A conference room has two analog (12-hour format) clocks,

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Veritas Prep

A conference room has two analog (12-hour format) clocks, one on the north wall and one on the south wall. The clock on the north wall loses 30 seconds per hour, and the clock on the south wall gains 15 seconds per hour. If the clocks begin displaying the same time, after how long will they next display the same time again?

A. 32 days
B. 36 days
C. 40 days
D. 44 days
E. 48 days

OA C.

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by deloitte247 » Sun Sep 23, 2018 8:50 am
The clock on the north wall loses twice as much time as the clock on the south wall gains,
Assuming that both clock starts from 12 o'clock,
The next time they will display the same time is 4 o'clock because by that time the clock on the north wall would have lost 8 hours and the clock on the south wall would have gained 4 hours.

South Wall Clock=
4 hours to seconds = 60 * 60 * 4 = 14,4000 seconds and the clock gains 15 seconds per hour. For it to get to 4 o'clock at the same time with the north clock on the north wall=
15 seconds = 1 hour
14,4000 seconds = x (cross match)
$$x\ =\ \frac{\left(14,400\right)}{15\ }=\ 960\ hours.$$
24 hours = 1 day
960 hours = x
$$x\ =\ \frac{960}{24}=\ 40\ days.$$

North Wall Clock=
8 hours to seconds = 60 * 60 * 8 = 28, 800s and this clock loses 30 seconds per hour for it to get to 4 o'clock at the same time with the clock on the south wall.
30 secounds = 1 hour
28,800 seconds = x
$$x\ =\ \frac{28800}{30}=960\ hours.$$
24 hours = 1 day
$$960\ hours=\ \frac{960}{24}=40\ days.$$
In either clock, the number of days needed = 40
Option C is CORRECT.

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AAPL wrote:Veritas Prep

A conference room has two analog (12-hour format) clocks, one on the north wall and one on the south wall. The clock on the north wall loses 30 seconds per hour, and the clock on the south wall gains 15 seconds per hour. If the clocks begin displaying the same time, after how long will they next display the same time again?

A. 32 days
B. 36 days
C. 40 days
D. 44 days
E. 48 days
This is an excellent example of a problem in which "blending" two of our most powerful techniques solves the exercise immediately (in just one line)!
Techniques: Relative Velocity AND Units Control
\[\left\{ \begin{gathered}
{\text{North}}:\,\,\frac{{ - 30\,\,{\text{s}}}}{{1\,\,{\text{h}}}} \hfill \\
{\text{South}}:\,\,\frac{{ + 15\,\,{\text{s}}}}{{1\,\,{\text{h}}}} \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\,\,\,\, \sim \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( * \right)\,\,\,\left\{ \begin{gathered}
{\text{North}}:\,\,{\text{regular}}\,\,{\text{clock}} \hfill \\
{\text{South}}:\,\,\frac{{ + 45\,\,{\text{s}}}}{{1\,\,{\text{h}}}} \hfill \\
\end{gathered} \right.\]
\[?\,\,{\text{in}}\,\,\left( * \right)\,\,:\,\,\, + 12{\text{h}}\,\,\,\,{\text{in}}\,\,\,{\text{?}}\,\,{\text{h}}\]

Once DATA and FOCUS were structurally presented, let´s connect them! (This is our method´s "backbone".)

\[?\,\,\,\, = \,\,\,\, + {\text{12h}}\,\,\,\left( {\frac{{60 \cdot 60\,\,\,{\text{s}}}}{{1\,\,{\text{h}}}}\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}} \right)\,\,\,\left( {\frac{{1\,\,{\text{h}}}}{{ + 45\,\,\,{\text{s}}}}\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}} \right)\,\,\,\,\left( {\frac{{1\,\,{\text{day}}}}{{24\,\,{\text{h}}}}\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}} \right)\,\,\,\,\, = \,\,\,\,\,\frac{{12 \cdot 60 \cdot 60}}{{45 \cdot 24}} = 40\,\,{\text{days}}\]
Obs.: arrows indicate licit converters.


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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by Scott@TargetTestPrep » Sun Sep 23, 2018 3:45 pm
AAPL wrote:Veritas Prep

A conference room has two analog (12-hour format) clocks, one on the north wall and one on the south wall. The clock on the north wall loses 30 seconds per hour, and the clock on the south wall gains 15 seconds per hour. If the clocks begin displaying the same time, after how long will they next display the same time again?

A. 32 days
B. 36 days
C. 40 days
D. 44 days
E. 48 days
We can see that the clock on the north wall loses twice as much time as the clock on the south wall gains. We can assume the time that both clocks display is 12 o'clock, i.e., the hour hand is on the number 12 on both clocks. The next time they will display the same time is 4 o'clock, since the clock on the south wall gains 4 hours and the clock on the north wall loses 8 hours.

From the south clock point of view:

Since 4 hours = 4 x 3600 = 14400 seconds and the clock on the south wall gains 15 seconds per hour, it needs 14400/15 = 960 hours or 40 days to strike 4 o'clock as the clock on the north wall strikes the same time.

Or, from the north clock point of view:

Since 8 hours = 8 x 3600 = 28800 seconds and the clock on the north wall loses 30 seconds per hour, it needs 28800/30 = 960 hours or 40 days to strike 4 o'clock as the clock on the south wall strikes the same time.

In either case, the number of days needed is 40.

Answer: C

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by [email protected] » Sun Sep 23, 2018 7:27 pm
Hi All,

We're told that a conference room has two analog (12-hour format) clocks, one on the north wall and one on the south wall; the clock on the north wall LOSES 30 seconds per HOUR, and the clock on the south wall GAINS 15 seconds per HOUR. We're asked, if the clocks begin displaying the SAME time, after how long will they next display the SAME time again. This question can be approached in a number of different way, but you can take advantage of the 'format' of the answer choices to do some minimal math to find the correct answer.

Since all five answers are in 'days', we can think of all of these calculations in terms of how the rates are calculated per day.
The North clock loses 30 seconds/hour; with 24 hours/day, that would be 12 minutes LOST per DAY.
The South clock gains 15 seconds/hour; with 24 hours/day, that would be 6 minutes GAINED per DAY.

There are 60 minutes/hour, so to make these calculations a little easier to think about, I'm going to think in terms of hours instead of minutes:
After 5 days....
The North clock will have LOST (12 minutes/day)(5 days) = 60 minutes in 5 days = 1 hour LOST every 5 days
The South clock will have GAINED (6 minutes/day)(5 days) = 30 minutes in 5 days = 1/2 hour GAINED every 5 days

Thus, every multiple of 5 days will give us a relatively nice fraction (either 1 or 1/2). Looking at the answer choices, there's only one answer that's a multiple of 5....
IN 40 DAYS.....
The North clock will have LOST (1 hour/5 days)(40 days) = 8 hours LOST every 40 days
The South clock will have GAINED (1/2 hour/5 days)(40 days) = 4 hours GAINED every 40 days.
On a 12-hour clock face, 8 hours lost combined with 4 hours gained would lead to the exact SAME time, so this MUST be the answer.

Final Answer: C

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