So – who stole the apple?

During lunch, 5 of Mr. Bryant’s students visit the supermarket.

One of the 5, stole an apple.

When questioned…

  Jim said: it was Hank or Tom.
  Hank said: neither Eddie or I did it.
  Tom Said: you’re both lying
  Don said: no one of them is lying, the other is speaking the truth.
  Eddie said: no Don, that’s not true.

When the shop owner asked Mr. Bryant, he said that three of the boys are always truthful, but two lie all the time.

So – who stole the apple?

And the name of the person who stole the apple is......! 

Tom is a Apple Thief!

What's the story behind the title? 

Let's recall who said what.

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

  Jim said: it was Hank or Tom.
  Hank said: neither Eddie or I did it.
  Tom Said: you’re both lying
  Don said: no one of them is lying, the other is speaking the truth.
  Eddie said: no Don, that’s not true.

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

Clearly, Don and Eddie are making contradicting statements. Hence, one of them must be liar.

So there must be 1 more liar among Jim, Hank and Tom (since there are 2 liars & 3 truth tellers).

Tom's statement - you're both lying points Jim and Hank as liars. But there are total 2 liars with one being from either or Eddie as deduced above.

Hence, Tom himself must be that other liar.

Now we are sure that Jim and Hank must be telling the truth & as told by Hank, Eddie or Hank himself is not thief.

Since, Hank is not the one who stole the apple, the statement made by Jim suggests that Tom is the person who stole that apple.

With that, since 1 liar found among Jim, Hank and Tom, the immediate statement made by Don must be lie and the statement made by Eddie is truth.

Divide 1 Cube into 20 Cubes!

From a 1987 Hungarian math contest for 11-year-olds:

How can a 3 × 3 × 3 cube be divided into 20 cubes (not necessarily the same size)?

Cut this way to get 20 cubes....

Division of 1 Cube into 20 cubes

What was the challenge?

Mark cube for cutting 3 x 3 x 3 = 27 cubes. Cut any section of 2x2x2 = 8 cubes & cut rest of 27-8 = 19 cubes. So these 19 cubes plus 1 cube of 2x2x2 give us total number of 20 cubes. 

Need of Speed For Average Speed

A man drives 1 mile to the top of a hill at 15 mph. How fast must he drive 1 mile down the other side to average 30 mph for the 2-mile trip?

Here is calculation of that speed needed!

Impossible Average Speed Challenge

What was average speed challenge?

A man drives 1 mile to the top of a hill at 15 mph. That means he took, 1/15 hours i.e.4 minutes to reach at the top of a hill.

To achieve average speed of 30 mph, the man has to complete 2 miles trip in 1/15 hours i.e. 4 minutes. But he has already taken 4 minutes to reach at the top of a hill, hence he can't achieve average speed of 30 mph over entire trip. 

MATHEMATICAL PROOF:

Let 'x' be the speed needed in the journey down the hill.

Average Speed = Total Distance/Total time

Average Speed = (1 + 1)/(1/15 + 1/x)

30 = 2/(1/15 + 1/x)

(1/15 + 1/x) = 2 / 30 = 1/15

1/x = 0

x = Infinity/Not defined.

To conclude, it's impossible to achieve average speed of 30mph in trip.

    
 

A Visit To Grandmother's Home!

A father wants to take his two sons to visit their grandmother, who lives 33 kilometers away. His motorcycle will cover 25 kilometers per hour if he rides alone, but the speed drops to 20 kph if he carries one passenger, and he cannot carry two. Each brother walks at 5 kph. 

Can the three of them reach grandmother’s house in 3 hours?

Do you think it's impossible? Click here!

Planning Journey Towards Grandmother's Home

What was the challenge in the journey?

Yes, all three can reach at grandmother's home within 3 hours. Here is how.

Let M be the speed of motorcycle when father is alone, D be the speed of motorcycle when father is with son and S is speed of sons.  Let A and B are name of the sons.

As per data, M = 25 kph, D = 20 kph, S = 5 kph.

1. Father leaves with his first son A while asking second son B to walk. Father and A drives for 24 km in 24/20 = 6/5 hours. Meanwhile, son B walks (6/5) x 5 = 6 km.

2. Now father leaves down son A for walking and drives back to son B. The distance between them is 24 -6 = 18 km.

3. To get back to son B, father needs 18/(M+S) = 18/(25+5) = 18/30 = 3/5 hours & in that time son B walks for another (3/5) x 5 = 3 km. Now, son B is 6 + 3 = 9 km away from source where he meets his father. While son A walks another (3/5) x 5 = 3 km towards grandmother's home.

4. Right now father and B are 24 km while A is 6 km away from grandmother's home. So in another 24/20 = 6/5 hours father and B drive to grandmother's home. And son B walks further (6/5) x 5 = 6 km reaching grandmother's home at the same time as father & brother B.

In this way, all three reach at grandmother's home in (6/5) + (3/5) + (6/5) = 3 hours.

In this journey, both sons walks for 9 km spending 9/5 hours and drives 24 km with father taking (6/5) hours. Whereas, father drives forward for 48 km (24 km + 24 km) in (6/5) + (6/5) hours and 15 km backward in 3/5 hours. 
  

Escape Safely to The Ground!

You find yourself trapped at top an 800 foot tall building. The surrounding land is completely flat, plus there are no other structures nearby. You need to get to the bottom, uninjured, and can only safely fall about 5feet.

You look down the four walls; they are all completely smooth and featureless, except that one of the walls has a small ledge 400feet above the ground. Furthermore, there are two hooks, one on this ledge, and one directly above it on the edge of the roof. The only tools you have are 600feet of rope, and a knife.

 How do you get to the bottom? 

This should be your strategy! 

Strategy To Land Safely On The Ground

Why strategy needed to be planned?

1.Tie one end of the rope to the to hook and climb down to the ledge.

2. Cut (without dropping) the rope that hangs below the ledge, then climb back to the roof carrying the extra rope that you cut. You now have two lengths of rope: one that is 400 feet long and one that is 200 feet long.

3.At the top, untie the rope from the hook.

 Now setup the ropes like : Tie a small loop at one end of the 200-foot long rope. String the 400-foot long rope through the loop so that half of its length is on either side of the loop. Make sure that the loop is snug enough that the 400-foot long rope won't fall out by itself, but loose enough that you can pull the rope out later.

4. Now, tie the end of the 200-foot rope without the loop to the first hook. The 200-foot long rope lets you climb halfway to the ledge. 

5.For the remaining 200 feet, you carefully climb down the 400-foot rope, which hangs down 200 feet from where it is held by the loop. 

6.Once you get to the ledge, pull the 400-foot rope out of the loop.

7. Finally, tie it to the second hook, and climb the rest of the way to the ground.

The Lightning Fast Addition!

A  story tells that, as a 10-year-old schoolboy, Carl Friedrich Gauss was asked to find the sum of the first 100 integers. The tyrannical schoolmaster, who had intended this task to occupy the boy for some time, was astonished when Gauss presented the correct answer, 5050, almost immediately.

How did Gauss find it?

Actually, he used this trick! 

 

Trick for The Lightneing Fast Addition!

Why lightning fast speed needed?

Gauss attached 0 to the series and made pairs of numbers having addition 100.

100 + 0 = 100

99 + 1 = 100

98 + 2 = 100

97 + 3 = 100

96 + 4 = 100

95 + 5 = 100
..
..
..
..
..
..
51 + 49 = 100

This way he got 50 pair of integers (ranging in between 1-100) having sum equal to 100.

So sum of these 50 pairs = 100x50 = 5000.

 
And the number 50 left added to above total to get sum of integers 1 - 100 as 5000 + 50 = 5050

Story of 7 Generous Dwarfs

The Seven Dwarfs are having breakfast, and Snow White has just poured them some milk. Before drinking, the dwarfs have a ritual. First, Dwarf #1 splits his milk equally among his brothers' mugs (leaving himself with nothing). Then Dwarf #2 does the same with his milk, etc. The process continues around the table, until Dwarf #7 has distributed his milk in this way. (Note that Dwarf #7 is named Dopey!) At the end, each dwarf has exactly the same amount of milk as he started with!

 

How much milk does each cup contain, if there were 42 ounces of milk altogether?

Finding difficult? Click here for answer! 

Behind the Story of 7 Generous Dwarfs

What was the story?

First thing is very clear that Dwarf 7 must have 0 ounces of milk at the start and end. Let's assume that 'a' be the maximum amount of milk (in ounces) that any dwarf has in his mug at any point of time. 

For a moment, let's assume Dwarf 1 himself has this 'a' amount of milk.

Now, when D1 distributes his 'a' amount of milk among 6 others, D7 receives 'a/6' amount of milk. At this point of time somebody else will be having maximum amount of milk 'a'. Let D2 be that person now having milk 'a'.

Next is D2's turn where he gives a/6 to all. So now D1 has a/6, D7 has 2a/6 and somebody else say D3 has maximum a. Continuing in this way, for each Dwarf's turn gives - 

Now, when we assumed D2 has maximum milk amount a after receiving a/6 from D1, then it's clear that he must had earlier 5a/6. Similarly, D3 had maximum amount of milk a after receiving a/6 from D1 and D2 indicates that he had 4a/6 milk initially. Continuing in this way, we can find the amount of milk each had initially like below.

So at any point of time, the milk distribution is like a, 5a/6, 4a/6, 3a/6, 2a/6, a/6, 0 where amounts are distributed among 7 dwarfs in cyclic order. But the total amount of milk available is 42 ounces.Hence, 

a + 5a/6 + 4a/6 + 3a/6 + 2a/6 + a/6 +  0 = 42

a + 15a/6 = 42

a + 5a/2 = 42

7a = 84

a = 12.

That means at any point of time the maximum amount of milk that one dwarf can have is equal to 12 ounces. And then others would have 10, 8, 6, 4, 2, 0.

To conclude, the 7 dwarfs had 12, 10, 8, 6, 4, 2, 0 ounces of milk initially.  

A Determined Cat on a Ladder!

A ladder is leaning against a wall. On the center rung is a cat. She must be a very determined cat, because she remains on that rung as we draw the foot of the ladder away from the tree until the ladder is lying flat on the ground. What path does the cat describe as she undergoes this indignity?

She follows this path!

A Path Followed by Determined Cat

What was the question?

 Interestingly, the cat follows the circular path whose center is at the foot of tree. 

Actually, as ladder is drawn out a series of right triangles with the same hypotenuse (the ladder) are created with respect to the foot of tree.

The point of hypotenuses where cat is sitting determinedly will be always at the same distance from all 3 vertices. So if all such point are joined then we get a circular path having center at the foot of tree. 

(Figure is for illustration purpose only & may not have accurate measurements.

Navigation Paths Between Two Points

Consider a rectangular grid of 4×3 with lower left corner named as A and upper right corner named B. Suppose that starting point is A and you can move one step up(U) or one step right(R) only. This is continued until B is reached. 

How many different paths from A to B possible ? 

Here is calculation of total number of paths. 

Combinations of Naviagation Paths

What is the question?

If the right move is represented as R and up move as U then, RRRUU is the one path to reach at B.

UURRR is one more path between points A and B.

URRRU is another way to reach at B.

Further, one can reach at B via RURRU.

So number of such paths are possible.

However, if all paths above are observed, we can conclude that total 5 moves are needed to reach from point A to B. Out of those 5, 3 have to be RIGHT and 2 have to be UP. 

That is, any combination having 3R and 2U in 5 moves will give a valid path to reach at B.

Now, number of ways 3R can be placed in 5 moves can be calculated as - 

C(5,3) = 5!/(5-3)! * 2! = 10.

To sum up, there are 10 paths available between points A and B.
 

Language Barrier in International Meeting

Of the 1985 people attending an international meeting, no one speaks more than five languages, and in any subset of three attendees, at least two speak a common language. Prove that some language is spoken by at least 200 of the attendees.

Here is the proof! 

For The Communication in International Meeting

What was the barrier in the communication?

For any attendee A and B, having no common language there must be C who know the language of either A or B to form a trio as mentioned.

Let's make assumption contradicting the statement made in question. Suppose there are only 198 people who can talk in particular language with A or B. Since A can communicate in 5 languages, there are 5 x 198 = 990 people who can talk with A.

That is 990 people are there who have sharing 1 common language with 1 of 5 languages known by A. Similarly, B also can communicate with 990 more people.

Now, if A and B have no common language then there are only 990 + 990 = 1980 people having potential to become C in the trio. This obviously doesn't cover total of remaining people i.e. 1985 - 2 (A and B) = 1983.

Hence, our assumption goes wrong there. So there must be at least 200 attendees knowing the same language .