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Showing posts from October, 2019

### "Hear & Identify What the Time is!"

Your grandma’s wall clock chimes the appropriate number of times at every whole hour, and also once every 15 minutes. If you hear the wall clock chime once, how much more time do you need to figure out what the time is, without looking at it?

This is the way to figure out the exact time!

### The Count Will Tell The Exact Time!

What was the challenge?

Since the clock chimes appropriate number of times at every whole hour, it's not difficult to predict the exact time after counting sounds but the exception of 1:00 AM/PM.

For example, if the first sound you heard at 4:30 AM then you will hear another sound at 4:45 AM and clock will strike 5 times at 5 AM by which you will easily know the exact time. So, even if you heard it at 15 minutes past any hour, you will need only 45 minutes to figure out the exact time.

But at 1:00 AM/PM the clock will strike only once & there the problem starts.

In worst condition, if you have heard first sound at 12:15 then clock will strike once for next 6 times at 12:30, 12:45, 1:00, 1:15, 1:30, 1:45. That is you will hear clock strike once for 7 consecutive times. So in worst condition, you need 1 hour and 30 minutes to figure out the exact time.

And if you hear the single strike for less than 7 consecutive times, then you can easily figure out the exact time.

### Arrange Positions Around The Round-Table

King Arthur and his eleven honorable knights must sit on a round-table. In how many ways can you arrange the group, if no honorable knight can sit between two older honorable knights?

Here are the possible combinations!

### Possible Positions Around The Round Table

What was the challenge?

If king K is sitting at the center then the youngest knight must sit to right or left of the king i.e. 2 possible positions for him.

The second youngest knight now can sit either left or right of the group of 2 made above.

The third youngest knight now can sit either left or right of the group of 3 made above.

And so on.

That is every knight has 2 possible positions except the oldest knight who will have only 1 position left.

This will be make sure each of the knight (except youngest one) has at least 1 younger neighbor (youngest one has king as one neighbor).

So after putting youngest in 2 possible ways the next youngest can be put another 2 possible ways. That is 4 possible combinations for 2.

Similarly, for arranging 3 knights' positions there can be 2^3 = 8 possible combinations.

This way for 10 knights (excluding the oldest which will have only 1 seat available at the end) there are 2^10 = 1024 possible combinations.

### Sharing The Driving Time!

John and Mary drive from Westville to Eastville. John drives the first 40 miles, and Mary drives the rest of the way. That afternoon they return by the same route, with John driving the first leg and Mary driving the last 50 miles.

Who drives the farthest, and by what distance?

### Uneven Sharing of Driving Time!

But actual how it was shared?

Let's assume for a moment, the distance between Westville and Eastville is 50 miles.

In this case, John drives only 40 miles while Mary drives 10 + 50 = 60 miles. For any distance beyond 50 miles, Mary drives east equal distance as John drives west.

So in any case the difference will remain same of 20 miles. So Mary drives the farthest by 20 miles.

### What is Color of His Hat?

There is a basket full of hats. 3 of them are white and 2 of them are black. There are 3 men Tom, Tim, and Jim. They each take a hat out of the basket and put it on their heads without seeing the hat they selected or the hats the other men selected.

The men arrange themselves so Tom can see Tim and Jim’s hats, Tim can see Jim’s hat, and Jim can’t see anyone’s hat.

Tom is asked what color his hat is and he says he doesn’t know.

Tim is asked the same question, and he also doesn’t know.

Finally, Jim is asked the question, and he does know.

What color is his hat?

### 'THIS' is The Color of His Hat!

What was the challenge?

Since there are only 2 black hats if Tom had saw 2 black hats (on heads of Tim & Jim) then he would have realized that he must be wearing white hat. Since, he says he doesn't know color of his hat that mean he can see either 2 white  hats or 1 black & 1 white hat on heads of other 2.

Now when Tim is saying he doesn't know color of his hat that means Jim must not be wearing black hat. If Jim had black hat then Tim would know that his color of hat is white but can't be black again as in that case Tom would have identified color of his hat in first attempt.

Hence, Jim must be wearing white hat!

### Plan an Unbeatable Strategy

Two people play a game of NIM. There are 100 matches on a table, and the players take turns picking 1 to 5 sticks at a time. The person who takes the last stick wins the game. (Both players has to make sure that the winner would be picking only 1 stick at the end)

Who has a winning strategy?

And what must be winning strategy in the person who takes the last stick looses?

### Planned The Unbeatable Strategy!

What is the game?

The first person can plan an unbeatable winning strategy.

CASE 1 : The person picking last stick is winner.

All that he has to do is pick 4 sticks straightaway at the start leaving behind 96 stick. Then, he has to make sure that the count of remaining stick will be always divisible by 6 like 96, 90, 84, 78......6.

So if the opponent takes away 2 sticks in his first turn, then first person has to take 6 - 2 = 4 sticks leaving behind 90 sticks there. That is if the opponent takes away X stick the first person need to pick 6 - X sticks.

Now, when there are 6 stick left, even if opponent takes away 5 sticks then 1 stick will be left for the first person.

And even if the opponent picks 4 sticks then first person will take 2 remaining sticks.

CASE 2 : The person picking last stick is looser.

Now the first person need to take away 3 sticks in first turn leaving behind 97. Next, he has to make sure the count of remaining sticks reduced by 6 after each of his turn. That is, the count should be like 91,85,79,72......7.

So if the opponent takes away 4 sticks in his first turn, then first person has to take 6 - 4 = 2 sticks leaving behind 91 sticks there. That is if the opponent takes away X stick the first person need to pick 6 - X sticks.

When there are 7 sticks are left then even if the opponent takes away 5 sticks then first person can force him to pick the last stick by picking only 1 stick of remaining 2.

And if the opponent takes away 4 sticks at this stage, the first person still can force him to pick last stick by picking 2 of remaining 3 sticks.

Conclusion : The first person always has a chance to plan a winning strategy.

### Fill in the Empty Boxes

Is it possible to fill each box in with an arithmetic operation so that this becomes a true equation?

### Correct Operators in Empty Boxes!

What wasn't looking possible?

Yes, it's possible. All you need to do is recall BODMAS (Brackets, Of, Division, Multiplication, Addition, Subtraction) rule in mathematics that we learned in school.

### What is the Weight of the Empty Jar?

A full jar of honey weighs 750 grams, and the same jar two-thirds full weighs 550 grams.

What is the weight of the empty jar in grams?

Find the correct way to find the weight here!

### Calculation of Weight of the Empty Jar

Collect the given data.

Let J be the weight of empty jar and H be the weight of honey when jar was full.

J + H = 750                                            ......(1)

And in second case,

J + (2/3)H = 550                                     .....(2)

Subtracting (2) from (1),

(1/3)H = 200

Hence, H = 600.

Putting this value in (1),

J = 750 - 600 = 150.

Therefore, the weight of the empty jar is 150 grams.

### Develop an Unbeatable Strategy!

Consider a two player coin game where each player gets turn one by one. There is a row of even number of coins, and a player on his/her turn can pick a coin from any of the two corners of the row. The player that collects coins with more value wins the game.

Develop a strategy for the player making the first turn, such he/she never looses the game.

Note : The strategy to pick maximum of two corners may not work. In the following example, first player looses the game when he/she uses strategy to pick maximum of two corners.

============================================================
Example :

18 20 15 30 10 14

First Player picks 18, now row of coins is
20 15 30 10 14

Second player picks 20, now row of coins is
15 30 10 14

First Player picks 15, now row of coins is
30 10 14

Second player picks 30, now row of coins is
10 14

First Player picks 14, now row of coins is
10

Second player picks 10, game over.

The total value collected by second player is more (20 +
30 + 10) compared to first player (18 + 15 + 14).
So the second player wins.

===================================================================

### The Unbeatable Strategy for a Coin Game!

Why strategy needed in the case?

Let's recall the example given in the question.
-------------------------------------------------------------------

Example

18 20 15 30 10 14

First Player picks 18, now row of coins is
20 15 30 10 14

Second player picks 20, now row of coins is
15 30 10 14

First Player picks 15, now row of coins is
30 10 14

Second player picks 30, now row of coins is
10 14

First Player picks 14, now row of coins is
10

Second player picks 10, game over.

The total value collected by second player is more (20 +
30 + 10) compared to first player (18 + 15 + 14).
So the second player wins.

---------------------------------------------------------------------------------

Here, it's very clear that the player who chooses coins numbered at even positions wins & the one who chooses odd position loses.

So first player who is going to choose coin first need to be smart. All that he/she need to do is make sum of values of all coin at even position, sum of values of all coins at odd positions and compare them.

If he finds the sum of values of coins at odd position greater then he should choose 1st coin (odd position) followed by 3rd,5th.....(odd positions) & force other to choose even coins

And if he finds the sum of values of coins at even positions greater then he should choose
last coin (which is at even position as number of coins are even).

For example, in above case,

18 20 15 30 10 14

first player calculates

Sum of odd coins = 18 + 15 + 10 = 43
Sum of even coins = 20 + 30 + 14 = 64.

Since sum of even coins is greater, he should choose 6th coin (which will be followed by 4th and 2nd) and force other player to choose 5th,3rd and 1st coin.

Even in case second player selects 1st coin after 14 at other corner is selected by first, still he can be forced to choose the coin at 3rd position (odd position) if first player selects 2nd coin.

In short, first player need to make sure that he collects all the coins that are at even positions or odd position whichever has greater sum!

Note that the total number of coins in the case can't be odd as then distribution of coins among 2 will be unequal.

### "What's The Area of The Triangle?"

If I place a 6 cm × 6 cm square on a triangle, I can cover up to 60% of the triangle. If I place the triangle on the square, I can cover up to 2/3 of the square. What is the area, in cm2, of the triangle?

(a) 22 4/5
(b) 24
(c) 36
(d) 40
(e) 60

Here is that mathematical calculation!

### Calculation of Area of Triangle

What was the question?

The most important thing to note here is the area that square overlaps on the triangle is equal to the area that triangle overlaps the square with the maximum contact area. That is both the areas are supposed to be equal in amount.

So if T is the area of the triangle & S is area of the square then,

0.6 T = 2/3  x S = 2/3 x (6x6) = 2/3 x 36 = 24

T = 40 Square cm.

Hence, answer is option (d) 40.

### The Dice Date Indicator!

How can you represent days of month using two 6 sided dice? You can write one number on each face of the dice from 0 to 9 and you have to represent days from 1 to 31, for example for 1, one dice should show 0 and another should show 1, similarly for 29 one dice should show 2 and another should show 9.

This is how it can be indicated!

### Making of The Dice Date Indicator

What was the challenge?

Dice 1: 0 1 2 3 5 7

Dice 2: 0 1 2 4 6 8

The number 0 has to be present on both the dice. The '0' on first die needed for dates from 1 to 9 to show them as 01,02,03......and the '0' on second die will be used for the dates 10,20,30.

The number 1 and 2 are repeated on the dates 11 and 22 so those 2 numbers has to be there on both dice.

Now, we are left with total 6 positions but 7 numbers - 3 to 9.

However, 6 and 9 can be represented by single die if it is written on one of the side of the die. In normal position it will represent 6 & in inverted position it will show 9 or say vice versa.

For example, the dates 6 and 9 can be indicated as -