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.