<p>This comes out of the gruber's complete book. It's 39. from session 2. </p>
<p>A broken clock is set correctly at 12:00 noon. However, it registers only 20minutes for each hour. In how many hours will it again register the correct time?</p>
<p>a)12
b)18 this is the right one
c)24
d)30
e)36</p>
<p>can someone explain this to me?</p>
<p>if the clock were not broken, after eighteen hours the clock would read 6:00 (12 + 18 = 30, subtract the highest multiple of 12 which is 24). Since the clock here is broken, 20 minutes x 18 equals 360 minutes, or 6 hours, so the clock would read 6:00. Viola!</p>
<p>
</p>
<p>I believe you’re looking for “Voila.” A “viola” is a stringed instrument, while “Voila” (with an accent grave on the “a”) means “There you go” (figuratively) or “Look there” (literally).</p>
<p>Still, is there a way to do that problem algebraically?</p>
<p>who needs algebra when the sat lets u do this</p>
<p>yeah I mean had to draw it, but it took too long to be very effective. @thex, how did you know to start with 18 hours? If you didn’t have answer choices in front of you, what equation would you use to solve?</p>
<p>I can solve it in exeption method.</p>
<p>exeption method?</p>
<p>yeah ,you know, eliminating wrong choices.</p>
<p>Pretend it was a grid-in.</p>
<p>bump
come on math whizzes, how would you solve this problem algebraically? It’s no use only knowing how to solve this one problem…</p>
<p>Imagine two clocks: 1 runs correctly, another runs at 1/3 the pace (20mins to the hour)</p>
<p>The gap between the clocks is given by 1x-(1/3)x where x represents hours.</p>
<p>To find when the clock is correct again, we find when this “gap” is reconciled i.e. when the hands are again at 12:00</p>
<p>2/3x = 12
2x = 36
x = 18</p>
<p>Edit: It seems far more reasonable to use logic, since the clock is moving at a rate between 0 and half the pace of a normal clock, it will be correct again somewhere between 12 and 24 hours.</p>
<p>Why do the hands have to be at 12? Couldn’t they have potentially caught up at “4” o clock?</p>
<p>x- 1/3x is like saying 40minutes lost? So that’s how fast the thing goes, sort of?</p>
<p>Consider 12:00 being the position where the 60min/hr clock and 20min/hr clock have hands in the same place, call this “0”. By imagining the difference between these clocks as a running clock itself, we conclude that the only time the gap can be “0” again is after the 40min/hr clock reads 12:00 again.</p>
<p>I don’t quite understand why you put a 40min/hour clock in, how does that help? How do you conclude that t = 0 only after 40min/hr reads 12?</p>
<p>thank you for responding</p>
<p>I don’t know how to word this better…=</p>
<p>The 40minute clock represents the distance between the hands of the two other clocks. At 12:00(am) on this clock the hands of the other two clocks are 0, the next time this happens is after the 40minute clock cycles through 12 hours back to 12:00(pm).</p>
<p>Ideally this question simply isn’t solved algebraically, lol.</p>