Assumptions:

1. None of the parrots and rabbits are ‘handicapped’
2. No parrots / rabbits have any anomalies such as more number of heads / feet than normal
3. The ‘number of heads’ and ‘number of feet’ in the cage are of the aforesaid parrots and rabbits only.

Let the number of parrots be ‘x’ and number of rabbits, ‘y’.

We have:

x + y = 72  (1)
2x + 4y = 200  (2)

Solving (1) and (2), we get x = 44 and y = 28. i.e., there are 44 parrots and 28 rabbits.

Assuming that the box has provision to support at least two locks at a time AND that delay in receiving the letter is not a matter of concern:

Use the ‘double lock’ method. ‘X’ should lock the box and send it to Y (X keeps the key with himself). Y, in turn, should add another lock on the box and send it back (key remains with Y). On receipt of the box, X should open and remove his lock and send it back to Y (with the other lock intact.) Once the box is received, Y can open the box and retrieve the letter by opening his own lock.

The guards were standing face-to-face, looking at opposite directions. So, they could see each other’s face.

Since both the trains travel at the same speed of 100 mph, and the tunnel is 200 miles, they would meet in the middle one hour from the point of entry. Since “Mr. Superbee” travels at 1000 mph, he would have covered 1000 miles in that much time.

Only THREE flowers (one of each) are there.

Since the puzzle specifies the presence of Roses, Tulips and Daisies, it is evident that there MUST be at least ONE piece of each of these. That makes three flowers, one of each kind. Now, if we add at least one flower of any kind, the ‘all but two’ condition will fail for all other kinds of flowers. e.g. suppose there are two roses. Then the ‘all but two’ condition will fail for tulip (because now there are three flowers – two roses + one daisy – other than tulip. Same for daisy as well. Hence, it is impossible to have even a single flower more than the three.

A modified version of the ‘Chocolate box’ puzzle (https://puzzlefry.com/puzzles/electronic-weighing-machine-to-fine-defective-chocolate-box-puzzle/)

Refer to the answer given to that puzzle. The same logic can be applied here. Number the bottles 1 to 10 and take as many marbles from each bottle as the bottle’s number and weigh. Since the ‘standard’ marbles are 1 g each, the total weight of the marbles, if all were standard, should be 55 g. (1 + 2 + 3 + … 10) The actual weight X obtained will be more than this (Because at least one marble selected weighs 1.1 g.) The difference (X – 55) divided by 0.1 will give the number of the bottle that contains the heavier marbles. e.g. if the weight obtained is 55.4 g, the bottle No. 4 contains heavier marbles (from which we had taken 4 marbles of 1.1 g each, resulting in 0.4 g excess weight.)

Theoretically, the company could have produced 84,958 flavours of soda…! Since the puzzle says the company produced “84,958 bottles” and “same number of bottles of each flavour” (with no other condition / restriction whatsoever on the number of bottles or flavours), it is possible that each bottle the company produced was of its own unique flavour, thus making 84,958 bottles of as many flavours (No. of bottles of each flavour = 1). Practical / commercial feasibility and / or any other considerations are beyond the scope of the puzzle.

Answer- 25 coconuts

At three coconuts per post (one each for each bag), after crossing the 10th Check Post, the trader will have exhausted thirty coconuts, which is equivalent of one bag.
So, thereafter, he will be carrying two bags only.

For the next fifteen posts, he will have to spend 30 coconuts (one for each bag) again, thus exhausting one more bag, leaving just one bag of thirty coconuts.

For the remaining 5 check posts, he will spend 5 more coconuts, leaving him with 25 coconuts.

If negative numbers can be included with prefix ‘Minus’: Smallest number could be: MINUS FORTY;

Largest would be: FIVE THOUSAND.
Otherwise, smallest would be ZERO.

(“Indian” style of counting with terms such as ‘LAKH’ is NOT considered.)

Assuming that the balance is a common two-pan one (rather than a digital one), here’s a possible solution with MAX. 3 attempts:

Step 1: Split the coins into two sets of three coins each.
We have TWO possibilities: One, BOTH counterfeit coins are in the SAME set (Case 1) OR both sets contain one counterfeit each (Case 2).
Step 2: Weigh the sets against each other. (W1).

Case 1: The set containing the counterfeits will be lighter. In this case, take this set for the next step (to identify the counterfeits). Let the coins be C11, C12 and C13.
Step 3(1): Keep one coin (C11) aside. Now, since two of the three coins are counterfeits, we have two possibilities here. P1: Both the remaining coins (C12 AND C13) are counterfeit; P2: C11 and one of the other two coins (C12 OR C13) are counterfeit.
Step 4(1) W2: Weigh C12 and C13. If they are equal, both are counterfeit. Otherwise, the lighter of the two and C11 the counterfeits. SOLVED.

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Case 2: Both the sets are of equal weight (because both contain one counterfeit each). Take any set. Let the coins be C21, C22, C23.
Step 3(2): Keep one coin (C21) aside. Again, we have two possibilities here: p1: the coin set aside (C21) is counterfeit; p2: One of the other two (C22 OR C23) is the counterfeit.

Step 4(2): W2: Weigh C22 and C23. If they are equal, both are good coins and C21 is the counterfeit. Otherwise, the lighter of the two (C22 / C23) is the counterfeit in the currently selected set. One of the counterfeits identified.

Step 5(2): W3: Repeat the Step 3(2) and 4(2) on the second set to determine the counterfeit in that set. BOTH Counterfeits identified; problem SOLVED.