## Jumps when it walks

What jumps when it walks and sits when it stands?

A kangaroo.

What jumps when it walks and sits when it stands?

A kangaroo.

You're walking down a path and come to two doors. One of the doors leads to a life of prosperity and happiness, and the other door leads to a life of misery and sorrow. You don't know which door is which.
In front of the door is ONE man. You know that this man either always lies, or always tells the truth, but you don't know which. The man knows which door is which.
You are allowed to ask the man ONE yes-or-no question to figure out which door to go through. To make things more difficult, the man is very self-centered, so you are only allowed to ask him a question about what he thinks or knows; your question cannot involve what any other person or object (real or hypothetical) might say.
What question should you ask to ensure you go through the good door?

You should ask: "If I asked you if the good door is on the left, would you say yes?"
Notice that this is subtly different than asking "Is the good door on the left?", in that you are asking him IF he would say yes to that question, not what his answer to the question would be. Thus you are asking a question about a question, and if it ends up being the liar you are talking to, this will cause him to lie about a lie and thus tell the truth. The four possible cases are:
The man is a truth-teller and the good door is on the left. He will say "yes".
The man is a truth-teller and the good door is on the right. He will say "no".
The man is a liar and the good door is on the left. He will say "yes" because if you asked him "Is the good door on the left?", he would lie and say "no", and so when you ask him if he would say "yes", he will lie and say "yes".
The man is a liar and the good door is on the right. Similar to the previous example, he'll say "no".
So regardless of whether the man is a truth-teller or a liar, this question will get a "yes" if the door on the left is the good door, and a "no" if it's not.

A teacher decides to give a pop quiz one day but all of her students refuse to take the quiz thinking that the teacher will call off the quiz.
She can give only one of these students a detention for skipping the quiz.
All of the students know each other's names and if a student knows he/she is getting a detention they take the quiz.
How can she threaten her students with the single detention so they all take the quiz?

She tells them that she will give the student who skips the quiz whose name comes first alphabetically a detention.
This student won't skip because they know they are getting a detention if they do.
The next person alphabetically will then know that they will get a detention so they won't skip either, and so on.

Hockey Stick and ball cost $50. If the Stick cost $49 more than the ball. What is the cost of each ?

Hockey Stick $49.50 & ball $0.50.

If you have two coins which total 35 cents and one of the coins is not a dime, what are the two coins?

A quarter and a dime. One coin is not a dime, but the other one is.

You have twelve balls, identical in every way except that one of them weighs slightly less or more than the balls.
You have a balance scale, and are allowed to do 3 weighings to determine which ball has the different weight, and whether the ball weighs more or less than the other balls.
What process would you use to weigh the balls in order to figure out which ball weighs a different amount, and whether it weighs more or less than the other balls?

Take eight balls, and put four on one side of the scale, and four on the other.
If the scale is balanced, that means the odd ball out is in the other 4 balls.
Let's call these 4 balls O1, O2, O3, and O4.
Take O1, O2, and O3 and put them on one side of the scale, and take 3 balls from the 8 "normal" balls that you originally weighed, and put them on the other side of the scale.
If the O1, O2, and O3 balls are heavier, that means the odd ball out is among these, and is heavier. Weigh O1 and O2 against each other. If one of them is heavier than the other, this is the odd ball out, and it is heavier. Otherwise, O3 is the odd ball out, and it is heavier.
If the O1, O2, and O3 balls are lighter, that means the odd ball out is among these, and is lighter. Weigh O1 and O2 against each other. If one of them is lighter than the other, this is the odd ball out, and it is lighter. Otherwise, O3 is the odd ball out, and it is lighter.
If these two sets of 3 balls weigh the same amount, then O4 is the odd ball out. Weight it against one of the "normal" balls from the first weighing. If O4 is heavier, then it is heavier, if it's lighter, then it's lighter.
If the scale isn't balanced, then the odd ball out is among these 8 balls.
Let's call the four balls on the side of the scale that was heavier H1, H2, H3, and H4 ("H" for "maybe heavier").
Let's call the four balls on the side of the scale that was lighter L1, L2, L3, and L4 ("L" for "maybe lighter").
Let's also call each ball from the 4 in the original weighing that we know aren't the odd balls out "Normal" balls.
So now weigh [H1, H2, L1] against [H3, L2, Normal].
-If the [H1, H2, L1] side is heavier (and thus the [H3, L2, Normal] side is lighter), then this means that either H1 or H2 is the odd ball out and is heavier, or L2 is the odd ball out and is lighter.
-So measure [H1, L2] against 2 of the "Normal" balls.
-If [H1, L2] are heavier, then H1 is the odd ball out, and is heavier.
-If [H1, L2] are lighter, then L2 is the odd ball out, and is lighter.
-If the scale is balanced, then H2 is the odd ball out, and is heavier.
-If the [H1, H2, L1] side is lighter (and thus the [H3, L2, Normal] side is heavier), then this means that either L1 is the odd ball out, and is lighter, or H3 is the odd ball out, and is heavier.
-So measure L1 and H3 against two "normal" balls.
-If the [L1, H3] side is lighter, then L1 is the odd ball out, and is lighter.
-Otherwise, if the [L1, H3] side is heavier, then H3 is the odd ball out, and is heavier.
If the [H1, H2, L1] side and the [H3, L2, Normal] side weigh the same, then we know that either H4 is the odd ball out, and is heavier, or one of L3 or L4 is the odd ball out, and is lighter.
So weight [H4, L3] against two of the "Normal" balls.
If the [H4, L3] side is heavier, then H4 is the odd ball out, and is heavier.
If the [H4, L3] side is lighter, then L3 is the odd ball out, and is lighter.
If the [H4, L3] side weighs the same as the [Normal, Normal] side, then L4 is the odd ball out, and is lighter.

A doctor and a bus driver are both in love with the same woman, an attractive girl named Sarah. The bus driver had to go on a long bustrip that would last a week. Before he left, he gave Sarah seven apples. Why?

An apple a day keeps the doctor away!

Finish the sequence:
7 8 5 5 3 4 4 ?

6 - the number of letters in the month august; (January has 7 letters, February has 8 etc.)

John Heysham Gibbon was most renowned surgeon of 1940-1970. More than 90% of his surgeries he performed are highly successful and still almost all of his patients die.

The surgery was performed way back, by now approx 90% of them have died by old age.

I know a number which when multiplied by multiple of 9 i.e 9 18 27 36 45 ... The output consist of number containing only one digit.
Can you identify the number?

12345679
12345679 × 9 = 111111111 (only 1s)
12345679 × 18 = 222222222 (only 2s)
12345679 × 27 = 333333333 (only 3s)
12345679 × 36 = 444444444 (only 4s)
12345679 × 45 = 555555555 (only 5s)