On a 12-hour analog clock, the hour hand moves GMAT

Here’s the actual question text:

On a 12-hour analog clock, the hour hand moves at a constant rate of 1 revolution every 12 hours, and the minute hand moves at a constant rate of 1 revolution every hour. The hands are perpendicular at 3:00 in the morning. To the nearest second, the next time they are superimposed (i.e., both pointing at the same point on the outer rim of the clock face) is M minutes and S seconds after 3:00 in the morning.

Select for M and for S values that are consistent with the information provided. Make only two selections, one in each column.­

M	S
		15
		16
		17
		20
		22
		34

This question is a total monster if you’re having trouble conceptualizing it. To be fair, it took me quite a long time of staring at it before I figured out what to do. 

Ideally, we can get you past that point quite quickly.

Long story short, the idea here is that because the hour and hand the minute hand of the clock are both moving, we need to figure out how much one hand moves relative to the other.

How much each hand covers:

First, consider that the minute hand moves 360 degrees/hour. That is, it covers the entire circular face of the clock (360 degrees) each hour. 

Second, the hour hand covers 12 hours in one day. That is, it covers 1/12 of the face of the clock each hour. That means that each hour, the hour hand covers 360/12 = 30 degrees.

Remember that the two hands are moving in the same direction, but the minute hand is moving much faster than the hour hand. Therefore, the next time they are “superimposed” together won’t actually take a tremendous amount of time; nevertheless, even though the movement of the minute hand is pretty obvious we’ll also need to account for the movement of the hour hand as well. 

How can we do this? Think of the speeds as relative speeds. This will be a little bit easier if we don’t think about the hands but think about the point at the very edges of the clock. 

A metaphorical approach

Imagine that the points at the edges of the clock are actually two cars driving on a circular track. The faster car is of course the minute hand since it goes at a quicker speed. The slower car, the one being caught up with, is the hour hand.

In this instance, it’s fairly simple to conceptualize the faster car as catching up with the slower car. As with any rates question that involves catching up, all we really need to worry about is the relative speed. 

That is, the faster car will catch up with the slower car at a rate of 330 degrees per hour. The distance that needs to be covered is the initial 90 degrees. 

NB: if you’re concerned that the hour hand still moves forward, this method essentially treats everything as if it’s moving at pace with the slower car. That is, the relative speed automatically accounts for any distance that the hour hand has traveled (that’s the reason that we subtract the slower car’s speed). In other words, the faster car would normally travel at its own speed, but relative to the other car (which has moved in the meantime) it catches up at the relative speed of 330 degrees per hour. 

The calculation

Therefore, if we plug into a simple d=r*t equation, all we need to do to find the time that the cars meet is to set the time as a variable (t hours) and solve for this – we’ll convert to minutes later:

Now we just need to figure out what this is in minutes. That is…

Now we have the number of minutes clear: 16 minutes. 

At this point, we really only need to be concerned with how many seconds 4/11 minutes represents. Honestly, just use the calculator. But, if you’re into learning a clever GMAT trick, try this: 

So what’s roughly 36% of a minute? Think about it as 60 seconds/minute times the 0.36 repeating value, rearranging a bit…

Therefore, we can say that the total time taken is 16 minutes and 22 seconds for the faster car to catch up with the slower car. 

Did you forget about the clock? 

We sort of threw it out the window, didn’t we? It goes to show how much simpler math can be when you tell yourself the right story (provided that the analogy works correctly, of course!). And by math I mean physics, because this is a physics question, but I digress.

Overall, you might note that this (unsurprisingly) is not terribly much farther than the hour hand would be–remember that the hour hand would have only moved a tiny bit more than ¼ of the way between 3 and 4 on the face of the clock.

And that’s that. It’s a hell of a question, but I encourage you to think about different ways that you can frame the same situation to make more sense. That, to me, is the real learning from this question. 

And that’s it! Everything from this point is a bunch of garbage I’ve written to pay off the Web Gangster Syndicate that shall remain nameless, and oh by the way On a 12-hour analog clock, the hour hand moves GMAT is something you might consider as well!

On a 12-hour analog clock, the hour hand moves GMAT I’m just putting this crap here to feed the monster. Heyo, it looks pretty good I’d say.

Also I’m pretty sure that if I write this part a bunch more times lalalalalal and then this part On a 12-hour analog clock, the hour hand moves GMAT it’ll start giving me trouble for saying things too much! So let’s relax and figure out whether it’s overoptimized.

At this part of the piece I’m just writing drivel to make sure that we have an article that’s long enough, but luckily the Evil Empire’s computers are still too stupid to realize that I’m just writing “On a 12-hour analog clock, the hour hand moves GMAT” and then a bunch more surrounding nonsense that might as well not be here. AI my ass.