Solution:

In one hour,

Soln 1:
the large inlet tube fills 1 / 3 of the tank;
the small inlet tube fills 1 / 6 of the tank;
the outlet tube empties 1 / 9 of the tank; and therefore
all three tubes together fill [ (1 / 3) + (1 / 6) − (1 / 9) ] of the tank.

Fraction of the tank that will be filled in 0.64 hours =
0.64 [ (1 / 3) + (1 / 6) − (1 / 9) ] = 0.25.


Soln 2:
Let V be the total volume of the tank. From equation (1),

Flow Rate (large inlet tube) = V / 3
Flow Rate (small inlet tube) = V / 6
Flow Rate (outlet tube) = V / 9.

Substituting in equation (2),

Rate of Accumulation in tank = (V / 3) + (V / 6) − (V / 9).

Using the above result in equation (1),

Time required to fill the complete tank = V / [ (V / 3) + (V / 6) − (V / 9) ].

Note that V cancels out on simplifying the above expression.

Fraction of the tank that will be filled in 0.64 hours =

0.64 [ (1 / 3) + (1 / 6) − (1 / 9) ] = 0.25.

Solution:

To do it, the marbles must be arranged as the vertices of an octahedron. If you are unsure, check the image below.

Solution:

One of the cop will move clockwise while the other will move anti-clockwise every day. In such a manner, they will catch the thief in minimum seven days.

Solution:

   9567
+ 1085
======
 10652

Solution:

Everyone will done one day or other.

Abraham Lincoln era is too old and we expect most of them(if not all) to be dead by now.

Solution:

The competition was held on 11 November 2013. The logic is simple. In America, the dates are mentioned as Month/Day/Year and in Britain, dates are written as Day/Month/Year. But on 11 November 2013, the date and month will be the same. And thus both will write in the same manner i.e. 11/11/2013. If it would have been any other date, the evaluators would have been able to identify it.

Solution:

Fred Odea won.

Just look at the announcement lines. The first sentence in 'Ok now everybody!' If you take out just the first letter of every word, it will form 'ONE'. In the same manner the second sentence 'The winner of…..' will give form 'TWO' and the third sentence 'The Hardest Riddle Ever Event.' Will form THREE.

Thus the final sentence must form FOUR. There is only one candidate who can suffice with the first two letter and he is Fred Odea. The final sentence of the mastermind must be, 'Fred Odea: Ultimate Riddler!'

Solution:

If you are thinking to hold one chocolate from each box in hand and then balancing weight in bare hands, you are thinking all wrong.

Let us begin by labelling boxes as 1, 2, 3 and so on till 10.

Now pick one chocolate from box 1, two chocolates from box 2, three from box 3 and so on. In total, you will have 55 chocolates now. (1 + 2 + 3 + ..... + 10)

The ideal weight of the chocolates should be 55 * 20 = 1100. However, somewhere in there are the defected chocolate/s.

You can judge that clearly by noting down the result of 1100 - total weight of chocolates. If the weight is less than 1 gram, the defected box is box 1, if the weight is less than 2 grams, the defected box is box 2 and so on.

Solution:

12 steps:

Steps 11 Liter jug 6 Liter Jug
1. 11 -
2. 5 6
3. 5 0
4. 0 5
5. 11 5
6. 10 6
7. 10 0
8. 4 6
9. 4 0
10. 0 4
11. 11 4
12. 9 6 ==> i found 9 liter in 11liter jug(as required)

Solution:

Q = 1
U = 9
I = 6
C = 8
K = 3.

Explanation:
Replace values in( Q + U + I + C + K)^3 = QUICK
(1+9+8+6+3)^3 = 19863
27^3 = 19863
19863 = 19863

Solution:

Shikhar hit a sixer and then got retired hurt.
Suresh Raina came but he got run out after he and Kohli had crossed halfway.
On the next ball, Kohli scored a six and was not out at 100.