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Suresh plans a trip from city A to city B which are ‘x’ km apart and returning to city A. There are two light traffic zones of 25 km length each where he goes at a speed of 20 km/hr and returns at a speed of 25 km/hr. There are two heavy traffic zones of 35 km length each where he goes with a speed of 14 km/hr and returns at a speed of 21 km/ hr. For other part of the journey he maintains his speed at 90 km/hr while going and 60 km/hr while returning. If it took 2 hrs extra to go to city B from city A than to return from city B to city A, what is the approximate average speed with he was moving during his return journey (in km/hr)?

Suresh plans a trip from city A to city B which are ‘x’ km apart and returning to city A. There are two light traffic zones of 25 km length each where he goes at a speed of 20 km/hr and returns at a speed of 25 km/hr. There are two heavy traffic zones of 35 km length each where he goes with a speed of 14 km/hr and returns at a speed of 21 km/ hr. For other part of the journey he maintains his speed at 90 km/hr while going and 60 km/hr while returning. If it took 2 hrs extra to go to city B from city A than to return from city B to city A, what is the approximate average speed with he was moving during his return journey (in km/hr)? Correct Answer 26

For the journey from city A to city B

Time = Speed/Distance

Total time taken = Time taken in light traffic zones + Time taken is heavy traffic zones + Time taken to cover rest of the distance

Ta = 2 × 25/20 + 2 × 35/14 + (x - 120)/90 = 7.5 + (x - 120)/90

For journey from city B to city A

Tb = 2 × 25/25 + 2 × 35/21 + (x - 120)/60 = 5.333 + (x - 120)/60

Given Ta - Tb = 2

0.167 = (x - 120)/60 - (x - 120)/90

0.167 = {3(x - 120) - 2(x - 120)}/180

⇒ x = 150 km

Time taken in reverse journey = 5.333 + (150 - 120)/60

= 5.833 hrs

Average speed = 150/5.8333 ≈ 26 km/hr

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