At its usual speed of 40 kmph, the bus covers its journey on schedule.

Question

At its usual speed of 40 kmph, the bus covers its journey on schedule. But when its speed reduces to 35 kmph, it takes 15 more minutes than scheduled time. Find the distance of the journey.

A. 50 km 
B. 60 km 
C. 70 km 
D. 75 km 
E. 80 km

navaneethmmb · Answer

When speed S = 40 kmph, Distance = D km, Time = t hrs

When speed = 35, Distance = D (no change) , but time = (t+15 mins) ==> (t+0.25) hrs

Solving above equations, we get

D = 40*t = 35(t+0.25)
40t = 35t + 8.75
5t = 8.75. ==> t = 1.75 hrs

D = 40* 1.75 = 70 km

Ans D

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brains · Answer

It can be done easily with proportionality. When anyone of the factor is constant , here in this case distance is constant ,then we can use other two relations to derive the proportionality

Since Distance is constant

Speed(S) is inversely proportional to time(t)

Ratio of speeds = 40:35 = 8 :7
So ratio of time will be = 7 :8

The difference between the ratio of time is 1 unit which is equivalent to 15 minutes

Hence 7 parts will be 7*15 = 105 minutes

Distance covered = 40*(105/60) = 70 Km . Answer is C

Jawad001 · Answer

Let the distance = x

x/35 -x/40 = 15/60

x/(7x5) -x/(8x5) = 15/60

x/(5 x 7 x8) = 1/4

x = 70(C)

ScottTargetTestPrep · Answer

We can create the distance equation:

40t = 35(t + 1/4)

40t = 35t + 35/4

5t = 35/4

20t = 35

t = 35/20 = 7/4

Thus, distance is 40 x 7/4 = 70 km.

Alternate Solution:

Let the distance of the journey be d kilometers and the scheduled time be t hours.

Since the bus completes the journey on the scheduled time when it travels at 40 km/h, we have d/40 = t.

Since the bus is 15 minutes = 1/4 hours late when it travels at 35 km/h, we have d/35 = t + 1/4.

Let’s substitute t = d/40 in the equation d/35 = t + 1/4:

d/35 = d/40 + 1/4

d/35 - d/40 = 1/4

5d/(35*40) = 1/4

d = (35*40)/(4*5) = 7 * 10 = 70 kilometers

Answer: C

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At its usual speed of 40 kmph, the bus covers its journey on schedule.

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