<p>When 60 minutes elapse on a correct clock, 62 minutes register on clock F (fast), and only 56 minutes register on clock S (slow). If later in the day clock F reads 8:00 and clock S reads 7:00, what was the correct time when the two clocks were originally set?</p>
<p>Please Help!</p>
<p>When 60 minutes elapse on a correct clock, 62 minutes register on clock F (fast), and only 56 minutes register on clock S (slow). If later in the day clock F reads 8:00 and clock S reads 7:00, what was the correct time when the two clocks were originally set?</p>
<p>Ok, its actually pretty simple:
Let w = the original time the clocks were set
Let h = the true number of hours past</p>
<p>Now, it states that clock F goes 2 minutes fast.
So, simply do: 62/60 = 1.033
And for Clock S: 56/60 = 0.933</p>
<p>Now, set up two equations:</p>
<p>w + 1.033h = 8:00
w + 0.933h = 7:00</p>
<p>Now, solve by using substitution and you get h = 10.</p>
<p>Now plug this value back into one of the equations: </p>
<p>w + 1.033(10) = 8</p>
<p>w = -2.333. Assume 12:00 is equal to 0. -2.33 translates to -2 hours and 20 minutes. So 2 hours and 20 minutes before 12 is 9:40.</p>
<p>SO, the original time is 9:40</p>
<p>Thank you so much guardiandevil!</p>
<p>was this a real SAT question??</p>
<p>bio_freak -
Many of us enjoy good math challenges, and there are tons of awesome places on the Internet where we can quench a math itch.
A couple of the best ones:
<a href="http://www.artofproblemsolving.com/Forum/index.php%5B/url%5D">http://www.artofproblemsolving.com/Forum/index.php</a>
<a href="http://www.unl.edu/amc/a-activities/a7-problems/problemdir.html%5B/url%5D">http://www.unl.edu/amc/a-activities/a7-problems/problemdir.html</a></p>
<p>As much as your "HUH-questions" are entertaining diversions, they might be bio_freaking out a lot of this forum participants.
Have mercy! SAT and AMC "are two big differences". </p>
<p>On your timely question.
I'd rather not use equations when I can do without them.</p>
<p>At the initial time when clocks S'n'F were set, they were
showing the same time.
1 hour later their readings differed by 62 - 56 = 6 min.
Correct time was (S clock) + 4 min. = (F clock) - 2 min.</p>
<p>2 hours later they differed by 2x62 - 2x56 = 2x(62-56) = 2x6 = 12 min.
Correct time: (S) + 2x4 min. = (F) - 2x2 min.</p>
<p>3 hours later: by 3x6 = 18 min.
Correct: (S) + 3x4 min. = (F) - 3x2 min.
....
When S and F struck 7 and 8 respectively, their readings differed by 60 min = 10x6 min.
Correct time at that moment:
(S) + 10x4 min. = 7:40 = (F) - 10x2 min.
It happened 10 hours past initial time.
10 hours back from 7:40 is 9:40.
THE END.</p>
<p>Yeah my first instinct was to do it the second way, without equations. Just personal preference I think</p>
<p>Since we do have a couple of suggestions, I'll add mine that is a mere variance. However, if this were on the SAT, this how I would do since I WANT to remember that lengthy problem most often have the simplest of solutions! </p>
<ol>
<li>Difference of rates between clocks is 2 - (-4) = 6 minutes</li>
<li>Total Difference at the end is 8 - 7 = 1 hour or 60 minutes</li>
<li>Numbers of hours needed to account for 60 minutes difference is 60/6 = 10.</li>
</ol>
<p>Now, we know that ten REAL hours are needed, so let's take a look at both clocks; </p>
<p>In ten real hours, the fast clock would have marked 10 x 62 or 620 minutes. So, the initial time was 8 - 10h20 or 9h40. </p>
<p>Time to solve this? about 30 seconds and 15 if you used military time! :)</p>
<p>PS The only numbers on my paper were 6 ... 60 ...10... 620...2000 ...1020 ... 940 </p>
<p>PPS You could have taken the slow clock and done 10 x 56 or 560 minutes. The initial time would be 7 - 9h20 or 9h40.</p>
<p><strong>applauds xiggi's ability to reduce math problems to their simplest form</strong></p>