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

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

manishgeorge already answered.

7:20 means smaller hand also moved (30*4/360)*30= 10 deg; So angle between hands= 30*3+10=100

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

SO answer is A

br,
Binu

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

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

@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)

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.

So the statements could be something like "He was out for more than 3 hr 10 minutes" etc...

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.

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.

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.

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

Yeah silly me
Any one should have said it is 32.xxx minutes just by dividing mentally

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

angle b/w long and short hand in clock is = 11/2 * ( min) - 30 * hr
plug in answer options
only option A is correct which is not divisible by 15.