A) 48
B) 36
C) 44
D) 32
Correct Answer: A
Solution :
Let the digits in the tens place and the ones place be x and y respectively. Then, according to the problem, \[\text{4}0={{\text{2}}^{\text{3}}}\times \text{5}\] \[\text{6}0={{\text{2}}^{\text{2}}}\times \text{3}\times \text{5}\] \[\therefore \] and \[={{2}^{2}}\times 3\times 5\times 2=120\] \[f(p)={{\text{p}}^{\text{3}}}+\text{6}{{\text{p}}^{\text{2}}}+\text{lip}+\text{6}\]\[f(p)\] If \[p(x)={{x}^{2}}+3x-2\],then\[\Rightarrow \] \[p(-1)={{(-1)}^{2}}+3(-1)-2=(-4)\]The number is 24. If\[p(-1)=-4\]then\[({{m}^{2}}+9){{x}^{2}}+13x+6m\]. \[\alpha \]The number is 48. Hence, the required number is 48.
You need to login to perform this action.
You will be redirected in
3 sec
