Four friends need to cross a dangerous bridge at night. Unfortunately, they have only one torch and the bridge is too dangerous to cross without one. The bridge is only strong enough to support two people at a time. Not all people take the same time to cross the bridge. Times for each person: 1 min, 2 mins, 7 mins and 10 mins. What is the shortest time needed for all four of them to cross the bridge?
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 mins. 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)
We all know that New Year occurs after a week from Christmas and thus it falls on the same day as of Christmas. But this will not happen in 2050. In 2050, Christmas will appear on Sunday while New Year will appear on Saturday.
Read the question carefully again. New Year do falls after Christmas but that happens if two different years. The question is put up against the year 2050 and thus there will be 51 weeks and 2 days in between them as New Year will appear on 1 January 2050 and Christmas will happen on 25 December 2050.
At first, there were 2 apples on the tree. After the wind blew, one apple fell on the ground. So there where no apples on the tree and there were no apples on the ground.
*apples => more than one apple
Another solution(non tricky one)
The wind blew so hard that the apples fell of the tree and blew along the ground.
An old man in the village feels that his end is near. He calls his two sons to discuss about the land he owns and other properties. He tells them to have a race on their horses till the city border. The one with the slower horse will be rewarded with the entire property.
Both of them keep wandering here or there without any result as no one is filling to reach the border. Then they visit the wise man of the village and seek his advice. The wise men tells them something listening to which they jump on the horses and race as fast as they can till the city border.
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
Suppose there is a Christmas tree and four angels are sitting on it amidst the other ornaments. Two of them have black halos and two of them have white halos. No body among them can see above their head. Angel A is sitting on the top branch and can see angels B and C sitting below him. B can see C who is sitting in a branch lower than his. Angel D is at the base of the tree and can't be seen due to the branches in between. Also, he can't see anybody as well.
If they are asked to guess the color of their own halo (they dont know that), who do you think will be able to deduce and speak up first with a right answer?
Now there can be two solutions to the given situation because there can be two situations:
Suppose if B and C have same colors, A will know that his color is the other one and he will be able to speak up the first.
Now if B and C do not have same colors, A will stay silent. This will tell B that his and C's colors are different. Thus he will speak up first.
So either A or B will speak up first depending on the situation.
We have 4 blue, 4 red, 4 green and 4 yellow hats. All the hats must be labelled with an arithmetic sign – ‘+’, ‘-‘, ‘x’ or ‘/’ in a manner that one sign is used on a particular color only once. Now these hats must be arranged in a 4x4 grid in a manner that no two rows or columns have a repeating color or sign.
To give you a kick start, we have arranged 4 hats in the grid. Can you place the rest of them ?
You have to place all the digits from 1 to 9 without repeating any number in a tic tac toe board. The condition is that the numbers should add up to 15 whether you add the numbers in each row, column or diagonally.