Calculation of Time For The Journey

What was the journey?

Let the speed of RED ant is R & that of BLACK ant is B

Let time taken by them to meet be T.

Now we will apply the basic formula of distance:

Distance = Speed * Time.

The RED ant travels R T distance before meeting and 20 R after the meeting.

The BLACK ant travels B T distance before meeting and 5 B after the meeting.

Now as per the question,The distance traveled by RED ant before they both meet will be equal to the distance covered by BLACK ant after they meet. We can say the same for the vice versa case as well.

Thus,

RT = 5B and BT = 20R i.e. B = 20R/T, putting in RT = 5B

R T = [20R/T] * 5

RT = 100R/T

T^2 = 100

T = 10.

Thus the RED ant will require 10 + 20 = 30 seconds to travel the distance.

And the BLACK ant will take 10 + 5 = 15 seconds to travel the distance. 

Flip The Triangle

Here you see a triangle formed by ten coins. The triangle points upwards. How can just three coins be moved to make the triangle point downwards?

Flip It By Moving 3

 Here are pointed those 3 coins 

Flipped The Triangle!

What's the challenge? 

Here should be those 3 coins that you need to move so that the triangle point towards down.

And after moving those coin, it will look like,

Distance Between Houses?

Four friends built a colony for themselves. They built their own houses at different
distances from each other.

Chris lived 60 km away from Alex.
Darren lived 40 km away from Bill.
Chris lived 10 km nearer to Darren than he lived to Bill.

Can you find out how far was Darren's house from Alex?

Click here to know the distance! 

Locations of 4 Friends

What was the given data & question? 

Let's represent houses of friends by their initial letter of the name i.e. C is the house of Chris, D is the house of Darren, B is Bill's house & A is name of Alex's house.

Now C & A are located at 60 km away from each other while D & B are 40 km apart i.e. CA = 60 & DB = 40.

As per given data, D is 10 km nearer than B from C. Hence, CB = CD + 10

There are 5 possibilities of the locations of these 4 friends.

Case 1. D & B are in between C & A.

In this case, 

CD + DB + BA = 60

but DB = 40,

CD + BA = 20,

CB - 10 + BA = 20

CB + BA = 30 i.e. CA = 30

But CA = 60, hence this combination is not possible. Other way, CB >= 40, since CD = CB-10, CD >= 30. That means CD + DB >= 70 for which B needs to be beyond A as CA = 60.

Case 2 : A is between D & B.

In this case, CD + DA  = 60, DA + AB = 40 . If we subtract second one from first, we get,

CD - AB = 20

CB - 10 - AB = 20

CB - AB = 30  

CA = 30 but CA = 60. Hence, this combination too is invalid.

This Coffee Puzzle Harder Than You Think!

With all credits to the creator of this picture puzzle, here it is just reproduced once again. 

So what do you think, which coffee cup will fill up first?

Hint : Don't conclude early!

 Coffee Cup Puzzle

This should be the first one!

 

Only This Cup Will Fill Up!

Purpose of the title of the post!

If you think that cup 4 will fill up first then you are deceived by the creator successfully.  You might have found the order of filling cups as 4,9,7 & 5. Congrats, your brain hasn't processed details! Didn't you notice the blocks? 

   Coffee Cup Puzzle

Finding The Average Speed

A man drives his car to the office at 20miles/hr. After reaching the office, he realizes that it's a new year holiday so he went back home at a speed of 30miles/hr.

Discounting the time spent in the stoppage what was his average speed of his journey?

Right way to calculate the average speed! 

Source 

Right Way To Find The Average Speed

What was the given data?

If you are finding average speed of 2 given speeds as (Speed 1 + Speed 2)/2 then you are getting tricked by questioner. The speed itself is a distance covered per unit i.e. Speed = Distance/Time. Calculating average speed like that means,you are doing like,

Average Speed = (Speed1 + Speed2)/2
                     
                      = (Distance1/Time1 + Distance 2/Time2)/2              
            
So you need to find the total distance traveled & the total time taken to complete the entire journey.It should be like,


Average Speed = Total Distance/Total Time

                       = (Distance1 + Distance2)/(Time1 + Time2)

In the given problem, let D be the distance traveled by car to the office. Let T1 be the time required to go to the office & T2 be the time to return back.

Since, Speed = Distance / Time, Time = Distance / Speed.

Hence, 

T1 = D / 20

T2 = D / 30

Now,

Total distance traveled =  D + D = 2D, Total Time taken = T1 + T2

Hence,

Average Speed = 2D / (T1 + T2)

Average Speed = 2D / (D/20 + D/30)

Average Speed = 2D / (50D/600)

Average Speed = 2D / (D / 12)

Average Speed = 24 miles/hr.  

Therefore, the average speed of the journey is 24 miles/hr not 25 miles/hr [(20 + 30)/2].

  
 

Guess The Order Of Cards

A man sitting opposite you has four cards in his hand facing him: 2, 3, 4 and 5 (but not in that order). He wants them placed in ascending order from his left to his right. To do this, he takes the leftmost card (from your perspective) and puts it last. He then takes the third card from the right (your right) and puts it in the last place.

What was the previous order of the cards?

Order Of Cards?

Here was the previous order! 

Source 

Identified The Order of The Cards

How cards were shuffled? 

Since he puts the leftmost card (from our point of view) to it's opposite location that card must be 2 (that's the smallest one & needs to be first in ascending order).

So the last card (from his point of view) must be 2. We still don't know exact positions of 3,4,5. Now 2 is in first position

Now what he does is that, moves third card from his left (our right) to the last position. Now to be in ascending order the last card must be 5. So the card that he used in this move must be 5.

Since these 2 moves completes the ascending order, rest must be in ascending from left to right already i.e as 3 & 4 but 5 in between them.

Hence the order before he moved cards must be as 3,5,4,2.

  Previous Order of The Cards

 

Time Of Arrival?

One day Rohit decided to walk all the way from city Bangalore to Tumkur. He started exactly at noon. And Samit in city Tumkur decided to walk all the way to Bangalore from Tumkur and he started exactly at 2 P.M. on the same day.

Both met on the Bangalore - Tumkur Road at five past four, and both reached their corresponding destination at exactly the same time.

At what time did we both arrive?

      Beginning of Journey

Find here their arrival time! 

Finding The Time Of Arrival

What was the puzzle? 

Let 'x' km/minute be the speed of Rohit's walk. He started to walk at 12 PM & met Samit at 4:05 PM. That means he has walked for 245 minutes. 

Distance traveled by Rohit = 245x km

Let 'y' km/minute be the speed of Samit's walk. He started to walk at 2 PM & met Samit at 4:05 PM. That means he has walked for 125 minutes.

Distance traveled by Samit = 125y km 

    Time of Meeting

Now after meeting each other they resumed their journey further. That means Rohit continues to Tumkur & covers distance of 125y km at his speed of x km/minute. Time taken by him further to complete the journey is 125y / x minutes (Time = Distance/Speed).

Journey From Top To Ground

Galileo dropped balls of various weights from the top of the Leaning Tower of Pisa to refute an Aristotelian belief that heavier objects fall faster than lighter objects. 

If the balls were dropped from a height of 54 meters, how long did it take for the balls to hit the ground?

Click here to know the answer! 

Time Needed To Reach The Ground

What was the problem? 

To solve the problem, we need to remember what we have learned in our early days of school. The gravity of the earth accelerates the falling object at the rate of 9.8 meter per second square. We know, the kinematic equation,

s = ut + (1/2) a t^2   .......(1)

where,

s = distance covered,
u = initial velocity,
a = acceleration,
t = time taken to travel distance s.

In this case, initial velocity must be 0 as it is dropped from height 54m. Again height here is the distance covered by the object. And the acceleration in this case is nothing but the acceleration due to gravity which is 9.8 m/s^2.

So putting s = 54 m, u = 0 m/s, a = 9.8 m/s^2 in equation (1) above,

54 = 0 x t + (1/2) x 9.8 x t^2

54 = 4.9 t^2

t^2 = 11.021

t = 3.3 seconds. 

So theoretically both balls should take 3.3 seconds to reach the ground. But resistance due to air will make the difference in time taken by balls to hit the ground. 

Round Table Coin Game

You are sitting with one opponent at an empty, round table. Taking turns, you should place one dollar on the table, in such a way that it touches none of the coins that are already on the table. The first player that is not able to place a dollar on the table has lost. By tossing a coin, it has been decided that you may start.

Which strategy will you follow to make sure you are guaranteed to win?

  
Trick to win this game always! 

Never Loose Round Table Coin Game

What was the game? 

There is little trick with which you will always end on winning side in this Round Table Coin Game. Since you have got first chance to place the coin you should place the coin right at the center of the round table. Now for every next coin placed by opponent you need to place coin in such a way that it 'mirrors' opponent's coin.

Imagine line from the center to opponent's coin. Place your coin exactly opposite to that coin at distance equal to distance between center & opponent's coin. Or imagine a circle (with the center fixed at the round table) with opponent's coin lying on it's border.  And place your coin at diagonally opposite point of point where opponent placed coin on that imaginary circle. (Assume imaginary circle though it's not appearing perfectly in the image above)

In this way for every move of your opponent, you will have 'answer' in form of space for placing the coin. This will continue until last place left on the table with your turn of placing the coin in the end. 

This is how to make sure you always on winning side in this 'Round Table Coin Game'!

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.