The toothpicks in the picture have been arranged to form a donkey shaped figure. You have to move two matchsticks in a way that the entire shape is rotated / reflected while being intact. Also, you can't change the tail's direction it should be pointing up.
Step1. Lets say y = x
Step2. Multiply through by x xy = x2
Step3. Subtract y2 from each side xy - y2 = x2 - y2
Step4. Factor each side y(x-y) = (x+y)(x-y)
Step5. Divide both sides by (x-y) y = x+y
Step6. Divide both sides by y y/y = x/y + y/y
Step7. And so... 1 = x/y + 1
Step8. Since x=y, x/y = 1 1 = 1 + 1
Step9. And so... 1 = 2
Step 5 is invalid, because we are dividing by (x-y), and since x=y, we are thus dividing by 0. This is an invalid mathematical operation (division by 0), and so by not following basic mathematical rules
In a recreational activity, you are given four different jars of 2 liters, 4 liters, 6 liters and 8 liters respectively with an unlimited water supply. Then you are asked to measure exactly 5 liters of water using them.
This is a famous probability puzzle in which you have to choose the correct answer at random from the four options below.
Can you tell us whats the probability of choosing correct answer in this random manner.
For my anniversary, I decided to surprise my wife. Since she is a voracious reader, I decided to collect a lot of books for her. On the first day of the month, I bought one book, on the second, I bought two and on the third, I bought three. This process went on till the anniversary and on the day, I had 276 books with me to gift her.
Can you calculate, on which day is our anniversary?
1. Fill 5 liters jar ( 5l-jar:5, 3l-jar:0)
2. Transfer to 3 liters pail (5l-jar:2, 3l-jar:3)
3. Empty 3 liters jar ( 5l-jar:2, 3l-jar:0)
4. Transfer 2q from 5 pail to 3 pail (5l-jar:0, 3l-jar:2)
5. Fill 5 liters pail(5l-jar:5, 3l-jar:2)
6. Transfer 1q from 5 pail to 3 pail(5l-jar:4, 3l-jar:3)
Three college toppers are summoned by the inspecting faculty. To identify the best from them, the faculty takes them into a room and places one hat on each of their heads. Now all of them can see the hats on other’s heads but can’t see his own. There are two colored hats – green and red.
Now the faculty announces that he had made sure that the competition is extremely fair to all three of them. He also gives them a hint that at least one of them is wearing a red hat. Now the first one who is able to deduce his own hat color will be awarded the most intelligent student of all award. After a few minutes, one of them raises his hand and is able to deduce the color correctly.
There are two things to keep in mind:
Firstly there is at least one red hat. (There can be two or three as well).
Secondly the competition is fair for everyone.
Thus if there is only one red hat, that person will see two green hats on other heads and will be able to deduce his own color as red. However the other students will see one red and one green hat and can never be sure. In such manner, the competition will prove to be partial for one student.
Suppose if there are two red hats. Then the students who are wearing red hats will see one red and one green hat on others. Now they must have deduced that there can’t be just one red hat. Thus they will know that they are also wearing a red hat. But the one who is wearing a green hat will see two red hats and can never be sure of his own color. In this case as well, the competition will not be fair.
Thus the only possible and fair means is if all of them are wearing a red hat. The one who is able to deduce the situation first, will raise his hand and will tell the correct answer.
Set the first switches on for abt 10min, and then switch on the second switch and then enter the room.
Three cases are possible
1.Bulb is on => second switch is the ans
2.Bulb is off and on touching bulb , you will find bulb to be warm
=>1st switch is the ans.
3.Bulb is off and on touching second bulb , you will find bulb to be normal(not warm)
=>3rd bulb is the ans.
* They have only one torch and the river is too risky to cross without the torch.
* If all people cross simultaneously then torch light wont be sufficient.
* Speed of each person of crossing the river is different.cross time for each person is 1 min, 2 minutes, 7 minutes and 10 minutes.
What is the shortest time needed for all four of them to cross the river ?
The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 minutes. Is that it? No. That would make this question too simple even as a warm up question.
Let's brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let's put all this together.
1 and 2 go cross
2 comes back
7 and 10 go across
1 comes back
1 and 2 go across (done)
Total time = 2 + 2 + 10 + 1 + 2 = 17 minutes
Submit your Email Address to get latest post directly to your inbox.