I won’t post an answer too quickly, but to give it a’nudge’ …

The Teaser can be modified so that the ‘discount’ on the meal is extremely large.

For example:

Three people enjoy a meal.

The waiter says the bill is £30, so each guest contributes £10. However, the manager/till-operator realizes the bill should only be £10. To rectify this, he gives the waiter £20 to return to the guests.

On his way back to the table, the waiter realizes that he cannot divide the money equally. Because the guests didn’t know the total of the revised bill, the waiter decides to give each guest just £6 and keep £2 as a tip for himself. Each guest therefore received £6 back.

So now, each guest has only paid £4, bringing the total paid for their meal to £12. The waiter has £2. And £12 + £2 = £14.

So, if the guests originally handed over £30, what happened to the remaining £16?

Now it’s a bit more obvious that the situation, as outlined, is quite unreasonable and hopefully it should be a little bit easier to resolve and explain !!.

Well, I guess that after four days, a further ‘nudge’ wouldn’t be out of place …

Using the extremely large discount on a meal, (hence leaving the original version intact),

after the waiter has refunded the diners, and pocketed his ‘Tip’ the £30 is located as follows

Till : £10
Each diner: £6 so for 3 diners = £18
Waiter : £2

The misleading part of the narrative is “… So now, each guest has only paid £4, bringing the total paid for their meal to £12. The waiter has £2.And £12 + £2 = £14. …”

Especially the text in Bold italics
That part of the narrative should have simply said " … including the £2 tip held by the waiter …"
The next bit of narrative could have added " … the balance of £18 is held as £6 by each of the three diners "

The original £30 is a red herring. The bill was £25 and a £2 tip was added, making £27. Each diner paid £9 pounds (made up of £10 each, less £1 each returned as change) = £27 and the waiter keep the £2 tip. The total transaction was £29.

He he. £30 was handed over, £3 was returned to the 3 diners, the waiter kept £2, and £25 stayed in the till to pay for the meal. There’s no missing £1.

Two gentlemen, let’s call them Richard and Nigel, both connoisseurs of elegant wristwatches, had agreed to meet at 10:00 am for coffee at The Watch Club in Piccadilly.

Richard believed that his watch was ten minutes slow, so planned to arrive ten minutes before the time showing on his watch.

Meanwhile, Nigel thought his watch was ten minutes fast, so he aimed to arrive ten minutes after the time shown on his watch.

However, both chaps were wrong. Richard’s watch was actually ten minutes fast, and Nigel’s was ten minutes slow.

Who was the first to arrive at The Watch Club ?

How many minutes difference, if any, were there between the arrival times of Richard and Nigel ?

My calculations suggest Richard arrives when his watch says 9:50 (thinking it’s 10.00) but it’s actually 9:40.
Nigel arrives when his watch says 10.10 (thinking it’s 10:00) but it’s actually 10:20.
So Richard arrives first at 9:40, with Nigel arriving 40 minutes later at 10:20.

A bit long-winded I’m afraid.
In an hour, the minute hand moves 360 degrees, but the hour hand only one twelfth of that, ie 30 degrees. So the hour hand moves at a twelfth of the speed of the minute hand.
At 10 o’clock, the hour hand is 60 degrees from vertical.
For each minute after 10 o’clock, the minute hand moves 6 degrees, so the hour hand will move 0.5 degrees.
Let’s assume that when the hands are inclined at equal angles, the minute hand will have moved 6x degrees clockwise from vertical, and the hour hand 0.5x degrees clockwise from 60 degrees.
The angle is equal when 60-0.5x=6x, so 60=6.5x, so x=9.23 minutes (or 9 minutes and 14 seconds).
As usual, I’m sure there is a neater solution.