Solution:

There are sixty marks in the given figure, thus it is safe to assume that each mark represents a minute.

Now, the hour hand is on a mark when the minute hand is on [0, 12, 24, 36, 48]. When the minute hand is on a mark, the second hand is on noon. But the orange hand cannot be the second hand. (It would be onnoon and it would be eighteen minutes past/before a full hour.)

a) Purple is on 12. Green indicates twenty four minutes. orange indicates forty two [minutes] = 8 + 2/5 [hours] = 8:24
b) Purple is on 12. Green indicates thirty six minutes. orange indicates eighteen [minutes] = 3 + 3/5 [hours] = 3:36
c) Green is on 12 = 3:36
d) Green is on 12 = 8:24.

Please note that the watch could be rotated upside down as well.

Solution:

9 - 8 + 76 - 5 + 4 + 3 + 21 = 100


Other ways of making number 100 (withiout using number 32)
98 + 7 + 6 - 5 - 4 - 3 + 2 - 1 = 100
98 - 7 - 6 - 5 - 4 + 3 + 21 = 100
9 - 8 + 76 - 5 + 4 + 3 + 21 = 100
98 - 7 + 6 + 5 + 4 - 3 - 2 - 1 = 100
98 + 7 - 6 + 5 - 4 + 3 - 2 - 1 = 100
98 + 7 - 6 + 5 - 4 - 3 + 2 + 1 = 100
98 - 7 + 6 + 5 - 4 + 3 - 2 + 1 = 100
98 - 7 + 6 - 5 + 4 + 3 + 2 - 1 = 100
98 + 7 - 6 - 5 + 4 + 3 - 2 + 1 = 100
98 - 7 - 6 + 5 + 4 + 3 + 2 + 1 = 100
9 + 8 + 76 + 5 + 4 - 3 + 2 - 1 = 100
9 + 8 + 76 + 5 - 4 + 3 + 2 + 1 = 100

Solution:

charles put the ring into the box, secure it with one of your locks, and send the box to the girl firend.
She should then attach one of his own locks and return it. When you receive it again, remove your lock and send it back with her lock. Now she can unlock his own lock and retrieve the object.

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.

Solution:

This problem can be solved with algebra as well as logical reasoning. We are going to tell you how it is logically possible.

Be it any quantity that was present in the beginning and any liquid as well. We know that we have taken a spoonful from one glass and put it into another. Then, we have taken a spoonful from the other glass and put it into the first glass. So, at the end of it, the quantity of liquid in either glass remains same as it was in the beginning.

Therefore, we can conclude the fact that any amount of coke that is missing in the glass with coke will be present in the lemonade glass. Also, the same quantity of lemonade will be missing from the lemonade glass and will be present in the coke glass.

Solution:

7, 4, 6, 5, 8

Solution:

First fill the 8 liters jug complete - 4, 8, 0
Fill the 5 liters jug with 8 liters jug - 4, 3, 5
Pour back the beer from 5 liters jug to 12 liters jug - 9, 3, 0
Pour the 3 liters from 8 liters jug to 5 liters jug - 9, 0, 3
Fill the 8 liters jug completely from 12 liters jug - 1, 8, 3
Fill the 5 liters jug from the 8 liters jug - 1, 6, 5
Pour the entire 5 liters jug back in 12 liters jug - 6, 6, 0

You have successfully split the beer into two equal parts.

Solution:

25 Fruits

Remember we told you that Birbal is a witty trader. So his sole motive is to get rid of the sacks as fast as he can.

For the first sack:
He must be able to fill fruits from one sack to other two sacks. Assume that he is able to do that after M check points. Now to find M,

(Space in first sack) M + (Space in second sack) M = (Remaining fruits in Third Sack) 30 – M
M = 10

Thus after 10 checkpoints, Birbal will be left with only 2 sacks containing 30 fruits each.

Now he must get rid of the second sack.
For that, he must fill the fruits from second sack to the first sack. Assume that he manages to do that after N checkpoints.

(Space in First Sack) N = (Remaining fruits in second sack) 30 – N
N = 15

Thus after he has crossed 25 checkpoints, he will be left be one sack with 30 fruits in it. He has to pass five more checkpoints where he will have to give five fruits and he will be left with twenty five fruits once he has crossed all thirty check points.

Solution:

he was genius.

Officials must have thought he was jumping with a hostage, they would never risk giving him a faulty parachute.

Solution:

1/36

Explanation:
The number of possibilities for the first as well as the second dice are 6 and 6 respectively. Therefore the total possibilities or outcome are 6 * 6 = 36.
Out of 36 outcomes, we need only one case i.e. the first gives 2 and the second gives 5.
Therefore, the probability will be 1 on 36 or 1/36.

Solution:

23rd day

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21 + 22 + 23 = 276

Thus our anniversary falls on the 23rd day of the month.

You can apply other formulas to shorten the process but that is the simplest way to do it.