Solution:

There is only one possible shape i.e. a cone.

Solution:

1:1 (approx)

Explanation:
We know that the probability of having a boy or having a girl is same and thus, half of the couples will stop after having a boy child. Half of the others will have a girl and will have a baby again. Out of those half of the couples, half will have a boy and will stop and half will have a girl again. This will keep on going like this.

Now, if there are X number of couples, there will be X boys.

1/2 have a boy and stop: 0 girls
1/4 have a girl, then a boy: X/4 girls
1/8 have 2 girls, then a boy: 2*X/8 girls
1/16 have 3 girls, then a boy: 3*X/16 girls
1/32 have 4 girls, then a boy: 4*X/32 girls
…
Total: X boys and
1X 2X 3X 4X
– + – + – + — +… = ~X

Therefore, the proportion of boys to girls will be extremely close to 1:1

Solution:

(0! + 0! + 0!)! = 6
(1 + 1 + 1)! = 6
2 + 2 + 2 = 6
3 x 3 - 3 = 6
√4 + √4 + √4 = 6
5 + 5/5 = 6
6 + 6 - 6 = 6
7 - 7/7 = 6
8 - √√(8 + 8) = 6
√(9 x 9) - √9 = 6

Solution:

What must be kept in mind before you start calculating through any method is that the weights can be put on both the pans.

So the weights - 1, 3, 7, 12, 43, 76, 102

The oldest man that can be weighed using these weights is a man aged 122 years

Solution:

fIvE thOUsAnd(5000)

Solution:

They can be expressed as the sum of the cubes of two different sets of numbers.

10^3 + 9^3 = 1729
and
12^3 + 1^3 = 1729

Solution:

The check was for dollars 31.63.

The bank teller gave her dollars 63.31
She spent .05, and then she had dollars 63.26, which is twice the check.

Let x be the dollars of the check, and y be the cent.

The check was for 100x + y cent
He was given 100y + x cent

Also
100y + x - 5 = 2(100x + y)
Expanding this out and rearranging, we find:
98y = 199x + 5

or 199x ≡ -5 (mod 98)
or 98*2*x + 3x ≡ -5 (mod 98)
3x ≡ -5 ≡ 93 (mod 98)
this quickly leads to x = 31

Solution:

11 / .11

Solution:

12

In general, with n+1 people, the number of handshakes is the sum of the first n consecutive numbers: 1+2+3+ ... + n.
Since this sum is n(n+1)/2, we need to solve the equation n(n+1)/2 = 66.
This is the quadratic equation n2+ n -132 = 0. Solving for n, we obtain 11 as the answer and deduce that there were 12 people at the party.

Since 66 is a relatively small number, you can also solve this problem with a hand calculator. Add 1 + 2 = + 3 = +... etc. until the total is 66. The last number that you entered (11) is n.

Solution:

The only lockers that remain open are perfect squares (1, 4, 9, 16, etc) because they are the only numbers divisible by an odd number of whole numbers; every factor other than the number's square root is paired up with another. Thus, these lockers will be 'changed' an odd number of times, which means they will be left open. All the other numbers are divisible by an even number of factors and will consequently end up closed.

So the number of open lockers is the number of perfect squares less than or equal to one thousand. These numbers are one squared, two squared, three squared, four squared, and so on, up to thirty one squared. (Thirty two squared is greater than one thousand, and therefore out of range.) So the answer is thirty one.