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Tuesday, March 8, 2011

Speed Distance Time Question TANCET

Speed, Distance and Time is an important topic from a TANCET quant perspective. You could expect to get one to two questions from this topic.

The core concept in this topic is the relation between these three parameters
Distance = speed * time.

The second point to keep in mind while solving questions from this topic is to check if the units of the three parameters match.
i.e., if distance is measured in kilometres and time in hours, speed should be in km/hr. Alternatively, if speed is in kilometres and time in minute, speed should be in km/min.

Here is an easy to moderate level difficulty question in this topic.

A bus travels at 40 kmph for the first 100 kms, 60 kmph for the next 100 kms and 48 kmph for the final 200 kms, what is the average speed over the total distance?
1. 48 kmph
2. 50 kmph
3. 49 kmph
4. 52 kmph
5. 50.4kmph

The concept tested in this question is that of average speeds.

The answer to this question is 48 kmph.

Explanatory Answer

Average speed for the first 200kms = 2ab/(a + b) = 2 * 40 * 60/(40 + 60) = 48 kmph.

Bus travels equal distances at two different speeds, average speed = 2ab/(a + b)

Now, if we break the overall journey into two stretches of 200 kms each, the bus travels the first 200kms at 48kmph, and the second at 48 kmph. Overall, average speed = 48 kmph.
Correct answer (1)

Alternatively, you could find out the total time taken to cover the entire distance of 400 kms

Time taken for the first 100 kms = 100 / 40 = 10/4 hours

Time taken for the second 100 kms = 100/60 = 10/6 hours

Time taken for the last 200 kms = 200/48 = 50/12.

Adding up all three, we get 10/4 + 10/6 + 50/12 = 100/12 hours.

Average speed = total distance / total time

= 400 /(100/12) = 4800/100 = 48 kmph

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