Solution:

Let's say the Mice are X and Y and Bottle are B1, B2....B9

Step1:
X drinks B1 B2 & B3
Y drinks B1 B4 & B5

Possible Outcomes
If both mice die, B1 is the poison bottle.
If only mice X dies, Either B2 or B3 is the poison bottle.
If only mice Y dies, Either B4 or B5 is the poison bottle.
If nobody dies, Poison is in B6 B7 B8 or B9


Step2: (If nobody dies)
X drinks B6 B7
Y drinks B6 B8

Possible Outcomes
If both mice die, B6 is the poison bottle.
If only mice X dies, B7 is the poison bottle.
If only mice Y dies, B8 is the poison bottle.
If nobody dies, B9 is the poison bottle.

Step2: (Only X dies - Either B2 or B3 is the poison bottle)
Y drinks B2, if Y dies B2 is the poison bottle else B3

Step2: (Only Y dies - Either B4 or B5 is the poison bottle)
X drinks B4, if X dies B4 is the poison bottle else B5

Solution:

Firstly: Policeman should take the thug across the river thames and escort the boat back with him. (Thug across, everyone else on this side)
Then, Policeman will take one cat along, leave it there, and on the way back, bring back the thug (since thug cannot be left with anyone without policeman's supervision).
Then Mrs. Mother will take the second cat on the other side. Drop it there and bring the boat back. For the 4th trip, Mrs. Mother And Mr. Father Will travel to the other side, and Mrs. Mother will stay there, and MR. Father will bring the boat back. (Now Mrs. Mother and 2 cats are on the other side.)
5th trip, the policeman takes the thug across. Both of them stay there and Mrs. Mother will come back. Then for the next trip, Mr. Father and Mrs. Mother will go on the boat, and Mrs. Mother will stay there, and Mr. Father will return back.
Then for the 7th trip, Mr. Father will travel with one dog. On the return, the policeman and the thug will return. Then in the next trip, the Policeman will take the second dog. (Now Mrs. Mother and her two cats; Mr. Father and his two dogs and the Policeman is here).
Now finally the policeman takes the boat on the other side and brings back the thug.
Tough one, right ?

Solution:

The old lady of course.
After helping the old lady into the car, I give the car keys to my best friend.
And then i go ib the bus with child love <3

Solution:

7 Races.

Solution:

Yes, you should switch the envelope. In the beginning when you picked up the envelope, you had a 1/3 probability of finding an offer letter in the envelope. There was 2/3 chance that the letter was there in the two envelopes on the table.

If you keep your selected envelope, you still have a 1/3 chance of finding an offer letter in that. However, since the interviewer has removed one empty envelope from the table, if you switch, you have a probability of 2/3 that the offer letter is inside that.

Solution:

He askes the god to beat him half to death.

Solution:

3,3,8

Lets break it down. The product of their ages is 72. So what are the possible choices?

2, 2, 18 sum(2, 2, 18) = 22
2, 4, 9 sum(2, 4, 9) = 15
2, 6, 6 sum(2, 6, 6) = 14
2, 3, 12 sum(2, 3, 12) = 17
3, 4, 6 sum(3, 4, 6) = 13
3, 3, 8 sum(3, 3, 8 ) = 14
1, 8, 9 sum(1,8,9) = 18
1, 3, 24 sum(1, 3, 24) = 28
1, 4, 18 sum(1, 4, 18) = 23
1, 2, 36 sum(1, 2, 36) = 39
1, 6, 12 sum(1, 6, 12) = 19

The sum of their ages is the same as your birth date. That could be anything from 1 to 31 but the fact that Torres was unable to find out the ages, it means there are two or more combinations with the same sum. From the choices above, only two of them are possible now.

2, 6, 6 sum(2, 6, 6) = 14
3, 3, 8 sum(3, 3, 8 ) = 14

Since the eldest kid is taking piano lessons, we can eliminate combination 1 since there are two eldest ones. The answer is 3, 3 and 8.

Solution:

192

Explanation:
I started with the number 192.
Adding the reversed 291 to 192 gave me 483.
Similarly,
483 + 384 = 867
867 + 768 = 1635
1635 + 5361 = 6996

Solution:

Flyball should pay Lavy $11 instead of accepting his counter offer as in that case the club will end up paying almost double the amount.

Solution:

She is near the trunk of the tree as shown below.

Solution:

13 numbers

Explanation:
Since we are looking to find odd reminder which implies that the remainder will be either 1 or 3
Therefore number must follow below two conditions:
A) We can list number 5n + 1 or 5n + 3. (where n = 0,1,2,3.....)
B) The number must end in 1 or 3.

List of such numbers less than 100 are:
1, 3, 11, 13, 21, 23, 31, 33, 41, 43, 51, 53, 61, 63, 71, 73, 81, 83, 91, 93

Let's remove the nonprime ones which are 1, 21, 33, 51, 63, 81 and 91

So, I will be left with 13 numbers which fit the bill.
3, 11, 13, 23, 31, 41, 43, 53, 61, 71, 73, 83 and 93