Simplification questions

$3 \frac{1}{2} \times \frac{7 \frac{2}{5}}{9 \frac{3}{5}} \times 8^{2} \times 60=2^{4} \times x$

$\frac{7}{2} \times\left(\frac{37}{5} \times \frac{5}{48}\right) \times 8^{2} \times 60=16\times x$

$\frac{7}{2} \times \frac{37}{48} \times 64 \times \frac{60}{16}=x$

$x=\frac{1295}{2}=647.5$

The given expression=

$2 \div \frac{3}{17} \text { of }\left(2 \frac{3}{4}+3 \frac{5}{8}\right)+\frac{2}{5} \div 2 \frac{1}{5}+\frac{2}{9}$
$\begin{aligned}=2 \div & \frac{3}{17} \text { of }\left(\frac{22+29}{8}\right)+\frac{2}{5} \div \frac{11}{5}+\frac{2}{9} \\ &=2 \div \frac{3}{17} \text { of } \frac{51}{8}+\frac{2}{5} \div \frac{11}{5}+\frac{2}{9} \\ &=2 \div \frac{3}{17} \times \frac{51}{8}+\frac{2}{5} \div \frac{11}{5}+\frac{2}{9} \\ &=2 \times \frac{9}{8}+\frac{2}{5} \div \frac{11}{5}+\frac{2}{9} \\ &=2 \times \frac{8}{9}+\frac{2}{5} \times \frac{5}{11}+\frac{2}{9} \\ &=\frac{16}{9}+\frac{2}{11}+\frac{2}{9}=\frac{176+18+22}{99} \\ &=\frac{216}{99}=\frac{24}{11} \end{aligned}$

$\sqrt{500}-\sqrt{125}$ =$\sqrt{100\times5}-\sqrt{25\times5}=10\sqrt5-5\sqrt5=5\sqrt5$

$1 \times \frac12 \times \frac13 \times \frac14 = \frac{1}{24}$

$\mathrm{A}=\frac{3}{4} \div \frac{5}{6}=\frac{3}{4} \times \frac{6}{5}=\frac{3}{2} \times \frac{3}{5}=\frac{9}{10}$

$\mathrm{~B}=3 \div\left[(4 \div 5) \div \frac{6}{1}\right]=3 \div\left[\frac{4}{5} \times \frac{1}{6}\right]$

$=3 \div\left[\frac{2}{15}\right]=3 \times \frac{15}{2}=\frac{45}{2}$

$\mathrm{C}=[3 \div(4 \div 5)] \div 6$

$=\left[3 \div \frac{4}{5}\right] \div 6$

$=\left[3 \times \frac{5}{4}\right] \div 6$

$=\frac{15}{4} \times \frac{1}{6}=\frac{5}{8}$

$D=3 \div 4(5 \div 6)=3 \div 4 \times \frac{5}{6}$

$=3 \div \frac{10}{3}=3 \times \frac{3}{10}=\frac{9}{10}$

Add the coefficients having same power of variables

$(5x^{2} -2x -1 )- (3x^{2}-5x+7)$

$(5x^{2} -2x -1 - 3x^{2}+5x-7)$

= $(2x^{2} +3x -8 )$

$(3 \div 8)$ of $16-[2 of 15 \div 25 \div\{(3 \div 5) \times(7 \div 2) \div(25 \div 4)\}]$ of $14 \div 25$

$\Rightarrow\frac{3}{8} \times 16-\left[\frac{30}{25} \div\left\{\frac{3}{5} \times \frac{7}{2} \times \frac{4}{25}\right\} \times \frac{14}{25}\right]$

$\Rightarrow 6-\left[\frac{30}{25} \times \frac{250}{84}\right] \times \frac{14}{25}$

$6-2=4$

Hence, option A is correct.

• We know that $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$

• $\frac{a^{3}+b^{3}+c^{3}-3 a b c}{a^{2}+b^{2}+c^{2}-a b-b c-c a}=a+b+c$

• $\therefore \frac{0.9 \times 0.9 \times 0.9+0.7 \times 0.7 \times 0.7+0.2 \times 0.2 \times 0.2-3 \times 0.9 \times 0.7 \times 0.2}{0.9 \times 0.9+0.7 \times 0.7+0.2 \times 0.2-0.9 \times 0.7-0.7 \times 0.2-0.2 \times 0.9}$

  $=0.9+0.7+0.2$ $=1.8$

$\sqrt50 + \sqrt18 - \sqrt8$

= $\sqrt{25\times2} + \sqrt{9\times2} - \sqrt{4 \times2}$

=$5\sqrt2 + 3\sqrt2 - 2\sqrt2$

= $6\sqrt2$

Calculation :

• $27^{\frac{1}{3} } \times 8^{\frac{2}{3} } \times 125^{-\frac{2}{3} } \times 16^{-\frac{1}{2} }$
  $= 3^{ 3 \times \frac{1}{3} } \times 2^{3 \times \frac{2}{3} } \times 5^{3 \times -\frac{2}{3} } \times 4^{2 \times -\frac{1}{2} }$
  $= 3^{1} \times 2^{2} \times 5^{-2} \times 4^{-1}$
  $= \frac{3 \times 4}{ 25 \times 4} = \frac{3}{25}$

Hence the answer is option C

$\left(\frac{7}{16} \div \frac{1}{2}\right.$ of $\left.\frac{1}{5}\right) \times \frac{4}{5}-\frac{1}{3} \times \frac{5}{8} \div \frac{1}{2}+\frac{3}{4}$

Using BODMAS:

$\Rightarrow\left(\frac{7}{16} \div \frac{1}{10}\right) \times \frac{4}{5}-\frac{1}{3} \times \frac{5}{8} \div \frac{1}{2}+\frac{3}{4}$

$\Rightarrow \frac{70}{16} \times \frac{4}{5}-\frac{1}{3} \times \frac{5}{8} \div \frac{1}{2}+\frac{3}{4}$

$\Rightarrow \frac{70}{16} \times \frac{4}{5}-\frac{1}{3} \times \frac{5}{4}+\frac{3}{4}$

$\Rightarrow \frac{7}{2}-\frac{5}{12}+\frac{3}{4}=\frac{42-5+9}{12}=\frac{46}{12}=\frac{23}{6}$

$80\times\sqrt p = 1120$

$\frac{3}{4}$( 1+$\frac{3}{4}$ )( 1+$\frac{2}{3}$ )( 1-$\frac{2}{5}$ )(1+$\frac{6}{7}$)(1-$\frac{12}{13}$)

$= \frac{3}{4}$ $\times$$\frac{7}{3} ×$ $\frac{5}{3} ×$ $\frac{3}{5} ×$ $\frac{13}{7} ×$$\frac{1}{13}$

$= \frac{1}{4}$

$\frac{x+5}{2}+x+\frac{3x+3}{6}=x+10$

$\Rightarrow$$\frac{3x+15+6x+3x+3}{6}=x+10$

$\Rightarrow$ $12x+18=6x+60$

$\Rightarrow$$6x=42$

$\therefore$$x=7$

$(p-q)^{2} = p^{2} +q{2} - 2pq$
substituting values,
$(6)^{2} =116 - 2pq$
$36 = 116 -2pq$
$2pq = 80$
$pq = 40$

Given:

$3 × 8 ÷ 9$ of $6 - 2 ÷ 3 × (5 - 2) × 2 + 18 ÷ 3$ of $3$

Calculation:

$3 × 8 ÷ 9$ of $6 - 2 ÷ 3 × (5 - 2) × 2 + 18 ÷ 3$of$3$

$⇒ 3 × 8 ÷ 9 of 6 - 2 ÷ 3 × 3 × 2 + 18 ÷ 3 of 3$

$⇒ 3 × \frac{8}{54} - \frac{2}{3} × 3 × 2 +\frac{18}{9}$

$⇒ \frac{4}{9} - 4 + 2$

$⇒ \frac{4 - 36 + 18}{9}$

$⇒ - \frac{14}{9}$

$\Rightarrow-1 \frac{5}{9}$

$\therefore$ Required answer is $-1 \frac{5}{9}$

$\frac{1}{2\frac{1}{3}}$ +$\frac{1}{1\frac{3}{4}}+\frac{1}{1\frac{2}{5}}$ = $\frac{1}{\frac{7}{3}}$ +$\frac{1}{\frac{7}{4}}+\frac{1}{\frac{7}{5}}$
$\frac{3}{7}$ +$\frac{4}{7}+\frac{5}{7}$
$1+\frac{5}{7}=\frac{12}{7}$

[Using : $(a + b) (a - b) = a^2 - b^2 ]$

$(2x + 3y) (2x - 3y) = (2x)^2 - (3y)^2$

$4x^2 - 9y^2$

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

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

$\frac{2}{3} \times \frac{3}{\frac{5}{4}\times\frac{4}{5}}$

$\frac{2}{3} \times 3$ = 2

$\frac{(3\frac{1}{2} - 2\frac{1}{3} -\frac{1}{3})}{(5\frac{1}{2} \times \frac{5}{11})- 1\frac{2}{3}}$$= \frac{\frac{7}{2}-\frac{7}{3}-\frac{1}{3}}{\frac{11}{2}\times\frac{5}{11}-\frac{5}{3}} = \frac{21-14-2}{6}\times\frac{6}{15-10}= 1$

$\begin{equation} \begin{aligned} &\text { (\( x \)) } =\frac{9|3-5|-5|4| \div 10}{-3(5)-2 \times 4 \div 2}\\ &=\frac{9 \times 2-5 \times 4 \div 10}{-15-2 \times 2}\\ &=\frac{18-2}{-19}=-\frac{16}{19} \end{aligned} \end{equation}$

$a-\frac{1}{a-3}=5$
$(a-3)-\frac{1}{(a-3)}= 2$
Cubing both sides
$((a-3)-\frac{1}{(a-3)})^{3}= 2^{3}$
$(a-3)^{3}-\frac{1}{(a-3)^{3}}-3 \times (a-3) \times \frac{1}{(a-3)}((a-3)-\frac{1}{(a-3)}) =8$
$(a-3)^{3}-\frac{1}{(a-3)^{3}}- 3 \times 2=8$
$(a-3)^{3}-\frac{1}{(a-3)^{3}}$=$8+6=14$

The expression is divisible by $x^2-x-6$, so $x^2-x-6=0, x=- 2/3$ <br> Substitute the values of x in the expression .<br> With x= 3, we get expression as $3p-q=3$<br> With x=-2, we get $2p+q=2$ <br>On solving, we get $p=1 , q=0$

$\frac{p}{q}= (\frac{2}{3}) ^3 \div (\frac{3}{2})^{-3}= (\frac{2}{3})^3 \times (\frac{3}{2})^3=1$
$\therefore (\frac{p}{q})^{-10}= 1^{-10}=1$

$\sqrt{20164}=142$

Now, $\sqrt{201.64}+\sqrt{2.0164}$

$\sqrt{\frac{20164}{100}}+\sqrt{\frac{20164}{10000}}=\frac{142}{10}+\frac{142}{100}=14.2+1.42=15.62$

Given: $$\frac{13 \frac{7}{9} \div \frac{2}{3} \text { of } \frac{1}{3}+\frac{2}{3} \times 3}{\frac{7}{5} \text { of } 325-20+7}$$ By using BODMAS rules- $$=\frac{\frac{124}{9} \div \frac{2}{9}+2}{7 \times 65-13}$$ $$=\frac{\frac{124}{9} \times \frac{9}{2}+2}{455-13}$$ $$=\frac{62+2}{442}$$ $$=\frac{64}{442}$$ $$=\frac{32}{221}$$

$(a+b)^2 = a^2+ b^2+2ab$

Here $= a= 2x, b= 3y$

so $(2x+3y)^{2}= (2x)^2+ (3y)^2+2 \times 2x \times 3y$

$=4x^2+ 9y^2+4x \times 6y$

$= 4x^{2} + 12xy + 9y^{2}$

$=\frac{3}{5}+\frac{12}{5} \div\left[\left(\frac{7-6}{8}\right) \times 4-\frac{20}{3} \div \frac{4}{3}\right] \times \frac{1}{3}$

$=\frac{3}{5}+\frac{12}{5}\div\left[\frac{1}{8} \times 4-\frac{20}{3} \times \frac{3}{4}\right] \times \frac{1}{3}$ $=\frac{3}{5}+\frac{12}{5} \div\left[\frac{1}{2}-5\right] \times \frac{1}{3}$

$=\frac{3}{5}+\frac{12}{5}\div\left(\frac{-5}{2}\right) \times \frac{1}{3}$

$=\frac{3}{5}+\frac{12}{5} \times\left(\frac{2}{-5}\right) \times \frac{1}{3}$

$=\frac{3}{5}-\frac{4}{5} \times \frac{2}{5}$

$=\frac{15-8}{25}$ $=\frac{9}{25}$

• Given that,

  $\frac{P}{Q}$= 5

• By taking Q$^{2}$ common in both numerator and denominator

  $\Rightarrow \frac{P^{2}+Q^{2}}{P^{2}-Q^{2}}= \frac{Q^{2}\left [ \left ( \frac{P^{2}}{Q^{2}} \right )+1 \right ]}{Q^{2}\left [ \left ( \frac{P^{2}}{Q^{2}} \right )-1 \right ]}$

• By cancelling Q$^{2}$term in both numerator and denominator

  = $\frac{\left [ \left ( \frac{P^{2}}{Q^{2}} \right )+1 \right ]}{\left [ \left ( \frac{P^{2}}{Q^{2}} \right )-1 \right ]}$

  $=\frac{\left [ \left ( \frac{P}{Q} \right ) ^{2}+1\right ]}{\left [ \left ( \frac{P}{Q} \right )^{2} -1\right ]}$

  $\Rightarrow \frac{5^{2}+1}{5^{2}-1} \ \left ( \because \frac{P}{Q} = 5\right )$

  $\Rightarrow \frac{25+1}{25-1}$

  $= \frac{26}{24}$

• Hence, the answer is $\frac{26}{24}$

$16 \div 8 \div 4 \div 2 =2 \div 4 \div 2= \frac{1}{2} \div 2= \frac{1}{4}$

$\sqrt {\sqrt{64} +248}$ =$\sqrt{248+8}$ = 16

2p + $\frac{1}{p}$ = 4

p + $\frac{1}{2p}$ = 2

$\therefore$ $(p+\frac{1}{2p})^{3}$ = $p^{3}$ + $\frac{1}{8p^{3}}$+ 3p $\times$ $\frac{1}{2p}$(p+$\frac{1}{2p}$=8)

($p^{3}$ + $\frac{1}{8p^{3}}$) = $p^{3}$ + $\frac{1}{8p^{3}}$ + $\frac{3}{2}$ $\times$ 2

= 8 - 3 = 5

$100 \div 10\div 5 \div 1$
=$100 \times \frac{1}{10} \times \frac{1}{5} \times\frac{1}{1}$
=$\frac{100}{10 \times5}$
=$\frac{10}{5}$
=2

Given:

$$5 \frac{1}{3} \div\left[7-3 \div\left(1-\frac{1}{4}\right) \times \frac{2}{3}+1\right]-3 \div 1+2$$

By using BODMAS rules

$$\begin{aligned} &=\frac{16}{3} \div\left[7-\frac{3}{\left(\frac{3}{4}\right)} \times \frac{2}{3}+1\right]-3+2 \\ &=\frac{16}{3} \div\left[7-\frac{8}{3}+1\right]-1 \\ &=\frac{16}{3} \div\left[\frac{21-8+3}{3}\right]-1 \\ &=\frac{16}{3} \times \frac{3}{16}-1 \\ &=1-1 \\ &=0 \end{aligned}$$

Now,

Calculations:

$(\frac{3}{2})× (\frac{5}{7}) ÷ (\frac{x}{21}) × (\frac{3}{5}) + (\frac{2}{7}) – 1 = \frac{16}{7}$

$⇒ (\frac{3}{2}) × (\frac{5}{7} × \frac{21}{x}) × (\frac{3}{5}) + (\frac{2}{7}) – 1 = \frac{16}{7}$

$⇒ (\frac{3}{2}) × (\frac{15}{x}) × (\frac{3}{5}) + (\frac{2}{7}) – 1 = \frac{16}{7}$

$⇒ \frac{27}{2x} = (\frac{16}{7}) – (\frac{2}{7}) + 1$

$⇒ \frac{27}{2x} = \frac{21}{7}$

$⇒ x = (7 × 27)/(2 × 21)$

$⇒ x =\frac{ 9}{2}$

$∴ x = \frac{9}{2}$

We know that: If $A+B+C=0$, then $A^{3}+B^{3}+C^{3}=3 A B C$

The value of $\frac{5\left(a^{6}-b^{6}\right)^{3}+5\left(b^{6}-c^{6}\right)^{3}+5\left(c^{6}-a^{6}\right)^{3}}{2\left(a^{3}-b^{3}\right)^{3}+2\left(b^{3}-c^{3}\right)^{3}+2\left(c^{3}-a^{3}\right)^{3}}$

$$=\frac{5\left(\left(a^{6}-b^{6}\right)^{3}+\left(b^{6}-c^{6}\right)^{3}+\left(c^{6}-a^{6}\right)^{3}\right)}{2\left(\left(a^{3}-b^{3}\right)^{3}+\left(b^{3}-c^{3}\right)^{3}+\left(c^{3}-a^{3}\right)^{3}\right)}$$

$$=\frac{5 \times 3\left(a^{6}-b^{6}\right)\left(b^{6}-c^{6}\right)\left(c^{6}-a^{6}\right)}{2 \times 3\left(a^{3}-b^{3}\right)\left(b^{3}-c^{3}\right)\left(c^{3}-a^{3}\right)}$$

$$=\frac{5}{2}\left(a^{3}+b^{3}\right)\left(b^{3}+c^{3}\right)\left(c^{3}+a^{3}\right)$$

Hence, option $\mathrm{C}$ is correct.

$\begin{aligned} \frac{\sqrt{5}}{\sqrt{3}+\sqrt{2}} &=\frac{\sqrt{5}(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})} \\ &=\frac{\sqrt{15}-\sqrt{10}}{3-2}=\sqrt{15}-\sqrt{10} \\ \frac{3 \sqrt{3}}{\sqrt{5}+\sqrt{2}} &=\frac{3 \sqrt{3}}{\sqrt{5}+\sqrt{2}} \times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}} \\=& \frac{3 \sqrt{3}(\sqrt{5}-\sqrt{2})}{5-2}=\sqrt{15}-\sqrt{6} \\ \frac{2 \sqrt{2}}{\sqrt{5}+\sqrt{3}} &=\frac{2 \sqrt{2}(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})} \\ &=\frac{2 \sqrt{2}(\sqrt{5}-\sqrt{3})}{5-3}=\sqrt{10}-\sqrt{6} \end{aligned}$
$\begin{aligned} \therefore & \text { } \\=&(\sqrt{15}-\sqrt{10})-(\sqrt{15}-\sqrt{6})+(\sqrt{10}-\sqrt{6}) \\=& \sqrt{15}-\sqrt{10}-\sqrt{15}+\sqrt{6}+\sqrt{10}-\sqrt{6} \\=& 0 \end{aligned}$

=$[5 (27^{\frac{1}{3}} +8^{\frac{1}{3}})^{3}]^{\frac{1}{4}}$

=$[5(\sqrt[3]{27} +\sqrt[3]{8})^{3}]^{\frac{1}{4}}$

=$[5(3+2)^{3}]^{\frac{1}{4}}$

=$[5\times5^{3}]^{\frac{1}{4}}$

=$[5^{4}]^{\frac{1}{4}}$

=$5$