400 puzzle -- Part 3

Problem 1:

A soldier looses his way in a thick jungle. At random he walks from his camp but mathematically in an interesting fashion.
              First he walks one mile East then half mile to North. Then 1/4 mile to West, then 1/8 mile to South and so on making a loop.
              Finally how far he is from his camp and in which direction?

Solution:

The soldier is 0.8944 miles away from his camp towards East-North.

It is obvious that he is in East-North direction.

Distance travelled in North and South directions
= 1/2 - 1/8 + 1/32 - 1/128 + 1/512 - 1/2048 + and so on... (a geometric series with r = (-1/4) )

   (1/2) * ( 1 - (-1/4)ⁿ )
= ---------------------------
         ( 1 - (-1/4) )

= 1 / ( 2 * ( 1 - (-1/4) ) )
= 2/5

Similarly in East and West directions
= 1 - 1/4 + 1/16 - 1/64 + 1/256 - and so on... (a geometric series with r = (-1/4) )

   (1) * ( 1 - (-1/4)ⁿ )
= ---------------------------
         ( 1 - (-1/4) )

= 1 / ( ( 1- (-1/4) )
= 4/5

So the soldier is 4/5 miles away towards East and 2/5 miles away towards North. So using right angled triangle, soldier is 0.8944 miles away from his camp.

Answer: 

                 0.8944 miles

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Problem 2:

Raj has a jewel chest containing Rings, Pins and Ear-rings. The chest contains 26 pieces. Raj has 2 1/2 times as many rings as pins, and the number of pairs of earrings is 4 less than the number of rings.

How many earrings does Raj have?

Solution :

Assume that there are R rings, P pins and E pair of ear-rings.

It is given that, he has 2 1/2 times as many rings as pins.
R = (5/2) * P or P = (2*R)/5

And, the number of pairs of earrings is 4 less than the number of rings.
E = R - 4 or R = E + 4

Also, there are total 26 pieces.
R + P + 2*E = 26
R + (2*R)/5 + 2*E = 26
5*R + 2*R + 10*E = 130
7*R + 10*E = 130
7*(E + 4) + 10*E = 130
7*E + 28 + 10*E = 130
17*E = 102
E = 6

Hence, there are 6 pairs of Ear-rings i.e. total 12 Ear-rings

Answer:  

                 12 earrings

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Problem 3:

How many ways are there of arranging the sixteen black or white pieces of a standard international chess set on the first two rows of the board?
                     Given that each pawn is identical and each rook, knight and bishop is identical to its pair.

Solution:

There are total 16 pieces which can be arranged on 16 places in ¹⁶P₁₆ = 16! ways.
(16! = 16 * 15 * 14 * 13 * 12 * ..... * 3 * 2 * 1)

But, there are some duplicate combinations because of identical pieces.

·   There are 8 identical pawn, which can be arranged in ⁸P₈ = 8! ways.

·   Similarly there are 2 identical rooks, 2 identical knights and 2 identical bishops. Each can be arranged in ²P₂ = 2! ways.

Hence, the require answer is
= (16!) / (8! * 2! * 2! * 2!)
= 6,48,64,800

Answer: 

                6,48,64,800 ways

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Problem 4:

A person with some money spends 1/3 for cloths, 1/5 of the remaining for food and 1/4 of the remaining for travel. He is left with Rs 100/-
                   How much did he have with him in the beginning?

Solution : 

Assume that initially he had Rs. X
He spent 1/3 for cloths =. (1/3) * X
Remaining money = (2/3) * X

He spent 1/5 of remaining money for food = (1/5) * (2/3) * X = (2/15) * X
Remaining money = (2/3) * X - (2/15) * X = (8/15) * X

Again, he spent 1/4 of remaining maoney for travel = (1/4) * (8/15) * X = (2/15) * X
Remaining money = (8/15) * X - (2/15) * X = (6/15) * X

But after spending for travel he is left with Rs. 100/- So
(6/15) * X = 100
X = 250

Answer:   

                    Rs. 250/-
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Problem 5:

Grass in lawn grows equally thick and in a uniform rate. It takes 24 days for 70 cows and 60 days for 30 cows to eat the whole of the grass.
                How many cows are needed to eat the grass in 96 days?

Solution : 

g - grass at the beginning
r - rate at which grass grows, per day
y - rate at which one cow eats grass, per day
n - no of cows to eat the grass in 96 days

From given data,
g + 24*r = 70 * 24 * y ---------- A
g + 60*r = 30 * 60 * y ---------- B
g + 96*r = n * 96 * y ---------- C

Solving for (B-A),
(60 * r) - (24 * r) = (30 * 60 * y) - (70 * 24 * y)
36 * r = 120 * y ---------- D

Solving for (C-B),
(96 * r) - (60 * r) = (n * 96 * y) - (30 * 60 * y)
36 * r = (n * 96 - 30 * 60) * y
120 * y = (n * 96 - 30 * 60) * y [From D]
120 = (n * 96 - 1800)
n = 20

Hence, 20 cows are needed to eat the grass in 96 days.

Answer:  

                   20 cows
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Problem 6:

There is a safe with a 5 digit number as the key. The 4th digit is 4 greater than the second digit, while the 3rd digit is 3 less than the 2nd digit. The 1st digit is thrice the last digit. There are 3 pairs whose sum is 11.
                   Find the number.

Solution : 

As per given conditions, there are three possible combinations for 2nd, 3rd and 4th digits. They are (3, 0, 7) or (4, 1, 8) or (5, 2, 9)

It is given that there are 3 pairs whose sum is 11. All possible pairs are (2, 9), (3, 8), (4, 7), (5, 6). Now required number is 5 digit number and it contains 3 pairs of 11. So it must not be having 0 and 1 in it. Hence, the only possible combination for 2nd, 3rd and 4th digits is (5, 2, 9)

Also, 1st digit is thrice the last digit. The possible combinations are (3, 1), (6, 2) and (9, 3), out of which only (6, 2) with (5, 2, 9) gives 3 pairs of 11. Hence, the answer is 65292.

Answer:

               65292
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Problem 7:

Four friends - Arjan, Bhuvan, Guran and Lakha were comparing the number of sheep that they owned.
             It was found that Guran had ten more sheep than Lakha.
             If Arjan gave one-third to Bhuvan, and Bhuvan gave a quarter of what he then held to Guran, who then passed on a fifth of his holding to Lakha, they would all have an equal number of sheep.
             How many sheep did each of them possess? Give the minimal possible answer

Solution :

Assume that Arjan, Bhuvan, Guran and Lakha had A, B, G and L sheep respectively. As it is given that at the end each would have an equal number of sheep, comparing the final numbers from the above table.

Arjan's sheep = Bhuvan's sheep
2A/3 = A/4 + 3B/4
8A = 3A + 9B
5A = 9B

Arjan's sheep = Guran's sheep
2A/3 = A/15 + B/5 + 4G/5
2A/3 = A/15 + A/9 + 4G/5 (as B=5A/9)
30A = 3A + 5A + 36G
22A = 36G
11A = 18G

Arjan's sheep = Lakha's sheep
2A/3 = A/60 + B/20 + G/5 + L
2A/3 = A/60 + A/36 + 11A/90 + L (as B=5A/9 and G=11A/18)
2A/3 = A/6 + L
A/2 = L
A = 2L

Also, it is given that Guran had ten more sheep than Lakha.
G = L + 10
11A/18 = A/2 + 10
A/9 = 10
A = 90 sheep

Thus, Arjan had 90 sheep, Bhuvan had 5A/9 i.e. 50 sheep, Guran had 11A/18 i.e. 55 sheep and Lakha had A/2 i.e. 45 sheep.

Answer: 

                Arjan, Bhuvan, Guran and Lakha had 90, 50, 55 and 45 sheep respectively.
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Problem 8:

Consider a number 235, where last digit is the sum of first two digits i.e. 2 + 3 = 5.
           How many such 3-digit numbers are there?

Solution:

The last digit can not be 0.

If the last digit is 1, the only possible number is 101. (Note that 011 is not a 3-digit number)

If the last digit is 2, the possible numbers are 202 and 112.

If the last digit is 3, the possible numbers are 303, 213 and 123.

If the last digit is 4, the possible numbers are 404, 314, 224 and 134.

If the last digit is 5, the possible numbers are 505, 415, 325, 235 and 145.

Note the pattern here - If the last digit is 1, there is only one number. If the last digit is 2, there are two numbers. If the last digit is 3, there are three numbers. If the last digit is 4, there are four numbers. If the last digit is 5, there are five numbers. And so on.....

Thus, total numbers are
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45

Altogether then, there are 45 different 3-digit numbers, where last digit is the sum of first two digits.

Answer:

There are 45 different 3-digit numbers. 
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Problem 9:

Find the smallest number such that if its rightmost digit is placed at its left end, the new number so formed is precisely 50% larger than the original number.

Solution :

If its rightmost digit is placed at its left end, then new number is 428571 which is 50% larger than the original number 285714.

The simplest way is to write a small program. And the other way is trial and error !!!

Answer: 

               The answer is 285714.
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Problem 10:

Two identical pack of cards A and B are shuffled throughly. One card is picked from A and shuffled with B. The top card from pack A is turned up. If this is the Queen of Hearts, what are the chances that the top card in B will be the King of Hearts?

Solution :

There are two cases to be considered.

CASE 1 : King of Hearts is drawn from Pack A and shuffled with Pack B

Probability of drawing King of Hearts from Pack A = 1/51 (as Queen of Hearts is not to be drawn)
Probability of having King of Hearts on the top of the Pack B = 2/53

So total probability of case 1 = (1/51) * (2/53) = 2 / (51 * 53)

CASE 2 : King of Hearts is not drawn from Pack A

Probability of not drawing King of Hearts from Pack A = 50/51 (as Queen of Hearts is not to be drawn)
Probability of having King of Hearts on the top of the Pack B = 1/53

So total probability of case 2 = (50/51) * (1/53) = 50 / (51 * 53)

Now adding both the probability, the required probability is
= 2 / (51 * 53) + 50 / (51 * 53)
= 52 / (51 * 53)
= 52 / 2703
= 0.0192378

Answer:  

                  52 / 2703
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Problem 11:

There are 3 ants at 3 corners of a triangle, they randomly start moving towards another corner.

What is the probability that they don't collide?

Solution :

Let's mark the corners of the triangle as A,B,C. There are total 8 ways in which ants can move.

1.  A->B, B->C, C->A

2.  A->B, B->C, C->B

3.  A->B, B->A, C->A

4.  A->B, B->A, C->B

5.  A->C, C->B, B->A

6.  A->C, C->B, B->C

7.  A->C, C->A, B->A

8.  A->C, C->A, B->C

Out of which, there are only two cases under which the ants won't collide :

·   A->B, B->C, C->A

·   A->C, C->B, B->A 
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Problem 12: Find all sets of consecutive integers that add up to 1000.Submitted by : James Barberousse

Solution : There are total 8 such series: 1.  Sum of 2000 numbers starting from -999 i.e. summation of numbers from -999 to 1000. (-999) + (-998) + (-997) + ..... + (-1) + 0 + 1 + 2 + ..... + 997 + 998 + 999 + 1000 = 1000 2.  Sum of 400 numbers starting from -197 i.e. summation of numbers from -197 to 202. (-197) + (-196) + (-195) + ..... + (-1) + 0 + 1 + 2 + ..... + 199 + 200 + 201 + 202 = 1000 3.  Sum of 125 numbers starting from -54 i.e. summation of numbers from -54 to 70. (-54) + (-53) + (-52) + ..... + (-1) + 0 + 1 + 2 + ..... + 68 + 69 + 70 = 1000 4.  Sum of 80 numbers starting from -27 i.e. summation of numbers from -27 to 52. (-27) + (-26) + (-25) + ..... + (-1) + 0 + 1 + 2 + ..... + 50 + 51 + 52 = 1000 5.  Sum of 25 numbers starting from 28 i.e. summation of numbers from 28 to 52. 28 + 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45 + 46 + 47 + 48 + 49 + 50 + 51 + 52 = 1000 6.  Sum of 16 numbers starting from 55 i.e. summation of numbers from 55 to 70. 55 + 56 + 57 + 58 + 59 +60 + 61 + 62 + 63 + 64 + 65 + 66 + 67 + 68 + 69 + 70 = 1000 7.  Sum of 5 numbers starting from 198 i.e. summation of numbers from 198 to 202. 198 + 199 + 200 +201 + 202 = 1000 8.  Sum of 1 number starting from 1000. 1000 = 1000  ------------------------------------------------------------------------------------------------- Problem 13:

There is a 4-character code, with 2 of them being letters and the other 2 being numbers.

How many maximum attempts would be necessary to find the correct code? Note that the code is case-sensitive.

Solution :

There are 52 possible letters - a to z and A to Z, and 10 possible numbers - 0 to 9. Now, 4 characters - 2 letters and 2 numbers, can be selected in 52*52*10*10 ways. These 4 characters can be arranged in ⁴C₂ i.e. 6 different ways - the number of unique patterns that can be formed by lining up 4 objects of which 2 are distinguished one way (i.e. they must be letters) and the other 2 are distinguished another way (i.e. they must be numbers).

Consider an example : Let's assume that @ represents letter and # represents number. the 6 possible ways of arranging them are : @@##, @#@#, @##@, #@@#, #@#@, ##@@

Hence, the required answer is
= 52*52*10*10*6
= 16,22,400 attempts
= 1.6 million approx. 

Answer:

The maximum number of attempts required are 16,22,400 

 Thanks to Tim Sanders for opening BrainVista's brain !!!
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Problem 14:

How many possible combinations are there in a 3x3x3 rubic cube?

In other words, if you wanted to solve the rubic cube by trying different combinations, how many might it take you (worst case scenario)?

How many for a 4x4x4 cube?Submitted

Solution :

Let's consider 3x3x3 Rubics first.

There are 8 corner cubes, which can be arranged in 8! ways.
Each of these 8 cubes can be turned in 3 different directions, so there are 3^8 orientations altogether. But if you get all but one of the corner cube into chosen positions and orientations, only one of 3 orientations of the final corner cube is possible. Thus, total ways corner cubes can be placed = (8!) * (3^8)/8 = (8!) * (3^7)

Similarly, 12 edge cubes can be arranged in 12! ways.
Each of these 12 cubes can be turned in 2 different directions, so there are 2^12 orientations altogether. But if you get all but one of the edge cube into chosen positions and orientations, only one of 2 orientations of the final edge cube is possible. Thus, total ways edge cubes can be placed = (12!) * (2^12)/2 = (12!) * (2^11)

Here, we have essentially pulled the cubes apart and stuck cubes back in place wherever we please. In reality, we can only move cubes around by turning the faces of the cubes. It turns out that you can't turn the faces in such a way as to switch the positions of two cubes while returning all the others to their original positions. Thus if you get all but two cubes in place, there is only one attainable choice for them (not 2!). Hence, we must divide by 2.

Total different possible combinations are
= [(8!) * (3^7)] * [(12!) * (2^11)] / 2
= (8!) * (3^7) * (12!) * (2^10)
= 4.3252 * 10^19


Similarly, for 4x4x4 Rubics total different possible combinations are
= [(8!) * (3^7)] * [(24!)] * [(24!) / (4!^6)] / 24
= 7.4011968 * 10^45

Note that there are 24 edge cubes, which you can not turn in 2 orientations (hence no 2^24 / 2). Also, there are 4 center cubes per face i.e. (24!) / (4!^6). You can switch 2 cubes without affecting the rest of the combination as 4*4*4 has even dimensions (hence no division by 2). But pattern on one side is rotated in 4 directions over 6 faces, hence divide by 24.

Answer: 

               There are 4.3252 * 10^19 possible combinations for 3x3x3 Rubics and 7.4012 * 10^45 possible combinations for 4x4x4 Rubics.
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Problem 15:

Substitute digits for the letters to make the following relation true.

               N  E  V  E  R

               L  E  A  V  E

            +           M  E

            -----------------

               A  L  O  N  E

Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter M, no other letter can be 3 and all other M in the puzzle must be 3.

Solution :

A tough one!!!

Since R + E + E = 10 + E, it is clear that R + E = 10 and neither R nor E is equal to 0 or 5. This is the only entry point to

solve it. Now use trial-n-error method.

     N  E  V  E  R               2  1  4  1  9

     L  E  A  V  E               3  1  5  4  1

  +           M  E            +           6  1

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

     A  L  O  N  E               5  3  0  2  1
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Problem 16:

One of the four people - Mr. Clinton, his wife Monika, their son Mandy and their daughter Cindy - is a singer and another is a dancer. Mr. Clinton is older than his wife and Mady is older than his sister.

1.  If the singer and the dancer are the same sex, then the dancer is older than the singer.

2.  If neither the singer nor the dancer is the parent of the other, then the singer is older than the dancer.

3.  If the singer is a man, then the singer and the dancer are the same age.

4.  If the singer and the dancer are of opposite sex then the man is older than the woman.

5.  If the dancer is a woman, then the dancer is older than the singer.

Whose occupation do you know? And what is his/her occupation?

Solution :

Cindy is the Singer. Mr. Clinton or Monika is the Dancer.

From (1) and (3), the singer and the dancer, both can not be a man. From (3) and (4), if the singer is a man, then the dancer must be a man. Hence, the singer must be a woman.

CASE I : Singer is a woman and Dancer is also a woman
Then, the dancer is Monika and the singer is Cindy.

CASE II : Singer is a woman and Dancer is also a man
Then, the dancer is Mr. Clinton and the singer is Cindy.

In both the cases, we know that Cindy is the Singer. And either Mr. Clinton or Monika is the Dancer.
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Problem 17:

There are 20 people in your applicant pool, including 5 pairs of identical twins.

If you hire 5 people randomly, what are the chances you will hire at least 1 pair of identical twins? (Needless to say, this could cause trouble ;))Submitted

Solution :

5 people from the 20 people can be hired in 20C5 = 15504 ways.

Now, divide 20 people into two groups of 10 people each :
G1 - with all twins
G2 - with all people other than twins

Let's find out all possible ways to hire 5 people without a single pair of indentical twins.

People from G1	People from G2	No of ways to hire G1 without a single pair of indentical twins	No of ways to hire G2	Total ways
0	5	10C0	10C5	252
1	4	10C1	10C4	2100
2	3	10C2 * 8/9	10C3	4800
3	2	10C3 * 8/9 * 6/8	10C2	3600
4	1	10C4 * 8/9 * 6/8 * 4/7	10C1	800
5	0	10C5 * 8/9 * 6/8 * 4/7 * 2/6	10C0	32
Total				11584

Thus, total possible ways to hire 5 people without a single pair of indentical twins = 11584 ways

So, total possible ways to hire 5 people with at least a single pair of indentical twins = 15504 - 11584 = 3920 ways

Hence, the probability to hire 5 people with at least a single pair of indentical twins

= 3920/15504

= 245/969
= 0.2528
= 25.28%

Answer:  

The probability to hire 5 people with at least 1 pair of identical twins is 25.28%
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