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.

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

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

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

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

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

=\([5^{4}]^{\frac{1}{4}}\)

=\(5\)