Time Speed Distance Tough Question | Arithmetic

Sometimes we come across questions in which the question stem itself is hard to explain (so no point worrying about the difficulty in explaining the solution as of now!).

Today we will take a look at one such question. It uses the concepts of relative speed and standardised tests could give you some pretty intimidating questions at higher levels. So what should be your strategy when you come across a question which takes a minute or more to sink in? After you understand the question, first of all you should congratulate yourself that the toughest part is already over. If the question is hard to understand, the solution would be cake walk (well, at least it will feel like it).

Of course, another approach is to skip such a question within 20 secs and move on but in the interest of this post, we will assume that you will not do that. Also, if you get such a question, chances are that you are ‘good’ at quant and that you would have performed quite well in the test till then. In that case, you would have plenty of extra time to challenge your intellect with such a question.

Let’s look at this question now:

Question: On a straight road, a biker noticed that every 12 minutes a bus overtakes him and every 4 minutes he meets an oncoming bus. If all buses leave the station at the same fixed time intervals and run at the same constant speed and the biker moves at a constant speed, what is the time interval between consecutive buses?

(A) 5 minutes

(B) 6 minutes

(D) 9 minutes

(E) 10 minutes

Solution:

First let’s review the information given in the question:

– All buses run at the same constant speed.

– They leave the station at fixed time intervals, say every $t$ mins. (We have to find the value of $t$ ) Had the biker been stationary, he would have met a bus every $t$ mins from both directions.

– The biker is moving at a constant speed which is less than the speed of the bus. We can infer that the biker’s speed must be less than the speed of the bus because buses overtake him from behind every 12 mins. Had his speed been equal to or more than the speed of the buses, the buses could not have overtaken him.

– Since the biker is moving too, (say going due east) he meets buses coming from one direction (say going from east to west) more frequently and buses coming from the other direction less frequently.

Imagine a scenario where the biker is stationary:

Bus –> Bus –> Bus –> Bus –> Bus –> Bus –>

Biker

Bus <– Bus <– Bus <– Bus <– Bus <– Bus <–

He will meet a bus coming from either direction every $t$ mins. Note that the distance between consecutive buses will stay the same.

Why? Let me explain this using an example:

Assume that starting from a bus station, all buses run at the same speed of 50 mph.

Say a bus starts at 12:00 noon. Another starts at 1:00 pm i.e. exactly one hr later on the same route. Can we say that the previous bus is 50 miles away at 1:00 pm? Yes, so the distance between the two buses initially will be 50 miles. The 1 o clock bus also runs at 50 mph. Will the distance between these two buses always stay the same i.e. the initial 50 miles? Since both buses are moving at the same speed of 50 mph, relative to each other, they are not moving at all and the distance between them remains constant. The exact same concept is used in this question.

Now imagine what happens when the biker starts moving too. Say, he is traveling due east.

Bus –> Bus –> Bus –> Bus –> Bus –> Bus –>

…………………………..Biker ->

Bus <– Bus <– Bus <– Bus <– Bus <– Bus <–

Say, he just met two buses, one from each direction. Now the Bus (in bold) is a fixed distance away from him. The biker and the Bus are traveling toward each other so they will cover the distance between them faster. Their relative speed is the sum of the speed of the bus and the speed of the biker. They take only 4 mins to meet up. $t$ must be more than 4.

On the other hand, the biker is moving away from the Bus (in italics) so the effective speed of Bus is only the difference between the speed of the bus and the speed of the biker. So Bus takes 12 mins to catch up with the biker. $t$ must be less than 12.

Now that we have understood the question, solving it is relatively easy.

Say the speed of the biker is $K$ and the speed of the bus is $B$ . The ratio of the relative speeds in the two cases will be the inverse of the ratio of time taken (Ratio of speed and time is covered in a previous post).

When the relative speed is $(B + K)$ , the time taken is 4 mins.

When the relative speed is $(B - K)$ , the time taken is 12 mins.

$\frac{B + K}{B -K} = \frac{12}{4}$ (inverse of $\frac{4}{12}$ )

This gives us $K = \frac{1}{2}B$

This means that the bus travelling at a relative speed which is half its usual speed $(B-K = \frac{B}{2})$ takes 12 minutes to meet the man. If it were travelling at its usual speed (i.e. if the biker were stationary), it would have taken half the time i.e. $\frac{12}{2} = 6$ mins to meet the biker. Had the biker been stationary, the time taken by the bus to cover the distance between them would be the same as the time interval between consecutive buses i.e. $t$ mins.