[Aces high]

Finally,I get an answer to question1:



Okey,.so he decides to walk.In the meantime his wife starts and after walking for x mins he meets her and they get back home at 6:00pm but if he hadnt walked he would have reached home at 6:00pm which is the normal time.


So total travelling time when he walks is:

<-xmins,walk-> <x-mins by car-> <-t->
station---------------pick up point---------------------------car when he starts walk-----home


therefore total time for car to travel: 2x+2t -eqn1


When he doesnt walk:

<-y-> <-x-> <-t->
station--------------if he walked--------------------car if he walked------home
he would wudve bin here
hv bin here

Where y is the time the car has to travel,cause the guy didnt walk

Total time=2y+2x+2t -eqn2

From the problem,the difference is 10 minutes between the two,hence


subteacting eqn1 from eqn2,we get

2y=10,y=5

Therefore the car covers y in 5 minutes,then the man would've covered this in 25 mins,cause car is five times faster.

Therefore he waited for 25 minutes,he would have been home at 6:10 minus 25 minutes

Answer 5:45pm


Damn why are all my diagrams getting squished!!!

[FreakazoiD]

Originally Posted by Aces high

Finally,I get an answer to question1:

Okey,he reaches station x minutes early.so he decides to walk.In the meantime his wife starts and walking for x mins he meets her and they get back home at 6:00pm but if he hadnt walked he would have reached home at 6:00pm which is the normal time.

Now time,speed,distance is not exactly my fav topic, infact I prefer to avoid these kind of questions. But I have a question about your solution.

If this guy reaches the station x minutes early, then after he starts walking he will meet his wife not after x minutes but less than that, since his wife now drives a shorter distance. (If he stayed at the station he would have to wait x minutes to meet his wife)

The other two questions were fairly easy and have already been solved.

2. I would solve this by drawing a venn diagram (3 circles intersecting). Each of the 7 separate regions takes 1. So you get required answer as 5.

3. For the question on number of scores, the way I would solve it is :

max score possible = 150, minimum score possible = -50. And between each 2 integers, 0.33 and 0.66 is possible. So number of scores = 200x3+1 = 601

But 149.66, 149.33 and 148.33 are not possible as they require more than 150 attempts.

So number of possible scores = 601-3 = 598

the freak

[surbsj]

gr8 help guys

[Aces high]

hey freak,point taken,I dont afffect the answer though(I think)....

[just08in]

Originally Posted by FreakazoiD

Now time,speed,distance is not exactly my fav topic, infact I prefer to avoid these kind of questions. But I have a question about your solution.

If this guy reaches the station x minutes early, then after he starts walking he will meet his wife not after x minutes but less than that, since his wife now drives a shorter distance. (If he stayed at the station he would have to wait x minutes to meet his wife)

The other two questions were fairly easy and have already been solved.

This question the imp thing to be noted is the distance the guy has travelled and the time the wife has saved...let him reach sation some xmin...and they meet after some z mins...its of no use...

the wife has saved 10mins ...cos he walked some distance....otherwise she had to travel that distance...i've explain this poblem in my previous post!..

[Cittarasu]

Hi Amit,

My Head is going to blast ... :-) .. Please tell the answer for the Red/Hat Puzzle ...

Thanks
Chitu

[Cittarasu]

Ans for Wife/Huspand pick up :

1) Let the total distance between station and home is "d", and speed of the husband by walk be "y" and the car is "5y" :

Then we know that by car they will reach home by 6:10 minutes , therefore when the husband reaches the station early and walks for "x" distance and then picked up by car then they save 10 minutes ...

so 2d/5y - 2(d-x)/5y = 10 minutes .... A

Solve the equation A and we will get 25 minutes ..that is the man walks for 25 minutes.

If he had reached the station 25 minutes before the usual time , then they will be saving 25 minutes.

They will reach the home at 5:45 ..

2)

1st hat - liked by A
2nd hat - liked by B
3rd hat - liked by C
4th Hat - liked by A&B
5th Hat - liked by B&C
6th Hat - liked by C&A
7th Hat - liked by A,B&C

C - 4 hats were grey , so from the above table we can see that all the combination were C involved will be grey Hat .

That is 7th Hat, 6th Hat, 5th Hat, and 3rd Hat. ( Total 4 grey Hat)

Now we can see that two hats of A are grey, and he is involved in the combinatioin 6th and 7th Hat.

Therefore the combination A&B will be a different colour Hat other then grey Hat.

B had 3 grey Hat , he was involved in 5th and 7th Hat combination , therefore the single Hat chosen by him only will also be Grey Hat.

There for there will be 5 Grey Hat.

3) I had saw others answer for q3 , is 598 or 595

[Amit@Career_Avenues]

Qn 2.
I would have looked for an answer using venn diagram. Whenever you see words like "two had but the third did not" and "all did" or "only one did", you should immediately identify it as venn diagram. Using that, it would be damn simple, and would work out to 5 out of 7 hats.

Qn 3.
Just make sure you avoid counting the three numbers 148.33, 149.33 and 149.66.
So we have -50 to 148 (a total of 148+50+1=199). Gaps between them are 198. Each gap has 2 more possibilities. So total from -50 to 148 = 199 + 2 * 198 = 595.
Add possibilities > 148, i.e. 148.66, 149 and 150
So ans = 598.

Qn 1. Let me get more responses. Answer tmrw.

Here are todays questions.

A dice game is played thus: On each turn, a normal pair of dice is rolled. The score is calculated by taking the product, rather than the sum, of the two numbers shown on the dice. On a particular game where the pair of dice is rolled 5 times, the score for the second roll is five more than the score for the first; the score for the third roll is six less than that of the second; the score for the fourth roll is eleven more than that of the third; and the score for the fifth roll is eight less than that of the fourth. In how many different ways of rolls of the dice is this possible ?
1] 1 2] 2 3] 5 4] > 5

This question is from an old CAT paper. Classical as well as elegant solutions would be entertained.

In table tennis the first player to score 21 points wins. Service alternates between the two players
every 5 points. A players can score points both during his service and his opponent’s service.
Doshi beat Venkat 21–16 in a game, 24 of the 37 points played were won by the player serving.
Who served first?
1] Doshi 2] Venkat 3] Indeterminate 4] Inconsistent data


Cheers
Amit

[coolkunal]

Hi,

I was trying the Red/Blue Hat problem. I think the answer should be 12 days. My reasoning goes something like this:

Case (i):

Consider a scenario wherein there were only 2 gauls (say R1 and R2) wearing Red Hats. In that case:

On day 1: R1 and R2 turn up at the party.

R1 thinks R2 is the only person wearing a Red Hat and will not turn up on Day 2.
R2 thinks R1 is the only person wearing a Red Hat and will not turn up on Day 2.

On Day 2: R1 and R2 turn up again at the party.

R1 knows that if R2 was the only person wearing a Red Hat, then he would not have seen anyone else wearing a Red Hat on Day 1 and would not have turned up on Day 2. Same is the case with R2.

So, they can both conclude that there are 2 people wearing a Red Hat. Now, they can see all the other gauls. None of them are wearing a Red Hat. So, it becomes obvious to them that they are the ones wearing a Red Hat, and they will not turn up for the party on Day 3.


Case (ii):

Consider a scenario wherein there were 3 gauls (say R1,R2 and R3) wearing Red Hats. In that case:

On day 1: R1,R2 and R3 turn up at the party.

R1 thinks (R2 and R3) are the only people wearing a Red at and he knows that it will take them 2 days to realize the situation (based on the reasoning above). So, they will not turn up from Day 3 onwards. Ditto is the case with R2 and R3.

On day 2: R1,R2 and R3 turn up at the party (as expected).

On day 3: R1,R2 and R3 turn up at the party (again).

Now, R1 knows that there are definitely more than 2 gauls wearing Red Hats, since R2 and R3 have come to the party contrary to his expectations. Since he can see all the other gauls wearing a Blue Hat, he can conclude that he is wearing a Red Hat.

Ditto is the case for R2 and R3. The 3 of them don't turn up for the party after day 3.


As can be seen; for 2 Gauls, it takes 2 days and for 3 gauls it takes 3 days. Extending the concept, I would go with the answer that for 12 gauls it would take 12 days to realize that they are wearing red hats.

Please let me know if i've missed something somewhere.

[coolkunal]

Originally Posted by Amit@Career_Avenues

Qn 2.
This question is from an old CAT paper. Classical as well as elegant solutions would be entertained.

In table tennis the first player to score 21 points wins. Service alternates between the two players
every 5 points. A players can score points both during his service and his opponent’s service.
Doshi beat Venkat 21–16 in a game, 24 of the 37 points played were won by the player serving.
Who served first?
1] Doshi 2] Venkat 3] Indeterminate 4] Inconsistent data


Cheers
Amit

I think the answer to this one has to be (1) - Doshi.
I considered 2 scenarios: First, where Doshi (D) starts the game and Second, wherein Venkat (V) starts the game.
Case (i): Doshi Starts the game

Serve By / Lost / Won / Doshi-Venkat Score at this point
D , 0 , 5 , 5-0
V , 0 , 5 , 5-5
D , 0 , 5 , 10-5
V , 0 , 5 , 10-10
D , 5 , 0 , 10-15
V , 5 , 0 , 15-15
D , 1 , 4 , 19-16
V , 2 , 0 , 21-16

For the second scenario, I could not find a combination which satisfies all the given criteria. I know that this is a very crude way of attacking a CAT problem, but at this odd hour couldn't think of anything else. I could possibly even be wrong, and the answer could be (3).