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

Question

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

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

Spartan85 · Accepted Answer

Answer: C

The clocks can show same time again only when their cumulative difference is 12 hours, since one is gaining and the other is losing.

For every 1 hour, the relative difference is 45 seconds (30 +15)

If 45 seconds difference is created in 1 hour
Then 12 hours difference is created in (12*3600) / 45 = 960 hours = 40 days.

RR88 · Answer

Option C

:N-clock (north clock) loses 30 seconds per hour & S-clock(south-clock) gains 15 seconds per hour.
:If both clocks started at 00:00:00, then -

After one hour - 
: N-clock = (1 - 30/3600) hour = (1 - 1/120) hour & S-clock = (1 + 15/3600) hour = (1 + 1/240) hour.
For every 12 hours N-clock loses = 12*1/120 = 1/10 hour & S-clock gains = 12*1/240 = 1/20 hour.

So, after 120 hours (5 days) N-clock would have lost 1 hours & after 240 hours (10 days) S-clock would have gained 1 hours.

Hence, after the  40  days N-clock would have lost 8 hours (will show 04:00:00) & S-clock would have gained 4 hours (will show 04:00:00).

BillyZ · Answer

Spartan85 . I could not understand the logic why we need to divide by 45s?

1 hr → 45 s lose
12 hrs → ?

(12 hr)(\frac{60 min}{hr})(\frac{60 s}{min}) = 12*60*60 s

\frac{12*60*60s}{45s}

Eventually, the unit cancel off each other.

Spartan85 · Answer

Unitary method,

If 45 seconds lost in 1 hour
then 1 second lost in 1/45 hour
then 12* 60 * 60 seconds lost in ? hours = (12*60 *60) *(1/ 45) hours= 960 hours = 40 days.
(the units dont really cancel each other in this case, but we maintain constant units on each side so that we dont make a mistake)

gracie · Answer

because north:south clock speed ratio is 2:1,
they will meet again--show the same time--
when north clock has lost 8 hours
and south clock has gained 4 hours
north clock loses 12 minutes each day
8 hours=480 minutes
480/12=40 days
C

va95 · Answer

One of the most frequent tricks the GMAT uses is an attention trick. Spent more than 10 minutes trying to solve this problem and then noticed that it was mentioned that the clocks are 12-hour format. I guess math requires as much attention as verbal does.

gracie · Answer

total distance=12 hours*60 minutes*60 seconds=43,200 seconds
combined rate=45 seconds per hour
total time=43,200/45=960 hours
960/24=40 days
40
C

ScottTargetTestPrep · Answer

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

fskilnik · Answer

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}
 {	ext{North}}:\,\,\frac{{ - 30\,\,{	ext{s}}}}{{1\,\,{	ext{h}}}} \hfill \
 {	ext{South}}:\,\,\frac{{ + 15\,\,{	ext{s}}}}{{1\,\,{	ext{h}}}} \hfill \ 
\end{gathered} ight.\,\,\,\,\,\,\,\,\,\,\,\, \sim \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( * ight)\,\,\,\left\{ \begin{gathered}
 {	ext{North}}:\,\,{	ext{regular}}\,\,{	ext{clock}} \hfill \
 {	ext{South}}:\,\,\frac{{ + 45\,\,{	ext{s}}}}{{1\,\,{	ext{h}}}} \hfill \ 
\end{gathered} ight.

?\,\,{	ext{in}}\,\,\left( * ight)\,\,:\,\,\, + 12{	ext{h}}\,\,\,\,{	ext{in}}\,\,\,{	ext{?}}\,\,{	ext{h}}

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

?\,\,\,\, = \,\,\,\, + {	ext{12h}}\,\,\,\left( {\frac{{60 \cdot 60\,\,\,{	ext{s}}}}{{1\,\,{	ext{h}}}}\begin{array}{*{20}{c}}
 
earrow \ 
 
earrow 
\end{array}} ight)\,\,\,\left( {\frac{{1\,\,{	ext{h}}}}{{ + 45\,\,\,{	ext{s}}}}\begin{array}{*{20}{c}}
 
earrow \ 
 
earrow 
\end{array}} ight)\,\,\,\,\left( {\frac{{1\,\,{	ext{day}}}}{{24\,\,{	ext{h}}}}\begin{array}{*{20}{c}}
 
earrow \ 
 
earrow 
\end{array}} ight)\,\,\,\,\, = \,\,\,\,\,\frac{{12 \cdot 60 \cdot 60}}{{45 \cdot 24}} = 40\,\,{	ext{days}}

Obs.: arrows indicate licit converters.

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

Regards, 
Fabio.

P.S.: many solutions presented before had exactly the same rationale, by the way.

EMPOWERgmatRichC · Answer

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 in 40 days 
The South clock will have GAINED (1/2 hour/5 days)(40 days) = 4 hours GAINED in 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

GMAT assassins aren't born, they're made, 
Rich

GianKR · Answer

I looked at this problem as I would look at any  relative speed problem that involves two bodies moving towards each other in opposite directions.

Considering the 12 hours clock format, the two clocks are moving in opposite directions towards each other and need together to close a gap of 12hours=720 minutes, in order to meet (display the same time).

The 1st clock rate is 30 seconds/hour, converted to minutes per day= 12minutes/day.
The 2nd clock rate is 15 seconds/hour, converted to minutes per day= 6 minutes/day.

Their relative speed, since they are moving in opposite directions, is the sum of their speeds= 12+6= 18 minutes/day.

From here,  we have got the relative speed= 18 min/day and the gap= 720 min , we have matching units and can get the time applying the rate, time, distance formula:  t= 720/18= 40 days.

Correct answer is C

Bunuel  have you get an official answer to share with us?

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