Explorers of the Deep

@YouKnowMe @HrishabhNayal The thing is, the only fault I know in clocks is that the speed of their minute hand is slower/faster than the usual $6^\circ/\mathrm{min}$. But, the speed of the minute hand affects that of the hour hand and they follow this relation, no matter the speed of the minute or the hour hand : one complete revolution for the minute hand means a $30^\circ$ revolution for the hour hand.

The point is, it's not possible for even a slower/faster clock to show a time that is not possible. And a time of $1:06$ where the minute and the hour hand coincide is not possible because…

What I think it means instead is that the clock takes 66 minutes when it is supposed to take $65\dfrac{5}{11}$ for the hands to coincide....

Rajdeep Sindhu

7:14 AM

@YouKnowMe Yo!

@YouKnowMe Is the answer to this question : gains/loses 12 minutes?

I think it's : loses 12 minutes...

@BLM $\downarrow$

qphs.fs.quoracdn.net/main-qimg-50ea75f446aef849cfce9812074765b2

BLM

@RajdeepSindhu man, he's right

Rajdeep Sindhu

qphs.fs.quoracdn.net/main-qimg-48572b262c6ab166eeab4358b7b8137c

qphs.fs.quoracdn.net/main-qimg-418873aa02e19ecc4faa117517b4594d

qphs.fs.quoracdn.net/main-qimg-60653126058ca30f719d3c008a1a8cf6

BLM

@RajdeepSindhu quora has the potential to turn into a Meme app

Rajdeep Sindhu

@BLM lol, yes it does :D

qphs.fs.quoracdn.net/main-qimg-918dfa10abe2e9f34093727c6c9d7f9b

BLM

7:30 AM

it's already one, just not officially

Rajdeep Sindhu

@BLM Right, for educational purposes, we use SE anyway...

BLM

@RajdeepSindhu yep :)

Rajdeep Sindhu

qphs.fs.quoracdn.net/main-qimg-1ebbc9e7a031c3cbc5fe8c9f21ec1174

BLM

@RajdeepSindhu call Narcotics

Rajdeep Sindhu

@BLM LoL

BLM

7:33 AM

Man, she had to reload and still missed

Rajdeep Sindhu

@YouKnowMe Is my answer correct?!?!?!?!?!?!?!

YouKnowMe

@RajdeepSindhu $11\dfrac{109}{121}$ minutes loss

Rajdeep Sindhu

@YouKnowMe Never mind

@YouKnowMe Did you get this $\downarrow$

42 mins ago, by Rajdeep Sindhu

@YouKnowMe @HrishabhNayal The thing is, the only fault I know in clocks is that the speed of their minute hand is slower/faster than the usual $6^\circ/\mathrm{min}$. But, the speed of the minute hand affects that of the hour hand and they follow this relation, no matter the speed of the minute or the hour hand : one complete revolution for the minute hand means a $30^\circ$ revolution for the hour hand.

The point is, it's not possible for even a slower/faster clock to show a time that is not possible. And a time of $1:06$ where the minute and the hour hand coincide is not possible because…

YouKnowMe

Why is everyone using "quoracdn"?

@RajdeepSindhu I didn't try.

Do you want me to read that?

Rajdeep Sindhu

@YouKnowMe No, no, I just linked your name to that out of fun...

@YouKnowMe Memes from Quora -_-

YouKnowMe

7:39 AM

-_-

I am reading that.

Rajdeep Sindhu

@YouKnowMe Good...

YouKnowMe

@RajdeepSindhu

"And a time of $1:06$ where the minute and the hour hand coincide is not possible because…"

Why not?

Rajdeep Sindhu

@YouKnowMe Just read what follows -_-

YouKnowMe

@RajdeepSindhu What?

@RajdeepSindhu Why are you wasting time on a wrong question?

Rajdeep Sindhu

@YouKnowMe Shut up, you brat!! (jk)

You don't have any idea why it's wrong, but still insist on the same.

You just don't understand the language.

And when I told you what it possibly means, you thought that it is something it could possibly mean but the long message I sent was to clarify to you that such a situation where a slower clock's minute and hour hand coincide at 1:06 is impossible -_-

@YouKnowMe So, can we agree that even in slower/faster clocks, a complete revolution of the minute hand means a $30^\circ$ revolution of the hour hand?

YouKnowMe

7:47 AM

Yes

Rajdeep Sindhu

Great, now, can we also agree on the fact that in that case, if the minute hand has moved by a certain amount, the corresponding movement in the hour hand will be unique, no matter the speed of the hands?

YouKnowMe

No

@RajdeepSindhu

Rajdeep Sindhu

@YouKnowMe Why?

YouKnowMe

@RajdeepSindhu Elaborate

Rajdeep Sindhu

@YouKnowMe -_-

I think it would be better to first know why you disagree...

YouKnowMe

7:54 AM

@RajdeepSindhu What does "will be unique" mean?

Rajdeep Sindhu

@YouKnowMe Only one possible position of the hour hand...

Basically, don't you think that it's safe to say that even a slower/faster clock can't show an impossible time?

The position of the minute hand determines the position of the hour hand, not the speed...

YouKnowMe

@RajdeepSindhu Impossible time -_- Do you mean the angle between hour hand and minute hand?

@RajdeepSindhu Yes

Rajdeep Sindhu

@YouKnowMe Right, for example, a time of 3:30 and he hands coincide is not possible even if the clock is slower/faster than usual

YouKnowMe

@RajdeepSindhu Yes

Rajdeep Sindhu

@YouKnowMe And we know that when the minute hand does a full revolution, the hour hand revolves by $30^\circ$.

Similarly, when the minute had does a half revolution, the hour hand revolves by $15^\circ$

YouKnowMe

7:57 AM

Wait

Rajdeep Sindhu

Alright...

YouKnowMe

@RajdeepSindhu Yes

Rajdeep Sindhu

@YouKnowMe Now, now, let's say that the time is 1:00

So, we have the minute hand at $0^\circ$ from the 12:00 point

and he hour hand at $30^\circ$ from the 12:00 point...

Now, let's say that the minute hand moves by $36^\circ$.

On a clock with a normal minute arm speed, that would mean that the time is $1:06$, by the way...

YouKnowMe

@RajdeepSindhu Yea

Rajdeep Sindhu

So, now, the minute hand is at the 6th minute mark

It has moved a 10th of the full revolution

YouKnowMe

8:01 AM

0

Q: The hands of a faulty clock meet every 66 minutes; which solution is correct?

The minute hand of a clock overtakes the hour hand at intervals of $66$ minutes of the correct time. How much time does the clock gain or lose in $4$ hours? Solution 1 The normal clock time it takes for the minute hand to overtake the hour hand is $65\cfrac{5}{11}$. So the time $\und...

algebra-precalculus

Rajdeep Sindhu

Which means that the hour hand will move a 10th of $30^\circ$ and hence, it will move by $3^\circ$

YouKnowMe

Hmm What?

Rajdeep Sindhu

@YouKnowMe This does not mean that 66 minutes have passed in the faulty clock when the arms coincide

But instead, it means that 66 minutes of our usual time have passed by the time the arms coincide...

YouKnowMe

The normal clock time it takes for the minute hand to overtake the hour hand is $65\cfrac{5}{11}$

Don't you think this line is wrong?

Rajdeep Sindhu

@YouKnowMe Right, if the time is measured with the conventional definition of a minute

@YouKnowMe I do...

What it's supposed to say is most probably that after $65\dfrac{5}{11}$ minutes from a give time, the arms are still at the same angle as they initially were...

@YouKnowMe Did you get what I was trying to say?

i.e. did you get that having a clock such that the arms on it coincide at 1:06 is impossible?

And that there are two possible meanings of the statement, one of which we just discarded?

YouKnowMe

8:08 AM

@RajdeepSindhu When do hands coincides in correct clock?

Rajdeep Sindhu

@YouKnowMe 12:00 and then with periods of $65\dfrac{5}{11}$ minutes of the usual time from then.

And in the case of a faulty clock, the time period might change, but the positions at which they coincide don't...

YouKnowMe

@RajdeepSindhu Yes

Rajdeep Sindhu

So, you get it, nice :D

YouKnowMe

@RajdeepSindhu

Who do iron fillings assemble in this position?

Rajdeep Sindhu

Dunno, tbh

Google "Doctor Strange : In the multiverse of madness". Look at the release date and cast...

YouKnowMe

8:20 AM

@RajdeepSindhu Just for information, Tenet will release on 15Dec digitally.

Rajdeep Sindhu

@YouKnowMe :D

YTS. where art thou, YTS?

YTS(dot)movie is working :D

YTS said that they know my IP address. They also said that I'm currently in Czechia, lol.

(I'm not even using a VPN)

Hrishabh Nayal

8:59 AM

@RajdeepSindhu seems legit

Rajdeep Sindhu

9:11 AM

@HrishabhNayal :D

@YouKnowMe People from ALLEN are coming to visit out home...

I wonder what they'll talk about.

They better not ask any chemistry related questions -_-

YouKnowMe

@RajdeepSindhu Why?

Rajdeep Sindhu

@YouKnowMe Dunno, tbh. They just said that they wanted to meet us...

YouKnowMe

Will you do one thing?

Rajdeep Sindhu

¯_(ツ)_/¯

@YouKnowMe Depends on what the thing is...

brb

YouKnowMe

@RajdeepSindhu Can you ask for the Tallentex question paper?

Rajdeep Sindhu

9:16 AM

@YouKnowMe bruh

of course

not

2

YouKnowMe

@RajdeepSindhu -_-

Why ?

Rajdeep Sindhu

Of course they won't give it, it's pointless to ask

Even pointless-er if they're coming just for talking...

YouKnowMe

@RajdeepSindhu Atleast give it a try.

Rajdeep Sindhu

@YouKnowMe No, thanks :)

YouKnowMe

@RajdeepSindhu They won't just talk, they would promote...

@RajdeepSindhu :-(

Rajdeep Sindhu

9:20 AM

@YouKnowMe Right, that's what they call talking...

@YouKnowMe Do you expect them to carry a hard copy with them at all time, lol?

YouKnowMe

Rajdeep Sindhu

Also, most probably, they aren't even teachers...

YouKnowMe

@RajdeepSindhu They can provide a softcopy after some day.

@RajdeepSindhu So...

Rajdeep Sindhu

@YouKnowMe So, never mind...

@YouKnowMe Sowwie, me won't ask...

Why do you need it so bad, though?

YouKnowMe

@RajdeepSindhu Nevermind :-(

Rajdeep Sindhu

9:22 AM

@YouKnowMe :(

Well, they're coming, gtg. Bye :D

YouKnowMe

12:14 PM

@RajdeepSindhu What happened in the meeting?

Rajdeep Sindhu

@YouKnowMe Yo!

So, they said that they chose to talk to the parents of top 10 kids from every city from class 10th...

@BLM They were from ALLEN Chandigarh.

The entrance exam for SRG is on 6th December, 2020 :O

And the classes for this advanced batch will begin on 15th December, 2020...

But I'm not sure if this SRG exam is just for the Chandigarh branches or for all branches, what do you think @BLM ?

YouKnowMe

@RajdeepSindhu What's SRG?

Rajdeep Sindhu

@YouKnowMe Special Rankers' Group aka star batch...

Rajdeep Sindhu

1:55 PM

@YouKnowMe Got any difficult questions?

YouKnowMe

2:17 PM

Not now :-(

Rajdeep Sindhu

2:40 PM

@YouKnowMe :(

YouKnowMe

4:33 PM

@RajdeepSindhu

Let $p(x)=a x^{2}+b x, q(x)=\left(x^{2}+m x+n .\right.$ If
$p(1)-q(1)=0, p(2)-q(2)=1$ and $p(3)-q(3)=4$ then $p(4)-q(4)=$

Not difficult

But I want to see your approach :-(

BLM

@RajdeepSindhu all

@RajdeepSindhu stats show the other way lmao

Rajdeep Sindhu

7:27 PM

@BLM lol...

@BLM I wonder why those people said then "Assume that almost all the children from North India who gave TALLENTEX will give the SRG exam"...

BLM

7:45 PM

@RajdeepSindhu just like all 10th graders give ntse

Transcript for

$Mathematics$ Explorers of the Deep