<p>A broken clock is set correctly at 12:00 noon. However it registers only 20 minutes for each hour. In how many hours will it again register the correct time?
a) 12
b) 18
c)24
d)30
e) 36</p>
<p>Does the correct time mean 12:00 again? And how to solve this problem?</p>
<p>Is the answer E?</p>
<p>Uhm no. I don’t know whether they use the standard time or the time has been cut off in this problem</p>
<p>Is it a real BB question?</p>
<p>Answer is (B).</p>
<p>Solve with following equation: x/3 = x-12, where x is the answer.</p>
<p>what’s 12?</p>
<p>I don’t know what you mean; it’s a number, but you probably know that.</p>
<p>This is definitely not a BB problem, hence it is flawed.</p>
<p>If the clock makes no distinction between AM and PM, then the answer is 18. The broken clock would display 6:00 [PM] and a properly functioning clock would display 6:00 <a href=“the%20brackets%20indicate%20that%20AM/PM%20would%20be%20irrelevant”>AM</a>. However, the fact that the problem says “12 noon” indicates that a distinction is should be made (then 36 would be the correct answer). Whether a distinction should be made is not made explicitly clear in the problem, which is a flaw of non BB problems.</p>
<p>Furthermore, when the problem says “In how many hours will it again register the correct time” is it referring to hours passed on the broken clock or on a working clock? Again, it’s not clear.</p>
<p>Bottom line: this problem is crap. Use college board materials ONLY.</p>
<p>EDIT: I see what the question is asking, but it’s not very good.</p>
<p>It’s saying when will the broken clock’s pointer be on the real time. (what I said is confusing, but I will explain). Since you already know the answer now, you could use 18 hours and add that to 12 noon, which would be 6 AM. For the broken clock in 18 hours it’s pointer will move 6 hours, which will be at 6.</p>
<p>If the algebra doesn’t come to mind, this problem can be solved by brute force. Make a table of “real time” and “Bad Clock time” and keep adding rows until the two times match. Then look to see how much real time has passed.</p>
<p>RT 12 BCT 12
RT 1 BCT 12:20
RT 2 BCT 12:40
RT 3 BCT 1
RT 4 BCT 1:20
RT 5 BCT 1:40
RT 6 BCT 2
RT 7 BCT 2:20
RT 8 BCT 2:40
RT 9 BCT 3
RT 10 BCT 3:20
RT 11 BCT 3:40
RT 12 midnight BCT 4
RT 1 BCT 4:20
RT 2 BCT 4:40
RT 3 BCT 5
RT 4 BCT 5:20
RT 5 BCT 5:40
RT 6 BCT 6 </p>
<p>There you go. After 18 hours the Bad Clock shows the right time.</p>
<p>Here’s a graphical version – albeit a long one. On the SAT, a clock should be presented as 12 hours “old style” clock. </p>
<p>Hours Clock 1 Clock 2
0 12 12
1 1 1220
2 2 1240
3 3 1
4 4 120
5 5 140
6 6 2
7 7 220
8 8 240
9 9 3
10 10 320
11 11 340
12 12 4
13 1 420
14 2 440
15 3 5
16 4 520
17 5 540
18 6 6
19 7 620
20 8 640
21 9 7
22 10 720
23 11 740
24 12 8
25 1 820
26 2 840
27 3 9
28 4 920
29 5 940
30 6 10
31 7 1020
32 8 1040
33 9 11
34 10 1120
35 11 1140
36 12 1200</p>
<p>However, you really can use a three hours’ jump </p>
<p>0 12 12
3 3 1
6 6 2
9 9 3
12 12 4
15 3 5
18 6 6
21 9 7
24 12 8
27 3 9
30 6 10
33 9 11
36 12 12</p>
<p>The algebraic relationship is as follows:</p>
<p>HGC = Hours Good Clock
HBC = Hours Bad Clock
HE = Hours Elapsed</p>
<p>HGC = HE [Mod[/url</a>] 12
HBC = (HE / 3) Mod 12 (where 1/3 = 20 minutes Bad Clock/60 minute Good Clock)</p>
<p>Solve for HGC = HBC:
HE Mod 12 = (HE / 3) Mod 12</p>
<p>Because of the nature of the Mod operator, you don’t need to concern yourself with values of HE less than 12 for the HGC = HBC equality. Then the question is, what value of HE will satisfy the equation such that HE divided by 3 (and then divided by 12) results in the same remainder when HE is divided by 12? </p>
<p>Because the [url=<a href=“http://mathforum.org/library/drmath/view/51619.html]reverse”>http://mathforum.org/library/drmath/view/51619.html]reverse</a> of the mod function](<a href=“http://en.wikipedia.org/wiki/Modulo_operation]Mod[/url”>Modulo - Wikipedia) gives several possible values, the strategy at this point is to start evaluating the relationship with values starting at 12. To save time, use the values from the multiple choice answers:</p>
<p>For HE = 12, HE Mod 12 = (HE / 3) Mod 12 : 0 <> 4
For HE = 18, HE Mod 12 = (HE / 3) Mod 12 : 6 = 6 (this satisfies the equation)
…
For HE = 36, HE Mod 12 = (HE / 3) Mod 12 : 0 = 0 (this also satisfies the equation) (see [url=<a href=“http://talk.collegeconfidential.com/1063262570-post9.html]#9[/url”>http://talk.collegeconfidential.com/1063262570-post9.html]#9[/url</a>] for discussion on possible interpretations of the question)</p>
<p>The time indicated by the bad clock should be equal to the actual time.
Let h be the number of hours passed.It cannot be less than 12 because then it would show a time of hX20/60 hrs =h/3 hours. </p>
<p>So if h> 12 then the correct hours shown by the clock should be h-12 and the actual hours that would be shown = (h X 20)/60 =h/3</p>
<p>Now clock will be correct when h-12 =h/3
3h - 36=h
h=18</p>
<p>I don’t think the question is flawed; I’ve never seen a clock that dinstinguishes between A.M. and P.M…</p>
<p>@silverturtle, I don’t understand your equation x/3= x+12. How do you get 12?</p>
<p>I did it in a more logical than algebraic way, I found the number of minutes it would take to get back to 12 which is 12(60) = 720. Then I divided that by 40 (the number of minutes skipped per hour). This gave me B) 18.</p>
<p>but how do you know that the next correct hour is also 12?</p>
<p>Interesting…but LOTS of clocks indicate AM or PM. For example, look at the lower right hand corner of your computer screen! Just sayin…</p>
<p>The clock on my computer can’t break in the way the question describes. But this argument is rather pointless. :)</p>