We need to compare the value of $\(\frac{0.9999}{1.0002}\)$ with $\(\frac{0.9998}{1.0001}\)$ i.e. to compare $\(\frac{9999}{10002}\)$ with $\(\frac{9998}{10001}\)$

Since the denominators of the fractions are two consecutive numbers i.e. 1002 and 10001, they must be co-prime (Co-prime numbers are those which do not have any factor in common except 1).

Making their denominators same, we get the fractions as $\(\frac{9999 \times 10001}{10002 \times 10001}=\frac{99 \ldots . \ldots 8 \text { times. } \cdot 9}{10002 \times 10001}\)$ and

$$
\(\frac{9998 \times 10002}{10001 \times 10002}=\frac{99 \ldots 7 \text { times } \cdot .96}{10002 \times 10001}\)
$$


With same denominator fractions the comparison remains between their numerator values only and the fraction with higher numerator is greater.

Now clearly $\(999 \ldots 8\)$ times ... 9 is greater than $\(99 \ldots . .7$ tines.. 96\) , so is greater.

Hence column A has greater quantity, so the answer is (A).