1) Start with ZBG and ZBGJ. It should be either "the/then" or "you/your" combination as they appear more.


B R W Q H L F K W H J K Q I B W K
o b s t a c l e s a r e t h o s e

Q I C E D W Z B G W K K M I K E
t h i n g s y o u s e e w h e n

Z B G Q H S K Z B G J K Z K W
y o u t a k e y o u r e y e s

B U U Z B G J D B H F W.
o f f y o u r g o a l s.


2) It is obvious that between 5 O'clock and 6 O'clock the hands will not be exactly opposite to each other. It is also obvious that the hands will be opposite to each other just before 5 O'clock. Now to find exact time:

The hour hand moves 1 degree for every 12 degrees that the minute hand moves. Let the hour hand be X degree away from 5 O'clock. Therefore the minute hand is 12X degree away from 12 O'clock.

Therefore solving for X

Angle between minute hand and 12 O'clock + Angle between 12 O'clock and 4 O'clock + Angle between 4 O'clock and hour hand = 180
12X + 120 + (30-X) = 180
11X = 30
Hence X = 30/11 degrees
(hour hand is X degree away from 5 O'clock)

Now each degree the hour hand moves is 2 minutes.

Therefore minutes are
= 2 * 30/11
= 60/11
= 5.45 (means 5 minutes 27.16 seconds)

Therefore the exact time at which the hands are opposite to each other is
= 4 hrs. 54 min. 32.74 seconds


3) They took their percentages from 40 and not from 39, so they got more than their share.

The oldest son got 1/2 of 40 = 20 which is 0.5 more
The second son got 1/4 of 40 = 10 which is 0.25 more
The third son got 1/8 of 40 = 5 which is 0.125 more
The youngest son got 1/10 of 40 = 4 which is 0.1 more

And the stranger got 1/40 of 40 = 1 which is 0.025 more (As he is not supposed to get anything)

All these fractions add to = 0.5 + 0.25 + 0.125 + 0.1 + 0.025 = 1 which stranger took away.


4) There are total 2 couples and a son. Grandfather and Grand mother, their son and his wife and again their son. So total 5 people.


Grandfather, Grandmother
|
|
Son, wife
|
|
Son


5) He lost Rs.600

First time restaurant has given food worth Rs.105 and Rs. 395 change. Similarly second time, food worth Rs.80 and Rs.20 change. Here, we are not considering food restaurant profits.


6) 3651 represents LENS.

Let's assign possible values to each letter and then use trial-n-error.

S must be 1.

Then D (under L) must be greater than 5. If D is 6, then L is 0. But then A must be 0 or 1 which is impossible. Hence, the possible values of D are 7, 8 or 9.

N must be E + 1. Also, D must be A + 5 as the possible values of D are 7, 8 or 9, D can not be (10+A) + 5.

Now using trial-n-error, we get S=1, I=2, L=3, A=4, N=5, E=6 and D=9


S L I D E 1 3 2 9 6


- D E A N - 9 6 4 5


----- -----


3 6 5 1 L E N S

Hence, 3651 represents LENS.


7) Clark is the landlord.

Assume each statement true, one at a time and see that no other statement is contradicted.

Let's assume that Statement (1) is true. Then, Adam is the landlord and lives in apartment 1. Also, other three's apartments will be on the right of his apartment - which contradicts Statement (4) i.e. If Edmund's apartment is right of Adam's apartment, then the landlord is Burzin. Thus, Adam is not the landlord.

Let's assume that Statement (2) is true. Then, Edmund is the landlord and lives in apartment 4. Also, other three's apartments will be on the left of his apartment - which again contradicts Statement (4) i.e. If Edmund's apartment is right of Adam's apartment, then the landlord is Burzin. Thus, Edmund is not the landlord either.

Let's assume that Statement (3) is true. Then, Clark is the landlord and lives in apartment 3. It satisfies all the statements for
(1) Adam - (2) Edmund - (3) Clark - (4) Burzin

Hence, Clark is the landlord.

Similarly, you can assume Statement (4) true and find out that it also contradicts.


8) J is the married man.

Note that a person's sister-in-law may be the wife of that person's brother or the sister of that person's spouse.

There are 2 cases:

If B's legal spouse is J, then J's sibling must be P and P's sister-in-law must be B.

If B's legal spouse is P, then P's sister-in-law must be J and J's sibling must be B.

It is given that B's legal spouse and J's sibling are of the same sex. Also, it is obvious that P's sister-in-law is female. Then, B's legal spouse and J's sibling both must be males.


B's spouse J's sibling P's sister-in-law


(male) (male) (female)




Case I J P B


Case II P B J


Case II is not possible as B & P are married to each other and both are male. Hence, J is the married man.


9) 44

36 of the cubes have EXACTLY 2 of their sides painted black, but because a cube with 3 of its sides painted black has 2 of its sides painted black, you must also include the corner cubes. This was a trick question, but hopefully the title of the puzzle tipped you off to this.


10) t takes 5 moves to make the triangle with 5 rows point the other way.

0 = a coin that has not been moved.
X = the old position of the moved coin
8 = the new position of the moved coin.

________X
_______X X
____8 0 0 0 8
_____0 0 0 0
____X 0 0 0 X
_______8 8
________8


For traingle of any number of rows, the optimal number of moves can be achieved by moving the vertically symmetrical coins i.e. by moving same number of coins from bottom left and right, and remaining coins from the top.


For a triangle with an odd number of rows, the total moves require are :
(N^2/4) - (N-4) Where N = 4, 6, 8, 10, ...

For a triangle with even number of rows, the total moves require are :
((N^2-1)/4) - (N-4) Where N = 5, 7, 9, 11, ...


11) Each envelope contains the money equal to the 2 raised to the envelope number minus 1. The sentence "Each envelope contains the least number of bills possible of any available US currency" is only to misguide you. This is always possible for any amount !!!

One more thing to notice here is that the man must have placed money in envelopes in such a way that if he bids for any amount less than $25000, he should be able to pick them in terms of envelopes.

First envelope contains, 2^0 = $1
Second envelope contains, 2^1 = $2
Third envelope contains, 2^2 = $4
Fourth envelope contains, 2^3 = $8 and so on...

Hence the amount in envelopes are $1, $2, $4, $8, $16, $32, $64, $128, $256, $512, $1024, $2048, $4096, $8192, $8617

Last envelope (No. 15) contains only $8617 as total amount is only $25000.

Now as he bids for $8322 and gives envelope number 2, 8 and 14 which contains $2, $128 and $8192 respectively.

Envelope No 2 conrains one $2 bill
Envelope No 8 conrains one $100 bill, one $20 bill, one $5 bill, one $2 bill and one $1 bill
Envelope No 14 conrains eighty-one $100 bill, one $50 bill, four $10 bill and one $2 bill


12) The minute and the hour hand meet 11 times in 12 hours in normal watch i.e. they meet after every
= (12 * 60) / 11 minutes
= 65.45 minutes
= 65 minutes 27.16 seconds

But in our case they meet after every 65 minutes means the watch is gaining 27.16 seconds.


13) The number is 45, simply because
45 = 5 * (4 + 5)
How does one find this number?

Let T be the digit in the tens place and U be the digit in the units place. Then, the number is 10*T + U, and the sum of its digits is T + U.

The following equation can be readily written:
10*T + U = 5*(T + U) or
10*T + U = 5*T + 5*U or
5*T = 4*U

Thus, T / U = 4 / 5

Since T and U are digits, T must be 4 and U must be 5.


14) Total no of balls = 5 + 7 + 14 + 16 + 18 + 29 = 89

Total number of balls are odd. Also, same number of red balls and blue balls are left out after selling one box. So it is obvious that the box with odd number of balls in it is sold out i.e. 5, 7 or 29.

Now using trial and error method,
(89-29) /2 = 60/2 = 30 and
14 + 16 = 5 + 7 + 18 = 30

So box with 29 balls is sold out.


15) 47 Chocolates and 16 Friends

Let's assume that there are total C chocolates and F friends.

According to first case, if she gives 3 chocolates to each friend, one friend will get only 2 chocolates.
3*(F - 1) + 2 = C

Similarly, if she gives 2 chocolates to each friends, she will left with 15 chocolates.
2*F + 15 = C

Solving above 2 equations, F = 16 and C = 47. Hence, Ekta got 47 chocolates and 16 friends


16) Esha's age is 45 years.

Assume that Esha's age is 10X+Y years. Hence, her hunsbands age is (10Y + X) years.

It is given that difference between their age is 1/11th of the sum of their age. Hence,
[(10Y + X) - (10X + Y)] = (1/11)[(10Y + X) + (10X + Y)]
(9Y - 9X) = (1/11)(11X + 11Y)
9Y - 9X = X + Y
8Y = 10X
4Y = 5X

Hence, the possible values are X=4, Y=5 and Esha's age is 45 years.


17) The fish is 128 inches long.

It is obvious that the lenght of the fish is the summation of lenghts of the head, the body and the tail. Hence,
Fish (F) = Head (H) + Body (B) + Tail (T)

But it is given that the lenght of the head is 4 inches i.e. H = 4. The body is three-quarters of its total length i.e. B = (3/4)*F. And the tail is its head plus a quarter the lenght of its body i.e. T = H + B/4. Thus, the equation is
F = H + B + T
F = 4 + (3/4)*F + H + B/4
F = 4 + (3/4)*F + 4 + (1/4)*(3/4)*F
F = 8 + (15/16)*F
(1/16)*F = 8
F = 128 inches

Thus, the fish is 128 inches long.


18) Everybody in the World will get to know the scandal in 8 hours.

You came to know of a scandal and you passed it on to 4 persons in 30 minutes. So total (1+4) 5 persons would know about it in 30 minutes.

By the end of one hour, 16 more persons would know about it. So total of (1+4+16) 21 persons would know about it in one hour.

Similarly, the other (1+4+16+64) persons would have know about it in one and a half hours. (1+4+16+64+256) persons would have know about it in two hours and so on...

It can be deduced that the terms of the above series are the power of 4 i.e. 4^0, 4^1, 4^2, 4^3 and so on upto (2N+1) terms. Also, the last term would be 4^2N where N is the number of hours.

Sum of the above mentioned series = [4^(2N+1)-1]/3