| (I) 6 | (II) 66 |
| (III) 125 | (IV) 6666 |
A) Both (I) and (II)
B) Both (III) and (IV)
C) (I), (II) and (III)
D) All of (I), (II), (III) and (IV)
Correct Answer: D
Solution :
(d): \[\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}\Rightarrow \frac{4}{p}\ne \frac{p}{-2}\ne \frac{21}{15}\Rightarrow {{p}^{2}}\ne -8\](always true). Also, \[p\ne \frac{-42}{15}\] \[\therefore \] All values in question satisfy the above.
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