CUET UG 2025 — General Test Arithmetic

Let the distance from the village to the post-office be $d$d km.

The distance from village to post-office is the same as the distance from post-office back to village.

Speed going to post-office = 25 km/hr

Speed walking back = 4 km/hr

Total time for both trips = 5 hours 48 minutes

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Converting the total time to hours:

5 hours 48 minutes = $5 + \dfrac{48}{60}$5+6048​ hours

$$= 5 + \dfrac{4}{5}$$=5+54​

$$= 5 + 0.8$$=5+0.8

$= 5.8$=5.8 hours

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Using Time = Distance ÷ Speed:

Time to go to post-office = $\dfrac{d}{25}$25d​ hours

Time to walk back = $\dfrac{d}{4}$4d​ hours

The total time is 5.8 hours:

$$\dfrac{d}{25} + \dfrac{d}{4} = 5.8$$25d​+4d​=5.8

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The LCM of 25 and 4 is 100.

$$\dfrac{4d}{100} + \dfrac{25d}{100} = 5.8$$1004d​+10025d​=5.8

$$\dfrac{29d}{100} = 5.8$$10029d​=5.8

$$29d = 580$$29d=580

$$d = 20$$d=20

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The distance of the post-office from the village is 20 km.