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
$(a+b)(a+b)=(a+b{)}^2=(a^2+b^2+2ab)$ and
$(a-b)(a-b)=(a-b{)}^2=(a^2+b^2-2ab)$
$\Rightarrow \frac{(7+\sqrt{5})(7+\sqrt{5})-(7-\sqrt{5})(7-\sqrt{5})}{(7{)}^2-(\sqrt{5}{)}^2}$
$\Rightarrow \frac{[(7{)}^2+(\sqrt{5}{)}^2+2(7)(\sqrt{5})]-[(7{)}^2+(\sqrt{5}{)}^2-2(7)(\sqrt{5})}{49-5}$
$\Rightarrow \frac{[49+5+14\sqrt{5}-[49+5-14\sqrt{5}}{44}$
$\Rightarrow \frac{54+14\sqrt{5}-54+14\sqrt{5}}{44}$
$\Rightarrow \frac{28\sqrt{5}}{44}$
$\Rightarrow \frac{7\sqrt{5}}{11}=a+7\sqrt{5}b$
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:
$(a+b)(a+b)=(a+b{)}^2=(a^2+b^2+2ab)$
$\Rightarrow \frac{(3{)}^2+(\sqrt{7}{)}^2+2(3)(\sqrt{7})}{(3{)}^2-(\sqrt{7}{)}^2}$
$\Rightarrow \frac{9+7+6\sqrt{7}}{9-7}$
$\Rightarrow \frac{16+6\sqrt{7}}{2}$
$\Rightarrow 8+3\sqrt{7}=a+b\sqrt{7}$
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 $a^2-b^2+ab$
$\Rightarrow {\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)}^2-{\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)}^2+\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)$
Using identity: $(a+b{)}^2=a^2+b^2+2ab$
$(a-b{)}^2=a^2+b^2-2ab$
$\Rightarrow \left(\frac{(\sqrt{3}{)}^2+(1{)}^2+2(\sqrt{3})(1)}{(\sqrt{3}{)}^2+(1{)}^2-2(\sqrt{3})(1)}\right)-\left(\frac{(\sqrt{3}{)}^2+(1{)}^2-2(\sqrt{3})(1)}{(\sqrt{3}{)}^2+(1{)}^2+2(\sqrt{3})(1)}\right)+1$
$\Rightarrow \left(\frac{3+1+2\sqrt{3}}{3+1-2\sqrt{3}}\right)-\left(\frac{3+1-2\sqrt{3}}{3+1+2\sqrt{3}}\right)+1$
$\Rightarrow \left(\frac{4+2\sqrt{3}}{4-2\sqrt{3}}\right)-\left(\frac{4-2\sqrt{3}}{4+2\sqrt{3}}\right)+1$
$\Rightarrow \left(\frac{(4+2\sqrt{3})(4+2\sqrt{3})-(4-2\sqrt{3})(4-2\sqrt{3})}{(4-2\sqrt{3})(4+2\sqrt{3})}\right))+1$
$\Rightarrow \frac{16+12+16\sqrt{3}-(16+12-16\sqrt{3})}{16+8\sqrt{3}-8\sqrt{3}-12}+1$
$\Rightarrow \frac{32\sqrt{3}}{4}+1$
$\Rightarrow 1+8\sqrt{3}$

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

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

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

$\Rightarrow \frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}$
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:
$4\frac{4}{5}÷\{2\frac{1}{5}-\frac{1}{2}(1\frac{1}{4}-\overline{\frac{1}{4}-\frac{1}{5}})\}$
$=4\frac{4}{5}÷\{\frac{11}{5}-\frac{1}{2}(\frac{5}{4}-\overline{\frac{1}{4}-\frac{1}{5}})\}$
$=\frac{24}{5}÷\{\frac{11}{5}-\frac{1}{2}(\frac{5}{4}-\frac{1}{20})\}$
$=\frac{24}{5}÷\{\frac{11}{5}-\frac{1}{2}\left(\frac{25-1}{20}\right)\}$
$=\frac{24}{5}÷\{\frac{11}{5}-\frac{1}{2}\times \frac{24}{20}\}$
$=\frac{24}{5}÷\{\frac{11}{5}-\frac{12}{20}\}$
$=\frac{24}{5}÷\left\{\frac{44-12}{20}\right\}$
$=\frac{24}{5}÷\frac{32}{20}$
$=\frac{24}{5}\times \frac{20}{32}$
$=\frac{3}{4}\times 4=3$

Given expression

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

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

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

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

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

$=\frac{15}{2}-[\frac{9}{4}÷\{\frac{5}{4}-\frac{2}{3}\}]$

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

$=\frac{15}{2}-[\frac{9}{4}÷\frac{7}{12}]$

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

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

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

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

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

$=\frac{25}{7}$

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

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

$=\frac{8}{2}$

$=4$

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

$=\frac{41}{25}$ 

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

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,
$\frac{a^2}{b}+\frac{b^2}{a}$
$\Rightarrow \frac{a^3+b^3}{ab}=\frac{(a+b)(a^2+b^2-ab)}{ab}$
$=\frac{[(5-2\sqrt{6})+(5+2\sqrt{6})][(49-20\sqrt{6})+(49+20\sqrt{6})-(5-2\sqrt{6})(5+2\sqrt{6})]}{(5-2\sqrt{6})(5+2\sqrt{6})}$
$=\frac{10[49+49-(25-24)]}{25-24}$
$=10\times 97$

$=970$

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

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

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

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

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