What word does this rebus represents?
Metaphor (Meta 4).
In a far away land, it was known that if you drank poison, the only way to save yourself is to drink a stronger poison, which neutralizes the weaker poison. The king that ruled the land wanted to make sure that he possessed the strongest poison in the kingdom, in order to ensure his survival, in any situation. So the king called the kingdom's pharmacist and the kingdom's treasurer, he gave each a week to make the strongest poison. Then, each would drink the other one's poison, then his own, and the one that will survive, will be the one that had the stronger poison. The pharmacist went straight to work, but the treasurer knew he had no chance, for the pharmacist was much more experienced in this field, so instead, he made up a plan to survive and make sure the pharmacist dies. On the last day the pharmacist suddenly realized that the treasurer would know he had no chance, so he must have a plan. After a little thought, the pharmacist realized what the treasurer's plan must be, and he concocted a counter plan, to make sure he survives and the treasurer dies. When the time came, the king summoned both of them. They drank the poisons as planned, and the treasurer died, the pharmacist survived, and the king didn't get what he wanted. What exactly happened there?
The treasurer's plan was to drink a weak poison prior to the meeting with the king, and then he would drink the pharmacist's strong poison, which would neutralize the weak poison. As his own poison he would bring water, which will have no effect on him, but the pharmacist who would drink the water, and then his poison would surely die. When the pharmacist figured out this plan, he decided to bring water as well. So the treasurer who drank poison earlier, drank the pharmacist's water, then his own water, and died of the poison he drank before. The pharmacist would drink only water, so nothing will happen to him. And because both of them brought the king water, he didn't get a strong poison like he wanted.logicshort
How far can you run into the woods?
Half way. If you go in any further, you’d be running OUT of the woods.cleanfunnyshort
What you call a witch at a beach?
Three people check into a hotel. They pay $30 to the manager and go to their room. The manager finds out that the room rate is $25 and gives $5 to the bellboy to return. On the way to the room, the bellboy reasons that $5 would be difficult to share among three people, so he pockets $2 and gives $1 to each person. Now, each person paid $10 and got back $1. So they paid $9 each, totalling $27. The bellboy has $2, totalling $29. )
Where is the remaining dollar?
Each person paid $9, totalling $27. The manager has $25 and the bellboy has $2. The bellboy's $2 should be added to the manager's $25 or substracted from the tenant's $27, not added to the tenants' $27.logic
You have just purchased a small company called Company X. Company X has N employees, and everyone is either an engineer or a manager. You know for sure that there are more engineers than managers at the company.
Everyone at Company X knows everyone else's position, and you are able to ask any employee about the position of any other employee. For example, you could approach employee A and ask "Is employee B an engineer or a manager?" You can only direct your question to one employee at a time, and can only ask about one other employee at a time. You're allowed to ask the same employee multiple questions if you want.
Your goal is to find at least one engineer to solve a huge problem that has just hit the company's factory. The problem is so urgent that you only have time to ask N-1 total questions.
The major problem with questioning the employees, however, is that while the engineers will always tell you the truth about other employees' roles, the managers may lie to you if they like. You can assume that the managers will do their best to confuse you.
How can you find at least one engineer by asking at most N-1 questions?
You can find at least one engineer using the following process:
Put all of the employees in a conference room. If there happen to be an even number of employees, pick one at random and send him home for the day so that we start with an odd number of employees. Note that there will still be more engineers than managers after we send this employee home.
Then call them out one at a time in any order. You will be forming them into a line as follows:
If there is nobody currently in the line, put the employee you just called out in the line.
Otherwise, if there is anybody in the line, then we do the following. Let's call the employee currently at the front of the line Employee_Front, and call the employee who we just called out of the conference room Employee_Next.
So ask Employee_Front if Employee_Next is a manager or an engineer.
If Employee_Front says "manager", then send both Employee_Front and Employee_Next home for the day.
However, if Employee_Front says "engineer", then put Employee_Next at the front of the line.
Keep doing this until you've called everyone out of the conference room. Notice that at this point, you'll have asked N-1 or less questions (you asked at most one question each time you called an employee out except for the first employee, when you didn't ask a question, so that's at most N-1 questions).
When you're done calling everyone out of the conference room, the person at the front of the line is an engineer. So you've found your engineer!
But the real question: how does this work?
We can prove this works by showing a few things.
First, let's show that if there are any engineers in the line, then they must be in front of any managers.
We'll show this with a proof by contradiction. Assume that there is a manager in front of an engineer somewhere in the line. Then it must have been the case that at some point, that engineer was Employee_Front and that manager was Employee_Next. But then Employee_Front would have said "manager" (since he is an engineer and always tells the truth), and we would have sent them both home. This contradicts their being in the line at all, and thus we know that there can never be a manager in front of an engineer in the line.
So now we know that after the process is done, if there are any engineers in the line, then they will be at the front of the line. That means that all we have to prove now is that there will be at least one engineer in the line at the end of the process, and we'll know that there will be an engineer at the front.
So let's show that there will be at least one engineer in the line. To see why, consider what happens when we ask Employee_Front about Employee_Next, and Employee_Front says "manager". We know for sure that in this case, Employee_Front and Employee_Next are not both engineers, because if this were the case, then Employee_Front would have definitely says "engineer". Put another way, at least one of Employee_Front and Employee_Next is a manager. So by sending them both home, we know we are sending home at least one manager, and thus, we are keeping the balance in the remaining employees that there are more engineers than managers.
Thus, once the process is over, there will be more engineers than managers in the line (this is also sufficient to show that there will be at least one person in the line once the process is over). And so, there must be at least one engineer in the line.
Put altogether, we proved that at the end of the process, there will be at least one engineer in the line and that any engineers in the line must be in front of any managers, and so we know that the person at the front of the line will be an engineer.logicmath
A women walks into a bank to cash out her check.
By mistake the bank teller gives her rupee amount in change, and her paise amount in rupees.
On the way home she spends 5 paise, and then suddenly she notices that she has twice the amount of her check.
How much was her check amount ?
The check was for Rupees 31.63.
The bank teller gave her Rupees 63.31
She spent .05, and then she had Rupees 63.26, which is twice the check.
Let x be the rupees of the check, and y be the paise.
The check was for 100x + y paise
He was given 100y + x paise
100y + x - 5 = 2(100x + y)
Expanding this out and rearranging, we find:
98y = 199x + 5
or 199x â‰¡ -5 (mod 98)
or 98*2*x + 3x â‰¡ -5 (mod 98)
3x â‰¡ -5 â‰¡ 93 (mod 98)
this quickly leads to x = 31logic
You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?
You need to conduct 7 races.
First, separate the horses into 5 groups of 5 horses each, and race the horses in each of these groups. Let's call these groups A, B, C, D and E, and within each group let's label them in the order they finished. So for example, in group A, A1 finished 1st, A2 finished 2nd, A3 finished 3rd, and so on.
We can rule out the bottom two finishers in each race (A4 and A5, B4 and B5, C4 and C5, D4 and D5, and E4 and E5), since we know of at least 3 horses that are faster than them (specifically, the horses that beat them in their respective races).
This table shows our remaining horses:
A1 B1 C1 D1 E1
A2 B2 C2 D2 E2
A3 B3 C3 D3 E3
For our 6th race, let's race the top finishers in each group: A1, B1, C1, D1 and E1. Let's assume that the order of finishers is: A1, B1, C1, D1, E1 (so A1 finished first, E1 finished last).
We now know that horse D1 cannot be in the top 3, because it is slower than C1, B1 and A1 (it lost to them in the 6th race). Thus, D2 and D3 can also not be in the to 3 (since they are slower than D1).
Similarly, E1, E2 and E3 cannot be in the top 3 because they are all slower than D1 (which we already know isn't in the top 3).
Let's look at our updated table, having removed these horses that can't be in the top 3:
A1 B1 C1
A2 B2 C2
A3 B3 C3
We can actually rule out a few more horses. C2 and C3 cannot be in the top 3 because they are both slower than C1 (and thus are also slower than B1 and A1). And B3 also can't be in the top 3 because it is slower than B2 and B1 (and thus is also slower than A1). So let's further update our table:
A1 B1 C1
We actually already know that A1 is our fastest horse (since it directly or indirectly beat all the remaining horses). So now we just need to find the other two fastest horses out of A2, A3, B1, B2 and C1. So for our 7th race, we simply race these 5 horses, and the top two finishers, plus A1, are our 3 fastest horses.logic
A man is found dead in the desert. He is wearing only his underwear. Half of a straw is found nearby.
How did this man die?
The man was flying in a hot-air balloon with another man over the desert. The balloon started to go down because of excess weight. Both men would die if they ended up stranded in the desert, so they stripped down to their underwear and threw their clothes off the balloon to try to reduce the weight. Unfortunately, that didn't work well enough. So they drew straws to decide who would jump. The dead man pulled the short straw and jumped out of the balloon.logic
Frank and some of the boys were exchanging old war stories. James offered one about how his grandfather (Captain Smith) led a battalion against a German division during World War I. Through brilliant maneuvers he defeated them and captured valuable territory. Within a few months after the battle he was presented with a sword bearing the inscription: "To Captain Smith for Bravery, Daring and Leadership, World War One, from the Men of Battalion 8." Frank looked at James and said, "You really don't expect anyone to believe that yarn, do you?" 7
What is wrong with the story?
It wasn't valled World War One until much later. It was called the Great War at first, because they did not know during that war and immediately afterward that there would be a second World War (WW II).