A new company has launched a range of juicy candies. There are two flavors of which these juice candies will be available for retail - Peach and Grape. The toffees are packed in three different boxes with one containing peach candies, one containing grape candies and one containing a mixture of both. It is informed that the packaging person has accidentally mislabeled all the three boxes.
If you are asked to label them correctly, how many candies will you have to pick and from which jar in order to label every box correctly?
There are many ways in which you can think a solution for the problem. Your approach should be to minimize the efforts. What if you took out one candy from the box labeled Peach and Grape?
What you have to consider is the fact that every box is labelled incorrectly. Thus if you pick a candy from the box labelled with Peach and Grape, it is evident that the box will not have a mixture of both (as it is labelled wrong). Thus you will get to know which candies are present in that box.
Suppose the candy you take out turns out to be peach flavored, then the box labeled as Grapes is the one which contains a mixture of both peach and grapes. This is because it is labelled incorrectly and since we have already figured out the box with Peach candies, we can be confident of the fact. And then the jar labelled as peach will be the one containing grape flavored candies.
If the candy you take out from the mixed labeled box is Grape flavored, you can use the same logic to find out the rest.
Consider picking a candy from the other two boxes instead of what is told in the solution and you will find out that you will have to take out a candy from each of the boxes before being sure.
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.
It is best if the person sits on the back of the train. After the stop, the train will be accelerating as it goes into the tunnel. Thus the train will be much master when the back of the train enters the tunnel. In this way, he will have to spend much less time in the tunnel.
There are three bags.The first bag has two blue rocks. The second bag has two red rocks. The third bag has a blue and a red rock. All bags are labeled but all labels are wrong.You are allowed to open one bag, pick one rock at random, see its color and put it back into the bag, without seeing the color of the other rock.
How many such operations are necessary to correctly label the bags ?
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.
David and Albert are playing a game. There are digits from 1 to 9. The catch is that each one of them has to cut one digit and add it to his respective sum. The one who is able to obtain a sum of exact 15 will win the game?
You are a friend of David. Do you suggest him to play first or second?
Let's suppose that David plays first and he picks 9. Then Albert will definitely pick 8. Now, David will have to pick 7 or Albert will pick 7 in his turn. But if David picks up 7, then he will score 16 that is beyond 15 and will lose. So one thing is for sure, no one will be willing to start with the highest digits.
Suppose David plays first and picks up 1, Albert will pick 2. Then David will pick 3 and Albert will pick 4. Now David will be forced to pick 9. The score is 6 to 13 and thus David will have no chance of winning.
If David Picks 9 after Albert has picked up 2, then Albert will pick 8 and the score will become 10 to 10. Thus David will pick 3 as picking 7 will send him past 15. Now Albert will pick 4 and David has nothing to pick for winning. Thus Albert wins.
Therefore, you should suggest David to play second.
Jonathan has three boxes containing milk chocolates and dark chocolates. The problem is that all of them have been labeled incorrectly as follows.
Box1: Dark Chocolates
Box2: Milk Chocolates
Box3: Dark Chocolates and Milk Chocolates
How will he label all the boxes correctly by just opening one box?
It has been clearly mentioned that all the boxes are labeled incorrectly. If he opens the Box3, then he will get either Dark Chocolates or Milk Chocolates as it is labeled incorrectly. Let us suppose he finds Dark Chocolates in there. Now since all are labeled incorrectly, Box B A must contain Milk Chocolates and Box B must contain Milk Chocolates and Dark Chocolates.
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?
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)
Total time = 2 + 2 + 10 + 1 + 2 = 17 mins
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