Hard riddles

funnylove

Three passengers

You are driving a car on one big stormy night. You pass a bus station. There are three people who are waiting for the bus: One old sick lady who is dying, One doctor who saved your life before, and one lady who is someone you have been dreaming to be with. You can only take one passenger, which one will you choose?
Give the car key to the doctor, let the doctor take the old lady to the hospital and stay to wait for the bus with the lady of your dreams!
90.74 %
62 votes

logic

Spring ahead

What is seen in the middle of March and April that can't be seen at the beginning or end of either month?
The letter "R"
90.67 %
45 votes

logicshort

Akbar and Birbal

One day, Emperor Akbar posed a question to Birbal. He asked him what Birbal would choose if he offered either justice or a gold coin. "The gold coin," said Birbal without hesitation. On hearing this, Akbar was taken aback. "You would prefer a gold coin to justice?" he asked, not believing his own ears. "Yes," said Birbal. The other courtiers were amazed by Birbal's display of idiocy. They were full of glee that Birbal had finally managed himself to do what these courtiers had not been able to do for a long time - discredit Birbal in the emperor's eyes! "I would have been disappointed if this was the choice made even by my lowliest of servants," continued the emperor. "But coming from you it's not only disappointing, but shocking and sad. I did not know you were so debased!" How did Birbal justify his answer to the enraged and hurt Emperor?
"One asks for what one does not have, Your Majesty." said Birbal, smiling gently and in quiet tones. "Under Your Majesty´s rule, justice is available to everybody. But I am a spendthrift and always short of money and therefore I said I would choose the gold coin." The answer immensely pleased the emperor and respect for Birbal was once again restored in the emperor's eyes.
90.67 %
45 votes

logic

Engineers and Managers

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.
90.47 %
44 votes

logicmath

In a bank

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 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
90.47 %
44 votes

logic

25 Horses

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 A2 B2 A3 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.
90.47 %
44 votes

cleanlogicshort

One word

Re-arrange the letters, O O U S W T D N E J R to spell just one word.
"Just one word".
90.47 %
44 votes

logic

Old war story

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).
90.26 %
43 votes

logic

Lost in the desert

Jack and Joe were on vacation and driving along a deserted country road from the town of Kaysville to the town of Lynnsville. They came to a multiple fork in the road. The sign post had been knocked down and they were faced with choosing one of five different directions. Since they had left their map at the last gas station and there was no one around to ask, how could Jack and Joe find their way to Lynnsville?
They need to stand the signpost up so that the arm reading Kaysville points in the direction of Kaysville, the town they had just come from. With one arm pointing the correct way, the other arms will also point in the right directions.
90.26 %
43 votes

cleanlogicmath

Mick and John

Mick and John were in a 100 meter race. When Mick crossed the finish line, John was only at the 90 meter mark. Mick suggested they run another race. This time, Mick would start ten meters behind the starting line. All other things being equal, will John win, lose, or will it be a tie in the second race?
John will lose again. In the second race, Mick started ten meters back. By the time John reaches the 90 meter mark, Mick will have caught up him. Therefore, the final ten meters will belong to the faster of the two. Since Mick is faster than John, he will win the final 10 meters and of course the race.
90.26 %
43 votes

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