Free Practice Questions for Simplification in Maths

$=\frac{\frac{1}{5}÷\frac{1}{25}}{\frac{1}{25}÷\frac{1}{5}}=\frac{\frac{25}{5}}{\frac{5}{25}}=\frac{25}{5}\times \frac{25}{5}$
$=25$

$\frac{(0.623{)}^3+(0.377{)}^3}{(0.623{)}^2-(0.623\times 0.377)+(0.377{)}^2}$
$=\frac{(0.623+0.377)({0.623}^3-0.623\times 0.377+{0.377}^2)}{(0.623{)}^2-0.623\times 0.377+(0.377{)}^2}$
$[\because a^3+b^3=(a+b)(a^2-ab+b^2)]$
$=0.623+0.377=1.000=1$

$4\frac{2}{3}\times \frac{9}{14}+5\frac{1}{6}\times 2\frac{2}{5}$

On solving the given question,

$\frac{14}{3}\times \frac{9}{14}+\frac{31}{6}\times \frac{12}{5}$

$\Rightarrow$3+$\frac{62}{5}$

$\Rightarrow$$\frac{62+15}{5}$

$\Rightarrow$$15\frac{2}{5}$

$\begin{array}{rl}243\times 124& -25340\\ =& 30132-25340=4792\end{array}$

$5-[\frac{3}{4}+\{2\frac{1}{2}-(0.5+\frac{\bar{1}}{6}-\frac{1}{7})\}]$
$=5-[\frac{3}{4}+\{\frac{5}{2}-(\frac{1}{2}+\frac{7-6}{42})\}]$
$=5-[\frac{3}{4}+\{\frac{5}{2}-(\frac{1}{2}+\frac{1}{42})\}]]$
$=5-[\frac{3}{4}+\{\frac{5}{2}-\frac{22}{42}\}]$
$=5-[\frac{3}{4}+\frac{105-22}{42}]=5-[\frac{3}{4}+\frac{83}{42}]$
$=5-\left[\frac{63+166}{84}\right]=5-\frac{229}{84}$
$=\frac{420-229}{84}=\frac{191}{84}$

$45-[28-\{37-(15-?)\}]=58$
$\begin{array}{lrl}\Rightarrow & -[28-\{37-(15-?)\}]& =58-45\\ \Rightarrow & -[28-\{37-(15-?)\}]& =13\\ \Rightarrow & 28-\{37-(15-?)\}& =-13\\ \Rightarrow & 28+13& =37-(15-?)\\ \Rightarrow & 41& =37-(15-?)\\ \Rightarrow & (15-?)& =37-41\\ \Rightarrow & (15-?)& =-4\\ \Rightarrow & (15+4)& =?\\ \Rightarrow & ?& =19\end{array}$

$\frac{(598+178{)}^2-(598-178{)}^2}{598\times 178}$
$=\frac{4\times 598\times 178}{598\times 178}[\because (a+b{)}^2-(a-b{)}^2=4ab]$
$=4$

$19587\times 637+19587\times 363$
$=19587(637+363)$
$=19587\times 1000$ $[ab+ac=a(b+c)]$
$=19587000$

$?+3699+1985-2047=31111$

$\Rightarrow$$?+3637=31111$

$\Rightarrow$$?=31111-3637$

$\Rightarrow$$?=27474$

Let $856=a$ and $167=b$

Then, by placing the value in the given expression,

$\frac{(a+b{)}^2+(a-b{)}^2}{a\times a+b\times b}$

$\frac{a^2+b^2+2ab+a^2+b^2-2ab}{a^2+b^2}$

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