Expect the Unexpected | Arithmetic

Most of us know that GMAT is a shrew, and is very hard to tame. It is well established that it is able to give a pretty accurate estimate of aptitude with just a few questions, and that the only way to “deceive” it is by actually improving your aptitude! It has numerous tricks up its sleeves to uncloak a rather basic player.

Let’s discuss one such trick today – a trick in which you need to realize that the situation calls for a complete U-turn of the usual.

Let’s take an example:

Question: Two cars run in opposite directions on a circular track. Car A travels at a rate of ${6 \times \pi}$ miles per hour and Car B runs at a rate of ${8 \times \pi}$ miles per hour. If the track has a radius of 6 miles and the cars both start from Point S at the same time, how long, in hours, after the cars depart will they again meet at Point S?

(A) ${\frac{6}{7}}$ hrs

(B) ${\frac{12}{7}}$ hrs

(D) ${6}$ hrs

(E) ${12}$ hrs

Solution: What would we usually do in such a question? Two cars start from the same point and run in opposite directions – their speeds are given. This would remind us of relative speed. When two objects move in opposite directions, their relative speed is the sum of their speeds. So we might be tempted to do something like this:

Perimeter of the circle = $2 \times \pi \times r = 2 \times \pi \times 6 = 12 \times \pi$ miles

Time taken to meet = $\frac{Distance}{Relative\ Speed} = \frac{12 \times \pi}{(6\times \pi + 8 \times \pi)} = \frac{6}{7}$ hrs

But take a step back and think – what does $\frac{6}{7}$ hrs give us? It gives us the time taken by the two of them to complete one circle together. In this much time, they will meet somewhere on the circle but not at the starting point. So this is definitely not our answer.

The actual time taken to meet at point S will be given by $\frac{12 \times \pi}{(8 \times \pi - 6 \times \pi)} = 6$ hrs

This is what we mean by unexpected! The relative speed should be the sum of their speeds. Why did we divide the distance by the difference of their speeds? Here is why:

For the two objects to meet again at the starting point, obviously they both must be at the starting point. So the faster object must complete at least one full round more than the slower object. In every hour, car B – the one that runs at a speed of $8 \times \pi$ mph covers $2 \times \pi$ miles more compared with the distance covered by car A in that time (which runs at a speed of $6 \times \pi$ mph). We want car B to complete one full circle more than car A. In how much time will car B cover $12 \times \pi$ miles (a full circle) more than car A?

$\frac{12 \times \pi}{2 \times \pi}$ hrs = 6 hrs.

Now we will keep the question the same but will change the figures a bit:

Modified Question: Two cars run in opposite directions on a circular track. Car A travels at a rate of ${3 \times \pi}$ miles per hour and Car B runs at a rate of ${5 \times \pi}$ miles per hour. If the track has a radius of 7.5 miles and the cars both start from Point S at the same time, how long, in hours, after the cars depart will they again meet at Point S?

So following the same logic as above,

Perimeter of the circle = $2 \times \pi \times r = 2 \times \pi \times 7.5 = 15 \times \pi$ miles

The time taken to meet at point S will be given by $\frac{15 \times \pi}{(5 \times \pi - 3 \times \pi)} = 7.5$ hrs

But note that the two cars will not even be at the starting point, S, in 7.5 hrs. So this answer is wrong. Why? It has something to do with the word “at least” used in the explanation above i.e. “So the faster object must complete at least one full round more than the slower object. “

Try to put it all together.

Meanwhile, let’s give you another method. This will not fail you no matter what the figures.

Using the original question:

Time taken by car A to complete one full circle = $\frac{12 \times \pi}{6 \times \pi} = 2$ hrs

Time taken by car B to complete one full circle $\frac{12 \times \pi}{8 \times \pi} = 1.5$ hrs

So every 2 hrs car A is at S and every 1.5 hrs, car B is at S. When will they both be together at S?

Car A at S -> 2 hrs, 4 hrs, 6 hrs, 8 hrs …

Car B at S -> 1.5 hrs, 3 hrs, 4.5 hrs, 6 hrs …

In 6 hrs – the first common time, both cars will be at the point S together. So answer is 6 hours.

Using the same method on the Modified Question,

Time taken by car A to complete one full circle = $\frac{15 \times \pi}{3 \times \pi} = 5$ hrs

Time taken by car B to complete one full circle = $\frac{15 \times \pi}{5 \times \pi} = 3$ hrs

So every 5 hrs, car A is at S and every 3 hrs, car B is at S. When will they both be together at S?

Car A at S -> 5 hrs, 10 hrs, 15 hrs, 20 hrs

Car B at S -> 3 hrs, 6 hrs, 9 hrs, 12 hrs, 15 hrs

In 15 hrs – the first common time (LCM of 3 and 5), both cars will be at the point S together.

This all makes sense now.

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