$(a)\;1\qquad(b)\;4.06\qquad(c)\;3\qquad(d)\;4.65$

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Answer: 0.333

Given 50% of the reaction of X was completed when 87.5% of the reaction of Y was completed, the time is the same in both cases..

$\Rightarrow$ Time $t = \large\frac{1}{k_1}$$ \ln $$\; \large\frac{100}{50}$$ = \large\frac{1}{k_2}$$ \ln $$\; \large\frac{100}{12.5}$

$\Rightarrow \large\frac{k_1}{k_2}$$ = \large\frac{\ln \large\frac{100}{50}}{\ln \large\frac{100}{12.5}}$$ = 0.333$

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