Clocks can measure time even when they do not show the right time. You just have to wind the clock up and…
We have to suppose that the journey to the friend and back lasts exactly the same time and the friend has a clock (showing the correct time) – it would be too easy if mentioned in the riddle.

Now there is no problem to figure out the solution, is there?

The wise man told them to switch camels.

The dish will be full at 12:44.

You can hang the iron rods on a string and watch which one turns to the north (or hang just one rod).
Gardner gives one more solution: take one rod and touch with its end the middle of the second rod. If they get closer, then you have a magnet in your hand.
The real magnet will have a magnetic field at its poles, but not at its center. So as previously mentioned, if you take the iron bar and touch its tip to the magnet’s center, the iron bar will not be attracted. This is assuming that the magnet’s poles are at its ends. If the poles run through the length of the magnet, then it would be much harder to use this method.
In that case, rotate one rod around its axis while holding an end of the other to its middle. If the rotating rod is the magnet, the force will fluctuate as the rod rotates. If the rotating rod is not magnetic, the force is constant (provided you can keep their positions steady).

Throw the ball straight up in the air.

The man is of short stature. He can’t reach the upper elevator buttons, but he can ask people to push them for him. He can also push them with his umbrella.

All the tools are random things that are not going to help you. All you have to do is pour some water into the pipe so that the ball swims up on the surface. And if you say that you don’t have any water, then think about what you drank today and if you can use that somehow.

Keep the first bulb switched on for a few minutes. It gets warm, right? So all you have to do then is … switch it off, switch another one on, walk into the room with bulbs, touch them and tell which one was switched on as the first one (the warm one) and the others can be easily identified 🙂

Daniel is 7 years old and Jessica is 4 years old.

The first half of the question involves Daniel (D) and Alan (A).

Daniel is currently one fifth of Alan’s age, so:

        A = D x 5           (1)

In 21 years time, Alan will be twice his age, so:

   A + 21 = (D + 21) x 2    (2)

Using (1) in (2) gives:

   A + 21 = (D + 21) x 2

  5D + 21 = (D + 21) x 2

  5D + 21 = 2D + 42

       3D = 21

        D = 7

So Daniel is 7 (and Alan is 35).

The second half of the question involves Jessica (J) and Betty (B).

Betty’s is exactly seven times the age of Jessica:

        B = J x 7           (3)

In 8 years time, Betty will be three times the age of Jessica, so:

    B + 8 = (J + 8) x 3     (4)

Using (3) in (4) gives:

    B + 8 = (J + 8) x 3

   7J + 8 = (J + 8) x 3

   7J + 8 = 3J + 24

       4J = 16

        J = 4

So Jessica is 4 (and Betty is 28). QED.

2,300 miles.

Take the alphabetic position of the first and last letters, add them together and multiply by 100. S = 19, D = 4, 19 + 4 = 23, 23 * 100 = 2,300.