They have property such that when you light the fire from one end , it will take exactly 60 seconds to get completely burn.
However they do not burn at consistent speed (i.e it might be possible that 40 percent burn in 55 seconds and next 60 percent can burn in 10 seconds).
There can be two possibilities for the given situation.
One is if I run on the perimeter. Then, the lion will eventually catch me as the lion can follow my radius and then his trajectory will be half of the circle and not a spiral. Therefore, the lion will subsequently catch me in a finite amount of time.
Secondly if I don’t run on the perimeter, then clearly, I have an infinite amount of time before the lion catches me.
* 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)
You can see the figure or draw one of your own. The scenario is as shown. There are three houses represented with the triangle over the square. There are three utilities: W, G and E representing water, gas and electricity respectively.
Can you draw a line and get each utility into every house (9) total lines without ever crossing any line?
A guy claims to do the following thing. He puts a coin in a glass bottle. Then, he shuts the mouth of the bottle with the help of a cork. Now he manages to remove the coin out from the bottle without taking out the cork or breaking the glass bottle.
There are three cars in a racing track. The track is made forming a perfect circle and is quite wide so that at one time, multiple cars can pass through it. The car which is leading is driving at 55 MPH and the slowest car is driving at 45 MPH. The car that is in middle of these two is driving in between the two speeds. For the time being you can say that the distance between the fastest car and the middle car is x miles and it is same between the middle car and the slowest car. Also, x is not equal to 0 or 1.
The car keeps running till the leading car catches up with the slowest car and then every car stops. Given the case, do you think that at any point, the distance between any two pairs will again become x miles?
In the question, the distance of x miles is given at a particular moment. The middle car is running at that time at 50 MPH. The distance will keep increasing with time as three are running at different speeds. It will increase by x miles every hour till a certain point of time and then it will start decreasing till the fastest car meets the slowest again.
As per question, all the cars stop then. Thus, the distance will never be x miles again.
A 12-year footballer Lavy was offered a contract with a club name "Flyball" that if he will practice 1 hour extra after school he will be offered 11$ after every week.
Lavy, however, suggested an alternative paid method in which
He will be paid just a penny on his first day.
Two pence will be paid on the second day,
Four pence will be paid on the third day.
And so on till 11th day.
You have ten boxes and an electronic weighing machine. In those ten boxes, you have chocolates. Each chocolate weigh 20 grams. But in one box the chocolates are defective and each weigh 19 grams exactly.
Now you can weigh in the electronic weighing machine but you can use that machine just once. How will you find out which box has the defected chocolates.
If you are thinking to hold one chocolate from each box in hand and then balancing weight in bare hands, you are thinking all wrong.
Let us begin by labelling boxes as 1, 2, 3 and so on till 10.
Now pick one chocolate from box 1, two chocolates from box 2, three from box 3 and so on. In total, you will have 55 chocolates now. (1 + 2 + 3 + ..... + 10)
The ideal weight of the chocolates should be 55 * 20 = 1100. However, somewhere in there are the defected chocolate/s.
You can judge that clearly by noting down the result of 1100 - total weight of chocolates. If the weight is less than 1 gram, the defected box is box 1, if the weight is less than 2 grams, the defected box is box 2 and so on.
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