Arithmetic

Time, Speed & Distance

Medium

A bus starts at 9 am and follows a fixed route every day. One day, it traveled at a constant speed of 60 km per hour and reached its destination 3.5 hours later than its scheduled arrival time. Next day, it traveled two-thirds of its route in one-third of its total scheduled travel time, and the remaining part of the route at 40 km per hour to reach just on time. The scheduled arrival time of the bus is

10:30 pm

7:00 pm

7:30 pm

9:00 pm

✅ Correct Option: 3

We have a bus that follows the same route every day, but on two different days, it behaves differently. We need to find when the bus is supposed to arrive according to its regular schedule.

Let's call the scheduled travel time $t$ hours and the total distance of the route $d$ km.

On Day 1, the bus traveled at 60 km/h but arrived 3.5 hours late.

Actual time taken: $t + 3.5$ hours (scheduled time + delay)

Speed: 60 km/h

Distance: $60 \times (t + 3.5)$ km

On Day 2, the bus completed the journey in exactly the scheduled time $t$ hours, but it did so in two parts:

Part 1: Two-thirds of the route in one-third of the scheduled time

Distance covered: $\tfrac{2d}{3}$ km

Time taken: $\tfrac{t}{3}$ hours

Speed: $\tfrac{\text{Distance}}{\text{Time}} = \tfrac{2d/3}{t/3} = \tfrac{2d}{t}$ km/h

Part 2: Remaining distance at 40 km/h

Distance remaining: $d - \tfrac{2d}{3} = \tfrac{d}{3}$ km

Time remaining: $t - \tfrac{t}{3} = \tfrac{2t}{3}$ hours

Speed: 40 km/h (given)

From Part 2 of Day 2, we can use the formula: Distance = Speed × Time

$\tfrac{d}{3} = 40 \times \tfrac{2t}{3}$

$\tfrac{d}{3} = \tfrac{80t}{3}$

$d = 80t$

Now we have two expressions for the same distance $d$ :

From Day 1: $d = 60(t + 3.5) = 60t + 210$

From Day 2: $d = 80t$

Since both expressions equal $d$ :

$80t = 60t + 210$

$20t = 210$

$t = 10.5$ hours

Bus starts at: 9:00 AM

Scheduled arrival time: 9:00 AM + 10 hours 30 minutes = 7:30 PM

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