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Showing posts from October, 2017

Treasure Seekers On The Mission

13 caves are arranged in a circle at the temple of doom. One of these caves has the treasure of gems and wealth. Each day the treasure keepers can move the treasure to an adjacent cave or can keep it in the same cave. The treasure keepers allowed to move only in 1 direction only i.e. right to the current position. Every day two treasure seekers visit the place and have enough time to enter any two caves of their choice.

How do the treasure seekers ensure that they find the treasure in minimum possible days?

Source

Tip For Treasure Seekers

What was the mission?

One of the treasure seeker should start moving clockwise & other should anti clockwise.

One starting from Cave C1 & other from Cave C13 make sure that treasures are not in those.

Now if we assume it was in C2 on day 1 & keeper moving it in clockwise then on Day 6 it would be in C7. At 7th day, seeker 1 should go to C8 & seeker 2 should go to C7. If keepers had kept it in same cave after Day 6, then seeker 2 would find it or if they had moved it to C8 then seeker 1 would find it for sure.

So we require minimum 7 days to make absolutely sure that seekers find the treasure. And if keeper starts from any other position it would require less number of days. For example, starting from C3 at Day1 , it would be in C8 on the 6th day & seeker by itself.

The Color Of The Last Ball?

You have 20 Blue balls and 10 Red balls in a bag. You put your hand in the bag and take off two at a time. If they’re of the same color, you add a Blue ball to the bag. If they’re of different colors, you add a Red ball to the bag. What will be the color of the last ball left in the bag?

Note: Assume you have a big supply of Blue and Red balls for this purpose. When you take the two balls out, you don’t put them back in, so the number of balls in the bag keeps decreasing.

Once you tackle that, what if there are 20 blue balls and 11 red balls to start with?

That's The Color Of Last Ball

There could be 3 possible combinations that we could get on each removal.

1. One is Red & other is Blue.

In this case, we are taking off Blue & Red but adding Red back though from other source. Effectively we are taking off only Blue keeping number of Red balls same.

2. Both are Red.

We are taking off 2 Red balls but adding 1 Blue.

3. Both are Blue.

Again we are taking off both Blue balls but adding 1 from other source. Effectively, we are keeping number of Blue intact in such cases.

What we observe from this is that the Red is always taken off in pair. And if it is taken off in single then other Red takes it's place as seen in case 1 above. On the other hand, Blue is added or taken off in single.

Since there are even number of Red balls i.e.10 which only can be taken off in pair, there won't be any single Red balls at the end. Hence, last ball must be Blue.

For odd number of Red balls i.e. 11 here, if taken off in pair then the last ball would be the Red always. Hence, the last ball must be Red in the case.

Who Did It: Murder in A Family of 4

One evening there was a murder in the home of married couple, their son and daughter. One of these four people murdered one of the others. One of the members of the family witnessed the crime.The other one helped the murderer.

These are the things we know for sure:

1. The witness and the one who helped the murderer were not of the same sex.

2. The oldest person and the witness were not of the same sex.

3. The youngest person and the victim were not of the same sex.

4. The one who helped the murderer was older than the victim.

5. The father was the oldest member of the family.

6. The murderer was not the youngest member of the family.

Who was the murderer?

Find out the murderer here!

Source

Roles of Family Members

What was the case?

From (6), we know the youngest one wasn't the murderer. Clue (3) suggests youngest wasn't the victim & (4) hints that youngest wasn't helper too. Hence, youngest must be witness.

So possibilities of others' role are listed in table below.

From (2), the witness had opposite sex than the oldest & from (5), we know father is the oldest. So witness must be daughter.

From (4), murderer was older than victim & since father is oldest he can't be victim. So last 2 possibilities are eliminated.

Challenge Of Crossing Desert

Mr. Rawat wishes to cross a the Sahara desert.

It requires 6 days to cross.

One man can only carry enough food and water for 4 days.

What is the fewest number of other men required to help carry enough food for Mr. Rawat to cross ?

He need only few helpers in the case!

Source

Efficient Way To Cross Desert

What's the challenge?

Mr. Rawat should take 2 helpers - let's name them as A & B.

Let 1 be the unit of food & water that is required for 1 day. So all are going to carry 4 units of food & water.

On the day 1, Mr. Rawat, B & A himself take the food & water from A. Then A will left with only 1 unit & to survive he should go back. In 1 day, he can consume that 1 unit of food & water & travel back to the origin.

On the day 2, Mr. Rawat & B consume 2 units food & water of helper B. Now helper B has to move back to origin with 2 units of food & water in 2 days. Again he can easily go back in 2 days covering distance traveled by him in 2 days of forward journey.

Now Mr. Rawat has 4 units of food & water which he can use in his 4 day's journey of crossing Sahara desert.