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# bhramarraj

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## Posts posted by bhramarraj

1. ### brain puzzle

fill both 500 ml bowls. Place one of those bowls in the one liter bowl. Now all three bowls contain the same amount of water?

2. ### brain puzzle

It looks impossible unless it is a tricky question. Or there has to be some outside aid like markings on the Jars....?

Or it has to do something by tilting of jars to certain angle so that a specific amount of water comes out.....?

Guys these are three empty bowls A bowL is 500mil , B bowl is 1 Litre, C bowl is 500mil and there is one jar of 1litre water aprat from this 3 bowls so we have to pore water in such a manner in all 3 bowl should have same level of water also once so how will you solve?

3. ### An irrational "aha!" puzzle

When a rectangle of dimensions g x 1 is cut into a square and another triangle, the square will be of dimensions 1 x 1 and another rectangle (g - 1) x 1

The proportions of the original rectangle = longer side g / shorter side '1'
The proportions of the second rectangle = longer side 1 / shorter side (g -1), Then
g/1 = 1/(g - 1) which can be written as g = squareroot of (g + 1)

Therefore g is an irrational number.

4. ### One Girl - One Boy

I think discussion so far might have not satisfied many learned friends. May be my explanation would clarify some doubts.

OP says that the couple has two kids, so obviously one will be younger and other elder.

Say couple sends elder kid to attend a function. Now what is the probability of this kid being a girl, if one kid is definitely a girl...?

I think you got it it now....!

Say younger is a girl, then there are two possibilities: elder may be a boy or a girl.

As bonanova stated correctly, OP does not specifies which kid is a girl. so younger kid could be a boy, so the third possibility is that elder could be a girl; here fourth possibility of elder being a boy, when younger is a boy, is ruled out because OP says that one kid is definitely a girl.

Now out of above three possibilities, there is only one possibility of both kids being girls.

5. ### Cockfit

Calculating from back:
Third arena:
You come out paying \$10 and balance with you becomes Zero.
So you had \$5 before the amount was doubled.
So before paying entrance fee you had \$15.
Second arena:
You had \$15 when you came out of second arena, so you before coming out you had \$25.
So you had \$12.5 before the amount was doubled.
So before paying entrance fee you had \$22.5.
First arena:
When you came out of first arena you had \$22.5.
So before leaving the first arena you had \$32.5.
So when you came at the first arena (before doubling the amount) you had \$16.25.
So before paying the entrance fee to enter the first arena you had \$26.25.

Above remains true whether we go for wording of OP in any way.

6. ### Mensa Math/Logic Problems...How to Solve?

*Last but one (3/20): 83

Multiply the 2nd and 3rd number of a row.

Reverse the digits of first number in this row. Subtract the number so obtained from the product obtained above. Result is the fourth number of this row.

7. ### Mensa Math/Logic Problems...How to Solve?

*

First 4/20: 75
Add first & last numbers in each row, reverse the digits of the sum if below 100, and resulting number occupies the center place in the row. If the sum of the first and the last number in a row is more than 100, then remove the first digit and add it to the remaining two digit number and then reverse the digits of the number so obtained.
*First 5/20: 3
Sum of all the numbers, on each side of the triangle formed by the circles, is equal to 21.
*First 8/20: Answer is the last pentagon in the options given.
When two pentagons are overlapped, those circles within pentagons disappear which overlap each other. The other circles occupy their respective places, and form the figure on top of these two circles.
Note: Second pentagon in the last row does not seem to follow this logic. This pentagon may occupy only three circles then the above logic follows perfectly.
* First 10/20: 67
Add last digits of first and last numbers in each row - say the result is X, then add first digits of the first and last numbers of the row – say the result is Y, then the middle number in the row is XY.
* First 11/20: 120
[This can be solved with the help of simple mathematics. I adopted rather longer procedure. If someone asks for it I will post it.]
* First 12/20: 30
The prime numbers are not included in the series.
* Second 5/20: 8
Product of the first and the third number in each row forms the second number but with the digits of product are reversed.
* First 15/20: 72
Next number is the number obtained by adding ‘1’ to the next prime number.
* Second 4/20: 25
Sum of all the numbers in one sector of the circle is three times the product of all the numbers in another sector just opposite to it.
* Second 8/20: 7
Sum of the two larger numbers is divided by 9, to get the third smaller number.
* First 9/20: Again Rectangle.
Writing the number of lines (contained by each figure) in words, next figure is seen to contain the number of lines equal to the number of letters in the previous word.
* Third 8/20: 20
1st Fig: 14 – 12 = 2 and 11 – 9 = 2, then 17 + 2 + 2 = 21 (the number in centre).
2nd Fig: 6 – 5 = 1 and 6 – 5 = 1, then 18 + 1 + 1 = 20 (the number in centre)
Similarly, 3rd Fig: 6 – 8 = (-) 2 and 5 – 2 = 3, then 19 - 2 + 3 = 20.
* Second 9/20: 62
Multiply the numbers in front of each other, in the outer circles. And deduct smaller product from the larger product to get the number in the center of the circle.
* First 14/20: First figure of the options given.
When two figures overlap each other, then if any circle (except the outermost), line or white dot overlaps another, it disappears, while the other ones occupy their respective places. The figure so formed is the figure on top of these two figures.
* Last 13/20: 31
Add all the numbers on the corners of the triangle, reverse the digits of the sum, number obtained occupies the center space of the triangle.

8. ### Not on your finger tips!

It seems that sometimes we take a puzzle lightly, as we think it has easy answer. Though this perticular puzzle looks like a math problem, but if we do not go for the depth of the question then we donot find the right answer. Important thing to notice is that the time interval between the snaps is from the 'Begining' The right answer is:

OP says that one snap is at the very begining and then second is after 1minute. then third is after double the time from begining i.e. after 2 minute. Then third snap is after 4 minute from the begining. So:

After Start – 1st snap

After 1 min from start– 2nd snap

After 2 min from start – 3
After 4 min from start – 4
After 8 min from start– 5
After 16 min from start– 6
After 32 min from start– 7
After 64 min from start– 8
After 128 min from start– 9
After 256 minfrom start – 10
After 512 minfrom start – 11
After 1024 min from start– 12
After 2048 min from start – 13
After 4096 min from start– 14
After 8192 min from start– 15
After 16384 min from start– 16
After 32768 min – 17
After 65536 min – 18
After 131072 min from start– 19
After 262144 min from start– 20
After 524288 min from start– 21 – Very close but still below 525,600 min total in a year.

There are more simpler ways to solve it.....

9. ### Not on your finger tips!

I came across a very good puzzle. I have reproduced the same as it was worded.!

Let's play a little game. Please snap your fingers. Then, 1 minute later please snap your fingers again. And so it goes on. Each time you double the time between the snaps, so you snap your fingers again after 2 minutes (from the beginning). Then snap your fingers after 4 minutes (from the beginning).

Then snap your fingers after 8 minutes (from the beginning).

How many snaps will you make in a year?

So the game starts ..Now.

19

60*24*365= minutes in a year= 525600

we are counting up by powers of 2, 2^19=524288 (largest power of 2 less than 525600)

Therefore the sum of 2^0-2^18=524287

After 19 clicks you are only 1313 minutes from the next year

We must use a geometric series with the initial value 1 and increasing at a factor of 2. The sum of such a series with N terms is 2^N-1. Like Nins said, the largest power of 2 less than 525600 (the number of minutes in a year) is 524288 (one less than that is the value of 2^N-1). There must have been 19 clicks after the first click (N=19). Thus, you make

20 clicks in one year.

Still needs more to think... you are very near...!

10. ### Not on your finger tips!

I came across a very good puzzle. I have reproduced the same as it was worded.!

Let's play a little game. Please snap your fingers. Then, 1 minute later please snap your fingers again. And so it goes on. Each time you double the time between the snaps, so you snap your fingers again after 2 minutes (from the beginning). Then snap your fingers after 4 minutes (from the beginning).

Then snap your fingers after 8 minutes (from the beginning).

How many snaps will you make in a year?

So the game starts ..Now.

19

60*24*365= minutes in a year= 525600

we are counting up by powers of 2, 2^19=524288 (largest power of 2 less than 525600)

Therefore the sum of 2^0-2^18=524287

After 19 clicks you are only 1313 minutes from the next year

Not correct. Sorry...!

Well, technically, I will have either snapped 0 times, because I'm a rebel who refuses to follow instructions, or way more than 19, because I snap my fingers at other times.

What? This is the Logic/Math section, isn't it?

Fine, then, I'll retreat back into the dark depths of the Word Riddle section...

11. ### Elevator

Actually, unless I have misread the solution, normdeplume's submission results in 4 correct floors (B, 4, 5, 7), not 3 as required by the author. One that might be a better fit would be:

8 -- 4 (x0.5)

7 -- 1 (simple mismatch)

6 -- 3 (x0.5)

5 -- 5 (OK)

4 -- 8 (x2.0)

3 -- 6 (x2.0)

2 -- 2 (OK)

1 -- 7 (simple mismatch)

B -- B (OK)

This solution also has the added benefit of being rather symmetrical (if that appeals to you...) since doubles and halves line up nicely.

Button 1 does not work correctly -> It will lead to 1/2 floor or 2nd floor. Must be 2nd floor.

Therefore button 2 does not work correctly -> It will lead to 1st floor or 4th floor. If it does not lead to 1st floor, no other button can. Must be 1st floor.

If so, how do we get to 4th floor? Can not be from button 2, cannot be from button 8 (no such button). It must be from button 4.

Out of 3 even button 2,4,6, button 4 works correctly, hence button 6 must lead to 3rd floor.

Therefore button 3 does not work correctly -> It will lead to 6th floor.

We come to the solution drawn by normdeplume.

Nice puzzle, thanks

OP clearly indicates ---- "Three of the seven buttons work correctly; two take one to a floor with a number twice that of the floor to which they should; and two take on to a floor with a number half that of the floor to which they should." Where is the confusion?

normdeplume's answer is correct & excelently explained by someids.

12. ### Not on your finger tips!

I came across a very good puzzle. I have reproduced the same as it was worded….!

Let's play a little game. Please snap your fingers. Then, 1 minute later please snap your fingers again. And so it goes on. Each time you double the time between the snaps, so you snap your fingers again after 2 minutes (from the beginning). Then snap your fingers after 4 minutes (from the beginning).
Then snap your fingers after 8 minutes (from the beginning).
How many snaps will you make in a year?

So the game starts …..Now….

13. ### Beggars can't be choosers

Out of 216 butts he made 216/ 6 = 36 cigarettes.

Then he got 36 butts from these cigarettes.
Out of these 36 butts he made 36/6 = 6 cigarettes.
Again he got 6 butts from these cigarettes; from these he made 1 cigarette.
Thus he could make and smoke 36 + 6 + 1 = 43 cigarettes.

14. ### Drawing the square root of 3

It seems to me as easy one, so I doubt following solution is correct.Draw a line at 30degree to AB at A, and make a perpendicular line at B. Say both lines meet at point C.The length BC = sqrt3.

Step 1 would be the problem.

If compass is given, there should be no problem, an equilateral triange has 60 degree angles. OR I am missing some thing?

Make a perpendicular line AD of length 3 at A and make an equilateral trangle ADE (in the first quadrant). Angle BAE will be 30 degree. Make a perpendicular to BA at point B, and extend AE to meet this perpendicular at C. The length BC = sqrt3.

15. ### Drawing the square root of 3

It seems to me as easy one, so I doubt following solution is correct.Draw a line at 30degree to AB at A, and make a perpendicular line at B. Say both lines meet at point C.The length BC = sqrt3.

Step 1 would be the problem.

If compass is given, there should be no problem, an equilateral triange has 60 degree angles. OR I am missing some thing?

16. ### Drawing the square root of 3

It seems to me as easy one, so I doubt following solution is correct.

Draw a line at 30degree to AB at A, and make a perpendicular line at B. Say both lines meet at point C.The length BC = sqrt3.
17. ### A stand up Logician

Sorry...! My last reply was posted in a hurry as I was called to attend an urgent meeting. There were mistakes in the post

As far as I know A truth teller always speak or act truthfully, and a liar always speak lie or act untruthfully, i.e. if he raised his hands first time and asked to raise his hands second timeif he raised his hands first time, then he will not raise his hands second time. Because he has to lie or act untruthfully like a robot.

So first time the logician must ask the crowd to raise their hands if they were truth tellers. Whole crowd will raise their hands.
Then second time logician should ask them to raise their hands who first time raised their hands, then only truth tellers will raise their hands.

Hope this time I am correct.

.

18. ### A stand up Logician

This method may be useful for identifying the number of always truth tellers and number of always liars.

Say first time truth tellers raised their hands are Xo (o stands for out of town), and liars Yi (i stands for in town).
So we know Xo + Yi = A...................................................[1]

And Xi = Yo......................................................................[2]

Then second time hands raised by Xo + Yo = B..............[3]

And Xi + Yo......................................................................[4]

Given Xo + Xi + Yo + Yi = 30............................................[5]
From the above equations equations we may know Yo + Yi; and Xo + Xi.

20. ### Which liar done it?

Arnold is always lying.

So Arnold is the murderer.

21. ### Two Ten Ton Tough Elephants

Let the initial weights of Tenny and Tonny were x ton & y ton respectively.

After 10 years Tenny will be (9^10*11^10x)/10^20 ton.

& after 10 years Tonny will be (11^10*9^10y)/10^20 ton.

Now they each weigh 10 ton (given).

So, x = y = 10^21/(11^10*9^10) ton

But this differs with dark_magician,s answer, as my value comes to approximately 11.17 Ton.

22. ### Two Ten Ton Tough Elephants

Let the initial weights of Tenny and Tonny were x ton & y ton respectively.

After 10 years Tenny will be (9^10*11^10x)/10^20 ton.
& after 10 years Tonny will be (11^10*9^10y)/10^20 ton.
Now they each weigh 10 ton (given).
So, x = y = 10^21/(11^10*9^10) ton
23. ### Two Ten Ton Tough Elephants

Let the initial weights of Tenny and Tonny were x ton & y ton respectively.

After 10 years Tenny will be 11^10x/10^10 ton.

& after 10 years Tonny will be 9^10y/10^10 ton.

Now they each weigh 10 ton (given).

So 11^10x/10^10 = 10 OR x = 10^11/11^10 ton

Similarly y = 10^11/9^10 ton

Let the initial weights of Tenny and Tonny were x ton & y ton respectively.

After 10 years Tenny will be 11^10x/10^10 ton.

& after 10 years Tonny will be 9^10y/10^10 ton.

Now they each weigh 10 ton (given).

So 11^10x/10^10 = 10 OR x = 10^11/11^10 ton

Similarly y = 10^11/9^10 ton

OOOPS....I missed the changes in summer....

I shall again post the solution.

24. ### Two Ten Ton Tough Elephants

Let the initial weights of Tenny and Tonny were x ton & y ton respectively.

After 10 years Tenny will be 11^10x/10^10 ton.
& after 10 years Tonny will be 9^10y/10^10 ton.
Now they each weigh 10 ton (given).
So 11^10x/10^10 = 10 OR x = 10^11/11^10 ton
Similarly y = 10^11/9^10 ton

25. ### Many More Discursive Judges

There are 9 judges. So thre will be 9 opinions about P and 9 opinions about Q, total 18 opinions.

CASE [A]: 6 opinions are that Q is true, 7 opinions are that P is true.

Majority of opinions (12 out of 18) is that P & Q are true. So Jacques is guilty.
CASE : Here 5 opinions are that P is true and 5 opinions are that Q is true. So majority of opinions are that P & Q both are true. Again Jacques is guilty.
CASE [C]: Here also majority of opinions are that P & Q both are true. Again Jacques is guilty.

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