Free Practice Questions for Simplification in Maths

Bodmas stands for

B - Brackets

O - Orders 

D - Division

M - Multiplication

A - Addition

S - Subtraction

So,The letter B in the BODMAS rule stands for Brackets.

$\frac{10}{21}÷1\frac{1}{14}$ of $\frac{5}{3}-\frac{5}{3}\times \frac{1}{10}+\frac{9}{28}÷\frac{3}{7}$ of $\frac{6}{5}$
$\frac{10}{21}÷\frac{15}{14}\times \frac{5}{3}-\frac{5}{3}\times \frac{1}{10}+\frac{9}{28}÷\frac{3}{7}\times \frac{6}{5}$
$\frac{10}{21}\times \frac{14}{15}\times \frac{3}{5}-\frac{5}{3}\times \frac{1}{10}+\frac{9}{28}\times \frac{7}{3}\times \frac{5}{6}$
$\frac{4}{15}-\frac{1}{6}+\frac{5}{8}$
$\frac{29}{40}$

$=36÷18$ of $3\times 2+8÷4$ of $2\times 5-5÷10\times 2$
$=36÷54\times 2+8÷8\times 5-\frac{5}{10}\times 2$
$=\frac{36}{54}\times 2+5-1$
$=5\frac{1}{3}$

$\frac{5}{7}$ of $\frac{2}{5}+\frac{3}{7}-\frac{1}{14}\times 2÷1$
$\frac{5}{7}\times \frac{2}{5}+\frac{3}{7}-\frac{1}{7}÷1$
$=\frac{2}{7}+\frac{2}{7}$
$=\frac{4}{7}$

$(2\frac{6}{7}$ of $4\frac{1}{5}÷\frac{2}{3})\times 5\frac{1}{9}÷(\frac{3}{4}\times 2\frac{2}{3}$ of $\frac{1}{2}÷\frac{1}{4})$
$(\frac{20}{7}\times \frac{21}{5}\times \frac{3}{2})\times \frac{46}{9}÷(\frac{3}{4}\times \frac{8}{3}\times \frac{1}{2}\times \frac{4}{1})$
$=18\times \frac{46}{9}÷4$
$=23$

$x^3+y^3=(\mathrm{x}+\mathrm{y})(x^2-\mathrm{xy}+y^2)$
Now, $\frac{(251+249)(251\times 251-62499+249\times 249)}{25.1\times 25.1-624.99+24.9\times 24.9}=5\times {10}^K$
$\Rightarrow (251+297)\times 100=5\times {10}^K$
$\Rightarrow 50000=5\times {10}^K$
$\Rightarrow 5\times {10}^4=5\times {10}^K$
$\Rightarrow \mathrm{K}=4$

$$\begin{array}{rl}& \Rightarrow \frac{33}{40}+\frac{1}{5}[\frac{4}{5}-\frac{1}{5}\times (\frac{7}{8}-\frac{5}{4})]-\frac{4}{5}\\ & =\frac{33}{40}+\frac{1}{5}[\frac{4}{5}-\frac{1}{5}\times \left(\frac{7-10}{8}\right)]-\frac{4}{5}\\ & =\frac{33}{40}+\frac{1}{5}[\frac{4}{5}-\frac{1}{5}\times \left(\frac{-3}{20}\right)]-\frac{4}{5}\\ & =\frac{33}{40}+\frac{1}{5}[\frac{4}{5}+\frac{3}{40}]-\frac{4}{5}\\ & =\frac{33}{40}+\frac{1}{5}\left[\frac{32+3}{40}\right]-\frac{4}{5}\\ & =\frac{33}{40}+\frac{1}{5}\left[\frac{35}{40}\right]-\frac{4}{5}\\ & =\frac{33}{40}+\frac{7}{40}-\frac{4}{5}\\ & =\frac{40}{40}-\frac{4}{5}\\ & =1-\frac{4}{5}=\frac{1}{5}\end{array}$$

Let If $x$ is added to $(\frac{5}{10}+\frac{5}{100}+\frac{5}{1000})$, it will be $\frac{5}{2}$. So
$\begin{array}{rl}& (\frac{5}{10}+\frac{5}{100}+\frac{5}{1000})+x=\frac{5}{2}\\ & \left(\frac{500+50+5}{1000}\right)+x=2.5\\ & \frac{555}{1000}+x=2.5\\ & 0.555+x=2.5\\ & x=2.5-0.555=1.945\end{array}$
Hence, $1.945$ should be added to $(\frac{5}{10}+\frac{5}{100}+\frac{5}{1000})$ to get $\frac{5}{2}$.

$\begin{array}{rl}& x=\frac{2}{5}\times \frac{-3}{7}-\frac{1}{14}-\frac{3}{7}\times \frac{3}{5}\\ & x=\frac{-6}{35}-\frac{1}{14}-\frac{9}{35}\\ & x=\frac{-12-5-18}{70}\\ & x=\frac{-35}{70}=\frac{-1}{2}\end{array}$
So, $3x=3\times \frac{-1}{2}=\frac{-3}{2}$
Hence, Reciprocal of $3x=\frac{1}{(-3/2)}=\frac{-2}{3}$

$\Rightarrow [6+3$ of $7-5]÷2$ of 11
$=[6+21-5]÷22$
$=[6+21-5]÷22$
$=22÷22=1$

$\Rightarrow 75÷30-\frac{3}{4}$ of $(\frac{5}{7}÷\frac{9}{14})-(3\frac{1}{3}-1\frac{2}{3})$
$=75÷30-\frac{3}{4}$ of $(\frac{5}{7}\times \frac{14}{9})-(\frac{10}{3}-\frac{5}{3})$
$=75÷30-\frac{3}{4}\times \left(\frac{10}{9}\right)-\left(\frac{5}{3}\right)$
$=75÷30-\frac{5}{6}-\left(\frac{5}{3}\right)$
$=\frac{5}{2}-\frac{5}{6}-\left(\frac{5}{3}\right)$
$=\frac{15-5-10}{6}=0$

$\Rightarrow \frac{312÷26+60÷(42-55\text{ of }4÷10)\times 2}{9+(12-2\text{ of }4)}$
$=\frac{12+60÷(42-220÷10)\times 2}{9+(12-8)}$
$=\frac{12+60÷20\times 2}{9+4}$
$=\frac{12+3\times 2}{13}$
$=\frac{12+6}{13}$
$=\frac{18}{13}=1\frac{5}{13}$

$\begin{array}{rl}& \Rightarrow 3\times 4÷3\text{ of }24-3÷2\times (3-4)\times 2+6÷4\text{ of }3\\ & =3\times 4÷72-3÷2\times (-1)\times 2+6÷12\\ & =\frac{1}{6}-\frac{3}{2}\times (-1)\times 2+\frac{1}{2}\\ & =\frac{1}{6}+3+\frac{1}{2}=\frac{1+18+3}{6}=\frac{22}{6}=3\frac{2}{3}\end{array}$

$(\frac{-5}{55}÷\frac{2}{11})=\frac{-1}{11}\times \frac{11}{2}=\frac{-1}{2}$
Hence, Reciprocal of $\frac{-1}{2}$ is $-2$.

$4\frac{3}{8}+\frac{5}{16}=\frac{35}{8}+\frac{5}{16}=\frac{70+5}{16}=\frac{75}{16}=4\frac{11}{16}$

$\begin{array}{rl}& \frac{5y}{2}=\frac{25}{4}\\ & y=\frac{50}{20}=\frac{5}{2}\end{array}$
So, $3y^2+2y=3\times \frac{5}{2}\times \frac{5}{2}+2\times \frac{5}{2}=\frac{75}{4}+5=\frac{75+20}{4}=\frac{95}{4}=23\frac{3}{4}$

$2\frac{1}{3}÷\frac{3}{5}=\frac{7}{3}\times \frac{5}{3}=\frac{35}{9}$
Hence, Reciprocal of $2\frac{1}{3}÷\frac{3}{5}=$ Reciprocal of $\frac{35}{9}=\frac{9}{35}$

${\left[{(-\frac{5}{6})}^{-5}\right]}^3={(-\frac{5}{6})}^{-5\times 3}={(-\frac{5}{6})}^{-15}$

${\left(\frac{2}{7}\right)}^{-3}\times {\left(\frac{4}{49}\right)}^6={\left(\frac{2}{7}\right)}^{2m-1}$
${\left(\frac{2}{7}\right)}^{-3}\times {\left({\left(\frac{2}{7}\right)}^2\right)}^6={\left(\frac{2}{7}\right)}^{2m-1}$
${\left(\frac{2}{7}\right)}^{-3}\times {\left(\frac{2}{7}\right)}^{12}={\left(\frac{2}{7}\right)}^{2m-1}$
${\left(\frac{2}{7}\right)}^{(-3+12)}={\left(\frac{2}{7}\right)}^{2m-1}(\because a^m\times a^n=a^{(m+n)})$
${\left(\frac{2}{7}\right)}^9={\left(\frac{2}{7}\right)}^{2m-1}$
$2m-1=9$
$2m=10$
$m=5$

$\Rightarrow [25+8÷2-\{16+(14$ of $7÷14)-(18÷12$ of $\frac{1}{2})\}]$
$=[25+8÷2-\{16+(98÷14)-(18÷6)\}]$
$=[25+8÷2-\{16+7-3\}]$
$=[25+8÷2-20]$
$=[25+4-20]$
$=9$