Discussion

Explanation:

As per the description,

AB = 20 km

AG1 = BG3

2G1G2 = G2G3

∴ x = 2.5 km

y + 2y = 20 − 2x

∴ y = 5 km

∴ Time to cover A to G3112+1560 = 20 min

∴ While coming back his speed is 60 km/hr.

∴ Time taken to cover the distance from G3 to A (i.e. 17.5 km) = 17.5 minutes

∴ Required time = 20 + 17.5 + 1 = 38.5 minutes

∴ The doctor will have 1.5 minutes to attend the patient.

Hence, option (c).

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