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.

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

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

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

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

$x^3+y^3=(\mathrm{x}+\mathrm{y})\left(x^2-\mathrm{xy}+y^2\right)$
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{aligned}
&\Rightarrow \frac{33}{40}+\frac{1}{5}\left[\frac{4}{5}-\frac{1}{5} \times\left(\frac{7}{8}-\frac{5}{4}\right)\right]-\frac{4}{5} \\
&=\frac{33}{40}+\frac{1}{5}\left[\frac{4}{5}-\frac{1}{5} \times\left(\frac{7-10}{8}\right)\right]-\frac{4}{5} \\
&=\frac{33}{40}+\frac{1}{5}\left[\frac{4}{5}-\frac{1}{5} \times\left(\frac{-3}{20}\right)\right]-\frac{4}{5} \\
&=\frac{33}{40}+\frac{1}{5}\left[\frac{4}{5}+\frac{3}{40}\right]-\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{aligned}

Let If $x$ is added to $\left(\frac{5}{10}+\frac{5}{100}+\frac{5}{1000}\right)$, it will be $\frac{5}{2}$. So
$
\begin{aligned}
&\left(\frac{5}{10}+\frac{5}{100}+\frac{5}{1000}\right)+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{aligned}
$
Hence, $1.945$ should be added to $\left(\frac{5}{10}+\frac{5}{100}+\frac{5}{1000}\right)$ to get $\frac{5}{2}$.

$
\begin{aligned}
&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{aligned}
$
So, $3 x=3 \times \frac{-1}{2}=\frac{-3}{2}$
Hence, Reciprocal of $3 x=\frac{1}{(-3 / 2)}=\frac{-2}{3}$

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

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

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

$
\begin{aligned}
&\Rightarrow 3 \times 4 \div 3 \text { of } 24-3 \div 2 \times(3-4) \times 2+6 \div 4 \text { of } 3 \\
&=3 \times 4 \div 72-3 \div 2 \times(-1) \times 2+6 \div 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{aligned}
$

$\left(\frac{-5}{55} \div \frac{2}{11}\right)=\frac{-1}{11} \times \frac{11}{2}=\frac{-1}{2}$
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{aligned}
&\frac{5 y}{2}=\frac{25}{4} \\
&y=\frac{50}{20}=\frac{5}{2}
\end{aligned}
$
So, $3 y^2+2 y=3 \times \frac{5}{2} \times \frac{5}{2}+2 \times \frac{5}{2}=\frac{75}{4}+5=\frac{75+20}{4}=\frac{95}{4}=23 \frac{3}{4}$

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

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

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

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