| Sajha.com Archives | |
| Username | Post |
| Biswo | Posted
on 08-May-04 11:51 PM
I hope some sajhaites will help me out in this.A nasty creature wanted to ruin my time by forwarding this thing:-) 5->52 6->63 7->? 8->46 9->18 Hope I didn't ruin your weekend. |
| nepalithhito | Posted
on 09-May-04 12:01 AM
the answer is quite simple ........... its 94 find the connection yourself |
| redstone | Posted
on 09-May-04 12:11 AM
94 it is! |
| Biswo | Posted
on 09-May-04 12:34 AM
Please, kindly explain. I have exhausted too much times thinking about this. |
| scatterbrain | Posted
on 09-May-04 12:45 AM
Biswo ji, to me it seems like the answer is 61. 5->52 6->63 7->61 8->46 9->18 Looking at it from a numerical perspective. :) |
| scatterbrain | Posted
on 09-May-04 12:46 AM
How about a graphical for an explanation! :) 
|
| scatterbrain | Posted
on 09-May-04 12:49 AM
should read "... graphical display for ..." |
| nepalithhito | Posted
on 09-May-04 01:13 AM
you are thinking in a very complicated way scatterbrain ........................ just square the number and reverse the digits you get your answer 5x5= 25 ............... 52 6x6= 36 ............... 62 7x7= 49 ............... 94 8x8= 64 ............... 46 9x9= 81 ............... 18 |
| niksnpl | Posted
on 09-May-04 03:09 AM
Where are you studying ScatterBrain?? Your knowledge of mathematics is good :u |
| An Indun Poet | Posted
on 09-May-04 07:16 AM
6x6=36....... 63 A man while looking at a photograph said, "Brothers and sisters have I none. That man's father is my father's son." Who was the person in the photograph? |
| rohini_a | Posted
on 09-May-04 07:25 AM
It's so damn easy. I figured out in 10 secs. The answer is clearly 94. |
| nepalithhito | Posted
on 09-May-04 07:44 AM
An Indun Poet thanks for pointing out my typo ........... well the answer to your puzzle is he is looking at the photograph of his son |
| scatterbrain | Posted
on 09-May-04 09:54 AM
Nepalithito, aha! Now it is that simple eh? I knew there was a simpler answer, but I didn't know what it was. So, I just thought I'd give it a different spin, travel a road where few would go. Beauty in simplicity is certainly the most appealing to me, but forgive me brother I may not be able to revel in it for I am a simple man with a complicated mind. :) |
| scatterbrain | Posted
on 09-May-04 09:57 AM
niksnpl. thank you for the compliment. i am studying in the DC metro area thesedays. :) |
| ajaybro | Posted
on 09-May-04 10:32 AM
Hi biswo ji I am Ajay from Saint louis.I mistakenly deleted your email ID and coulnt e-mail you.hope you are in good condition there.well,please forward me your ID with ph.no,if possible. last time i got from Amina who is in nepal now. My id is ajaypsr@yahoo.com ok,take care. Ajay |
| ajaybro | Posted
on 09-May-04 10:40 AM
The problem was so difficult to me till i got the answer.good for 'em who got good I Q brain.I Think my analytical power has gone down. |
| Badmash | Posted
on 09-May-04 10:54 AM
What scatterbrain demonstrated in the graphic and how he came up with 61 is beyond my comprehension. Anyone? |
| Biswo | Posted
on 09-May-04 11:16 AM
Nepalithitoji, Thanks for the answer. It seems so easy now.. Scatterbrainji, yours was really praiseworthy attempt to find the answer. Appreciate that. Ajay, please check for my email. AIPji, father and son. |
| EdHunter | Posted
on 09-May-04 11:22 AM
scaterbrain's answer's the most complex and creative one.. nice one scatterbrain... logical and explainable... :P nice one to the others as well.. me aws thinking too much abt it and juz gave it up... :P |
| An Indun Poet | Posted
on 09-May-04 03:27 PM
Try This! A man wanted to enter an exclusive club but did not know the password that was required. He waited by the door and listened. A club member knocked on the door and the doorman said, "twelve." The member replied, "six " and was let in. A second member came to the door and the doorman said, "six." The member replied, "three" and was let in. The man thought he had heard enough and walked up to the door. The doorman said ,"ten" and the man replied, "five." But he was not let in. |
| An Indun Poet | Posted
on 09-May-04 03:28 PM
What should have he said to ENTER: |
| redstone | Posted
on 09-May-04 03:42 PM
THE ANSWER IS THREE :) |
| Dananah | Posted
on 09-May-04 06:00 PM
hehe...i go with red with this one too..;) 3 it is :D... |
| nsshrestha | Posted
on 09-May-04 06:15 PM
Biswo, Got the answer now? :) |
| Badmash | Posted
on 09-May-04 07:33 PM
3 kasari? |
| redstone | Posted
on 09-May-04 08:08 PM
3 kasari re? Dorman took the answer -not the half of the number given, but the number of letters in the given number!!! yesari twelve." The member replied, "six " ---the number of letters in twelve is six. six." The member replied, "three --the number of letters in six is three. The doorman said ,"ten" and the man replied, "five."--the number of letters in ten is three not five. |
| An Indun Poet | Posted
on 09-May-04 09:38 PM
Another One Kevin is 14 inches taller than George. The difference between Kevin and Richard is two inches less than between Richard and George. Kevin at 6'6" is the tallest. How tall are Richard and George ? |
| nsshrestha | Posted
on 09-May-04 10:14 PM
1. K = G+ 14; 2. (K-R) = (R-G) - 2; 3. K = 78 inches Solvking eqn 1 2 and 3. K = 78 inches R = 72 inches G = 64 inches |
| An Indun Poet | Posted
on 09-May-04 10:42 PM
Three men go fishing on a camping trip. They collect all the fish they caught and put them in a cooler. They agree that when they leave they will divide up the fish equally. In the middle of the night the first man gets sick and decides to leave. He opens up the cooler and counts the fish. He throws one fish away and then takes his third of the fish. Later that night the second man has a family emergency and decides to leave. He opens up the cooler and counts the fish. He throws one fish away and then takes his third of the fish. The third man gets up very early the next morning and decides to leave. He opens up the cooler and counts the fish. He throws one fish away and then takes his third of the fish. None of the men know that the other men had taken any fish. What is the smallest number of fish that would make this possible? |
| nepalithhito | Posted
on 09-May-04 10:53 PM
the answer is 25 ............ calculations left for other people out there .......... |
| An Indun Poet | Posted
on 09-May-04 11:01 PM
Lots of smart guys here... Another one 1 11 21 1211 111221 ______ ________ __________ Fill in the blanks. |
| anonymous | Posted
on 09-May-04 11:12 PM
nsjresta,mine was Kevin - 6'6" Richard- 6' George - 5 '4" other question 25..too..dyam nepallithito beatme to the ans...:P last one seems tough...will try later..:D cheers..:D |
| Biswo | Posted
on 10-May-04 12:22 AM
A bit late, but I think the flaw in scatterbrain's original answer (7,61) was that he was assuming a lot of things in addition to the given four pairs. For example, he was assuming something for 5.1, 5.2, 6.4, and they all make up a nice quadratic line, from which he got (7,61). But the question gives only four such value pairs. |
| Brook | Posted
on 10-May-04 12:45 AM
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 ...? |
| An Indun Poet | Posted
on 10-May-04 01:43 AM
Some months have 30 days, some months have 31 days, how many months have 28 days..... |
| Brook | Posted
on 10-May-04 01:48 AM
Exactly 28 days? 1 or 0 Loosely, 12 months have 28 + days. |
| nepalithhito | Posted
on 10-May-04 01:50 AM
ya !!! brook you got it. i was baffaled too about the problem. had to ask for a senir dai's help on it. great problem. looking forward to seeing more of those problems. to those who are still baffled by the problem here are next two numbers of the sequence ............ try to figure out the relation 1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 132113111231132211 111212211231121321132221 |
| nepalithhito | Posted
on 10-May-04 01:51 AM
all 12 months have 28 days |
| scatterbrain | Posted
on 10-May-04 04:29 AM
Biswo ji: Good point. However, I was not assuming the data to be coming from a continuous quadratic function. Rather, I was thinking of it more in terms of a descrete function. A finite sequence, to be more precise, since what you had put forth is a finite sequence defined by the quadratic function over the domain of a finite subset of natural numbers, namely {5, 6, 7, 8, 9}. Hope it clears your doubts. :) |
| scatterbrain | Posted
on 10-May-04 04:32 AM
* discrete Oh boy, it's too early in the morning for me to spell correctly. |
| scatterbrain | Posted
on 10-May-04 05:00 AM
If you are still curious, Biswo ji, the discrete quadratic function f(x) = -6.5x^2 +82.5x-198 over the domain {5, 6, 7, 8, 9} gives the desired result of {52, 63, 61, 46, 18}. For the finite sequence, translate the above function to the left to get g(x)=-6.5(x+4)^2 + 82.5(x+4) - 198, which over the domain of {1, 2, 3, 4, 5} gives you the sequence {52, 63, 61, 46, 18}. Hope we beaten the horse to death by now. :) |
| An Indun Poet | Posted
on 10-May-04 05:21 AM
Now Lets See the Real Smart Ones Figure This Out! You have 12 coins, one weighs slighlty less orslightly more than the others. Using an equal arm balance and only making3 weighings determine which one is different and whether it is slightly less or slightly more. Warning: the fact that you don't know whether it is less or more is a major problem in the solution. |
| nsshrestha | Posted
on 10-May-04 06:24 AM
6-6 Keep heavy one. 3-3 Keep heavy one. 1-1, You know which heavy out of these two. if not these two, it will be the last one left. |
| nsshrestha | Posted
on 10-May-04 06:25 AM
didn't see less or more ..:) |
| nsshrestha | Posted
on 10-May-04 06:36 AM
Its too easy, answer is in Google..:) |
| rbaral | Posted
on 10-May-04 06:46 AM
nsshrestha ji The premise of your analysis is that the *one* is heavier than others. What do you do if the *one* is lighter than the remainings? |
| Dr. Strangelove | Posted
on 10-May-04 06:56 AM
Confounding Compounding!!! "Usually when the subject of a sentence is compound, and the components are connected by "and," the verb takes the plural form. For example, we say, "Bob and his wife ARE planning to drive to Florida"... not, "Bob and his wife IS planning to drive to Florida." Likewise, we say, "The vase and the book ARE on the table," not, "The vase and the book IS on the table." But, can you think of a situation where the components of a compound subject are connected by "and," yet the form of the verb must be singular, and not plural? |
| rbaral | Posted
on 10-May-04 07:17 AM
There are several - one is - AT and T is a good (or bad) telephone company. |
| anonymous | Posted
on 10-May-04 07:20 AM
separate them to 3 groups of 4 coins each 4 4 4 (A) (B) (C) Step 1:----1st weighing - A Vs B if balances--> means Different coin is atC...go to Step 2a if doesnt balance it means Different coin is either at A or Cgo to Step 2b.(ps will post 2b later its a little bit tricky and im tired typing :P) Step 2a: ---2nd weighing take 2 coins from C and weigh it with any 2 coins from A or B(since we know they are all weigh the same) if it balances -->means Different coin is one of the 2 coins fromC that was not weighed--go to step 2aa if it does not balance we know the Different coin is one of the 2 coins we just took from C and weighed--go to step 2ab Step 2aa:----3rd weighing take one 1 coin from that untouched C and weigh it with any 1 coin of the previous weighed coin... if it balances--> means the last coin left from Cwhich was not weighed is the Different Coin we were looking for.< if it doesnt balance, that coin u just took from C and weighed is the Different Coin.< Step 2ab:---3rd weighing take one of the coins from the 2(the one u just weighed from C) and weigh it with another one..any coin from A or B. if it balances, it means the coin that was left ou, is the Different coin we are looking for.< if it doesnt balance, current coin you weighed is the Different Coin you are looking for.< |
| nsshrestha | Posted
on 10-May-04 07:27 AM
1. Peanut butter and Jelly IS my favorite sandwich 2. Marconi and cheese tastes good. |
| Dr. Strangelove | Posted
on 10-May-04 07:30 AM
Baral, good one! But, AT&T is not quite a compund subject! |
| scatterbrain | Posted
on 10-May-04 07:51 AM
Dr. Strangeglove, did you know one AND one EQUALS two? :) |
| Dr. Strangelove | Posted
on 10-May-04 07:56 AM
Scatterbrain....is matterbrain. Good job! My closest friend and college roommate was the best man at my wedding." Because it's the same person, right? What common mathematical symbol, when placed between the numbers 4 and 5, will result in a number that is greater than 4 but less than 5? |
| Sonika_NY | Posted
on 10-May-04 08:00 AM
Strangelove, even kid can answer that. its dot. 4.5 is greater than 4 and less than 5. Riddles should something that makes you think and think and brain scattered.lol |
| Dr. Strangelove | Posted
on 10-May-04 08:14 AM
Sonika_NY, you are soooooooooo smart!! Since you drive around in a Porsche 911 GT, here's a car riddle for you! There are two guys sitting around, and one of them is looking at a piece of paper. He says, "Hah! This is interesting." The other guy says, "What's interesting?" And, the first guy says, "Well, this is a bunch of names of cars, past, present, and future, and there's something interesting about it. I'll read you some of them: AMC Eagle Ford Falcon Dodge Charger Ford Bronco Plymouth Colt And, my personal favorite...Daihatsu Rocky." The other guy says, "So what?" But, there is a big deal. What is it that's unique to the names of these cars? |
| DWI | Posted
on 10-May-04 08:15 AM
I guess I am an idiot then. I was thinking the symbol '>' as in 4>5. For Indun Poet's puzzle, that is a classic puzzle. I got stuck in step 2b of Anonymous as well. Will try more, guarantee the answer. |
| An Indun Poet | Posted
on 10-May-04 08:21 AM
There are 4 men who want to cross a bridge. They all begin on the same side. You have 17 minutes to get all of them across to the other side. It is night. There is one flashlight. A maximum of two people can cross at one time. Any party who crosses, either 1 or 2 people, must have the flashlight with them. The flashlight must be walked back and forth, it cannot be thrown, etc. Each man walks at a different speed. A pair must walk together at the rate of the slower man Man 1: 1 minute to cross Man 2: 2 minutes to cross Man 3: 5 minutes to cross Man 4: 10 minutes to cross |
| Sonika_NY | Posted
on 10-May-04 08:25 AM
Strangelove, You are funny, they all have football team name on it. You are wasting my time. Do you mind me driving a Porsche? duh |
| nsshrestha | Posted
on 10-May-04 08:32 AM
A compund interest problem: S E N D + M O R E ---------------- M O N E Y |
| nsshrestha | Posted
on 10-May-04 08:38 AM
1,2 goes, 1 returns time = 2 + 1 = 3 minuts 5,10 goes, 2 returns time = 3 + 10 + 2 = 15 minuts 1,2 goes time = 3 + 10 + 2 = 17 minuts |
| scatterbrain | Posted
on 10-May-04 08:39 AM
Trip across the bridge:1 2, 1, 3 4, 2, 1 2 |
| scatterbrain | Posted
on 10-May-04 08:50 AM
A man leaves home running, turns left, runs forward, turns left again, runs forward, turns left again, and returns home to find a masked man. Who was the masked man? |
| RBaral | Posted
on 10-May-04 08:59 AM
The Catcher and Umpire! (Lindsay, age 8) http://halife.com/family/kids/kackles.html |
| scatterbrain | Posted
on 10-May-04 09:14 AM
Rbaral ji, you certainly sound resourceful. Some more: -------------- A man got up in the middle of the night and went to the toilet. When he'd finished what he was there for, he flushed the toilet, turned off the light, and went back to bed. When he awoke in the morning, he listened to the radio and promptly committed suicide. Why did he commit suicide? -------------- A woman with her wedding ring finger missing and a man get off of a train, and are met by a man that they haven't seen for thirty years. He pulls out a gun and kills her. The police take no action. Why? -------------- |
| RBaral | Posted
on 10-May-04 09:32 AM
Answer to the second question - Andha Kanoon estyal. The man probably was her ex-husband, who had served an imprisonment for murdering her (upon her disappearance). |
| Biswo | Posted
on 10-May-04 10:56 AM
scatterbrainji, I agree with your solution in the AIP's problem about crossing bridge. I think nsji's solution has a problem that the second trip people (3 and 4) doesn't have a torch to carry with them. In earlier answer, I still stick with my contention and I think I need to make my argument clearer. My point was , when you try to fit points in a quadratic equation, we are actually assuming a lot of other pairs. In fact, infinitely many other pairs that makes up the quadratic equation. But the original question gave only four such pairs, and asked us to find the fifth one. So, although the equation gives correct values for the four other points, it is not correct in terms of fifth. I hope I am clear here. Other than that, I am enjoying everything here. |
| nsshrestha | Posted
on 10-May-04 11:22 AM
Biswo, SB and mine are identical answers. After first trip, 1 returns with the torch. :) |
| Biswo | Posted
on 10-May-04 01:35 PM
oops, sorry ,nsji. Yep. You're right. |
| scatterbrain | Posted
on 10-May-04 02:16 PM
Biswo ji: You are right about there being infinitely many ordered pairs that make up the quadratic function, but that is when the domain of the function is over all real numbers. If the function domain is RESTRICTED to just the set of {5, 6, 7, 8, 9}, it will be a discrete function, instead of being continuous as in the case you mention. Maybe the curve that I fitted in the graph was misleading. I was trying to show that the points mentioned (which were plotted with filled circles) were a subset of the underlying quadratic function (the curve). Had I just plotted the points, the relationship wouldn't have been as obvious and therefore as a guideline the curve was drawn. Just focus on the dots plotted and hopefully that will clear up your doubts. :) |
| SITARA | Posted
on 10-May-04 02:37 PM
Scatts still plotting? :P |
| scatterbrain | Posted
on 10-May-04 02:57 PM
Yeah, it was just a scatterplot. How are you twinkling? :) |
| SITARA | Posted
on 10-May-04 03:04 PM
Not very bright, I'm afraid! This thread's bombasted my thunk-ability! |
| DWI | Posted
on 10-May-04 07:50 PM
I don't think anybody solved IndusPoet's riddle, and I am still working on it. I got as far as anonymous did (I also tried making groups of 3 marbles instead of 4), but you get stuck at one point (in his case, Step 2b). Step 2b is the toughest one. Will anybody attempt? You have 12 coins, one weighs slighlty less orslightly more than the others. Using an equal arm balance and only making3 weighings determine which one is different and whether it is slightly less or slightly more. Warning: the fact that you don't know whether it is less or more is a major problem in the solution |
| Neural | Posted
on 10-May-04 08:20 PM
Answer to DWI: Make three groups of 4 coins Group 1: * * * * Group 2: * * * * Group 3: * * * * 1] Weigh two groups: If both groups of coins are equal in weight, the lighter coin must be in the group you didn't weigh. On the other hand, if one of the groups you weighed is lighter, the lighter coin must be in that group. 2] Grab the lighter group of 4 coins: Weigh two coins. If one of them is lighter, there it is! If they are equal in weight, the lighter coin must be the one you didn't weigh. And go for the third attempt. That's it!!! |
| anonymous | Posted
on 10-May-04 09:09 PM
hey DWI :D took me quite a long time to solve that bloody thingy for 2b hehe it is complicated and too complicated to type it here..dunno how im going to explain it...sems pretty long..:| buts heres some hint hope it helps..:D if A and B does not balance.(problem 2b) Hint-you have to now weigh using 3 coins on each side ;)(maybe this is a give away hehe..dunno what other hints i can give...) anyways have fun..:D) cheers :D |
| Brook | Posted
on 10-May-04 10:41 PM
Oh I thought this was solved by somebody already. Anyway. Here's the solution: Divide the 12 coins into 3 groups: A, B and C. For simplicity, let's label each coin as A1, A2, A3, and A4 and similarly for those in groups B and C. Step 1: Compare sum of A with sum of B. There are two possibilities (the third being trivial), Case 1: A = B; => the coin we require is in group C. Step 2: Compare B1+B2+B3 with C1+C2+C3. Again, there are two scenarios - Scenario 1: B1+B2+B3 = C1+C2+C3 => The problem coin is C4. Step 3: Compare any coin with C4 to determine whether . C4 is lighter or heavier. Scenario 2: B1+B2+B3<> C1+C2+C3 => It's either C1 or C2 or C3! If B1+B2+B3 > C1+C2+C3, then our coin is LIGHTER, Step 3: B1+C2 and C3 and C4, case (a) if B1+C2 > C3 + C4, => C3 is lighter. case (b) if B1+C2 < C3+C4, => C2 is lighter. case (c) if B1 + C2 = C3+C4, => C1 is lighter. Similarly if B1+B2+B3 Case 2: Coming up :) |
| Brook | Posted
on 10-May-04 11:41 PM
Case 2: This is slightly trickier. Step 1: Compare A and B. A <> B. Say A> B or, A1+A2+A3+A4 > B1+B2+B3+B4 Step 2: Compare, A1+A2+C1+C2 with B1+B2+C3+C4 Scenario 1: A1+A2+C1+C2 > B1+B2+C3+C4 => either one A1 or A2 is the heavier coin or B1 or B2 is the lighter coin. then, Step 3: Compare, A1+B2 with A2+C1 if (a) A1 + B2> A2 + C1, then A1 is the heavier coin. if (b) A1 +B2 < A2 + C1, then B2 is the lighter coin. if (c) A1 +B2 = A2 +C1, then B1 is the lighter coin. Scenario 2: A1+A2+C1+C2 < B1+B2+C3+C4 => either A3 or A 4 is the heavier coin or B1 or B2 is the lighter coin. Step 3, solve as above. Scenario 3: A1+A2+C1+C2 = B1+B2+C3+C4 => either A3 or A4 is the heavier coin or B3 or B4 is the lighter coin. Step 3, solve as above. The case when B> A is solved similarly. --*-- Now I can happily enjoy lunch. |
| noname | Posted
on 11-May-04 03:22 AM
Okay here is one simple riddle from this late comer. (Modified from original version from a text-book to keep googlers at bay :)) A boy wakes up at 6:00 in the morning and begins a long winding road up a mountain to his uncle's home at the summit. He spends all the day way up; he stops several times along the way at random and rests for a random amount of time at each stoppage. He reaches the top at 6:00 in the evening. The next day he starts from his uncle's home at the top exactly at 6:00 in the morning, and starts the journey to the base. Again he rests several times along the way at random and for random amounts of time, and reaches the bottom exactly at 6:00 in the evening. The question is, at any point is he at exactly the same place at the exact same time on the two days? |
| arch119 | Posted
on 11-May-04 04:50 AM
ya there should be one and only one of such points provided that the boys motion is monotonically upwards or downwards. Just imagine that when the boy starts climbing up the hill on a fine sunday morning , his alter ego starts descending the hill . Now if they don't engage themselves in playing hide and seek , they will meet at some point and the time at which they both will arrive at that point will be the same. It won't be difficult now to apprehend that such a point will exist even if the two motions occur in two different days rather than the same day. |
| noname | Posted
on 11-May-04 05:09 AM
:) |
| rbaral | Posted
on 11-May-04 05:11 AM
>The question is, at any point is he at exactly the same place >at the exact same time on the two days? For sure, there were two instances when he was at the same place - the beginning, and the end of his journey. |
| Neural | Posted
on 11-May-04 05:28 AM
>>The question is, at any point is he at exactly the same place at the exact same time on the two days? rbaral: " For sure, there were two instances when he was at the same place - the beginning, and the end of his journey" beginning was 6 AM and end of journey was 6 PM == is not it?? Time for the same place, if it is for Base = 6AM (first day) and 6PM (II day) or Time for the same place, if it is for Top = 6 PM (I day) and 6 AM (II day) 6 AM not equal to 6PM Noname has asked : same place at the exact same time. The best riddle till now- by Noname Ji :))) |
| Neural | Posted
on 11-May-04 05:34 AM
.Noname ji, is there any condition like : same climate on both days? Cuz many uncertainty like rain, sudden weather changes will surely affect the BOY during his journey, is not it? |
| anonymous | Posted
on 11-May-04 05:53 AM
i think arch119 ans is right.. and well in some ways r baral ans are logical too...but hehe if i had to choose i choose arch119 ans hehe... anyways heres smthing i had in my fwd..try it if u havent seen this before.. cheers :D 
|
| EdHunter | Posted
on 11-May-04 06:15 AM
me answers topah… and dun tell me it’s wrong.. :P 1. MTWTFSS.. go figure out… urself.. easy peasy.. :P 2. 5 + 5 + 5 NOT EQUAL to 550 duh!!! 3. go figure 4. 3 lines.. and I did it… what yah say?? ;0) 
|
| anonymous | Posted
on 11-May-04 06:19 AM
Ed or should i call u EH??(EH sounds better dun u reckon..:P) hmmm good ans....all sounds logical...but since this is an A level paper...we have to mark it strictly... according to the ans..and sorry to say though i like to give u full marks..(just for trying..:P)...u did manange to pass and got 50%...go find out which ones are right and which ones are wrong... cheers..:D |
| nsshrestha | Posted
on 11-May-04 06:47 AM
Here comes another one. google safe version. A monkey without eyes sees mangos on a mango tree, he does not take mangos and he does not leave mangos, How can it be? |
| An Indun Poet | Posted
on 11-May-04 06:58 AM
Three men were on a business trip and had to stay in a hotel over night. The price of the room was $30.00, so the men decided to split one room, three ways. Each one paid $10.00. Well after they paid, the manager realized that he overcharged them on their room. The room only cost $25.00, so he gave the bellboy five one dollar bills to give to the three men. On his way up to the room the bell boy was trying to think of a way to split $5.00 three ways. After thinking about it awhile, he decided to keep $2.00 for himself and give each man $1.00 back. Now, if each man (who paid $10) gets $1 back that means they each paid $9.00 ($10 - $1 = $9 ). $9.00 multiplied by 3 (because there are three men) equals $27.00 plus the $2.00 the bell boy kept equals $29.00! What happened to the missing dollar????? |
| EdHunter | Posted
on 11-May-04 07:16 AM
wrong..!!! actually… the men are paying tips to the bell boy plus the room charge.. which totals up to $27 ($25 for the room + $2 for the tips) so the guys pay $9 each for this…. the one dollar they are receiving back is the left over from this. $30 - $27 = $3. which is then divided by the 3 of em… so each guys gets back 1 dollar. so, in effect, they are receiving back what they are owed… :P makes sense??? if not go figure urself.. :P |
| RBaral | Posted
on 11-May-04 07:18 AM
Here, 
|
| DWI | Posted
on 11-May-04 10:23 AM
Brook Thanks for the brilliant answer. The thought of mixing the weights came to me, just didn't find the end. Good job. Neural, did you realize your answer wasn' t adequate? |
| noname | Posted
on 11-May-04 03:42 PM
Neural Since the boy is making random stoppage spending random amount of time, rain or sun it doesn't matter, provided, as arch 119 wittingly mentioned, the boy doesn't play hide and seek. |
| An Indun Poet | Posted
on 11-May-04 04:44 PM
Two people, named S and P, are talking about two numbers x and y. (note: x and y are both integers greater than or equal to 2). S knows their sum (x+y), while P knows their product (xy); however, initially NEITHER knows x and y. S: Hey P! I don't know what the numbers are. P: I'm not surprised. I KNEW you didn't know. However, I too don't know. S: You don't? Really! Then I know what the numbers are! What are the two numbers? |
| Neural | Posted
on 11-May-04 07:02 PM
Noname, "The question is, at any point is he at exactly the same place at the exact same time on the two days?" I guess no... because at any places he can be for different amount of time .. |
| noname | Posted
on 11-May-04 07:26 PM
AIP, That Sum guy looks really smart. It is easy for the Product guy to say that 'i knew you didn't know' coz there are lots of possible combinations for the Sum guy. Had it been the Sum guy saying something to that effect, the matter would have been a little easier. Wasted 2 hrs.........:( Anyone, already with the solution ? Neural, Let's try this riddle from AIP... |
| Neural | Posted
on 11-May-04 07:50 PM
Answer to AIP, the two nos. are 6 and 2. |
| niksnpl | Posted
on 12-May-04 03:34 AM
T= knows; F= doesn’t know X Y S = X + Y P = XY S P S 2 2 4 4 T 2 3 5 6 T 2 4 6 8 F T 2 5 7 10 F T T 2 6 8 12 F F ? 2 7 9 14 F T T 2 8 10 16 F F 2 9 11 18 F F 2 10 12 20 F F 2 11 13 22 F T 3 3 6 9 F T T 3 4 7 12 F F ? 3 5 8 15 F T T 3 6 9 18 F F 3 7 10 21 F T 3 8 11 24 F F 3 9 12 27 F T 3 10 13 30 F F 4 4 8 16 F F 4 5 9 20 F F From the table above, the minimum number that P doesn't know is for 12.... So the numbers can be 2 and 6 or 3 and 4.... But, Now S know the numbers.. :O Neural, How can you claim that the numbers are 2 and 6, they can also be 3 and 4.. Hope you will post your reason................. Cheers!!! |
| niksnpl | Posted
on 12-May-04 03:36 AM
Hey...where did my table disapper??????????????/ |
| Neural | Posted
on 12-May-04 04:02 AM
.niksnpl well start from 2+2, which is not possible... as anyone can know from sum that its 2,2 then 2+3 is also not possible with same logic. 2+4 is also not possible; 9 could tell the possibilities of 3+3 just try many possibilities, and 6 and 2 is only possibilities as 12 could give possibilities of 6,2 and also 4,3 But 7 is not multiplication of any number. Hence, I go for 6, 2. (May be i am wrong) |
| nepalithhito | Posted
on 12-May-04 04:10 AM
since S knows the sum of two number only he will know whether two numbers were 6&2 or 4&3. if this answer is taken correct than other two number that fit in this category is 6&3 or 9&2 |
| niksnpl | Posted
on 12-May-04 04:48 AM
Nepali thito !! There can be infinite combinations like that..So, we should limit it in the minimum, as it doesn't have maxima.... so the only possibility remain over here is 2 , 6 or 3, 4.......... Hope you understood that>>>>>> |
| EdMorah | Posted
on 12-May-04 06:11 AM
Another riddle.. I got it off somewhere… Once upon a time… (nice way to start ALL stories :P) There was a T-junction on a stretch of road. These T-junctions were very special. One side led to the gates of hell (bad sh*T) and the other to a lifetime of happiness and eternity.. (Good sh*T) On each side of this T-junction, 2 giant twins guarded the junction. One ALWAYS told the truth… The other ALWAYS told a lie… Now the qn is... as u are approaching the junction, depending on where u wanna go (i.e. good Sh*T or bad Sh*T), what SINGLE and IDENTICAL question must u ask each of these twins to go to ur specific destination (as I said. Some want good sh*T other’s the bad sh*T)?? The rule of this game is that these giant will only answer ONE qn each from yah and no more. And u must ask em both the SAME qn.. if u try asking difft qns, they will kick ur ass so hard u will be screaming for ur mommy... (hey.. dun blame me..!!! they’re twins.. telepathy and all that sh*T) So what is that ONE, single, sole, solitary (u get my point??) same, similar, identical (u got my other point as well din’t yah>>??) question u must ask these twins so that u’d be able to go to ur destination… (bad sh*T or good sh*T.. it’s up to u) rem.. one ALWAYS tells the truth the other ALWAYS lies.. thing is u dunno who lies and who tells the truth… they’re twins anyway, so u won’t know.. Ans anyone?? P.S: I hope I have made the riddle clear enough… if u deun understand.. well juz ask me :P |
| Dr. Strangelove | Posted
on 12-May-04 06:29 AM
Tim and Jethro were happy to have their jobs at the new self-serve gas station in town. And, since the Farmer's Almanac had predicted this to be the coldest winter since the last ice age, they were happy to be working indoors, while the customers pumped their own gas. This station was so modern that it had a video camera for each of the pumps, and a TV monitor that would show the rear of everyone's vehicles as soon as they pulled up to the pumps. When the boredom of their jobs finally set in, Tim and Jethro began playing a little game. The game involved trying to figure out which customers had pulled up to a pump with the fuel door on the wrong side-- that is, facing away from the pump. Now, they couldn't see the cars pull in to the gas station. The video cameras were only aimed at the back of the vehicles. So, there was no time during which they could see the side of a vehicle where the fuel door was located. They could only see the vehicle after it was in position to refuel. They had to make their bets before the driver shut off the key and exited the vehicle-- before he dope slapped himself for pulling in on the wrong side. Jethro was correct 99 percent of the time. Tim was correct about 50 percent of the time, because he was just guessing. What did Jethro know that enabled him to tell when a driver had pulled up to the pump with the fuel door facing the wrong way? |
| nepalithhito | Posted
on 12-May-04 06:47 AM
this is a classic puzzle: if i ask the other guard if that road leads to good shit would he say yes well here are four scenarios 1) you ask the honest one gaurding the good sh*t........... liar is gaurding the bad shit so he would answer yes ............. so yes will be the answer 2) you ask the honest one gaurding the bad sh*t........... liar is gaurding the good shit so he would answer no ............. so no will be the answer 3) you ask the liar one gaurding the good sh*t........... honest one is gaurding the bad shit so he would answer no ............. so yes will be the answer 4) you ask the liar one gaurding the bad sh*t ........... honest one is gaurding the good shit so he would answer yes ............. so no will be the answer so in any case you can safely take the road which is gaurded by the guard who says yes |
| llkathmandu | Posted
on 12-May-04 07:09 AM
.Answer for EdMorah's riddle is (ask any guard...): "If I ask the other guard for the HELL's (or Heaven's) gate, which gate will he show me ?" And go to the opposite gate for HELL (or Heaven) ;) |
| An Indun Poet | Posted
on 12-May-04 10:49 PM
Use 1, 5, 6 and 7 only once to come up with the answer: 21. While 1,2,6, and 7 can only be used once, you can use multiple operands. Another Old One: A man conducting a survey visits the home of a woman. He knocks on the door and sees a woman with her three children. He asks for the ages of her children. And she tells him, "The product of their ages is 72. And the sum of their ages is the number on the door." So the man goes outside and checks the number of the door. He comes back and says, "I need more information." She replies: "The oldest one likes strawberries." Immediately, the man figures out their ages. What are their ages? |
| sukuti | Posted
on 21-Jun-04 06:33 AM
What's the asnwer to the gas-station puzzle or riddle? |