Ananyaa travels a certain distance at a speed of 3 kmph and reaches her destination late by 15 minutes. If she travels at a speed of 4 kmph, she reaches her destination 15 minutes earlier. Find the distance that Ananyaa is traveling to reach her destination.

Let $d$ = distance (in km)

Let $t$ = the correct time to reach (in hours)

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When traveling at 3 kmph, Ananyaa is 15 minutes late.

Time taken = $\frac{d}{3}$ hours

15 minutes = $\frac{15}{60} = \frac{1}{4}$ hour

$\frac{d}{3} = t + \frac{1}{4}$ ... (Equation 1)

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When traveling at 4 kmph, Ananyaa is 15 minutes early.

Time taken = $\frac{d}{4}$ hours

$\frac{d}{4} = t - \frac{1}{4}$ ... (Equation 2)

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Subtracting Equation 2 from Equation 1:

$\frac{d}{3} - \frac{d}{4} = \left(t + \frac{1}{4}\right) - \left(t - \frac{1}{4}\right)$

$\frac{d}{3} - \frac{d}{4} = t + \frac{1}{4} - t + \frac{1}{4}$

$\frac{d}{3} - \frac{d}{4} = \frac{1}{2}$

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$\frac{4d}{12} - \frac{3d}{12} = \frac{1}{2}$

$\frac{d}{12} = \frac{1}{2}$

$d = 6$

Therefore, the distance is 6 km.