Journey of The Dog

Jessica, Warner decided to meet & left their home. And a puppy starts walking down a road. They started at the same time. Their homes are located at 33 KM away from each other.

• Warner walks at 5 miles/hour.
• Jessica walks at 6 miles/hour.

  

The puppy runs from Warner to Jessica and back again with a constant speed of 10 miles/hour.

 
The puppy does not slow down on the turn. How far does the puppy travel in till Jessica and Warner meet?

Know here the distance traveled by puppy! 

Distance in The Dog's Journey!

What was the puzzle? 

First thing on which we need to focus on in how much time Jessica and Warner would meet. Since they are moving towards each other the distance of 33 KM is being covered at 5 + 6 = 11 KM/h.  So they are going to meet each other in 33/11 = 3 hours. Now everything else here can deceive you to find distance covered by puppy. All you need to do is stick to the basics.

Speed = Distance / Time

Distance = Speed * Time 

Distance covered by Puppy = Speed of Puppy * Time for which it traveled.

Distance covered by Puppy = 10 * 3 = 30 KM 

So Puppy travels 30 KM to & fro until Jessica and Warner meets. 

 

Equate Number of Heads or Tails

You are blindfolded and 10 coins are placed in front of you on the table. You are allowed to touch the coins but can't tell which way up they are by feel. You are told that there are 5 coins head up, and 5 coins tail up but not which ones are which.

How do you make two piles of coins each with the same number of heads up? You can flip the coins any number of times.

This is how it can be done! 

Trick To Equate Number of Heads or Tails

What was the task? 

Without thinking too much we need to make 2 piles of 5 coins each. Now there are 3 possibilities here depending on number of heads in either pile. One of the pile might have either 0 or 1 or 2 heads (other having 5 or 4 or 3 heads).

Case 1 : 

P1 : T T T T T
P2 : H H H H H

Case 2 : 

P1 : H T T T T
P2 : H H H H T

Case 3 :

P1 : H H T T T
P2 : H H H T T

Now just flipping all the coins from single pile will make number of heads (or say tails) in both piles equal. So we can flip coins of either P1 or P2. Let's flip all coins of P2.


Case 1 : 

P1 : T T T T T         Number of heads - 0
P2 : T T T T T         Number of heads - 0

Case 2 : 

P1 : H T T T T         Number of heads - 1
P2 : T T T T H         Number of heads - 1

Case 3 :

P1 : H H T T T         Number of heads - 2
P2 : T T T H H         Number of heads - 2

Challenge of Grouping The Coins

You are given a unlimited number of coins and 10 pouches. Now, you have to divide these coins in the given pouches in a manner that if someone asks you for any number of coins between 1 to 1000, you should be able to give the amount by just giving the pouches. You are not allowed to open pouches for that.

How will you do it? 

Know here the only efficient way to do that! 

Source 

Grouping The Coins in Binary Numbers

What was the challenge? 

Once again here binary number system comes in handy. Similar kind of use of binary system in day to day life is here! Another intelligent use is here!  We are already provided 10 pouches which is exactly equal to the number of bits required to represent any number from 1 to 1000. Let's number the pouch as Pouch 0 to Pouch 9. So we need to group coins in 10 pouches like below.

Pouch 0 : 1
Pouch 1 : 2
Pouch 2 : 4
Pouch 3 : 8
Pouch 4 : 16
Pouch 5 : 32
Pouch 6 : 64
Pouch 7 : 128
Pouch 8 : 256
Pouch 9 : 512


Now if somebody asks us for 30 coins then we should give Pouch 4, Pouch 3, Pouch 2, Pouch 1. (11110) That's the binary representation of 30 if we assume Pouches as a bits. If another asks for 828 (binary - 1100111100) then we should give Pouch 9, Pouch 8, Pouch 5, Pouch 4, Pouch 3,
Pouch 2.     

Identify The Cards

From a pack of 52 cards ,I placed 4 cards on the table.

I will give you 4 clues about the cards:

Clue 1: Card on left cannot be greater than card on the right.
Clue 2: Difference between 1st card and 3rd card is 8.
Clue 3: There is no card of ace.
Clue 4: There is no face cards (queen,king,jacks).
Clue 5: Difference between 2nd card and 4th card is 7.

Identify four cards ?

Cards identified here!

Source 

Identified Cards From Clues

What was the task given? 

Let's list the clues once again here for our convenience.

Clue 1: Card on left cannot be greater than card on the right.
Clue 2: Difference between 1st card and 3rd card is 8.
Clue 3: There is no card of ace.
Clue 4: There is no face cards (queen,king,jacks).
Clue 5: Difference between 2nd card and 4th card is 7.

From Clue 4, it's very clear that there is no King, Queen or Jack card.

From Clue 2 & Clue 3, we have combinations of either 1,9 or 2,10 at first & third place. But Clue 3 eliminates the combination of 1,9.So at first place we have 2 & at third we have 10.

Again from Clue 5 & Clue 3, possible combinations at second & fourth place are 2,9 & 3,10. If it was 2,9 then 4 cards would have been like 2,2,10,9. But according to Clue 1 the card on left can't be greater than that at right. Here, the card at third place (10) is greater than that at fourth place (9) placed at right. Hence, this would be invalid combination.

Hence the correct combination for the second & fourth place is 3,10.

So we have 4 cards as 2,3,10,10.

Correct The Equation

Assume these lines as matchsticks though they don't look like by any angle. You need not to be maths expert to tell the equation is totally wrong as RHS is not equal to LHS.

The challenge here is that you are provided only 1 matchstick. You need to correct the above equation by putting that matchstick at right place. Can you?

Find the right place here! 

Correction in Wrong Equation

What was the challenge? 

So did you get the correct place for the matchstick to correct the wrong equation? Don't worry, if you didn't get it. All you need to place the matchstick near '+' sign so that it look like '4' as shown below.

Remember I had asked to assume lines as matchsticks in the question itself. 

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? 

This should be your advice!

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?

This should be the color of last ball! 

Source 

That's The Color Of Last Ball

What was the challenge? 

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.

 
Sub question's Answer :

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.

100m Running Race

Lavesh, Bolt, and Lewis race each other in a 100 meters race. All of them run at a constant speed throughout the race.

Lavesh beats Bolt by 20 meters.
Bolt beats Lewis by 20 meters.

How many meters does Lavesh beat Lewis by ? 

Know here the answer! 

Source 

Winner Beats Second Runner Up by...

What was the question? 

Let Lavesh's speed be 10 m/s. Then he must have taken 10 seconds to finish the race. Since Bolt was beaten by Lavesh by 20 m he must have run 80m when Lavesh finished race in 10 seconds (t=10). So his speed would be 8 m/s.

Now Bolt requires 100/0.8 = 12.5 seconds to finish the race. When he finished, Lewis was 20m behind i.e. 80m from starting point at t = 12.5. So Lewis speed is 80/12.5 = 6.4 m/s

At t = 10 seconds, when Lavesh finished his race Lewis must be at 6.4 x 10 = 64 m from starting point. Hence Lavesh beats Lewis by 100 - 64 = 36 m.

Another method.

Let L be the speed of Lavesh, B be the speed of Bolt & W be that of Lewis. Then,

L/B = 100/80 = 5/4

L = (5/4) B  .......(1)

Similarly,

B/W = 100/80 = 5/4

B = (5/4) W .......(2)

Putting (2) into (1),

L = (5/4) x (5/4) W

L/W = 25/16

L = (25/16) W

For given time t, when Lavesh finished the race,

Distance by L/ Distance by W = 25 / 16

100 m/ Distance by W = 25 / 16

Distance by W = (16 x 100) / 25 = 64.

  
So when Lavesh finished cross line at 100 m, Lewis was at 64m i.e. 36m behind. In other words, Lavesh beats Lewis by 100 - 64 = 36 m.