On a 12-hour analog clock, the hour hand moves at a constant rate of

Question

On a 12-hour analog clock, the hour hand moves at a constant rate of 1 revolution every 12 hours, and the minute hand moves at a constant rate of 1 revolution every hour. The hands are perpendicular at 3:00 in the morning. To the nearest second, the next time they are superimposed (i.e., both pointing at the same point on the outer rim of the clock face) is  M  minutes and  S  seconds after 3:00 in the morning.

Select for  M  and for  S  values that are consistent with the information provided. Make only two selections, one in each column.

ID: 700268

KarishmaB · Accepted Answer

The clock questions are easily done using relative speed. Consider this: an hour hand covers 30 degrees in 1 hr A minute hand covers 360 degrees (full circle) in 1 hr Relative speed of minute hand = 330 degree per hour. When the two hands are perpendicular at 3:00, they have angle of 90 degrees between them. To overlap, they need to cover this relative distance of 90 degrees Time Taken = \frac{Distance}{Speed} =\frac{90}{330}*60 mins = 16\frac{4}{11} min \frac{4}{11} mins = \frac{4}{11} * 60 secs = 21.8 secs ANSWER 16 mins, 22 secs Check this video for a discussion on relative speed in clocks: https://youtu.be/dGqC_6DIKkY

egmat · Answer

Wow! What a question this is. On the surface, it appears to be simple – why? Since it is based on an equipment that we see on a daily basis – a clock. In fact, it looks simple enough more so because the author has explained how a clock works – something that is general information. But the real complexity of this question lies in what it asks and the kind of processing that it requires.

Watch the video solution and realize these complexities in the question and how each of these are places where a test taker can falter.

https://www.youtube.com/watch?v=Nkesqt92wto

Visualizing what it means when the author says that the two hands superimpose – And draw inference to set the stage for this question.
 
 Hour hand would have traveled 90 degrees more than the minute hand. 
 If one is unable to draw this inference, then they might as well randomly mark some answer and move to the next question, in the test environment. This is how critical this step is.   
 Translating the inference into mathematical terms is the next critical step. This is where you need to apply your understanding of Geometry concepts pertaining to angles in a circle. See how practical application of Geometry is tested so beautifully here. This is where you set the approach to solve this question. 
 Once you have the approach established, the execution is also extensive. And there are places where you could falter – not doing the unitary conversion correctly and hence arriving at incorrect equation. 
 Lastly, not doing the correct conversion to find minutes and seconds.

Beautiful question – one that is packed with learnings!

chetan2u · Answer

The question just asks us when will the two hands of clock meet after 3:00.

For sure, the distance travelled will be different. The minute hand will travel 90 degree extra. So, let us equate the angles they would travel.

Say both meet after x minutes.

Hour hand: It travels 360 in 12 hours or 30 degree per hour, that is 1/2 degree per minute. So x minutes would mean x*(1/2) or x/2 degree movement.

Minute hand: It travels 360 in 1 hour, that is 6 degrees per minute. So x minutes would mean x*6 degree movement.
Thus should be equal to 90 degree plus (x/2) degree done by hour hand.

=>  6x=90+\frac{x}{2} 
 12x=180+x......11x=180......x=\frac{180}{11}=\frac{16*11+4}{11}=16+\frac{4}{11}

now 4/11 minutes would mean 60*4/11 or 240/11 or 22 seconds approx.

The answer is 16 minutes 22 seconds

Engineer1 · Answer

chetan2u Bunuel MartyMurray : Can anyone please help me solve this? This is from the DI 2023-24 Review. Thanks. https://gmatclub.com/forum/download/file.php?id=81537 GMAT-Club-Forum-xpylap48.png

MartyMurray · Answer

On a 12-hour analog clock, the hour hand moves at a constant rate of 1 revolution every 12 hours, and the minute hand moves at a constant rate of 1 revolution every hour. The hands are perpendicular at 3:00 in the morning. To the nearest second, the next time they are superimposed (i.e., both pointing at the same point on the outer rim of the clock face) is M minutes and S seconds after 3:00 in the morning.

Select for M and for S values that are consistent with the information provided. Make only two selections, one in each column.

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Official Answer

M: 16 S: 22

To determine when they will be superimposed, we must determine when they will both have completed the same fraction of a revolution.

At 3:00, the minute hand has completed none of a revolution, and the hour hand has completed 1/4 of a revolution.

The minute hand travels at the rate of 1 revolution/hour.

The hour hand travels at the rate of 1/12 revolution/hour.

So, we can determine how long they will need to take to arrive at the same spot using the following formula, in which T represents a quantity of time in hours:

1/4 + T × 1/12 = T × 1

1/4 + T/12 = T

1/4 = 11T/12

(1/4)/(11/12) = T

3/11 = T

Since there are 60 minutes in an hour, the number of minutes is 3/11 × 60 ≈ 16.36 minutes ≈ 16 minutes and 60 × 0.36 seconds ≈ 16 minutes and 22 seconds.

15
16
17
20
22
34

For M, select 16.

For S, select 22.

KushagraKirtiman · Answer

Hi Marty,
Just wanted an advise, I solved this question in approx 8 mins. Do you think that it would advisable to solve question which would take more than 3 mins?

GMATGuruNY · Answer

Alternate approach:
At 12am, the hour hand and the minute hand are superimposed over 12.
For the two hands to meet again, the minute hand must make 1 more revolution than the hour hand.
For the two hands to be superimposed after 3am, the minute hand must make 3 more revolutions than the hour hand.

Since the hour hand makes 1/12 revolution per hour, the number of revolutions made by the hour hand in t hours = (rate)(time) = (1/12)t
Since the minute hand makes 1 revolution per hour, the number of revolutions made by the minute hand in t hours = (rate)(time) = 1*t = t
Snce the number of revolutions for the minute hand must be 3 greater than the number of revolutions for the hour hand, we get:
t = (1/12)t + 3
12t = t + 36
11t = 36
t = 36/11 --> 3 and 3/11 hours --> 3/11 hour after 3am

(3/11 hour)(60 minutes per hour) = 16.36 minutes
Since 1/3 minute = 20 seconds, 0.36 minutes = a bit more than 20 seconds
Among the answer choices, only 22 is viable for the number of seconds.
Thus:
M = 16 minutes
S = 22 seconds

saynchalk · Answer

Hey Chetan, can you please explain the highlighted part, I got everything else

MartyMurray · Answer

Hi  KushagraKirtiman .

When you're practicing, taking 8 minutes to solve a question is fine. By taking all the time you need to answer a practice question, you give yourself time to learn. As you develop skill, you'll speed up and answer questions in less time.

On the other hand, in general, when you're taking the GMAT, you should not spend 8 minutes on any one question unless you're pretty sure you'll be super fast on the rest of the questions in the section or you're near the end of the section and you have extra time that you can use to spend 8 minutes on one question without geting behind on time. Spending 3 or 4 minutes on a question might make sense sometimes, but spending 8 or more rarely works out.

ruis · Answer

For thoise like me who struggle to visualize the clock´s hands. Perpendicular hands: https://gmatclub.com/forum/download/file.php?id=81810 GMAT-Club-Forum-4x3jjd3w.png

Akshaynandurkar · Answer

we know 1 revolution of hr hand requires 12 hr so x revolutions will require 12x hr
similarly 1 revolution of min hand requires 1 hr means for 15+x revolution requires 15+x hr (as we know hr and min hands will meet after 15 minutes only
equate these two 15+x = 12x 
x = 15/11 min = 1 min 21.8 sec

since we know min and hr hand after 15 min
it will take 16 min 22 sec to overlap each other

Gemmie · Answer

................................Speed

Hour hand (h).......... \frac{R}{60 * 12 min}

Minute hand (m)....... \frac{R}{60 min}

(R = revolution)

Distance between the end point of hour hand and the end point of minute hand:  \frac{R}{4}

m - h = \frac{R}{60} - \frac{R}{60*12} = \frac{11R}{60*12}

Time for the end point of hour hand to catch the end point of the minute hand
 \frac{R}{4} \div \frac{11R}{60*12} = \frac{12 * 60}{4 * 11} = \frac{180}{11}= 16.36  (min)

0.36 min = 0.36 * 60 = 22 (sec)

min: 16
sec: 22

Msk666 · Answer

Hi, how would the relative speed be 330 since both hands move in the same direction?

KarishmaB · Answer

Since they both move in the same direction, and minute hand covers 360 degrees in 1 hour while the hour hand covers 30 degrees is one hour, their relative speed is the difference of their speed i.e. 360 - 30 = 330 degrees per hour. I talk about it in detail here: https://www.youtube.com/watch?v=dGqC_6DIKkY

On a 12-hour analog clock, the hour hand moves at a constant rate of

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