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# In a circular clock, the long hand is the radius of the circ

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In a circular clock, the long hand is the radius of the circ [#permalink]  06 Jul 2011, 11:46
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In a circular clock, the long hand is the radius of the circle. At what time is the smaller angle between the hands of the clock NOT divisible by 15?

I 7:20
II 9:00
III 4:30

(A) Only I
(B) Only III
(C) I + II
(D) II+III
(E) I+II+III
[Reveal] Spoiler: OA
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Re: Clock hands [#permalink]  06 Jul 2011, 12:13
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Moderators : help needed here

as you can see from the diagram every division 12-1, 1-2, 2-3 etc etc makes a sector angle of 30 deg.
this can be mathematically achieved by = total central angle in the circle/total number of sector divisions = 360/12 =30 (because clock divisions make 12 sectors in the clock)

now see :
I says 7:20
so here one hand points at 7 and the other at 20 (ie 4). Now count number of divisions from 7 to 4. div = 3
convert div to angle. each div = 30 deg so div(from 7 to 4) = 3*30 = 90 which is divisible by 15

II says : 9:00 ie angle made between 9 and 00(or 12) which is 3 divisions = 3*30 =90 divisible by 15

III says : 430 ie 2 divisions ie 60deg = divisible by 15

so IMO the options provided are debatable. please check and confirm.

whats the OE and OA?
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Re: Clock hands [#permalink]  06 Jul 2011, 16:23
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in one minute hand take a 360 Degree circle and the hour hand take a 30 degree circle

For the hour hand 30 degree movement is accomplished in 1 hour,(60 Min) so in one minute 30/60 = .5 degree (hour hand travels that much in one minute)
The minute hand accomplish 6 degree rotation in one minute

No to the problem at 7.20

Position of hour hand from 12(0 degree) is = 30*7 + 20*.5 = 210 + 10 = 220

Position of minute hand is 20*6 = 120 degrees from 12(0 degree)

the difference in angles is 100 degrees Not divisible by 15...

Employing the same priciple on the rest I get answer A
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Re: Clock hands [#permalink]  06 Jul 2011, 17:04
fivedaysleft wrote:
Moderators : help needed here

as you can see from the diagram every division 12-1, 1-2, 2-3 etc etc makes a sector angle of 30 deg.
this can be mathematically achieved by = total central angle in the circle/total number of sector divisions = 360/12 =30 (because clock divisions make 12 sectors in the clock)

now see :
I says 7:20
so here one hand points at 7 and the other at 20 (ie 4). Now count number of divisions from 7 to 4. div = 3
convert div to angle. each div = 30 deg so div(from 7 to 4) = 3*30 = 90 which is divisible by 15

II says : 9:00 ie angle made between 9 and 00(or 12) which is 3 divisions = 3*30 =90 divisible by 15

III says : 430 ie 2 divisions ie 60deg = divisible by 15

so IMO the options provided are debatable. please check and confirm.

whats the OE and OA?

7:20 means smaller hand also moved (30*4/360)*30= 10 deg; So angle between hands= 30*3+10=100
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Re: Clock hands [#permalink]  28 Jul 2011, 07:15
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Formula for angle between Hour ( H) and minute ( m ) hand is

(60H-11m)/2

using this : 7.20 ->60*7 -11*20 /2 = 100 not divisible by 15
9:00 -> 180 divisible by 15
4:30 -> 105 divisible by 15

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Re: Clock hands [#permalink]  28 Jul 2011, 12:22
7:20
1. Hours hand = 7* 30 + 1/3 (30) = 210 + 10 = 220
2. Mins hand = 4* 30 = 120

diff = 100 deg. NOT divisible by 15..

9:00 and 4:30 are pretty easy
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Re: Clock hands [#permalink]  28 Jul 2011, 13:19
dimri10 wrote:
In a circular clock, the long hand is the radius of the circle. At what time is the smaller
angle between the hands of the clock NOT divisible by 15?
I 7:20
II 9:00
III 4:30
(A) Only I
(B) Only III
(C) I + II
(D) II+III
(E) I+II+III

I just do not get it!

The clock/circle has 360". Nowto figure out the SMALL angle between
hands of the clock I asigned a 30" value between the numbers on its face (360/12=30)
Now, at 7.20 one hand is on 7 and the other is on 4, thus, the small angle is 90" while the outside angle is 270". As follows, at 9.00 one hand is at nine while the other is on 12 - We have an angle formed by 3 intervals of 30". Small angle is 90".
In the last scenario, 4.30, One hand is at 4 and the other at 6 so the angle is 60. Everything divisible by 15...
Where do I go wrong
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Re: Clock hands [#permalink]  28 Jul 2011, 20:20
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Expert's post
katealpha wrote:
dimri10 wrote:
In a circular clock, the long hand is the radius of the circle. At what time is the smaller
angle between the hands of the clock NOT divisible by 15?
I 7:20
II 9:00
III 4:30
(A) Only I
(B) Only III
(C) I + II
(D) II+III
(E) I+II+III

I just do not get it!

The clock/circle has 360". Nowto figure out the SMALL angle between
hands of the clock I asigned a 30" value between the numbers on its face (360/12=30)
Now, at 7.20 one hand is on 7 and the other is on 4, thus, the small angle is 90" while the outside angle is 270". As follows, at 9.00 one hand is at nine while the other is on 12 - We have an angle formed by 3 intervals of 30". Small angle is 90".
In the last scenario, 4.30, One hand is at 4 and the other at 6 so the angle is 60. Everything divisible by 15...
Where do I go wrong

At 7:20, is the hour hand at 7 or a third between 7 and 8?
At 4:30, is the hour hand at 4 or mid way between 4 and 5?

You have to account for the little bit of distance covered by the hour hand too.
Employ Relative Speed here.
Minute hand covers 360 degrees in an hour.
Hour hand covers 360/12 = 30 degrees in an hour.
Speed of minute hand relative to hour hand is 360 - 30 = 330 degrees per hour.

At 7 o clock, the minute hand is 210 degrees behind the hour hand. In 20 minutes (at 7:20), it makes up 330/3 = 110 degrees. Now it will be 100 degrees behind the hour hand. The smaller angle between them is 100 degrees.

At 4 o clock, the minute hand is 120 degrees behind the hour hand. In half an hour, it covers 330/2 = 165 degrees to get 45 degrees ahead of the hour hand. The smaller angle between them is 45 degrees.

At 9 o clock, the angle between the two hands is 90 degrees.

(Such clock questions are applications of Relative Speed. Do you remember 'A cop is running after a crook at a speed of ... Initial distance between them is ... When will he catch up?' kind of questions? The principle employed is exactly the same here.)
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Re: Clock hands [#permalink]  30 Jul 2011, 18:31
@Karishma +1 Kudos

Loved the approach. Yes it is a relative speed problem and I like your way to approach the problem. Do post any variant of the question popping up because I am too lazy to create a variant.
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Re: Clock hands [#permalink]  01 Aug 2011, 01:30
Expert's post
krishp84 wrote:
@Karishma +1 Kudos

Loved the approach. Yes it is a relative speed problem and I like your way to approach the problem. Do post any variant of the question popping up because I am too lazy to create a variant.

Another type of clock questions that I have come across:
Mr. A leaves his house between 3 and 4 pm and notices that the angle between the hour and minute hands is x degrees. He returns between 6 and 7 pm same day and notices that the angle is again x degrees. For how much time was Mr A away from home? (The value of x is given. Try putting different values (180 degrees, 170 degrees, 45 degrees etc) and you can make PS/DS questions)
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Re: Clock hands [#permalink]  01 Aug 2011, 16:19
VeritasPrepKarishma wrote:
Another type of clock questions that I have come across:
Mr. A leaves his house between 3 and 4 pm and notices that the angle between the hour and minute hands is x degrees. He returns between 6 and 7 pm same day and notices that the angle is again x degrees. For how much time was Mr A away from home? (The value of x is given. Try putting different values (180 degrees, 170 degrees, 45 degrees etc) and you can make PS/DS questions)

Is the answer 3 hours 18 mns ?

okay - My thought process....
Distance between Hour hand and Minute hand between 3 and 4 PM is x degree - So Time will be 3:y, where y is the distance of the minute hand from 12 position

Distance between Hour hand and Minute hand between 6 and 7 PM is x degree - So Time will be 6:z, where z is the distance of the minute hand from 12 position

y = ()+x
z = ()+x+30*3

So the difference between 6:z and 3:y is (6-3) hours + 30*3/5 minute (Minute hand covers 30 degree in 5 minutes)
= 3 hours 18minutes....

PS - I did not substitute, But I will definitely substitute the answer choices to confirm.

Do let me know your thoughts.....
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Re: Clock hands [#permalink]  01 Aug 2011, 21:24
Expert's post
krishp84 wrote:
VeritasPrepKarishma wrote:
Another type of clock questions that I have come across:
Mr. A leaves his house between 3 and 4 pm and notices that the angle between the hour and minute hands is x degrees. He returns between 6 and 7 pm same day and notices that the angle is again x degrees. For how much time was Mr A away from home? (The value of x is given. Try putting different values (180 degrees, 170 degrees, 45 degrees etc) and you can make PS/DS questions)

Is the answer 3 hours 18 mns ?

okay - My thought process....
Distance between Hour hand and Minute hand between 3 and 4 PM is x degree - So Time will be 3:y, where y is the distance of the minute hand from 12 position

Distance between Hour hand and Minute hand between 6 and 7 PM is x degree - So Time will be 6:z, where z is the distance of the minute hand from 12 position

y = ()+x
z = ()+x+30*3

So the difference between 6:z and 3:y is (6-3) hours + 30*3/5 minute (Minute hand covers 30 degree in 5 minutes)
= 3 hours 18minutes....

PS - I did not substitute, But I will definitely substitute the answer choices to confirm.

Do let me know your thoughts.....

y and z are the distances from 12 in minutes in your solution, not in degrees. I am not sure how you made these equations. The question I gave is a generic example. The actual question would have a value for x. e.g.
Mr. A leaves his house between 3 and 4 pm and notices that the angle between the hour and minute hands is 45 degrees. He returns between 6 and 7 pm same day and notices that the angle is again 45 degrees. For how much time was Mr A away from home?
Even this would be a DS question with 2 statements giving further information because multiple answers are possible. The diagram below will explain why.
Attachment:

Ques3.jpg [ 14.19 KiB | Viewed 5736 times ]

So the statements could be something like "He was out for more than 3 hr 10 minutes" etc...
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Re: Clock hands [#permalink]  02 Aug 2011, 02:58
Karishma,
Shouldn't this question be like what will the maximum / minimum time was Mr A away from home, considering that we are going to have 2 possibilities.
Also, how can we solve this types of ques.
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Re: Clock hands [#permalink]  02 Aug 2011, 09:19
dimri10 wrote:
In a circular clock, the long hand is the radius of the circle. At what time is the smaller
angle between the hands of the clock NOT divisible by 15?
I 7:20
II 9:00
III 4:30
(A) Only I
(B) Only III
(C) I + II
(D) II+III
(E) I+II+III

The way I would solve this Q is this:
look at the answers - we want to check answer II first of all bc its easy to check and if ill cancel it, it will leave me with 2 answers. We know easily that at 0900 - the angel is 90 and therefore we can remove all the questions with answer II - meaning C,D,E - OUT.

now - bc we have no answer as "none" - we need to check only one of the answers. I or III. For me, 0430 is easier bc its dealing with half and not thirds.

we know that the big one is at 180, the small one is 4.5*30 = 135.
180-135=45. Done. III is out.
I dont need to check answer I bc i made sure all the others are wrong.
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Re: Clock hands [#permalink]  02 Aug 2011, 20:44
Expert's post
varunmaheshwari wrote:
Karishma,
Shouldn't this question be like what will the maximum / minimum time was Mr A away from home, considering that we are going to have 2 possibilities.
Also, how can we solve this types of ques.

Yes, if you want to make it a PS question, you can definitely ask for the maximum or minimum time. There are going to be 4 possibilities (2C1 * 2C1). 2 possibilities for the time at which he leaves. 2 possibilities for the time at which he returns.

How to solve it?
Mr. A leaves his house between 3 and 4 pm and notices that the angle between the hour and minute hands is 45 degrees. He returns between 6 and 7 pm same day and notices that the angle is again 45 degrees. What is the minimum time for which Mr A was away from home?

Let's focus on the minimum time that he was away. For that he should have left later and arrived earlier. So he should have left at around 3:25 and arrived at around 6:25 (Look at the diagram above to see the case we are talking about)

At 3 o clock, the angle between the hour and minute hand is 90 degrees.
The minute hand should cover the 90 degrees and then create an angle of 45 degrees between itself and the hour hand.
Relative speed of minute hand is 330 degrees/hour.
To cover 135 degrees relative to the hour hand, it will take 60/330 * 135 minutes = 270/11 minutes = 24(6/11) minutes
He must have left at 24(6/11) minutes past 3.

At 6 o clock, the angle between the hour and minute hand is 180 degrees.
The minute hand should cover 135 degrees relative to the hour hand to create an angle of 45 degrees.
Relative speed of minute hand is 330 degrees/hour.
To cover 135 degrees relative to the hour hand, it will take 60/330 * 135 minutes = 270/11 minutes = 24(6/11) minutes.
He must have arrived at 24(6/11) minutes past 6.

This means he was out for exactly 3 hours.

Now try the case of 'maximum time' and put it up.
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Re: Clock hands [#permalink]  08 Aug 2011, 18:42
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VeritasPrepKarishma wrote:
varunmaheshwari wrote:
Karishma,
Shouldn't this question be like what will the maximum / minimum time was Mr A away from home, considering that we are going to have 2 possibilities.
Also, how can we solve this types of ques.

Yes, if you want to make it a PS question, you can definitely ask for the maximum or minimum time. There are going to be 4 possibilities (2C1 * 2C1). 2 possibilities for the time at which he leaves. 2 possibilities for the time at which he returns.

How to solve it?
Mr. A leaves his house between 3 and 4 pm and notices that the angle between the hour and minute hands is 45 degrees. He returns between 6 and 7 pm same day and notices that the angle is again 45 degrees. What is the minimum time for which Mr A was away from home?

Let's focus on the minimum time that he was away. For that he should have left later and arrived earlier. So he should have left at around 3:25 and arrived at around 6:25 (Look at the diagram above to see the case we are talking about)

At 3 o clock, the angle between the hour and minute hand is 90 degrees.
The minute hand should cover the 90 degrees and then create an angle of 45 degrees between itself and the hour hand.
Relative speed of minute hand is 330 degrees/hour.
To cover 135 degrees relative to the hour hand, it will take 60/330 * 135 minutes = 270/11 minutes = 24(6/11) minutes
He must have left at 24(6/11) minutes past 3.

At 6 o clock, the angle between the hour and minute hand is 180 degrees.
The minute hand should cover 135 degrees relative to the hour hand to create an angle of 45 degrees.
This is because 180-45 = 135 degrees and not 90+45 degrees - Highlighted to see that evryone gets the reason. Anyway it will become clear in the Maximum example
Relative speed of minute hand is 330 degrees/hour.
To cover 135 degrees relative to the hour hand, it will take 60/330 * 135 minutes = 270/11 minutes = 24(6/11) minutes.
He must have arrived at 24(6/11) minutes past 6.

This means he was out for exactly 3 hours.

Now try the case of 'maximum time' and put it up.

Mr. A leaves his house between 3 and 4 pm and notices that the angle between the hour and minute hands is 45 degrees. He returns between 6 and 7 pm same day and notices that the angle is again 45 degrees. What is the maximum time for which Mr A was away from home?

Now we need -
Earliest time that Mr. A left b/w 3 and 4 PM
Latest time when Mr. A returned b/w 6 and 7 PM

Relative speed of Minute hand w.r.t Hour hand = 330 degrees per hour

At 3 PM, the angle b/w hour and minute hand = 90 degree
So earliest time when Mr. A leaves will be 90-45 = 45 degrees to be covered by the minute hand relative to Hour hand = (60/330)45 minutes past 3

At 6 PM, the angle b/w hour and minute hand = 180 degree
So latest time when Mr. A returns will be 180+45 = 225 degrees to be covered by the minute hand relative to Hour hand = (60/330)225 minutes past 6

So maximum time for which Mr A was away from home
= (60/330)225 minutes past 6 - (60/330)45 minutes past 3
= 3 hrs (60/330)180minutes
= 3 hours (360/11) minutes
= $$3 hours 34\frac{6}{11} minutes$$

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Re: Clock hands [#permalink]  09 Aug 2011, 02:43
Expert's post
krishp84 wrote:
Mr. A leaves his house between 3 and 4 pm and notices that the angle between the hour and minute hands is 45 degrees. He returns between 6 and 7 pm same day and notices that the angle is again 45 degrees. What is the maximum time for which Mr A was away from home?

Now we need -
Earliest time that Mr. A left b/w 3 and 4 PM
Latest time when Mr. A returned b/w 6 and 7 PM

Relative speed of Minute hand w.r.t Hour hand = 330 degrees per hour

At 3 PM, the angle b/w hour and minute hand = 90 degree
So earliest time when Mr. A leaves will be 90-45 = 45 degrees to be covered by the minute hand relative to Hour hand = (60/330)45 minutes past 3

At 6 PM, the angle b/w hour and minute hand = 180 degree
So latest time when Mr. A returns will be 180+45 = 225 degrees to be covered by the minute hand relative to Hour hand = (60/330)225 minutes past 6

So maximum time for which Mr A was away from home
= (60/330)225 minutes past 6 - (60/330)45 minutes past 3
= 3 hrs (60/330)180minutes
= 3 hours (360/11) minutes
= $$3 hours 34\frac{6}{11} minutes$$

Yes, that's right. Though, in the last step, 360/11 minutes would be 32(8/11) minutes
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Re: Clock hands [#permalink]  09 Aug 2011, 03:41
VeritasPrepKarishma wrote:
krishp84 wrote:
Mr. A leaves his house between 3 and 4 pm and notices that the angle between the hour and minute hands is 45 degrees. He returns between 6 and 7 pm same day and notices that the angle is again 45 degrees. What is the maximum time for which Mr A was away from home?

Now we need -
Earliest time that Mr. A left b/w 3 and 4 PM
Latest time when Mr. A returned b/w 6 and 7 PM

Relative speed of Minute hand w.r.t Hour hand = 330 degrees per hour

At 3 PM, the angle b/w hour and minute hand = 90 degree
So earliest time when Mr. A leaves will be 90-45 = 45 degrees to be covered by the minute hand relative to Hour hand = (60/330)45 minutes past 3

At 6 PM, the angle b/w hour and minute hand = 180 degree
So latest time when Mr. A returns will be 180+45 = 225 degrees to be covered by the minute hand relative to Hour hand = (60/330)225 minutes past 6

So maximum time for which Mr A was away from home
= (60/330)225 minutes past 6 - (60/330)45 minutes past 3
= 3 hrs (60/330)180minutes
= 3 hours (360/11) minutes
[color=#FF0000]= $$3 hours 34\frac{6}{11} minutes$$

[/color]

Yes, that's right. Though, in the last step, 360/11 minutes would be 32(8/11) minutes

Yeah silly me
Any one should have said it is 32.xxx minutes just by dividing mentally
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Re: Clock hands [#permalink]  09 Aug 2011, 03:50
VeritasPrepKarishma wrote:
varunmaheshwari wrote:
Karishma,
Shouldn't this question be like what will the maximum / minimum time was Mr A away from home, considering that we are going to have 2 possibilities.
Also, how can we solve this types of ques.

Yes, if you want to make it a PS question, you can definitely ask for the maximum or minimum time. There are going to be 4 possibilities (2C1 * 2C1). 2 possibilities for the time at which he leaves. 2 possibilities for the time at which he returns.

How to solve it?
Mr. A leaves his house between 3 and 4 pm and notices that the angle between the hour and minute hands is 45 degrees. He returns between 6 and 7 pm same day and notices that the angle is again 45 degrees. What is the minimum time for which Mr A was away from home?

Let's focus on the minimum time that he was away. For that he should have left later and arrived earlier. So he should have left at around 3:25 and arrived at around 6:25 (Look at the diagram above to see the case we are talking about)

At 3 o clock, the angle between the hour and minute hand is 90 degrees.
The minute hand should cover the 90 degrees and then create an angle of 45 degrees between itself and the hour hand.
Relative speed of minute hand is 330 degrees/hour.
To cover 135 degrees relative to the hour hand, it will take 60/330 * 135 minutes = 270/11 minutes = 24(6/11) minutes
He must have left at 24(6/11) minutes past 3.

At 6 o clock, the angle between the hour and minute hand is 180 degrees.
The minute hand should cover 135 degrees relative to the hour hand to create an angle of 45 degrees.
Relative speed of minute hand is 330 degrees/hour.
To cover 135 degrees relative to the hour hand, it will take 60/330 * 135 minutes = 270/11 minutes = 24(6/11) minutes.
He must have arrived at 24(6/11) minutes past 6.

This means he was out for exactly 3 hours.

Now try the case of 'maximum time' and put it up.

One more important thing to note here -
At 3 o clock, the angle between the hour and minute hand is 90 degrees.
The minute hand should cover the 90 degrees and then create an angle of 45 degrees between itself and the hour hand.

Relative speed of minute hand is 330 degrees/hour.
To cover 135 degrees relative to the hour hand, it will take 60/330 * 135 minutes = 270/11 minutes = 24(6/11) minutes

When minute hand moves, Hour hand will also move, but the distance in degree b/w both the hands will remain same.

Let Hour hand move by h degrees when minute hands moves ...
The minute hand would be 135+h degrees from 12'o clock position.....
But here we ae speaking about relative speed and
NOT ABSOLUTE SPEEDcolor=#0000BF], so relative to hour hand , the speed will remain constant (135 degrees)[/color]

These are simple points but make the difference to understand the core concepts
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Re: Clock hands [#permalink]  21 Sep 2013, 17:29
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Re: Clock hands   [#permalink] 21 Sep 2013, 17:29
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