+h b^{2}: m c^{2}+h d^{2}: m a^{2}-h b^{2}: m c^{2}-h d^{2}$

$\frac{m a^{2}+h b^{2}}{m c^{2}+h d^{2}}=\frac{m a^{2}-h b^{2}}{m c^{2}-h d^{2}}$

$\frac{m n^{2}+h b^{2}}{m a^{2}-h b^{2}}=\frac{m c^{2}+h d^{2}}{m c^{2}-h d^{2}}$ (by alternate)

$\frac{m a^{2}+h b^{2}+m a^{2}-h b^{2}}{m a^{2}+h b^{2}-\left(m a^{2}-h b^{2}\right)}=\frac{m c^{2}+h d^{2}+m c^{2}-h d^{2}}{m c^{2}+h d^{2}-\left(m c^{2}-h d^{2}\right)}$

$\frac{m a^{2}+h b^{2}+m a^{2}-h b^{2}}{m a^{2}+h b^{2}-ma^{2}+h b^{2}}=\frac{m c^{2}+n d^{2}+m c^{2}-b d^{2}}{mc^{2}+n d^{2}-m c^{2}+h d^{2}}$

$\frac{m a^{2}}{h b^{2}}=\frac{m c^{2}}{h d^{2}}$

multiply both sides by $\frac{h}{m}$

$\frac{m  a^{2}}{h b^{2}} \times \frac{h}{m}$= $\frac{m  c^{2}}{h d^{2}} \times \frac{h}{m}$

$\frac{a^{2}}{b^{2}}=\frac{c^{2}}{d^{2}}$

Talking square root both sides 

$\begin{aligned} \therefore \quad \frac{a}{b} &=\frac{c}{d} \\ a : b: & \therefore c=d \end{aligned}$

Hence a , b , c and d are in proportion 

(ii)

$\begin{aligned}&(a+b+c+d)(a-b-c+d)=(a+b-c-d)(a-b+c-d) \\&\frac{a+b+c d}{a+b-c-d}=\frac{a-b+c-d}{a-b-c+d}\end{aligned}$

$\frac{a+b+c d}{a+b-c d}=\frac{a-b+c-d}{a-b-c+d}$

$\frac{a+b+c+d+a+b-c-d}{a+b+c+d-(a+b-c-d)}$ =$\frac{a-b+c-d+a-b-c+d}{a-b+c-d-(a-b-c+d)}$ 

(By component and dividers)

$\frac{2 a+2 b}{2 c+2 d}=\frac{2 a-2 b}{2 c-2 d}$

$\frac{2 a+2 b}{2 a-2 b}=\frac{2 c+2 d}{2 c-2 d}$(By Alternate)

$\frac{2 a+2 b+2 a-2 b}{2 a+2 b-(2 a-2 b)}=\frac{2 c+2 d+2 c-2 d}{2 c+2 d-(2 c-2 d)}$ (By component and dividends)

$\frac{2 a+2 b+2 a-2 b}{2 b+2 b-2 b+2 b}=\frac{2 c+2 b+2 c-2 c}{2 c+2 d-2 c+2 d}$

$\frac{4_{a}}{4_{b}}=\frac{4_{c}}{4_{d}}$

$a=b :: c: d .$

∴ a , b , c and d are in proportion 

Question 5

Ans:5

Given,

a: b=c: d  and c: f=g: h

$\frac{a}{b}=\frac{c}{d}$ and $\frac{c}{f}=\frac{g}{h}$

Multiplying each other 

$\frac{a}{b} \times \frac{e}{f}=\frac{c}{d} \times \frac{g}{h}$

$\frac{a e}{b y}=\frac{c g}{d h}$

$\frac{a e+b f}{a e-b f}=\frac{c g+d h}{c g-d h}$ (By component and dividends)

∴ $a e+b f=$ $a e-b y=$ $c g+d h= c y-d h$

Hence proved, 

Question 6

Ans:6

Given,

$\begin{aligned}x &=\frac{10 p q}{p+q} \\x &=\frac{5 p \times 2 q}{p+q} \\\frac{x}{5 p} &=\frac{2 q}{p+q} \\\frac{x+5 p}{x-5 p} &=\frac{2 q+p+q}{2 q-(p+q)}\end{aligned}$  (By component and dividends)

$\frac{x+5 p}{x-5 p}=\frac{3 q+p}{2 q-p-q}$

$\frac{x+5 p}{x-5 p}=\frac{3 q+p}{q-p}$............(i)

Again , 

$x=\frac{10 p q}{p+q}$

$x=\frac{2 p \times 5 q}{p+q}$

$\frac{x}{5 q}=\frac{2 p}{p+q}$

$\frac{x+5 q}{x-5 q}=\frac{2 p+p+q}{2 p-(p+q)}$   (By component and dividends)

$\frac{x+5 q}{x-5 q}=\frac{3 p+q}{2 p-p-q}$

$\frac{x+5 q}{x-5 q}=\frac{3 p+q}{p-q}$ ..........(ii)

Add the eqn (i) and (ii)

$\begin{aligned} \frac{x+5 p}{x-5 p}+\frac{x+5 q}{x-5 q} &=\frac{3 q+p}{q-p}+\frac{3 p+q}{p-q} \\ &=\frac{3 q+p}{q-p}-\frac{3 p+q}{q-p} \\ &=\frac{3 q+p-(3 p+q)}{q-p} \\ &=\frac{3 q+p-3 p-q}{q-p} \\ &=\frac{2 q-2 p}{q-p} \end{aligned}$

$=\frac{2(q-p)}{q-p}$

=2 

Hence the value of $\frac{x+5 p}{x-5 p}+\frac{x+5 q}{x-5 q}$ is 2 .

Question 7

Ans:7

Given,

$x=\frac{6 p q}{p+q}$

$x=\frac{3 p \cdot x 2 q}{p+q}$

$\frac{x}{3 p}=\frac{2 q}{p+q}$

$\frac{x+3 p}{x-3 p}=\frac{2 q+p+q}{2 q-(p+q)}$  (By component and dividends)

$\frac{x+3 p}{x-3 p}=\frac{3 q+p}{2 q-p-q}$

$\frac{x+3 p}{x-3 p}=\frac{3 q+p}{q-p}$..........(i)

Again,

$\begin{aligned}&x=\frac{6 p q}{p+q} \\&x=\frac{2 p \times 3 q}{p+q} \\&\frac{x}{3 q}=\frac{2 p}{p+q}\end{aligned}$

$\frac{x+3 q}{x-3 q}=\frac{2 p+p+q}{2 p-(p+q)}$ (By component and dividends)

$\frac{x+3 q}{x-3 q}=\frac{3 p+q}{2 p-p-q .}$

$\frac{2 c+3 q}{x-3 q}=\frac{3 p+q}{p-q}$............(ii)

On adding eq (i) and (ii)

$\begin{aligned} \frac{x+3 p}{x-3 p}+& \frac{x+3 q}{x-3 q}=\frac{3 q+p}{q-p}+\frac{3 p+q}{p-q} \\ &=\frac{3 q+p}{q-p}-\frac{3 p+q}{q-p} \\ &=\frac{3 q+p-(3 p+q)}{q-p} \end{aligned}$

$=\frac{3 q+p-3 p-q}{q-p}$

$=\frac{2 q-2 p}{q-p}$

=2

Hence the value of $\frac{x+3 p}{x-3 p}+\frac{x+3 q}{x-3 q}=2$

Question 8

Ans:8

(i) Given

$\frac{3 x+\sqrt{9 x^{2}-5}}{3 x-\sqrt{9 x^{2}-5}}=\frac{5}{1}$

$\frac{3 x+\sqrt{9 x^{2}-5}+3 x-\sqrt{9 x^{2}-5}}{3 x+\sqrt{9 x^{2}-5}-\left(3 x-\sqrt{9 x^{2}-5}\right)}=\frac{5+1}{5-1}$   (By component and dividends)

$\frac{3 x+3 x}{3 x+\sqrt{9 x^{2}-5}-3 x+\sqrt{9 x^{2}-5}}=\frac{6}{4} .$

$\frac{6 x}{2 \sqrt{9 x^{2}-5}}=\frac{6}{4}$

$\frac{3 x}{\sqrt{9 x^{2}-5}}\frac{3}{2}$

$3 x \times 2=3 \sqrt{9 x^{2}-5}$

$6 x=\frac{3 \sqrt{9 x^{2}-5}}{6}$

$x=\frac{3 \sqrt{9 x^{2}-5}}{62}$

$x=\frac{\sqrt{9 x^{2}-5}}{2}$

$2 x=\sqrt{9 x^{2}-5}$

Squaring both side 

$\begin{aligned} \therefore \quad &(2 x)^{2}=\left(\sqrt{9 x^{2}-5}\right)^{2} \\ & 4 x^{2}=9 x^{2}-5 \\ & 4 x^{2}-9 x^{2}=-5 \\ &-5 x^{2}=-5 \end{aligned}$

$x^{2}=1$

$x=\sqrt{1}$

$x=1$

Hence the value of x is 1

(ii)Given,

$\frac{\sqrt{x+12}+\sqrt{x-3}}{\sqrt{x+2}-\sqrt{x-3}}=\frac{5}{1}$

$\frac{\sqrt{x+2}+\sqrt{x-3}+(\sqrt{x+2}-\sqrt{x-3})}{\sqrt{x+2}+\sqrt{x-3}-(\sqrt{x+2}-\sqrt{x-3})}=\frac{5+1}{5-1}$ (By component and dividends)

$\frac{\sqrt{x+2}+\sqrt{x-3}+\sqrt{x+2}-\sqrt{ x} -3}{\sqrt{x+2}+\sqrt{x-3}-\sqrt{x+2}+\sqrt{x-3}}=\frac{6}{4}$

$\frac{2 \sqrt{x+2}}{2 \sqrt{x-3}}$ $=\frac{6}{4}$

$\frac{\sqrt{x+2}}{\sqrt{x-3}}=\frac{3}{2}$

Squaring both sides 

$\begin{aligned}\left(\frac{\sqrt{x+2}}{\sqrt{x-3}}\right)^{2} &=\left(\frac{3}{2}\right)^{2} \\ \frac{x+2}{x-3} &=\frac{9}{4} \\ 4(x+2) &=9(x-3) \\ 4 x+8 &=9 x-27 \\ 4 x-9 x &=-8-27 \\-5 x &=-35 \end{aligned}$

x=7

Hence, the value of $x$ is 7 .

(iii) $\frac{x^{3}+3 x-341}{3 x^{2}+1}=\frac{31}{91}$

$\frac{x^{3}+3 x+\left(3 x^{2}+1\right)}{x^{3}+3 x-\left(3 x^{2}+1\right)}=\frac{341+91}{341-91}$  (By component and dividends)

$\frac{x^{3}+3 x+3 x^{2}+1}{x^{2}+3 x-3 x^{2}-1}=\frac{432}{250}$

$\frac{(x+1)^{3}}{(x-1)^{3}}$ = $\frac{216}{125}$

$\frac{(x+1)^{3}}{(x-1)^{3}}=\frac{(6)^{3}}{(5)^{3}}$

On taking cube root 

$\frac{x+1}{x-1}=\frac{6}{5}$

$5(x+1)=6(x-1)$

$5 x+5=6 x-6$

$5 x-6 x=-5-6$

$+x=+11$

$x=11$

Hence the value of x is 11

 

(iv) $\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=\frac{4 x-1}{2}$

$\frac{\sqrt{x+1}+\sqrt{x-1}+(\sqrt{x+1}-\sqrt{x-1})}{\sqrt{x+1}+\sqrt{x-1}-(\sqrt{x+1}-\sqrt{x-1})}=\frac{4 x-1+2}{4 x-1-2}$  (By component and dividends)

$\frac{\sqrt{x+1}+\sqrt{x-1}+\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}-\sqrt{x+1}+\sqrt{x-1}}=\frac{4 x-1}{4 x-3}$

$\frac{2 \sqrt{x+1}}{2 \sqrt{x-1}}=\frac{4 x+1}{4 x-3}$

$\frac{\sqrt{x+1}}{\sqrt{x-1}}=\frac{4 x+1}{4 x-3}$

Squaring both sides 

$\frac{(\sqrt{x+1})^{2}}{(\sqrt{x-1})^{2}}=\frac{(4 x+1)^{2}}{(4 x-3)^{2}}$

$\frac{x+1}{x-3}=\frac{(4 x)^{2}+(1)^{2}+2 x 4 x x 1}{(4 x)^{2}+(3)^{2}-2 \times 4 x+3}$

$\frac{x+1}{x-1}=\frac{16 x^{2}+1+8 x}{16 x^{2}+9-24 x}$

$(x+1)\left(16 x^{2}+9-24 x\right)=(x-2)\left(16 x^{2}+1+8 x\right)$

$16 x^{3}+9 x-24 x^{2}+16 x^{2}+9-24 x=16 x^{3}+x+8 x^{2}-16 x^{2}-1-8 x$

$\begin{aligned} 16 x^{3}-15 x &-8 x^{2} 9=16 x^{3}-7 x-8 x^{2}-1 \\ 16 x^{3}-15 x &=8 x^{2}+9-16 x^{3}+7 x+8 x^{2}+1=0 \\ &-8 x+10=0 \\ &=8 x=0-10 \\ &-8 x=-10 \end{aligned}$

$x=\frac{5}{4}$

Hence the value of x is  $=\frac{5}{4}$

Question 9

Ans:9

Given,

$16\left[\frac{a-x}{a+x}\right]^{3}=\frac{a+x}{a-x} \text {. }$

Multiply both sides by $\frac{a-x}{a+x}$

$16\left[\frac{a-x}{a+x}\right]^{3} \times \frac{a-x}{a+x}=\frac{a+x}{a-x} \times \frac{a-\pi}{a+x}$

$16\left[\frac{a-x}{a+x}\right]^{4}=1$

$\left[\frac{a-x}{a+x}\right]^{4}=\frac{1}{16}$

$\left[\frac{a-x}{a+x}\right]^{4}=\left[\frac{1}{2}\right]^{4}$

$\begin{aligned} \therefore \quad & \frac{a-x}{a+x}=\frac{1}{2} \\ & 2(a-x)=a+x \\ & 2 a-2 x=a+x \\ & 2 a-a=2 x+x \\ & a=3 x \\ & \therefore \quad x=\frac{a}{3} \end{aligned}$

Hence the value of x is $\frac{a}{3}$

Question 10

Ans-10:

$\begin{gathered}\frac{1+x+x^{2}}{1-x+x^{2}}=\frac{171(1+x)}{172(1-x)} \\\frac{(1-x)\left(1+x+x^{2}\right)}{(1+x)\left(1-x+x^{2}\right)}=\frac{171}{172} \\\frac{1-x^{3}}{1+x^{3}}=\frac{171}{172} \\172\left(1-x^{3}\right)=171\left(1+x^{3}\right) \\172-172 x^{3}=171+171 x^{3} \\172-171=172 x^{3}+171 x^{3} \\1=343 x^{3}\end{gathered}$

$\begin{aligned} \therefore x^{3} &=\frac{1}{343} \\ x &=\sqrt[3]{\frac{1}{343}} \\ x &=\sqrt[3]{\frac{1 \times 1 \times 1}{7 \times 7 \times 7}} \\ x &=\frac{1}{7} \end{aligned}$

Hence the value of x is  $\frac{1}{7}$

Question 11

Ans: 

Given

$\frac{x}{1}=\frac{\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}-b^{2}}}{\sqrt{a^{2}+b^{2}}

\sqrt{a^{2}b^{2}}}$

$\frac{x+1}{x-1}=\frac{\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}-b^{2}}+\left(\sqrt{a^{2}+b^{2}}-\sqrt{a^{2}-b^{2}}\right)}{\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}-b^{2}}-\left(\sqrt{a^{2}+b^{2}}-\sqrt{a^{2}-b^{2}}\right)}$

$\frac{x+1}{x-1}=\frac{\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}+b^{2}}-\sqrt{a^{2}-b^{2}}}{\sqrt{a^{2}+b}+\sqrt{a^{2}-b^{2}}-\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}-b^{2}}}$

$\frac{x+1}{x-1}=\frac{2 \sqrt{a^{2}+b^{2}}}{2\sqrt{a^{2}-b^{2}}}$

$\frac{x+1}{x-1}=\frac{\sqrt{a^{2}+b^{2}}}{\sqrt{a^{2}-b^{2}}}$

Squaring both sides 

$\frac{\left(x+b^{2}\right.}{(x-1)^{2}}=\left(\frac{\sqrt{a^{2}+b^{2}}}{\sqrt{a^{2}-b^{2}}}\right)^{2}$

$\frac{x^{2}+1+2 x x+1}{x^{2}+1-2 x^{2}(x)}=\frac{a^{2}+b^{2}}{a^{2}-b^{2}}$

$\frac{x^{2}+1+2 x}{x^{2}+1-2 x}=\frac{a^{2}+b^{2}}{a^{2}-b^{2}}$

$\frac{x^{2}+1+2 x+\left(x^{2}+1-2 x\right)}{x^{2}+1+2 x-\left(x^{2}+1-2 x\right)}=\frac{a^{2}+b^{2}+\left(a^{2}-b^{2}\right)}{x^{2}+b^{2}-\left(a^{2}-b^{2}\right)}$ (By components and dividends)

$\frac{2 x^{2}+2}{4 x}=\frac{2 a^{2}}{2 b^{2}}$

$\frac{2\left(x^{2}+1\right)}{4 x}=\frac{a^{2}}{b^{2}}$

$\frac{x^{2}+1}{2 x}=\frac{a^{2}}{b^{2}}$

$b^{2}\left(x^{2}+1\right)=2 a^{2} x$

$b^{2} x^{2}+b^{2}=2 a^{2} x$

$b^{2} x^{2}-2 a^{2} x+b^{2}=0$

Hence proved .

Question 12

Ans: 

Given,

$\frac{y}{1}=\frac{\sqrt{a+3 b}+\sqrt{a-3 b}}{\sqrt{a+3 b}-\sqrt{a-3 b}}$

$\frac{y+1}{y-1}=\frac{\sqrt{a+3 b}+\sqrt{a-3 b}+(\sqrt{a+3 b}-\sqrt{a-3 b})}{\sqrt{a+3 b}+\sqrt{a-3 b}-(\sqrt{a+3}-\sqrt{a-3 b})}$  (By components and dividends)

$\frac{y+1}{y-1}=\frac{\sqrt{a+3 b}+\sqrt{a-3 b}+\sqrt{a+3 b}-\sqrt{a-3 b}}{\sqrt{a+3 b}+\sqrt{a-3 b}-\sqrt{a+3 b}+\sqrt{a-3 b}}$

$\frac{y+1}{y-1}=\frac{2 \sqrt{a+3 b}}{2 \sqrt{a-3 b}}$

$\frac{y+1}{y-1}=\frac{\sqrt{a+3 b}}{\sqrt{a-3 b}}$

On squaring both sides 

$\frac{(y+1)^{2}}{(y-1)^{2}}=\frac{(\sqrt{a+3 b})^{2}}{(\sqrt{a-3 b})^{2}}$

$\frac{y^{2}+1+2 x y x 1}{y^{2}+1-2 x y \times 1}=\frac{a+3 b}{a-3 b}$

$\frac{y^{2}+1+2 y}{y^{2}+1-2 y}=\frac{a+2 b}{a-3 b}$

$\frac{y^{2}+1+2 y+\left(y^{2}+1-2 y\right)}{y^{2}+1+2 y-\left(y^{2}+1-2 y\right)}=\frac{a+3 b+(a-3 b)}{a+3b-(a-3 b)}$

$\frac{2 y^{2}+2}{4 y}=\frac{2 a}{6 b}$

$\begin{aligned} \frac{y^{2}+1}{2 y} &=\frac{a}{3 b} \\ 3 b\left(y^{2}+1\right) &=2 a y . \\ 3 b y^{2}+3 b &=2 a y \end{aligned}$

$3 b y^{2}-2 a y+3 b=0$

Hence, proved.

Question 13

Ans: 

given,

$\frac{a^{3}+3 a b^{2}}{3 a^{2} b+b^{3}}=\frac{x^{3}+3 x y^{2}}{3 x^{2} y+y^{3}}$

$\frac{a^{3}+3 a b^{2}+\left(3 a^{2} b+b^{3}\right)}{9^{3}+3 a b^{2}-\left(3 a^{2} b+b^{3}\right)}=\frac{x^{3}+3 x y^{2}+\left(3 x^{2} y+y^{3}\right)^{3}}{x^{3}+3 x y^{2}-\left(3 x^{2} y+y^{3}\right)}$

$\frac{9^{3}+3 a b^{2}+3 a^{2} b+b^{3}}{a^{3}+3 a b^{2}-3 a^{2} b-b^{3}}=\frac{x^{3}+3 x y^{2}+3 x^{2} y+y^{3}}{x^{3}+3 x y^{2}-3 x^{2} y-y^{3}}$

$\frac{a^{3}+b^{3}+3 a b^{2}+3 a^{2} b}{a^{3}-b^{3}+3 a b^{2}-3 a^{2} b}=\frac{x^{3}+y^{3}+3 x^{2} y+3 x y^{2}}{x^{3}-y^{3}+3 x y^{2}-3 x^{2} y}$

$\frac{(a+b)^{3}}{(a-b)^{3}}=\frac{(x+y)^{3}}{(x-y)^{3}}$

On taking cube both sides 

$\sqrt[3]{(a+b)^{3}}=\sqrt[3]{(a-y)^{3}}$

$\frac{a+3}{9-b}=\frac{x+y}{x-y}$

$\frac{a+b+(a-b)}{a+b-(a-b)}=\frac{x+y+(x-y)}{x+y-(x-y)}$

$\frac{a+b+a-b}{c+b-d+b}=\frac{x+y+x-y}{x+y-x+y}$

$\frac{ 2a}{2b}=\frac{2 x}{2 y}$

$\frac{a}{b}=\frac{x}{y}$

$\frac{y}{b}=\frac{x}{a} \quad$

$: \quad \frac{x}{a}=\frac{y}{b}$

Hence proved 

S Chand CLASS 10 Chapter 6 Ration and proportion Exercise 6A

  Exercise 6A

Question 1

Ans: 1

(i) $8: 14:: x: 28$

$\frac{8}{14}=\frac{x}{28} .$

$\therefore 8 \times 28=14 \times x$

$\frac{8 \times 28}{14}=x$

$8 \times 2=x$

$16=x$

$x=16 $

(ii) 

$\begin{aligned} x &=9=5: 3 \\ \frac{x}{9} &=\frac{5}{3} . \\ 3 x &=9 \times 5 \end{aligned}$

$x=\frac{9}{3} \times 5$

$x=15$

(iii) $12: x:=4: 15$.

$\begin{aligned}&\frac{12}{x}=\frac{4}{15} \\&12 \times 15=4 \times x\end{aligned}$

$x=\frac{12}{15} \times 4$

$3 \times 15=x$

$45=x$

$x=45$

Question 2

Ans: 

(i) 25,15,40

Let the fourth proportional be x

$\begin{aligned}\therefore \quad 25: 15 &: 40: x \\\frac{25}{15} &=\frac{40}{x} .\end{aligned}$

$25\times x =40 \times 15$

$x=\frac{40 \times 15}{25}$

$x=8 \times 3$

x=24

Hence , the fourth proportional is 24. 

(ii) $3 a^{2} b^{2}, a^{3}, b^{3}$.

Let the fourth proportional be x

$\begin{aligned} \therefore & 3 a^{2} b^{2}: a^{3}:: b^{3}: x \\ & \frac{3 a^{2} b^{2}}{9^{3}}=\frac{b^{3}}{x} \end{aligned}$

$3 a^{2} b^{2} \times x=b^{3} \times a^{3}$

$x=\frac{b^{3} \times a^{3}}{3 a^{2} b^{2}}$

$x=\frac{b a}{3}$

$x=\frac{a b}{3}$

Hence the fourth proportional is $\frac{a b}{3}$

(iii) $a^{2}-5 a+6, a^{2}+a-6, a^{2}-9$.

let the fourth proportional be x.

$\therefore a^{2}-5 a+6: a^{2}+a-6: a^{2}-9: x$.

$\frac{a^{2}-5 a+6}{a^{2}+a-6}=\frac{a^{2}-9}{x}$

$\left(a^{2}-5 a+6\right) \times x=\left(a^{2}-9\right) \times\left(a^{2}+a-6\right)$

$x=\frac{\left(a^{2}-9\right) \times\left(a^{2}+a^{2}-6\right)}{\left(a^{2}-5 a+6\right)}$

$x=\frac{\left((a)^{2}-(3)^{2}\right)\left(a^{2}+3 a-2 a-6\right)}{\left(a^{2}-3 a-2 a+6\right)}$

$x=\frac{(a+3)(a-3) \cdot(a(a+3)-2(a+3))}{(a(a-3)-2(a-3))}$

$x=(a+3)(a+3)$

$x=(a+3)^{2}$

Hence the fourth proportional $(a+3)^{2}$

Question 3

Ans: 

(i) 16 and 36 .

When, $a: b:: 3: c$

Let the third proportional be x 

$\therefore \quad 16: 36: 36: x$

$\frac{16}{36}=\frac{36}{x} $

$16 \times x=36 \times 36$

$x=\frac{36 \times 36}{16}$

$x=9 \times 9$

$x=81 $

Hence the third proportional is 81

(ii) $\frac{x}{y}+\frac{y}{x}$ and $\frac{x}{y}$

When a: b:: b:c 

Let the third proportional be x, 

$\therefore \frac{x}{y}+\frac{y}{x}: \frac{x}{y}:: \frac{x}{y}: x$

$\frac{\frac{x}{y}+\frac{y}{x}}{\frac{x}{y}}=\frac{x}{y}$

$\frac{\frac{x^{2}+y^{2}}{x y}}{\frac{x}{y}}=\frac{x}{\frac{y}{x}} .$

$\frac{x^{2}+y^{2}}{x y} \times x=\frac{x}{y} \times \frac{x}{y}$

$x=\frac{\frac{x^{2}}{y^{2}}}{\frac{x^{2}+y^{2}}{x y}}$

$x=\frac{x^{2} \times x y}{y^{2} \times x^{2}+y^{2}}$

$x=\frac{x^{3} y}{x^{2} y^{2}+y^{4}}$

$x=\frac{x^{3}}{x^{2} y +y^{3}}$

Hence , the third proportional is $\frac{x^{3}}{x^{2} y + y^{3}}$

(ii) $a^{2}-b^{2} a+b$. 

when, $a: b:: b: c$.

Let the third proportional be x, 

$\therefore a^{2}-b^{2}: a+b: a+b: x .$

$\frac{a^{2}-b^{2}}{a+b}=\frac{a+b}{x}$

$\left(a^{2}-b^{2}\right) \times x=(a+b) \times(a+b)$

$x=\frac{(a+b) \times (a+b)}{\left(a^{2}-b^{2}\right)}$

$x=\frac{(a+b) \times (a+b)}{(a+b)(a-b)}$

x = $\frac{a+b}{a-b}$

Hence , the third proportional is $\frac{a+b}{a-b}$

Question 4

Ans: 

(i) 5 and 80

According to question 

a: b:: b:c 

5:b::b:80

$\frac{5}{b}=\frac{b}{80}$

$5 \times 80=b \times b $

$\begin{aligned}&400=b^{2} . \\&b^{2}=400 \\&b=\sqrt{400} \\&b=\sqrt{20 \times 20}\end{aligned}$

$b=20$

Hence, the mean proportional is 20

(ii) $360a_{4}$ and $250 a^{2} b^{2}$

According to question, 

$\begin{aligned} & a: b:: b: c \\ & 360 a^{4}: b:: b: 250 a^{2} b^{2} \end{aligned}$

$360 a^{4} \times 250 a^{2} b^{2}=b^{2}$

$b^{2}=360 a^{4} \times 250 a^{2} b^{2}$

$b=\sqrt{360 a^{4} \times 250 a^{2} b^{2}}$

$b=\sqrt{360 \times 250 \times a^{4}+^{2} \times b^{2}}$

$b=\sqrt{360 \times 250 \times a^{6} \times b^{2}}$

$b=\sqrt{90000 \times 9^{6} \times b^{2}}$

$b=300 \times a^{3} \times b .$

$b=300 . a^{3} b $

Hence the mean proportional is $300 a^{3} b$

(iii) $(x-y) and \left(x^{3}-x^{2} y\right)$.

According to question,

$\begin{aligned} \therefore a: b: b &: c \\(x-y) &: b: b:\left(x^{3}-x^{2} y\right) . \\ \frac{(x-y)}{b} &=\frac{b}{x^{3}-x^{2} y .} \\(x-y) & x\left(x^{3}-x^{2} y\right)=b^{2} \\ b^{2} &=(x-y)\left(x^{3}-x^{2} y\right) . \\ b &=\sqrt{(x-y)\left(x^{3}-x^{2} y\right)} \\ b &=\sqrt{(x-y) x^{2}(x-y)} \\ b &=\sqrt{(x-y)^{2} x^{2}} \\ b &=(x-y) x \\ b &=x(x-y) \end{aligned}$

Hence, the mean proportional is x (x-y)

Question 5

Ans: 

(i) Given, 

x, 16,48,y are in continued proportion 

$\therefore \quad \frac{x}{16}=\frac{16}{48}=\frac{48}{y} $

$\begin{aligned} \frac{x}{16} &=\frac{16}{48} \\ 48 x &=16 \times 16 \end{aligned}$

$x=\frac{16 \times 16}{48}$

$x=\frac{16}{3}$

and.

$\begin{aligned}&\frac{16}{48}=\frac{48}{y} \\&16 y=48 \times 48\end{aligned}$

$y=48 \times 3 .$

$y=144 .$

Hence, $x=\frac{16}{3}$ and $y=144$.

(ii) Given,

$x, 9, y, 16$ are in continue proportion,

$\begin{aligned} \frac{x}{9} &=\frac{9}{y}=\frac{y}{16} \\ \frac{9}{y} &=\frac{y}{16} . \\ 9 \times 16 &=y^{2} . \\ y^{2} &=1.44 . \\ y &=\sqrt{144} \\ y &=\sqrt{12 \times 12} \\ y &=12 .\end{aligned}$

and. $\frac{x}{9}=\frac{9}{y} $

$\frac{x}{9}=\frac{9}{12} \quad(y=12)$

$12 x=9 \times 9 .$

$x=\frac{27}{4}$

Hence, $x=\frac{27}{4}$ and $y=12$.

Question 6

Ans: 

Let x be added to each number 

3+x, 5+x,7+x and 10+x are in proportion 

$\frac{3+x}{5+x}=\frac{7+x}{10+x}$

$(3+x)(10+x)=(7+1) x)(5+x)$

$30+3 x+16 x+x^{2}=35+7 x+5 x+x^{2}$

$30+13 x+x^{2}=35+12 x+x^{2}$.

$30+13 x+ x^{2}-35-12 x-x^{2}=0 .$

$13 x-12 x+30-35=0$

$x-5=0$

$x=0+5$

$x=5$

Hence 5 is added to each number 

Question 7

Ans: Let x be subtracted from each of the number 

$\therefore \quad 28 x: 53-x:: 19-x: 35-x$.

$\frac{28-x}{53-x}=\frac{19-x}{35-x} \text {}$

$(28-x)(35-x)=(19-x)(53-x) .$

$980-28 x-35 x+x^{2}=1007-19 x-53 x+x^{2}$

$980-28 x-35 x+ x^{2}-1007+19 x+53 x-x^{2}=0$

$980-63 x-1007+72 x=0 .$

$980-1007-63 x+72 x=0 .$

$-27+9 x=0$

$9 x=0+27$

$9 x=27$

x=3

Hence, 3 is subtracted from each number 

Question 8

Ans: 

(i) Given 

Mean proportional = 14 and third proportional= 122 

Let a and b be the two numbers

$\begin{aligned} a &: 14: 14: b \\ \frac{a}{14} &=\frac{14}{b} \\ a b &=14 \times 14 \\ a b &=196 . \end{aligned}$

$a=\frac{19 6}{b}$.............(i)

and a: b::b:112

$\begin{aligned} \frac{a}{b} &=\frac{b}{112} . \\ 112 a &=b^{2} . \\ 112 \times \frac{196}{b} &=b^{2} \end{aligned}$

$\begin{aligned} 21952 &=b^{3} \\ b^{3} &=21952 \\ b &=\sqrt[3]{21952} . \\ b &=\sqrt{28 \times 28 \times 28} . \\ b &=28 . \end{aligned}$

put the value of bin equation (i)

$a=\frac{196}{28}$

a=7 

Hence the two number a and b is 7 and 28 respectively 

(ii) Given,

Mean proportional =18 and 

Third proportional =144

Let a and b be the two number 

$\begin{aligned} \because a &: 18:: 18: b \\ \frac{a}{18} &=\frac{18}{b} \\ a b &=18 \times 18 \\ a b &=324 \end{aligned}$

a = $\frac{324}{b}$ ...........(i)

and a: b :: b:144 

$\frac{a}{b}=\frac{b}{144}$

$144 a=b^{2}$

$144 \times \frac{324}{b}=b^{2}$

$b^{2}=\frac{144 \times 324}{b^{\circ}}$

$b^{2}=\frac{46656}{b}$

$b^{3}=46656$

$b=\sqrt[3]{46656}$

$b=\sqrt{36 \times 36 \times 36}$

$b=36$

put the value of bin equation (i)

$a=\frac{324}{36}$

$a=9$

Hence, the two number a and b is 9 and 36 respectively.

Question 9

Ans:  Given 

$p+r=2 q$..........(i)

$\frac{1}{q}+\frac{1}{s}=\frac{2}{r}$..........(ii)

From eq (ii)

$\frac{s+q}{qs}=\frac{2}{r }$

$r(s+q)=2 q s$

$r(s+2)=(p+r) s$ (from (i))

rs +rq =ps +rs 

rq = ps +rs -rs

rq = ps 

$\frac{r}{s}=\frac{p}{q} .$

$r: s=p: q .$

$p: q=r: s$

Hence proved 

Question 10

Ans:  

Given

B is the mean proportional between a and c,

$\begin{aligned} \therefore \quad a &=b:: b: c \\ \frac{a}{b} &=\frac{b}{c} . \\ a c &=b^{2} . \end{aligned}$

According to the question 

$a, c, a^{2}+b^{2}$ and $b^{2}+c^{2}$

$\frac{a}{c}=\frac{a^{2}+b^{2}}{b^{2}+c^{2}}$

$\frac{a}{c}=\frac{a^{2}+a c}{a c+c^{2}} \quad\left(b^{2}=a c\right)$

$\frac{a}{c}=\frac{a(a+c)}{c(a+c)}$ 

$\frac{a}{c}=\frac{a}{c}$ 

Hence proved 

Question 11

Ans:  

Given 

x+ 7 is the mean proportional between (x+3) and (x+12)

∴ $x+3: x+7: \therefore x+7: x+12$

 $\frac{x+3}{x+7}=\frac{x+7}{x+12}$

$\left(x^{2}+12 x+2 x+12\right)=(x+7)(x+7)$

$\left(x^{2}+12 x+3 x+36\right)=(x+7)^{2}$

$x^{2}+15 x+36=x^{2}+(7)^{2}+2 \times x \times 7$

$x^{2}+15x+36= $x^{2}+49+14x

$x^{2}+15 x+36$$-x^{2}-49-4 x=0$

$15 x-14 x+36-49=0$

$x=13=0$

$x=0+13$

$x=13$

Hence, the value of x is 13

Question 12

Ans:  

Given 

$\frac{a^{2}+c^{2}}{a b+c d} = \frac{a b+c d}{b^{2}+d^{2}}$

$\left(a^{2}+c^{2}\right)\left(b^{2}+d^{2}\right)=(a b+c d)(a b+c d)$

$\left(a^{2} b^{2}+a^{2} d^{2}+b^{2} c^{2}+c^{2} d^{2}\right)=\left(a b+(d)^{2}\right.$

$\left(a^{2} b^{2}+a^{2} d^{2}+b^{2} c^{2}+c^{2} d^{2}\right)=(a b)^{2}+(c d)^{2}+2 x a b \times c d$

$a^{2} b^{2}+a^{2} d^{2}+b^{2} c^{2}+c^{2} d^{2}=a^{2} b^{2}+c^{2} d^{2}+2 a b c d $

$a^{2} b^{2}+a^{2} d^{2}+b^{2} c^{2}+c^{2} d^{2}-y^{2} b^{2}-c^{2} d^{2}-2 a b c d=0$

$a^{2} d^{2}+b^{2} c^{2}-2 a b c d=0 $

$(a d-b c)^{2}=0$

$a d-b c=0$

a d=b c

$\frac{a}{b}=\frac{c}{d}$, Hence Proved.