Riddles Options

[divz]

Arrey, bahooooooot hee asaaan sa sawaal poochha reyyyyyy...

Since they are working together at the same desk, they decide on a mutually acceptable basis that every week one of them will come 10 mins late to office.

Now whoever comes late, peeks in, checks what the other fellow is wearing, and if he is wearing a similar shirt, he goes to the men's room and changes into the spare one he's decided to carry for such an eventuality

Maths ko maaro goli rey, yahaan to chaalugiri chalti hai

Better still stop wearing shirts Sallu khan ki tarah

M and N u two are bherry funny........... vaise good answers ...

37 and 73 Divvz???

I will try for easy ones:) Though I just pass on the ones I get deariee

lagta hai you already knew the answers...

[Mandrake]

M and N u two are bherry funny........... vaise good answers ...

Chalo koi toh kadardaan mila

[Nimii]

Mandrake ji tik se padhyi .. she said M first and then N

Isi se kya pata chalta hai .. you are the best!!!!!!!!!!!!!!!

I always look with awe at you for all the info you post! Phew !!!!!!!!!!!

N :salutes Mandrake:

[vivekpm]

MJ .. aap to rehne hi dijiye.. dont tease us with such talk.. u r master of trade
and Hits.. please contribute.. appreciation aint enough

ok.. here is the next one..
Two Associates who went to school together start work on the same trading desk on the same day. They each know that the other only get his shirts from Thomas Pink, and most likely has 5 such shirts, three of which they both have, one which they probably both have, and another which the other probably doesn't have.

It is a top priority for both of them over their first 65 working days (after which they will get their first bonus and buy more shirts) that the two of them are never caught wearing the same shirt, lest they face a day of endless hazing by all the traders and salespeople. These are grown men, not high school girls, so of course they will never ask each other what they are going to be wearing. Each also knows that the other will almost never wear the same shirt two days in a row, and will also try to avoid easily identifiable patterns (such as always wearing the blue herringbone on Monday).

What simple strategy can one of the associates follow to minimize the chance of any of the above unpleasant conditions (in the order stated)? Can this be generalized to an arbitrary number of associates, known and unknown common shirts?

I tried to put my solution in as simple language as possible and this is the best I came up with. Pardon me if I have complicated things (which I think I have).

I believe correct solution to this problem is already provided by Mandrake and Nimii which I think makes it 100% certain that none of the two will wear same shirt on same day . For decency, I go with Mandrake's solution

Jokes apart, here is one possible approach (there can be multiple approaches for this problem):

Let us do some naming first:

Shirt1= Unique shirt for both associates
Shirt2 = Shirt which is "probably" unique.
Shirt3,Shirt4,Shirt5 = Common shirts

1. The associate who formulates the strategy should wear his Shirt1 on Monday.
2. On other days, he should wear the shirt worn by other person on previous days if associate 2 has worn either Shirt3,Shirt4,Shirt5
3. If Associate2 wears his Shirt1 or Shirt2 on monday (probability 2/5) as against other three shirts (probability 3/5), Associate1 should wear his Shirt2 on tuesday. In long run Associate1 will probably learn whether or not their Shirt2 are unique and luckily if they are, there is very less chance that they will end up wearing same shirt on same day.

For example, consider Associate1 to be the one who comes up with a strategy:
Associate 1 wears his Shirt1 on Monday.
Probability that Associate 2 does not wear his Shirt1 is 4/5 which is more as compared to the probability that he too will his Shirt1 (which is 1/5).

Now if, on Monday, Associate 2 wears Shirt2, then Associate 1 should wear his Shirt2 (remember that he doesn't know if Shirt2 was indeed Shirt2 for Associate2 if Shirt2 for both are different) and on other days he can either wear his unique shirt or wear shirt worn by Associate2 on previous day intermittently to avoid pattern.

Cheers,

[shivani]

Sigh
I tried to read it thrice and all I could read was shirt shirt shirt.. and suddenly I like Sallu's style better
Vivekah.. aapne likha hai to theek hi hoga

[vivekpm]

Sigh
I tried to read it thrice and all I could read was shirt shirt shirt.. and suddenly I like Sallu's style better
Vivekah.. aapne likha hai to theek hi hoga

Yehi to trick thi Complex likho to log apne aap sahi samaj lete hai...

Anyways, here is one more attempt to simplify things. Am leaving out probability here and writing only steps:

1. Associate1 wears unique shirt on monday.
2. On other days Associate1 wears shirt worn by Associate2 on previous day.

Exceptions:
1. If Associate2 wears his unique shirt on Monday, Associate1 wears his "probably same" shirt on tuesday.
2. If Associate2 wears his "probably same" shirt on Monday, Associate1 wears his "probably same" shirt on tuesday.

Does this help?

Cheers,

[shivani]

Thx a lot Vivekaah
(though I actually did not understand. guess its a side effect of conversing a lot with rommji)
go ahead and post something new : )

[vivekpm]

Ok... Here is the problem...

A banker interchanged the Rs. and Ps. when he cashed a cheque for some person X giving him Rs. instead of Ps. and vice-versa. (Ex. if cheque is for 40 Rs 70 Ps., banker gave him 70 Rs. 40 Ps.)

After buying a newspaper for 50 Ps., X noticed that he had exactly three times as much amount as the original cheque. What was the original amount of the cheque?

Cheers,

[hits]

Ok... Here is the problem...

A banker interchanged the Rs. and Ps. when he cashed a cheque for some person X giving him Rs. instead of Ps. and vice-versa. (Ex. if cheque is for 40 Rs 70 Ps., banker gave him 70 Rs. 40 Ps.)

After buying a newspaper for 50 Ps., X noticed that he had exactly three times as much amount as the original cheque. What was the original amount of the cheque?

Cheers,

The original amount was Rs. 18 and 56 paise.

I got the answer to this question by peering into the post sufficiently hard. Here's how the actual solution should go -

Let original value = x + y (x being Rs and y being paise)
Value post error = 100y + (x/100)

We are told that

100y + (x/100) - 0.5 = 3*(x + y).

This resolves to

9700y = 50 + 299x.

Add the following common-sensical constraints -

y < 1 (since y is paise in Rs denomination)
x > 1 (since x is discrete Rs).

The only realistic numbers which fit this scenario are x = 18 and y = 0.56.

[hits]

I tried to put my solution in as simple language as possible and this is the best I came up with. Pardon me if I have complicated things (which I think I have).

I believe correct solution to this problem is already provided by Mandrake and Nimii which I think makes it 100% certain that none of the two will wear same shirt on same day . For decency, I go with Mandrake's solution

Jokes apart, here is one possible approach (there can be multiple approaches for this problem):

Let us do some naming first:

Shirt1= Unique shirt for both associates
Shirt2 = Shirt which is "probably" unique.
Shirt3,Shirt4,Shirt5 = Common shirts

1. The associate who formulates the strategy should wear his Shirt1 on Monday.
2. On other days, he should wear the shirt worn by other person on previous days if associate 2 has worn either Shirt3,Shirt4,Shirt5
3. If Associate2 wears his Shirt1 or Shirt2 on monday (probability 2/5) as against other three shirts (probability 3/5), Associate1 should wear his Shirt2 on tuesday. In long run Associate1 will probably learn whether or not their Shirt2 are unique and luckily if they are, there is very less chance that they will end up wearing same shirt on same day.

For example, consider Associate1 to be the one who comes up with a strategy:
Associate 1 wears his Shirt1 on Monday.
Probability that Associate 2 does not wear his Shirt1 is 4/5 which is more as compared to the probability that he too will his Shirt1 (which is 1/5).

Now if, on Monday, Associate 2 wears Shirt2, then Associate 1 should wear his Shirt2 (remember that he doesn't know if Shirt2 was indeed Shirt2 for Associate2 if Shirt2 for both are different) and on other days he can either wear his unique shirt or wear shirt worn by Associate2 on previous day intermittently to avoid pattern.

Cheers,

Vivek and Shivani,

The problem is posed in a manner that the solution will be sub-optimal. So, without conclusively knowing the probability of the uniqueness of a shirt to each associate, the solution proposed above will break down as the days run by. The solution proposed above will give you the best possible result for a maximum of 8 days. Under such scenarios, the first week should be used as a "data-gathering" phase. What really needs to happen is to do what Vivek says above for exactly 6 days. At that point, you will have a first crack at the unknown probabilities. Then, you can frame the problem more accurately, and use this extra data to dress accordingly the next week, and so on.

As you would know, the probability of both wearing the same shirt will go up as time passes by. What the above situation will allow you to do is to actually manage "time" into the equation, thereby increasing your chances of success.

Thanks
Hits

[visuja]

It's a nice thing Im not regular here !!

The only questions that made sense to me so far were Anil-ji's and Mandrake's questions ! Were they answered ??

Heres my attempt to Anilji's question:
Short of halving total heat content ,
Since its 0 C ----- > 273 K,
twice as cold would imply ------ > 273/2 = 136.5 K ----- > (negative) 136.5 C ! (That shd be a new record !!)
Interestingly, twice as hot would imply ------> 546 K ------> 273 C ! (another record !!)
Does that make sense ?? Converting into degree F certainly isnt correct

Mandrake's question --- heres what I think happens:

The ice that freezes leaves out all salt behind, since natural freezing is quite a 'slow' process. The frozen ice is probably just pure water. Ice as such is lighter than water; even so than salt water. This sets up a density gradient, densest being at the bottom. Now salt also lowers the freezing point of water. (Thats why salt is sprinkled on roads after a snow shower). Sea water also has a higher heat capacity than ice --- ie more heat needs to be removed from sea water to cool it, than from ice. Basically its easier to cool ice ! So even though ice has a higher thermal conductivity (than sea water), its lower heat capacity and the fact that it floats on the surface, "shields" the water below from freezing over. and only a thin layer of ice is formed on the surface. The surface ice simply cools all the way down to the air temperature above.

I hope thats correct, Mandrake Im quite prone to making stupid mistakes

[vivekpm]

Ok... Here is the problem...

A banker interchanged the Rs. and Ps. when he cashed a cheque for some person X giving him Rs. instead of Ps. and vice-versa. (Ex. if cheque is for 40 Rs 70 Ps., banker gave him 70 Rs. 40 Ps.)

After buying a newspaper for 50 Ps., X noticed that he had exactly three times as much amount as the original cheque. What was the original amount of the cheque?

Cheers,

The original amount was Rs. 18 and 56 paise.

I got the answer to this question by peering into the post sufficiently hard. Here's how the actual solution should go -

Let original value = x + y (x being Rs and y being paise)
Value post error = 100y + (x/100)

We are told that

100y + (x/100) - 0.5 = 3*(x + y).

This resolves to

9700y = 50 + 299x.

Add the following common-sensical constraints -

y < 1 (since y is paise in Rs denomination)
x > 1 (since x is discrete Rs).

The only realistic numbers which fit this scenario are x = 18 and y = 0.56.

Spot on!!! You are correct Hitesh...

One more way to solve this is as follows:

Say the original amount is x Rs. and y Ps.:

Then either of following set of simultaneous equations should be true:

1. 3x = y, 3y = x - 50 ( If x > 50 )
2. 3x = y - 1, 3y = x + 50 ( If x < 50 )

1st equation gives negative value for x. so this is ruled out.
2nd equation gives x = 47/8 which is not a positive integer.
Now rewriting 2nd equation as follows:
3x = y - 2 (Subtracting one more rupee here) and 3y = x + 50 + 100 (Adding 100 Ps for subtracted rupee here

3x = y - 2, 3y = x + 150
This gives x = 18 and y = 56...

Good to see more participation here

Cheers,

[vivekpm]

I tried to put my solution in as simple language as possible and this is the best I came up with. Pardon me if I have complicated things (which I think I have).

I believe correct solution to this problem is already provided by Mandrake and Nimii which I think makes it 100% certain that none of the two will wear same shirt on same day . For decency, I go with Mandrake's solution

Jokes apart, here is one possible approach (there can be multiple approaches for this problem):

Let us do some naming first:

Shirt1= Unique shirt for both associates
Shirt2 = Shirt which is "probably" unique.
Shirt3,Shirt4,Shirt5 = Common shirts

1. The associate who formulates the strategy should wear his Shirt1 on Monday.
2. On other days, he should wear the shirt worn by other person on previous days if associate 2 has worn either Shirt3,Shirt4,Shirt5
3. If Associate2 wears his Shirt1 or Shirt2 on monday (probability 2/5) as against other three shirts (probability 3/5), Associate1 should wear his Shirt2 on tuesday. In long run Associate1 will probably learn whether or not their Shirt2 are unique and luckily if they are, there is very less chance that they will end up wearing same shirt on same day.

For example, consider Associate1 to be the one who comes up with a strategy:
Associate 1 wears his Shirt1 on Monday.
Probability that Associate 2 does not wear his Shirt1 is 4/5 which is more as compared to the probability that he too will his Shirt1 (which is 1/5).

Now if, on Monday, Associate 2 wears Shirt2, then Associate 1 should wear his Shirt2 (remember that he doesn't know if Shirt2 was indeed Shirt2 for Associate2 if Shirt2 for both are different) and on other days he can either wear his unique shirt or wear shirt worn by Associate2 on previous day intermittently to avoid pattern.

Cheers,

Vivek and Shivani,

The problem is posed in a manner that the solution will be sub-optimal. So, without conclusively knowing the probability of the uniqueness of a shirt to each associate, the solution proposed above will break down as the days run by. The solution proposed above will give you the best possible result for a maximum of 8 days. Under such scenarios, the first week should be used as a "data-gathering" phase. What really needs to happen is to do what Vivek says above for exactly 6 days. At that point, you will have a first crack at the unknown probabilities. Then, you can frame the problem more accurately, and use this extra data to dress accordingly the next week, and so on.

As you would know, the probability of both wearing the same shirt will go up as time passes by. What the above situation will allow you to do is to actually manage "time" into the equation, thereby increasing your chances of success.

Thanks
Hits

I agree. The solution does not guarantee that two people will not end up with same shirt on same day. It just maximizes chances of wearing different shirts.

As you pointed out Hitesh, the problem is not deterministic in nature. It uses words like "almost", "probably they both have" etc..

I too thought of "collecting data" approach for first few days but then it may happen that one or more shirts are never worn by the associates on a single day. Or worse still, the problem states that associates "most likely" has 5 shirts, seems even that is not certain.

Cheers,

[hits]

I too thought of "collecting data" approach for first few days but then it may happen that one or more shirts are never worn by the associates on a single day. Or worse still, the problem states that associates "most likely" has 5 shirts, seems even that is not certain.

Cheers,

Correct. The true mathematical solution to this problem would be to assign variable probabilities (p1, p2, etc..) to each of these non-deterministic variables, and then solve for a general solution. I can see it building up into a messy expression.

[shivani]

muniwar/ vivekah .. go ahead and post next riddle