Question

Evaluate
$\frac{(-1{)}^{13}}{2^3}+\frac{2^3-1^{10}+3^2}{3^2-2^2}+{\left(\frac{7}{11}\right)}^3÷\frac{98}{121}$

• A. $3\frac{173}{440}$
• B. $3\frac{17}{44}$
• C. $3\frac{137}{440}$
• D. $3\frac{21}{440}$

Answer 1

Answer: (A)

$3\frac{173}{440}$

$\frac{(-1{)}^{13}}{2^3}+\frac{2^3-1^{10}+3^2}{3^2-2^2}+{\left(\frac{7}{11}\right)}^3÷\frac{98}{121}$

$\Rightarrow \frac{-1}{8}+\frac{8-1+9}{9-4}+\frac{343}{11\times 11\times 11}÷\frac{98}{121}$

$\Rightarrow \frac{-1}{8}+\frac{7+9}{5}+\frac{7\times 7\times 7}{11\times 11\times 11}\times \frac{121}{98}$

$\Rightarrow \frac{-1}{8}+\frac{16}{5}+\frac{7}{11}\times \frac{1}{2}$

$\Rightarrow \frac{-1}{8}+\frac{16}{5}+\frac{7}{22}$

$\Rightarrow \frac{-55+16\times 88+20\times 7}{440}$

$\Rightarrow \frac{-55+1408+140}{440}$

$\Rightarrow \frac{-55+1548}{440}$

$\Rightarrow \frac{1493}{440}$

Hence, this is the answer.