Question : If $p=\frac{5}{18}$, then $27p^{3}–\frac{1}{216}–\frac{9}{2}p^{2}+\frac{1}{4}p$ is equal to:
Option 1: $\frac{4}{27}$
Option 2: $\frac{5}{27}$
Option 3: $\frac{8}{27}$
Option 4: $\frac{10}{27}$
Correct Answer: $\frac{8}{27}$
Solution :
Given: $p=\frac{5}{18}$
So, $27p^{3}–\frac{1}{216}–\frac{9}{2}p^{2}+\frac{1}{4}p$
Putting the value of $p=\frac{5}{18}$ in the expression,
= $27×(\frac{5}{18})^{3}–\frac{1}{216}–\frac{9}{2}×(\frac{5}{18})^{2}+\frac{1}{4}×\frac{5}{18}$
= $27×(\frac{125}{18×18×18})–\frac{1}{216}–\frac{9}{2}×(\frac{25}{18×18})+\frac{1}{4}×\frac{5}{18}$
= $\frac{125}{216}–\frac{1}{216}–\frac{25}{72}+\frac{5}{72}$
= $\frac{124}{216}–\frac{20}{72}$
= $\frac{124–60}{216}$
= $\frac{64}{216}$
= $\frac{8}{27}$
Hence, the correct answer is $\frac{8}{27}$.
Related Questions
Know More about
Staff Selection Commission Sub Inspector ...
Result | Eligibility | Application | Selection Process | Cutoff | Admit Card | Preparation Tips
Get Updates Brochure
Your Staff Selection Commission Sub Inspector Exam brochure has been successfully mailed to your registered email id “”.
