A person reaches his destination 32 minutes late if his speed is 6 km h and he reaches 18 minutes be...........

Quantitative Aptitude (MTS) Time & distance

A person reaches his destination 32 minutes late if his speed is 6 km/h, and he reaches 18 minutes before time if his speed is 7

km.h. Find the distance of his destination from his starting point.

This questions was previously asked in

SSC MTS 11th October 2021 Shift-2

28km

30km

32km

35km

- Let distance be D km.

- At 6 km/h, he is 32 mins late (that’s 32/60 = 8/15 hours late).

- At 7 km/h, he is 18 mins early (so he would have taken 18/60 = 3/10 hours less than on time).

- So, time difference = 32 + 18 = 50 mins = 5/6 hrs

Set up equations:

- Time at 6 km/h = D/6 (late by 8/15 hrs)

- Time at 7 km/h = D/7 (early by 3/10 hrs)

So, D/6 - D/7 = 8/15 + 3/10 = (16+9)/30 = 25/30 = 5/6

- Find D:

D/6 - D/7 = 5/6

(7D - 6D)/42 = 5/6

D/42 = 5/6

D = (5/6) × 42 = 35 km

Option 4 (35 km) is correct.

- Option 1 (28km), Option 2 (30km), and Option 3 (32km) do not satisfy the time difference.

- 35 km matches all given conditions and equations.

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