"A Puzzle A Day" - Flowers, and Venn Diagrams

The next puzzle can be solved longhand, but the best tool I would recommend is a Venn diagram, which solves the puzzle and elegantly shows your working (which my maths teachers said was always a good thing).

"Betty was making paper flowers for the local carnival.  11 flowers had red in them, seven had yellow in them, and five had red and yellow in them.  How many flowers did she make?"

The short answer (11+7+5 = 23) is incorrect; the point here is that the 11 flowers with red in them includes the five that had both red and yellow in them.  Similarly, the seven yellow flowers includes the five that had red in as well.

The quickest way to solve this is with a Venn diagram - see below.  Each circle contains the number of flowers of that colour, and the overlap shows the number of flowers which contain both red and yellow.

We know that there are five flowers which have both colours in them, so we can write 5 in the intersection (overlap) point of the two circles.  We also know that the red circle in total must contain 11 flowers, so the area outside the overlapping section is 11-5.  Finally, for the yellow the circle, the total needs be 7.

Simplifying this, we can see that there are 6 flowers that are red-only, 2 that are yellow-only, and five that are both red and yellow:

And hence the total is 13.  The initial "23" (11+7+5) counts the overlapped area twice more than it should be (i.e. it's counted as a red flower, a yellow flower and as a red-and-yellow flower).  By drawing out a simple Venn diagram, it's easy to see what the correct solution is.

"A Puzzle A Day" - add nine, reverse digits

This puzzle is one of those which is almost as simple to solve as it is to state:

Find a two-digit number that reverses its digits when you add nine to it.

Knowing that the nine-times table contains numbers which have two series of ascending tens and descending units, this should be a case of just identifying a pair:

9, 18, 27, 36, 45, 54, 63, etc.

However, there are many more pairs of numbers that fit the requirement in the question:

23 and 32
34 and 43
45 and 54 we've already mentioned
56 and 65
and so on

Any two digit number of the form (10x + (x+1)) or 11x+1 will reverse its digits when adding nine (up to 89 + 9 = 98).

The situation changes above 100, and its not possible to reverse digits by adding just nine (and we have three-digit numbers).

"A Puzzle A Day" - How Old Is Aunt Tabitha (age in days)

Next puzzle - another interesting one: 

Aunt Tabitha was extremely touch about her age.  When an impudent nephew was brave enough to ask her, she cunningly replied that she was 35 years old, not counting Saturdays or Sundays.  How old was she?

So Aunt Tabitha has given her age in weekdays, and weekdays only account for five-sevenths of the total week.  Therefore, Aunt Tabitha's stated age of 35 is only five-sevenths of her actual age; 35 / (5/7) = 49.

So she's 49 years old.  But I'll leave it to the impudent nephew tell her that.

It might have been more interesting if she'd given her age as 14 years old in weekends - which is also a more exciting and appealing way of describing her age.

Having said all that, I can say that I'm 30 years old (excluding weekends), or, even better, 12 years old in weekends.  I'm not touchy about my age, but the thought of being 12 years old in weekends (and not even a teenager) is definitely more appealing than stating my actual age!

"A Puzzle A Day" - A Three Horse Race

This puzzle comes from the Puzzle A Day pad (it's meant to last a year; at this rate I'll be completing it in 2025).

The question is simple (but with a slight twist):  "How many ways can a three horse race finish, including ties?"

The simple answer (excluding ties) can be found by looking at all the possible combinations; if we call the horses A, B and C, then we have:

ABC
ACB
BAC
BCA
CAB
CBA

Six ways.

Now, refreshing the list and including the ties (which will be shown in brackets):

ABC
A(BC)
(AB)C
ACB
A(CB)
(AC)B
BAC
B(AC)
(BA)C
BCA
B(CA)

(BC)A

CAB
(CA)B
C(AB)

CBA
(CB)A
C(BA)
(CAB)

13 in total - there are three variations for each of the previous combinations, so ABC now becomes ABC, (AB)C and A(BC) - but we must now deduplicate A(BC) and A (CB), and so on.  The red text above shows a duplicate of a combination which has already been seen above it in the list.

And there's one where all three tie together (ABC).  The need to deduplicate the ties makes this question more complex than it appears at first sight, and so (as is often the case) care is taken to fully understand the question.