Suppose you have the following problem: if the hour and minute hands are opposite to each other at 6 o'clock, when will they coincide?
My attempt at a solution:
Let $x$ = the number of minute spaces moved by the hour hand.
Let $y$ = the number of minute spaces moved by the minute hand.
Since the minute hand moves twelve times as fast as the hour hand, $y=12x$
Now, this is where it can get a little tricky. According to the problem, the hands are currently opposite each other; or, in other words, $y=60+x$. This is because, to reach the hour hand, the minute hand has to move 60 minute spaces (since that is how long an hour is), but then, to be opposite the other hand, move the same amount of minute spaces as the hour hand has (i.e., in this problem, 30 minute spaces, which is half an hour).
So, to coincide, the hour and minute hands must have the same number of minute spaces. I think that this should be represented by $y=60-x$, since, in the current scenario, the minute hand is $x$ minute spaces ahead of the hour hand, and thus it must move back $x$ spaces to coincide with the other hand.
So, the two equations are $y=12x$ and $y=60-x$.
After substituting $12x$ for $y$ in the second equation, we get:
$12x=60-x$
$13x$=60
$x=\frac{60}{13}$
Substituting this value back in the 1st equation, we obtain $\frac{720}{13}=y$. After converting that to a mixed number, we get $55 \frac{5}{13}$ minutes.
My Questions
To me, it doesn't make sense how I should convert this value into an answer; I have a number of minutes, but not a specific time on the clock.
I also don't know if my second equation for when the minute and hour hands coincide is correct, and therefore I have an overall doubt with regards to the accuracy of my solution. If my solution is wrong, could someone please explain my error so that I can get the right answer?
Thanks
Best Answer
The clock is now at 6 o'clock, so the minute hand points up and the hour hand points down. I'll use your definitions but make them a bit more precise:
Let x be the number of minute spaces moved by the hour hand between now and the moment the hands coincide.
Let y be the number of minute spaces moved by the minute hand between now and the moment the hands coincide.
As you say, the minute hand moves twelve times as fast as the hour hand, so y=12x.
The minute hand needs to travel 30 minute spaces around the clock to get where the hour hand starts off at, and then needs to travel as much as the hour hand to catch up. Basically the hour hand has a head start of 30 minute spaces that the minute hand needs to catch up on. This gives the second equation y=30+x.
Solving these two equations gives y=12*30/11 = 32.7272. This is the number of minute spaces that the minute hand has to travel, i.e. the number of minutes that have elapsed. This is 32 minutes and 43.6363 seconds after 6 o'clock.