[Jolts of volts]

Press buttons: T R F

If
Truth-teller: N J X
Liar: N N J
Random: N N N
If truth teller he need not press the button further. If liar, pressing third button F will give Jolt, but in case he is Random, he will get No jolt.

[Ice slice]

Could it be cut in circular shape, or is it restricted to straight slice?

[Whether to switch]

Here the probabilities on both occaisons seems to be equal as 50-50.

But you have already used 50 % chances to select right envelop having bigger amount.
Now if you take further chance Probility to win is reduced to half i.e. 25%. So switching is not advisable.

[Oh no I lost my baby!]

I am not well acquainted with the names in OP.

So before advancing further towards solution may any body tell whether the following is correct or not.
The Surnames mentioned in the OP: Wilson, Smith, Thompson, Larson, Miller, and George.
First name of Fathers: Chuck, Tom, Jacobs, Bob, Jim, Gary.
First names of Mothers: Betty, Judy, Susan, Mary, Trudy, carol.
Baby's Names: Amy, Jane, Ann, jason, Andrew, Rechard, Jennifer.

George is first name or Gary is surname.....?

I am little in confusion....

[How Far]

bhramarraj, we differ by an interpretation of OP, which is ambiguous.

If "2 minutes" means "every 2 minutes" or "a particular 2-minute interval" the answer could be what jordge and I posted or your answer or anything in between. That is, in the second case the first pole is passed (sometime) in the first minute and the second pole is passed (sometime) during the second minute. That would permit your answer if pole 1 happened at the beginning of minute 1, and the second pole happened at the end of minute 2.

Or am I missing something?

So, is John Wilson still around to clarify?

You are right bonanova.

Interestingly there would be muliple answers with the conditions given in OP, as can be seen by the formula:

30d/x = 30d - 1 where x = no. of poles or speed of the car in mph, and d is the distance between the poles in miles.

We can select value of x as from 1, 2, 3, 4,.........etc. and can have different values for d in feet.

[How Far]

Car's speed is say x miles per hour.

So it passes x lampposts in 2 minutes.
Say the distance between two lamposts is d miles.
So the distance between 30 lampposts is d(x-1).
So the car passes d(x-1)miles in 2 minutes.
Therefore, the car's speed is 30d(x-1) miles per hour
Therefore x = 30d(x-1) which can be written as x = 30d/(30d-1)
Value of x in the above eq. should be a whole number,
So the value acceptable for d is 2/30 miles.
So d = 2/30 miles = 2/30 * 5280 Feet = 352 feet

I hope this is the reasonable answer.

[Looking at the sun during an eclispe]

I agree to Bonanova's first post. Since the third person is equidistant from the other two and these three persons are colliniar, but they are assumed to be in the same direction to the sun, then the angle, third person will be making with the sun, will be smaller than 45 degree and greater than 30degree, if the three persons are on the same plane and collinear .

For second person say the horizontal distance from the sun is x, and vertical distance is say y. Since cot 45 degree = x/y = 1, so x = y.
Say the distance between second and third person is z, and since third person is in the middle of first and second person, so distance between first & second person will be 2z.
Then cot 30 degree 1.732 = (x + 2z)/y and since x = y, so
1 + 2z/y = 1.732 or z/y = 0.366........................[1]
Say the angle made by the third person is D degree, then cot D degree = (x+z)/y, and since x = y, so cot D = 1 + z/y........[2]
From [1] & [2], cot D = 1.366.
So D = 36.206 degree.

But the puzzles BMAD is presenting are not so easy.
The three persons might be collinear and on the same plane but their directions might be diffrent from the sun.

In that case vertical distance will remain same for each person say 'y', but the relation between the horizontal distances of the three persons from the sun is not so easy for me to calculate. I may come to some answer say tomorrow...

[The always liar]

Let us indicate always liar by 'L' and always truth teller by 'T'.

There are four sets of conversation:

Mike- Tell me something interesting.
Adam-one of us is always a liar.
Mike-Really?
Eric- Yes. They always lie, but one of us is also always a truth teller.
Mike- Oh, cool, well then …that would mean that Sam is the ‘always liar.’
Sam- Yes, yes I am.

Above tells us:

[1] Adam can't be 'L' if one of them is 'L'. He cold be 'T'.

[2] Eric can't be 'L', if one of them is 'T', but he could be 'T'.

[3] Mike's interpretation from the conversation is not correct, as the conversation which took so far between them could not ascertain that Sam is 'L'. So Mike could be 'L'.

[4] Sam can't be 'L', he is not 'T' also.

Second set of conversation:

Mike-…What? Wait a minute. I thought you were always a liar.

Sam- Yes I am.
Eric-He is.
Mike-He is?
Eric-yes
Sam-yes
Mike-yes?
Adam-no.
Mike-no.
Sam-no.

Above tells us:

[5] Mike could be 'L'.

[6] Sam can't be 'L' or 'T'.

[7] Eric can't be 'T'. Though here he is lying but if one of them was 'T', then he can't be 'L' as he could be 'T' in First set of conversation.

[8] Adam is right here, so he could be 'T'. So he can't be 'L'.

Third set of conversation:

Mike-wait….then does that make Eric the liar?
Eric- Yes
Sam- No
Adam-yes
Mike- What??

Above tells us:

[9] Mike could not be put into any catagoury.

[10] Eric could not be 'T'.

[11] Sam could be speaking truth.

[12] Adam could be telling truth here.

So if one of them is 'L', then it has to be Mike only. And then Adam will be 'T'.

[Funny Numbers]

[1] Say the first number is ‘abcde’

Then putting 1 before and after we get 1abcde * 3 = abcde1
Here 3 * e = x1, so e = 7
Then we have 1abcd7 * 3 = abcd71
Here 3 * d + carryover 2 = x7, therefore d = 5
Then we have 1abc57 * 3 = abc571
Now 3 * c + carryover 1 = x5, therefore c = 8
Then we have 1ab857 * 3 = ab8571
Now we have 3 * b + carryover 2 = x8, therefore b = 2
Then 1a2857 * 3 = a28571
Here 3 * a = x2, so a = 4,
Then finally we have 142857 * 3 = 428571
[2] Second smallest odd number which is perfect cube and perfect square.
Say it is a^3 which is equal to b^2. Here a & b both should be odd numbers to get required odd number.
Then a = (b/a)^2 = c^2 (say)
Then first smallest odd number which is greater than 1, and is a square of an odd number is 9, and second smallest odd number which is a square of an odd number is 25,
So 25^3 = b^2 = 125^2 = 15625

Third funny number will be worked out tomorrow...

[Execution Table]

It is clear that in first round all odd numbers will be executed.

Then on second round: dividing remaining even numbers by 2, the odd numbers we get will be executed.

The same will be repeated third time and so on.

In case of 13 persons, 6 persons will remain after first round, and three persons after second round and on third round only one will survive.

So 2*2*2 = 8 will be the survivor’s seat.

In case of 1 to 3 persons safe seat is no. 2.

In case of 4 to 7 persons, safe seat is no. 4.

In case of 8 to 15 persons, safe seat is no. 8.

In case of 16 to 31 persons, safe seat is no. 16.

In case of 32 to 63 persons, safe seat is no. 32.

The safe seat pattern is 2,4,8,16,32,64,128 and so on, corresponding to max. no. of persons as 3,7,15,31,63,127,255 and so on.

i.e. safe seat is:

2^1 for 2^1 to 2^0 + 2^1 Persons.

2^2 for 2^2 to 2^0 + 2^1 + 2^2 Persons,

2^3 for 2^3 to 2^0 + 2^1 + 2^2 + 2^3 persons.

2^4 for 2^4 to 2^0 + 2^1 +2^2 +2^3 +2^4 persons.

And so on………..

If n are the number of persons, and

n = 2^a to 2^0 + 2^1 + 2^2 + 2^3 + ..... + 2^a

Then 2^a is the safe seat for such a group of persons.

[Execution Table]

It is clear that in first round all odd numbers will be executed.

Then on second round: dividing remaining even numbers by 2, the odd numbers we get will be executed.
The same will be repeated third time and so on.
In case of 13 persons, 6 persons will remain after first round, and three persons after second round and on third round only one will survive.
So 2*2*2 = 8 will be the survivor’s seat.
In case of 1 to 3 persons safe seat is no. 2.
In case of 4 to 7 persons, safe seat is no. 4.
In case of 8 to 15 persons, safe seat is no. 8.
In case of 16 to 31 persons, safe seat is no. 16.
In case of 32 to 63 persons, safe seat is no. 32.
The safe seat pattern is 2,4,8,16,32,64,128 and so on, corresponding to max. no. of persons as 3,7,15,31,63,127,255 and so on.
i.e. safe seat is:
2^1 for 2^1 to 2^0 + 2^1 Persons.
2^2 for 2^2 to 2^0 + 2^1 + 2^2 Persons,
2^3 for 2^3 to 2^0 + 2^1 + 2^2 + 2^3 persons.
2^4 for 2^4 to 2^0 + 2^1 +2^2 +2^3 +2^4 persons.
And so on………..

[Date of the Dinner]

If Brenda was wrong, then the date was not greater than 13.

If Edward was wrong, then date was greater than 17.
Both are contradictory.
So one of them must be correct & other must be wrong.
Let us assume Brenda was right, and every one else was wrong. Then:
Brenda was right, so date was greater than 13.
Alan was wrong, so date was not an odd number; So possible dates are 14,16,18,20,22,24,26,28,30.
Carly was wrong so the date was a perfect square; only date 16 is the perfact square in the above list of possible dates.
Dara was wrong, so date was not a perfect cube. The date 16 is not a perfect cube.
Edward was wrong, so the date was greater than 17. But 16 is less than 17, so 16 is not the date.
Conclusion of above is that Brenda is not right; Therefore Edward must be right & every one else was wrong. Then:
Edward was right, so the date was less than 17.
Alan was wrong, so the date was not an odd nuber. So the possible dates are 16,14,12,10,8,6,4,2.
Brenda was wrong so the date was not greater than 13; so remaining possible dates are 12,10,8,6,4,2.
Carly was wrong so the date was a perfect square; only 4 is the date which is a perfect square.
Dara was wrong so the date was not a perfect cube; the date 4 is not a perfect cube.

Answer: The dinner date is 4th December.

[Two wrongs might be OK?]

The one error is title itself, since there is no error at all.

Another error is to look for the error even after reading the story of the boy.

[Is the name of this island Oahu? That depends on who is asking.]

Let us explore following possibilities;

(i) Sam – Type B, Janet – Type A
(ii) Sam – Type B, Janet – Type B
(iii) Sam – Type A, Janet – Type B
(iv) Sam – Type A, Janet – Type A

Say Sam is type B and Janet is type B, then the answer to the Q will be yes. but type B can’t ask such question. So Sam isn’t type B if Janet is type B.
Similarly another possibility of Janet and Sam both being Type A, OR Sam being type A and Janet being Type B isn’t possible, as in both cases, the answer to the Q would be NO, and such question can’t be asked by Type A.
Therefore only possibility Is that Sam is Type B and Janet is Type A.
Answer: Janet is type A.

[Who is Sylvia?]

Yesterday I was reading Bonanova's head explosion theory....! I think it could be solved by this method.

So a question is to be constructed such that Sylvia whether true or liar, may not be able to answer, while the other who isn't Sylvia is always able to answer.
I hope the question (directed to whomsoever) "Aren't you Sylvia?" is enough to find who is Sylvia.

[At Book Store]

Two men, David and Clifton, and their wives, Kim and Allison, go out shopping for books. Each person paid for each book a number of dollars equal to their number of books. David bought 1 more book than Kim, while Allison bought only 1 book. If each couple spent the same two-digit sum, who is Kim's husband?

Say Kim bought x books for x^2 dollars.
David bought x+1 books for (x+1)^2 dollars.
Allison bought only 1 book for 1 dollar.
Clifton bought y books for y^2 dollars.
If David and Kim were the couple, then they spent x^2 + (x + 1)^2 = two digit number = say z…....[1]
So x < 7 and x > 1
Then remaining couple Allison and Clifton would spend 1 + y^2 = z…………………………..…......[2]
So y can’t be greater than 9, i.e. z is not greater than 82.
Then from [1], x < 6 or z cannot be greater than 61.
Then from [2], y can’t be greater than 7, i.e. z can’t be greater than 50.
Then from [1], x < 5 or Z is not greater than 41.
And so on…
But we do not find suitable values of X and Y which satisfy both [1] & [2].

Note this can be prooved by equation y = SQRT 2x(x + 1), where we get only suitable values of x as 1 to get a whole number y. But from [1], X can't be lesser than 2, to get double digit number Z.
Therefore David and Kim are not the couple, so David and Allison are the husband & wife.
And Clifton is husband of Kim.
Then we have the two equations
(x + 1)^2 + 1 = z………………………………………………………………………………………..[3]
X^2 + y^2 = z…………………………………………………………………………………………….[4]
From [3] & [4], y^2 = 2x + 2 OR y = √(2x + 2)..…………………………………………………….[3]
[3] can be satisfied only when x= 7, or 1 to get y as a whole number.
But from [1], to get Z a double digit number X must be greater than 1.
Therefore X = 7, and y = 4.
Answer: David bought 8 books, his wife Allison bought 1 book. The spent total (64 + 1) = 65 dollars.
Clifton bought 4 books and his wife Kim bought 7 books, and they also spent 16 + 49 =65 dollars.

[1 wire]

As shown,one more piece of wire in a form of number 1

is to be bend and curved to produce the projections

(silhouette or shadow) of the 3 numerals on different

viewing angles. Find a way to do it.

3.JPG

At the top bend the wire to make a circle, and make the lower portion straight at 90 degree to it, bending at the bottom to give it J shape.From Top its projection will be Zero; but when prjected slightly at an angle, it will take a shape of Nine, at another angle It will project like Eight.

[Ships passing in the night]

15 ships.

It takes 7 days and 7 nights for a ship to reach Newyork to London and from London to Newyork.
Each day at Noon I ship leaves Newyork. So today when a ship X leaves London, 7 ships are already in the sea including a ship touching London.
The ship X will take 7 days and 7 nights to reach Newyork, so by the time X touches Newyork

7 more ships will sail into the sea, and eighth ship will be leaving Newyork.

Therefore, X will meet 7 + 8 ships during its course of journey.

[Road trip]

1) Doris ride for 15miles (say from A), and leaves the bicycle (say at B), starts on foot from there.and walks on foot for 10 miles,

By this time (3 hrs), Forrest reaches bicycle (at B) and ride
15 miles for 1hr (say up to C). Doris also walks further 5miles and both meet at C.
2) Above is repeated from this point, Forrest leaves bicycle, starts on foot; Doris rides for 15 miles, leaves the bicycle (say up to D), which is finally 15 miles away from destination, and starts on foot. Forrest reaches bicycle rides and reaches the destination, simultaneously Doris reaches the destination on foot

So Finally the bicycle is left 15 miles behind the destination.

[Where there's a will, there's a ... puzzle]

According to the OP wife gets 1/2 of what the son gets. and daughter gets half of what the wife gets.So the property will be divided in the ratio 2:4:1; i.e. if the property is 'a', wife will get 2a/7, son will get 4a/7, and daughter will get a/7.

[Of dogs and men]

Since speeds of the dog and soldiers are not given, it may be assumed that the dog has completed one round of the square formed by the soldiers, and followed the return path same as it took for starting its journey i.e. it has completed 400ft + 50ft + 50ft = 500ft.

[How much drink?]

Volume of Cylinder= pi x r^2 x h = 12 ounce

So pi x r^2 = 2 (Note: h = 6")
Considering half cone of height 4" and cylinder of height 2", the volume of liquid comes to

1/2[1/3 x pi x 4r^2 x 4"] + pi x r^2 x 2"
Putting the value of pi x r^2 as 2
Volume of liquid = 28/3
So 28/3 ounce is the answer.

[Who am I?]

2 possiblities giving same answer

all lying - 7-bill 8-susan

1 true rest lying 7- bill, 8-susan

All can't lie, because Statement 1 becomes true as soon as seven are considered lie. It is really amazing isn't it...?

But there are 8 people. If all of them lie, then statement 1 is false.

Please think again. There are 8 statements given in OP. If first statement is also considered to be a lie then there will be 8 false statements, but we can not make forced opinion. We have to go by statements available with us.

More clearly to say is that even if first person wanted to lie, he could not lie by giving this statement.

[Who am I?]

2 possiblities giving same answer

all lying - 7-bill 8-susan

1 true rest lying 7- bill, 8-susan

All can't lie, because Statement 1 becomes true as soon as seven are considered lie. It is really amazing isn't it...?

[Dog freedom?]

If I go for the wordings of OP, then the question was not asked about the area, but the 'yard' which is unit of length. If it is true then dog has excess to square root of (80^2 - T^2)/3 yards, where 'T' is the height of the tie in feet.