CUET UG 2025 — General Test Arithmetic

Let $d$d = distance (in km)

Let $t$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}$3d​ hours

15 minutes = $\frac{15}{60} = \frac{1}{4}$6015​=41​ hour

$\frac{d}{3} = t + \frac{1}{4}$3d​=t+41​ ... (Equation 1)

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

Time taken = $\frac{d}{4}$4d​ hours

$\frac{d}{4} = t - \frac{1}{4}$4d​=t−41​ ... (Equation 2)

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Subtracting Equation 2 from Equation 1:

$$\frac{d}{3} - \frac{d}{4} = \left(t + \frac{1}{4}\right) - \left(t - \frac{1}{4}\right)$$3d​−4d​=(t+41​)−(t−41​)

$$\frac{d}{3} - \frac{d}{4} = t + \frac{1}{4} - t + \frac{1}{4}$$3d​−4d​=t+41​−t+41​

$$\frac{d}{3} - \frac{d}{4} = \frac{1}{2}$$3d​−4d​=21​

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$$\frac{4d}{12} - \frac{3d}{12} = \frac{1}{2}$$124d​−123d​=21​

$$\frac{d}{12} = \frac{1}{2}$$12d​=21​

$$d = 6$$d=6

Therefore, the distance is 6 km.