Puzzles For Freshers [1]

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

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.


Consider a number 235, where last digit is the sum of first two digits i.e. 2 + 3= 5.

How many

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.
Answer

The answer is 285714.

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 !!!
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?

Answer 52 / 2703

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




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.
Answer

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

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.


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.


Answer 6,48,64,800 ways

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



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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 begining?

Answer Rs. 250/-

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


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?

Answer 20 cows

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.


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.

Answer 65292

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.