Question

If $x=y^3+4y$ and $y=\frac{7}{k}$, find the approximate value of $x$ when $k=21$.

• A. $1.37$
• B. $1.44$
• C. $1.67$
• D. $4.04$
• E. $4.11$

Answer 1

The correct option is A $1.37$

Given $x=y^3+4y$ and $y=\frac{7}{k}$

$\Rightarrow x={\left(\frac{7}{k}\right)}^3+4\left(\frac{7}{k}\right)$

$\Rightarrow x=\frac{343}{k^3}+\frac{28}{k}$

$\Rightarrow x=\frac{343+28k^2}{k^3}$

When $k=21$

$\Rightarrow x=\frac{343+28\times {21}^2}{{21}^3}$

$\Rightarrow x=\frac{343+12348}{9261}$

$\Rightarrow x=\frac{12691}{9261}=1.37$