Free Practice Questions for Simplification in Maths

$\frac{3}{4} \div \frac{5}{8} \div \frac{3}{5}$ of $\frac{5}{6}$ of $\frac{7}{10}$

$=\frac{3}{4} \div \frac{5}{8} \div \frac{7}{20}$

$=\frac{3}{4} \times \frac{8}{5} \times \frac{20}{7}$

$=\frac{24}{7}$

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

$2^{32}-(1)(2+1)\left(2^{2}+1\right)\left(2^{4}+1\right)\left(2^{8}+1\right)\left(2^{16}+1\right)$

$2^{32}-(2-1)(2+1)\left(2^{2}+1\right)\left(2^{4}+1\right)\left(2^{8}+1\right)\left(2^{16}+1\right)$

$2^{32}-\left(2^{2}-1\right)\left(2^{2}+1\right)\left(2^{4}+1\right)\left(2^{8}+1\right)\left(2^{16}+1\right)$

$=2^{32}-\left(2^{4}-1\right)\left(2^{4}+1\right)\left(2^{8}+1\right)\left(2^{16}+1\right)$

$=2^{32}-\left(2^{8}-1\right)\left(2^{8}+1\right)\left(2^{16}+1\right)$

$=2^{32}-\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-2^{32}+1=1$

$\left(\frac{81}{16}\right)^{\frac{-3}{4}} \times\left[\left(\frac{25}{9}\right)^{\frac{-3}{2}} \div\left(\frac{5}{2}\right)^{-3}\right]$
$\left(\frac{3^4}{2^4}\right)^{\frac{-3}{4}} \times\left[\left(\frac{5^2}{3^2}\right)^{\frac{-3}{2}} \div\left(\frac{5}{2}\right)^{-3}\right]$
$\left(\frac{3}{2}\right)^{-3} \times\left[\left(\frac{5}{3}\right)^{-3} \div\left(\frac{5}{2}\right)^{-3}\right]$
$\left(\frac{2}{3}\right)^3 \times\left[\left(\frac{3}{5}\right)^3 \div\left(\frac{2}{5}\right)^3\right]$
$\left(\frac{2}{3}\right)^3 \times\left[\frac{3^3}{5^3} \times \frac{5^3}{2^3}\right]$
$\frac{2^3}{3^3} \times \frac{3^3}{2^3}=1$

$40 \div 5 \text { of } 2 \times[18 \div 6 \times(12-9) \text { of } 5-(3-8)] \div 25$

Using BODMAS,

$ 40 \div 10 \times[3 \times 15-(-5)] \div 25 $

$ =4 \times[45-(-5)] \div 25 $

$ =4 \times[45+5] \div 25 $

$ =4 \times 50 \div 25 $

$=4 \times 2=8$

$\frac{63.5 \times 63.5 \times 63.5+36.5 \times 36.5 \times 36.5}{6.35 \times 6.35+3.65 \times 3.65-6.35 \times 3.65}$

We know that,

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

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

Here, $a=63.5$ and $b=36.5$

$=63.5 + 36.5 = 100$

$\Rightarrow 60 \div 5$ of $2 \times[18 \div 6 \times(12-9)$ of $5-(3-8)] \div 50$

$=60 \div 10 \times[18 \div 6 \times 3$ of $5-(-5)] \div 50$

$=6 \times[18 \div 6 \times 15+5] \div 50$

$=6 \times[3 \times 15+5] \div 50$

$=6 \times[45+5] \div 50$

$=6 \times 50 \div 50$

$=6 \times 1$

$=6$

$\Rightarrow \frac{5-30 \div 5 \times 15+5}{12-2}$
$=\frac{5-6 \times 15+5}{10}$
$=\frac{10-90}{10}$
$=\frac{-80}{10}$
$=-8$

$\begin{aligned}
&\Rightarrow 60^3+20^3-90^3 \\
&=216000+8000-729000 \\
&=224000-729000 \\
&=-505000
\end{aligned}
$

$ \frac{2 \frac{1}{2}+4 \frac{1}{3} \div 2 \frac{1}{3} \times 2 \frac{1}{4}-4 \frac{1}{4}}{1 \frac{1}{2} \times 2 \frac{1}{3}-5 \frac{1}{2}} \div 6 \times 2 $

$=\frac{\frac{5}{2}+\frac{13}{3} \div \frac{7}{3} \times \frac{9}{4}-\frac{17}{4}}{\frac{3}{2} \times \frac{7}{3}-\frac{11}{2}} \div 6 \times 2$

$=\frac{\frac{5}{2}+\frac{13}{3} \times \frac{3}{7} \times \frac{9}{4}-\frac{17}{4}}{\frac{3}{2} \times \frac{7}{3}-\frac{11}{2}} \div 6 \times 2$

$=\frac{\frac{5}{2}+\frac{117}{28}-\frac{17}{4}}{\frac{7}{2}-\frac{11}{2}} \div 6 \times 2$

$=\frac{\frac{70+117-119}{28}}{\frac{-4}{2}}\div 6 \times 2$

$=\frac{68}{28} \times \frac{2}{-4} \times \frac{1}{6}\times2$

$=\frac{-17}{42}$

$40 \%$ of $210+\sqrt{ } 289\times 15-51 \times \sqrt{ } 16=?$

$=84+255-204$

$=135$

$\Rightarrow(a+b)^{-1} \times\left(a^{-1}+b^{-1}\right)$
$=\frac{1}{(a+b)} \times\left(\frac{1}{a}+\frac{1}{b}\right),\left\{\because x^{-1}=\frac{1}{x}\right\}$
$=\frac{1}{(a+b)} \times\left(\frac{a+b}{a b}\right)$
$=\frac{1}{a b}$
$=\frac{1}{\frac{5}{11} \times \frac{3}{7}}=\frac{1}{\frac{15}{77}}=\frac{77}{15}$

$\frac{\sqrt{450}}{\sqrt{338}} \div \frac{\sqrt{125}}{\sqrt{245}}=\frac{\sqrt{450}}{\sqrt{338}} \times \frac{\sqrt{245}}{\sqrt{125}}=\frac{\sqrt{2 \times 225}}{\sqrt{2 \times 169}} \times \frac{\sqrt{5 \times 49}}{\sqrt{5 \times 25}}=\frac{15 \sqrt{2} \times 7 \sqrt{5}}{13 \sqrt{2} \times 5 \sqrt{5}}=\frac{21}{13}$

$
(0.102+0.25+0.111+0.1)=0.563
$
Let $x$ is added to $0.563$ to get 1 , then
$
\begin{aligned}
&0.563+x=1 \\
&x=1-0.563=0.437
\end{aligned}
$

$\Rightarrow\left[\left(\frac{4}{3}\right)^2\right]^3 \times\left(\frac{2}{3}\right)^{-4} \times\left(\frac{27}{16}\right)^2 \div 3^3$
$=\left(\frac{2^2}{3}\right)^{2 \times 3} \times\left(\frac{3}{2}\right)^4 \times\left(\frac{3^3}{2^4}\right)^2 \div 3^3$
$=\left(\frac{2^2}{3}\right)^6 \times\left(\frac{3}{2}\right)^4 \times\left(\frac{3^3}{2^4}\right)^2 \times \frac{1}{3^3}$
$=\frac{2^{12}}{3^6} \times \frac{3^4}{2^4} \times \frac{3^6}{2^8} \times \frac{1}{3^3}$
$=\frac{2^{12}}{3^{(6+3)}} \times \frac{3^{(4+6)}}{2^{(4+8)}}$
$=\frac{2^{12}}{3^9} \times \frac{3^{10}}{2^{12}}$
$=3$

$3 A B+A B 1=C 18$
$B+1=8$, So $B=7$
Now, $3 A 7+A 71=C 18$
$A+7=1$, So $A=4$
Now, $347+471=C 18$
$34+47=C 1$
$C 1=81$
$C=8$

$
2 \frac{1}{3} \div \frac{2}{5}=\frac{7}{3} \times \frac{5}{2}=\frac{35}{6}
$
Hence, $\frac{35}{6}$ is the correct answer.

$
-25 \times 7+(-25) \times 3=-25 \times(7+3)=-25 \times 10=-250
$
Hence, $-25 \times 7 \times 3$ is not same as $-25 \times(7+3)$.

$\begin{aligned}
&\Rightarrow 36 \frac{2}{5} \div\left[27+\frac{3}{8} \text { of } 24+\left(45 \div 9-4 \frac{3}{5}\right)\right] \\
&=\frac{182}{5} \div\left[27+9+\left(\frac{45}{9}-\frac{23}{5}\right)\right] \\
&=\frac{182}{5} \div\left[36+\left(\frac{225-207}{45}\right)\right] \\
&=\frac{182}{5} \div\left[36+\frac{18}{45}\right] \\
&=\frac{182}{5} \div\left[36+\frac{2}{5}\right] \\
&=\frac{182}{5} \div\left[\frac{182}{5}\right] \\
&=\frac{182}{5} \times \frac{5}{182} \\
&=1
\end{aligned}$

Using BODMAS formula ,
$
=\frac{67+\frac{2}{5} \text { of } 75}{5+\frac{2}{3} \times(58-37)}
$
$
\begin{aligned}
&=\frac{67+30}{5+\frac{2}{3} \times 21}=\frac{67+30}{5+14} \\
&=\frac{97}{19}=5 \frac{2}{19}
\end{aligned}
$

\begin{aligned}
&\Rightarrow \frac{5-\left[\frac{3}{4}+\left\{2 \frac{1}{2}-\left(\frac{1}{2}+\frac{1}{6}-\frac{1}{7}\right)\right\}\right]}{6} \\
&=\frac{5-\left[\frac{3}{4}+\left\{\frac{5}{2}-\left(\frac{1}{2}+\frac{7-6}{42}\right)\right\}\right]}{6} \\
&=\frac{5-\left[\frac{3}{4}+\left\{\frac{5}{2}-\left(\frac{1}{2}+\frac{1}{42}\right)\right\}\right]}{6} \\
&=\frac{5-\left[\frac{3}{4}+\left\{\frac{5}{2}-\left(\frac{21+1}{42}\right)\right\}\right]}{6} \\
&=\frac{5-\left[\frac{3}{4}+\left\{\frac{5}{2}-\left(\frac{22}{42}\right)\right\}\right]}{6} \\
&=\frac{5-\left[\frac{3}{4}+\left\{\frac{105-22}{42}\right\}\right]}{6} \\
&=\frac{5-\left[\frac{3}{4}+\left\{\frac{83}{42}\right\}\right]}{6} \\
&=\frac{5-\left[\frac{63+166}{84}\right]}{6} \\
&=\frac{\left[\frac{420-229}{84}\right]}{6} \\
&=\frac{191}{504}
\end{aligned}