A) 10
B) 2
C) 5
D) 15
Correct Answer: A
Solution :
If a and P are the zeros of the polynomial\[\therefore \] \[=\left( \frac{144}{48}+\frac{384}{48}+\frac{240}{48} \right)=3+8+5=16\]and\[\frac{a}{b}\] Given that\[\frac{c}{d}=\frac{L.C.M.(a,c)}{H.C.F.(b,d)}\], solving\[\Rightarrow \]and\[L.C.M.\], we get\[\frac{6}{14}and\frac{2}{7}\]and\[\Rightarrow \]. \[\frac{L.C.M.(6,2)}{H.C.F.(14,7)}=\frac{6}{7}\]\[\text{7}\times \text{13}+\text{13}=\text{1}0\text{4}=\text{23}\times \text{13}\]and\[\therefore \] \[\text{7}\times \text{13}+\text{13}\]You need to login to perform this action.
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