Free Practice Questions for Fraction in Maths

$\Rightarrow \frac{(7+\sqrt{5})(7+\sqrt{5})-(7-\sqrt{5})(7-\sqrt{5})}{(7-\sqrt{5})(7+\sqrt{5})}$
Using identity:
$(a+b)(a-b)=a^{2}-b^{2}$ and
$(a+b)(a+b)=(a+b)^{2}=\left(a^{2}+b^{2}+2 a b\right)$ and
$(a-b)(a-b)=(a-b)^{2}=\left(a^{2}+b^{2}-2 a b\right)$
$\Rightarrow \frac{(7+\sqrt{5})(7+\sqrt{5})-(7-\sqrt{5})(7-\sqrt{5})}{(7)^{2}-(\sqrt{5})^{2}}$
$\Rightarrow \frac{\left[(7)^{2}+(\sqrt{5})^{2}+2(7)(\sqrt{5})\right]-\left[(7)^{2}+(\sqrt{5})^{2}-2(7)(\sqrt{5})\right.}{49-5}$
$\Rightarrow \frac{[49+5+14 \sqrt{5}-[49+5-14 \sqrt{5}}{44}$
$\Rightarrow \frac{54+14 \sqrt{5}-54+14 \sqrt{5}}{44}$
$\Rightarrow \frac{28 \sqrt{5}}{44}$
$\Rightarrow \frac{7 \sqrt{5}}{11}=a+7 \sqrt{5} b$
On comparing, we get
$\Rightarrow a=0, b=\frac{1}{11}$

$
\Rightarrow(36)^{1 / 4} \times(6)^{-1 / 3}
$
LCM of 3,4 is 12
$\Rightarrow(36)^{\frac{1}{4} \times \frac{3}{3}} \times(6)^{-\frac{1}{3} \times \frac{4}{4}}$
$\Rightarrow(36)^{\frac{3}{12}} \times(6)^{-\frac{4}{12}}$
$\Rightarrow \sqrt[12]{36^{3} \times 6^{-4}}$
$\Rightarrow \sqrt[12]{\frac{36^{3}}{6^{4}}}$
$\Rightarrow \sqrt[12]{\frac{36 \times 36 \times 36}{6 \times 6 \times 6 \times 6}}$
$\Rightarrow \sqrt[12]{6 \times 6}$
$\Rightarrow \sqrt[12]{2^{2} \times 3^{2}}$
$\Rightarrow \sqrt[12]{6^{2}}$
$\Rightarrow 6^{2 / 12}$
$\Rightarrow 6^{1 / 6}$
$\Rightarrow \sqrt[6]{6}$

$\Rightarrow \frac{3+\sqrt{7}}{3-\sqrt{7}} \times \frac{3+\sqrt{7}}{3+\sqrt{7}}$
Using identity:
$(a+b)(a-b)=a^{2}-b^{2}$
$\Rightarrow \frac{(3+\sqrt{7})(3+\sqrt{7})}{(3)^{2}-(\sqrt{7})^{2}}$
Using identity:
$(a+b)(a+b)=(a+b)^{2}=\left(a^{2}+b^{2}+2 a b\right)$
$\Rightarrow \frac{(3)^{2}+(\sqrt{7})^{2}+2(3)(\sqrt{7})}{(3)^{2}-(\sqrt{7})^{2}}$
$\Rightarrow \frac{9+7+6 \sqrt{7}}{9-7}$
$\Rightarrow \frac{16+6 \sqrt{7}}{2}$
$\Rightarrow 8+3 \sqrt{7}=a+b \sqrt{7}$
On comparing we get $\Rightarrow \mathrm{a}=8, \mathrm{~b}=3$

$\Rightarrow a=\frac{\sqrt{3}+1}{\sqrt{3}-1}$ and $b=\frac{1}{a}=\frac{\sqrt{3}-1}{\sqrt{3}+1} \Rightarrow$ To find $a^{2}-b^{2}+a b$
$\Rightarrow\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)^{2}-\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)^{2}+\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)$
Using identity: $(a+b)^{2}=a^{2}+b^{2}+2 a b$
$(a-b)^{2}=a^{2}+b^{2}-2 a b$
$\Rightarrow\left(\frac{(\sqrt{3})^{2}+(1)^{2}+2(\sqrt{3})(1)}{(\sqrt{3})^{2}+(1)^{2}-2(\sqrt{3})(1)}\right)-\left(\frac{(\sqrt{3})^{2}+(1)^{2}-2(\sqrt{3})(1)}{(\sqrt{3})^{2}+(1)^{2}+2(\sqrt{3})(1)}\right)+1$
$\Rightarrow\left(\frac{3+1+2 \sqrt{3}}{3+1-2 \sqrt{3}}\right)-\left(\frac{3+1-2 \sqrt{3}}{3+1+2 \sqrt{3}}\right)+1$
$\Rightarrow\left(\frac{4+2 \sqrt{3}}{4-2 \sqrt{3}}\right)-\left(\frac{4-2 \sqrt{3}}{4+2 \sqrt{3}}\right)+1$
$\left.\Rightarrow\left(\frac{(4+2 \sqrt{3})(4+2 \sqrt{3})-(4-2 \sqrt{3})(4-2 \sqrt{3})}{(4-2 \sqrt{3})(4+2 \sqrt{3})}\right)\right)+1$
$\Rightarrow \frac{16+12+16 \sqrt{3}-(16+12-16 \sqrt{3})}{16+8 \sqrt{3}-8 \sqrt{3}-12}+1$
$\Rightarrow \frac{32 \sqrt{3}}{4}+1$
$\Rightarrow 1+8 \sqrt{3}$

$\Rightarrow \frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}} \times \frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\sqrt{35}$
Using identity:
$(a+b)(a-b)=a^{2}-b^{2}$
and
$(a-b)(a-b)=(a-b)^{2}=a^{2}+b^{2}-2 a b$
$\Rightarrow \frac{(\sqrt{7})^{2}+(\sqrt{5})^{2}-2(\sqrt{7})(\sqrt{5})}{(\sqrt{7})^{2}-(\sqrt{5})^{2}}+\sqrt{35}$

$\Rightarrow \frac{7+5-2 \sqrt{35}}{7-5}+\sqrt{35}$
$\Rightarrow \frac{12-2 \sqrt{35}}{2}+\sqrt{35}$
$\Rightarrow \frac{12-2 \sqrt{35}+2(\sqrt{35})}{2}$
$\Rightarrow \frac{12-2 \sqrt{35}+2 \sqrt{35}}{2}$
$\Rightarrow \frac{12}{2}$
$\Rightarrow 6$

$\Rightarrow \frac{(\sqrt{5}-2)(\sqrt{5}-2)-(\sqrt{5}+2)(\sqrt{5}+2)}{(\sqrt{5}+2)(\sqrt{5}-2)}$
Using identity:
$(a+b)(a-b)=a^{2}-b^{2}$
and
$(a-b)(a-b)=(a-b)^{2}=a^{2}+b^{2}-2 a b$
and
$(a+b)(a+b)=(a+b)^{2}=a^{2}+b^{2}+2 a b$
$\Rightarrow \frac{\left.\left[(\sqrt{5})^{2}+(2)^{2}-2(\sqrt{5})(2)\right]-\left[(\sqrt{5})^{2}+(2)^{2}+2(\sqrt{5})(2)\right)\right]}{(\sqrt{5})^{2}-(2)^{2}}$
$\Rightarrow \frac{5+4-4 \sqrt{5}-[5+4+4 \sqrt{5}]}{5-4}$
$\Rightarrow 9-4 \sqrt{5}-9-4 \sqrt{5}$
$\Rightarrow-8 \sqrt{5}$

$\Rightarrow \frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}$
Using identity:
$(a+b)(a-b)=a^{2}-b^{2}$
$\Rightarrow \frac{2+\sqrt{3}}{2^{2}-\sqrt{3}^{2}}$
$\Rightarrow \frac{2+\sqrt{3}}{4-3}$
$\Rightarrow 2-\sqrt{3}$

Given expression:
$4 \frac{4}{5} \div\left\{2 \frac{1}{5}-\frac{1}{2}\left(1 \frac{1}{4}-\overline{\frac{1}{4}-\frac{1}{5}}\right)\right\}$
$=4 \frac{4}{5} \div\left\{\frac{11}{5}-\frac{1}{2}\left( \frac{5}{4}-\overline{\frac{1}{4}-\frac{1}{5}}\right)\right\}$
$=\frac{24}{5} \div\left\{\frac{11}{5}-\frac{1}{2}\left(\frac{5}{4}-\frac{1}{20}\right)\right\}$
$=\frac{24}{5} \div\left\{\frac{11}{5}-\frac{1}{2}\left(\frac{25-1}{20}\right)\right\}$
$=\frac{24}{5} \div\left\{\frac{11}{5}-\frac{1}{2} \times \frac{24}{20}\right\}$
$=\frac{24}{5} \div\left\{\frac{11}{5}-\frac{12}{20}\right\}$
$=\frac{24}{5} \div\left\{\frac{44-12}{20}\right\}$
$=\frac{24}{5} \div \frac{32}{20}$
$=\frac{24}{5} \times \frac{20}{32}$
$=\frac{3}{4} \times 4=3$

Given expression

$=7 \frac{1}{2}-\left[2 \frac{1}{4} \div\left\{1 \frac{1}{4}-\frac{1}{2}\left(\frac{3}{2}-\overline{\frac{1}{3}-\frac{1}{6}}\right)\right\}\right]$

$=\frac{15}{2}-\left[\frac{9}{4} \div\left\{\frac{5}{4}-\frac{1}{2}\left(\frac{3}{2}-\overline{\frac{1}{3}-\frac{1}{6}}\right)\right\}\right]$

$=\frac{15}{2}-\left[\frac{9}{4} \div\left\{\frac{5}{4}-\frac{1}{2}\left(\frac{3}{2}-\frac{1}{6}\right)\right\}\right]$

$=\frac{15}{2}-\left[\frac{9}{4} \div\left\{\frac{5}{4}-\frac{1}{2}\left(\frac{9-1}{6}\right)\right\}\right]$

$=\frac{15}{2}-\left[\frac{9}{4} \div\left\{\frac{5}{4}-\frac{1}{2} \times \frac{4}{3}\right\}\right]$

$=\frac{15}{2}-\left[\frac{9}{4} \div\left\{\frac{5}{4}-\frac{2}{3}\right\}\right]$

$=\frac{15}{2}-\left[\frac{9}{4} \div\left\{\frac{15-8}{12}\right\}\right]$

$=\frac{15}{2}-\left[\frac{9}{4} \div \frac{7}{12}\right]$

$=\frac{15}{2}-\left[\frac{9}{4} \times \frac{12}{7}\right]$

$=\frac{15}{2}-\frac{27}{7}$

$=\frac{105-54}{14}$

$=\frac{51}{14}$$=3 \frac{9}{14}$

Given expression:
$5 \frac{1}{7}-\left\{3 \frac{3}{10} \div\left(2 \frac{4}{5}-\frac{7}{10}\right)\right\}$
$=\frac{36}{7}-\left\{\frac{33}{10} \div\left(\frac{14}{5}-\frac{7}{10}\right)\right\}$
$=\frac{36}{7}-\left\{\frac{33}{10} \div\left(\frac{28-7}{10}\right)\right\}$
$=\frac{36}{7}-\left\{\frac{33}{10} \div \frac{21}{10}\right\}$
$=\frac{36}{7}-\left\{\frac{33}{10} \times \frac{10}{21}\right\}$
$=\frac{36}{7}-\frac{11}{7}$
$=\frac{36-11}{7}$

$=\frac{25}{7}$

$=3 \frac{4}{7}$

The given expression
$=7 \frac{1}{3} \div \frac{2}{3}$ of $2 \frac{1}{5}+1 \frac{3}{8} \div 2 \frac{3}{4}-1 \frac{1}{2}$
$=\frac{22}{3} \div \frac{2}{3}$ of $\frac{11}{5}+\frac{11}{8} \div \frac{11}{4}-\frac{3}{2}$
$=\frac{22}{3} \div \frac{2}{3} \times \frac{11}{5}+\frac{11}{8} \div \frac{11}{4}-\frac{3}{2}$
$=\frac{22}{3} \div \frac{22}{15}+\frac{11}{8} \div \frac{11}{4}-\frac{3}{2}$
$=\frac{22}{3} \times \frac{15}{22}+\frac{11}{8} \div \frac{11}{4}-\frac{3}{2}$
$=5+\frac{11}{8} \times \frac{4}{11}-\frac{3}{2}$
$=5+\frac{1}{2}-\frac{3}{2}$
$=\frac{10+1-3}{2}$

$=\frac{8}{2}$

$=4$

Given expression:
$=4 \frac{4}{5} \div \frac{3}{5}$ of $5+\frac{4}{5} \times \frac{3}{10}-\frac{1}{5}$
$=\frac{24}{5} \div \frac{3}{1}+\frac{4}{5} \times \frac{3}{10}-\frac{1}{5}$
$=\frac{8}{5}+\frac{4}{5} \times \frac{3}{10}-\frac{1}{5}$
$=\frac{8}{5}+\frac{6}{25}-\frac{1}{5}$
$=\frac{40+6-5}{25}$

$=\frac{41}{25}$ 

$=1 \frac{16}{25}$

Given : $a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$\Rightarrow a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \times\left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right)$
$\Rightarrow a=\frac{(\sqrt{3}-\sqrt{2})^{2}}{3-2}$
$\Rightarrow a=5-2 \sqrt{6}$
Squaring both sides, we get :
$
a^{2}=49-20 \sqrt{6}
$
Similarly,
$
b=5+2 \sqrt{6} \text { and } b^{2}=49+20 \sqrt{6}
$
Now,
$\frac{a^{2}}{b}+\frac{b^{2}}{a}$
$\Rightarrow \frac{a^{3}+b^{3}}{a b}=\frac{(a+b)\left(a^{2}+b^{2}-a b\right)}{a b}$
$=\frac{[(5-2 \sqrt{6})+(5+2 \sqrt{6})][(49-20 \sqrt{6})+(49+20 \sqrt{6})-(5-2 \sqrt{6})(5+2 \sqrt{6})]}{(5-2 \sqrt{6})(5+2 \sqrt{6})}$
$=\frac{10[49+49-(25-24)]}{25-24}$
$=10 \times 97$

$=970$

Given : $x=\frac{2 \sqrt{24}}{\sqrt{3}+\sqrt{2}}$
$=x=\frac{2 \sqrt{24}}{\sqrt{3}+\sqrt{2}} \times\left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right)$
$=x=\frac{2 \sqrt{24}(\sqrt{3}-\sqrt{2})}{3-2}$
$x=2 \sqrt{72}-2 \sqrt{48}$
$x=6 \sqrt{8}-4 \sqrt{12}$
To find : $\frac{x+\sqrt{8}}{x-\sqrt{8}}+\frac{x+\sqrt{12}}{x-\sqrt{12}}$
$=\frac{6 \sqrt{8}-4 \sqrt{12}+\sqrt{8}}{6 \sqrt{8}-4 \sqrt{12}-\sqrt{8}}+\frac{6 \sqrt{8}-4 \sqrt{12}+\sqrt{12}}{6 \sqrt{8}-4 \sqrt{12}-\sqrt{12}}$
$=\frac{7 \sqrt{8}-4 \sqrt{12}}{5 \sqrt{8}-4 \sqrt{12}}+\frac{6 \sqrt{8}-3 \sqrt{12}}{6 \sqrt{8}-5 \sqrt{12}}$
$=\frac{(336-35 \sqrt{96}-24 \sqrt{96}+240)+(240-15 \sqrt{96}-24 \sqrt{96}+144)}{240-25 \sqrt{96}-24 \sqrt{96}+240}$
$=\frac{960-98 \sqrt{96}}{480-49 \sqrt{96}}$
$=\frac{2(480-49 \sqrt{6})}{480-49 \sqrt{6}}$

$=2$

Given : $x=\frac{2 \sqrt{6}}{\sqrt{3}+\sqrt{2}}$
$\Rightarrow x=\frac{2 \sqrt{6}}{\sqrt{3}+\sqrt{2}} \times\left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right)$
$\Rightarrow x=6 \sqrt{2}-4 \sqrt{3}$
Now: $\frac{x+\sqrt{2}}{x-\sqrt{2}}+\frac{x+\sqrt{3}}{x-\sqrt{3}}$
$=\frac{6 \sqrt{2}-4 \sqrt{3}+\sqrt{2}}{6 \sqrt{2}-4 \sqrt{3}-\sqrt{2}}+\frac{6 \sqrt{2}-4 \sqrt{3}+\sqrt{3}}{6 \sqrt{2}-4 \sqrt{3}-\sqrt{3}}$
$=\frac{7 \sqrt{2}-4 \sqrt{3}}{5 \sqrt{2}-4 \sqrt{3}}+\frac{6 \sqrt{2}-3 \sqrt{3}}{6 \sqrt{2}-5 \sqrt{3}}$
$=\frac{(84-35 \sqrt{6}-24 \sqrt{6}+60)+(60-15 \sqrt{6}-24 \sqrt{6}+36)}{60-25 \sqrt{6}-24 \sqrt{6}+60}$
$=\frac{240-98 \sqrt{6}}{120-49 \sqrt{6}}$
$=\frac{2(120-49 \sqrt{6})}{120-49 \sqrt{6}}=2$

$2 \sqrt{y}=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$
$\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} \times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{(\sqrt{5}+\sqrt{3})^{2}}{2}$
similarly, $\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{(\sqrt{5}-\sqrt{3})^{2}}{2}$
$\frac{(\sqrt{5}+\sqrt{3})^{2}}{2}+\frac{(\sqrt{5}-\sqrt{3})^{2}}{2}=2 \sqrt{(} y)$
$8=2 \sqrt{(} y)$
$y=16$

$\left(\frac{1728}{729}\right)^{1 / 3} \times \frac{3^{2}}{\sqrt{144}} \times \sqrt{5}$
$\Rightarrow\left[\frac{12^{3}}{9^{3}}\right]^{1 / 3} \times \frac{9}{12} \times \sqrt{5}$
$\Rightarrow\left[\frac{12}{9}\right]^{3 \times \frac{1}{3}} \times \frac{9}{12} \times \sqrt{5}$
$\Rightarrow \frac{12}{9} \times \frac{9}{12} \times \sqrt{5}$
$\Rightarrow \sqrt{5}$

$\left[\left(\sqrt{\frac{16}{25}}\right)^{4 x-9}\right]^{3}=\frac{64}{125}$
$\Rightarrow\left(\frac{4}{5}\right)^{4 x-9}=\sqrt[3]{\frac{64}{125}}$
$\Rightarrow\left(\frac{4}{5}\right)^{4 x-9}=\left(\frac{4}{5}\right)^{3 \frac{1}{3}}$
$\Rightarrow\left(\frac{4}{5}\right)^{4 x-9}=\left(\frac{4}{5}\right)^{1}$
$\Rightarrow 4 x-9=1$
$\Rightarrow 4 x=10$
$x=5 / 2$

$\left(\frac{2 a}{b}\right)^{2 x-4}=\left(\frac{b}{2 a}\right)^{2 x-4}$
$\Rightarrow\left(\frac{2 a}{b}\right)^{2 x-4}=\left(\frac{2 a}{b}\right)^{-2 x+4}$
$\Rightarrow 2 x-4=-2 x+4$
$\Rightarrow 4 x=4+4$
$\Rightarrow 4 x=8$
$\Rightarrow x=2$

$\Rightarrow \frac{14^{n+4}+7^{n+3} \times 2^{n+3}}{13 \times 14^{n}+14^{n}}$
$\Rightarrow \frac{14^{n+4}+14^{n+3}}{14^{n}(13+1)}$
$\Rightarrow \frac{14^{n+3}(14+1)}{14^{n}(13+1)} \Rightarrow \frac{14^{n+3-n} \times 15}{14}$
$\Rightarrow \frac{14 \times 14 \times 14 \times 15}{14} \Rightarrow 2940$