| If \[P=\frac{1}{25},\]then value of \[125{{p}^{3}}-\frac{1}{64}-\frac{75}{4}{{p}^{2}}+\frac{15}{16}p\]is equal to |
A) \[\frac{-1}{8000}\]
B) \[\frac{1}{8000}\]
C) \[\frac{1}{2000}\]
D) \[\frac{-1}{2000}\]
Correct Answer: A
Solution :
| \[P=\frac{1}{25}\] |
| Now,\[125{{p}^{3}}-\frac{1}{64}-\frac{75}{4}{{p}^{2}}+\frac{15}{16}p\] |
| \[={{(5p)}^{3}}-{{\left( \frac{1}{4} \right)}^{3}}-\frac{3\times 25}{4}{{p}^{2}}+\frac{3\times 5}{{{4}^{2}}}p\] |
| \[={{\left( 5p-\frac{1}{4} \right)}^{3}}\] |
| On putting \[p=\frac{1}{25},\] we get |
| \[{{\left( 5\times \frac{1}{25}-\frac{1}{4} \right)}^{3}}={{\left( \frac{1}{5}-\frac{1}{4} \right)}^{3}}\] |
| \[=\left( \frac{-1}{20} \right)=\frac{-1}{8000}\] |
You need to login to perform this action.
You will be redirected in
3 sec
