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The Obtuse Family takes a vacation

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When the airline ticket attendant asked Mr Obtuse how many tickets he and his family would need, he responded "well, let's see. We have 2 grandfathers, 3 fathers (with their sons), 3 cousins, a pair of twins, an uncle, a nephew, a niece and an aunt. Yup, that's the entire living family. Oh, and Auntie is, um, well, a little overweight, so we need 3 seats for her".

"Not a problem, sir. Okay, that will be $6098.81"

"Oh no", said Mr Obtuse. "I worked it out at home, before leaving. It wasn't anywhere near that high an amount".

How much money did Mr. Obtuse eventually pay for the tickets (after a lot of explaining)?

Clarifications:

  • No relationship is mentioned unless both parties are living (e.g. though all males are sons, the 3 sons mentioned each have a father that is on the flight).
  • "a pair of twins" means 2 people (one set of twins), not 4.

edit - I added the clarifications and the fact that the mentioned people represent the entire living family

edit - OMG I goofed. I overstated the number of uncles. I have adjusted the question. Instead of 2 uncles, there is only one, and instead of 2 seats, Auntie needs 3. Everything else remains the same. Sorry if this caused anyone an aneurysm.

edit - Again I goofed. I have changed the reference to Mr. Cheap to Mr. Obtuse (though this did not affect the answer).

Answer:

There are only six people on the flight. Their relationships are shown in the following diagram. Each person is lettered A to F. Letters within the brackets denote [g]randfathers, [f]athers, ons, [c]ousins, [t]wins, ncles, ne[p]hews, [n]ieces, and [a]unts. x's refer to deceased family members.

.......x

....../.\

...../...\

..../.....\

.../.......\

..x......A[gfu]

..|......../.\

..|......./...\

..|....../.....\

..|...../.......\

D[cn].B[gfsct].E[cta]

.........|

.........|

.........|

.......C[fps]

.........|

.........|

.........|

........F

D and E are female.

An initial count of Mr Obtuse's seating needs (ignoring relationships) yields (17 people + 2 extra seats =) 19 tickets. Therefore tickets must $320.99 each since the only factors of 609881 are 19 and 32099.

6 tickets @ $320.99 = $1,925.94

Note: It is also possible to have D be the daughter of A, if E becomes the sister of C.

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Posted · Report post

I think the travelling party would consist of:

----------Mr Obtuse

Mr Obtuse Jr ----Mrs Acute (nee Obtuse)

Mr Obtuse III Mr Acute and Miss Acute (twins)

Young master Obtuse

in the above family tree the 2 grandfarthers are Mr Obtuse and Mr Obtuse Jr

The 3 fathers are Mr Obtuse and Mr Obtuse Jr and Mr Obtuse III

The three cousins are Mr obtuse III , Mr Acute and Miss AcuteThe twins are Mr Acute and Miss Acute

The uncle is Mr Obtuse Jr

There Are 2 nephews Mr Obtuse III and Mr Acute (the OP says 1 but I can not see how you can have 3 cousins and only one each nephew and niece. I could be proved wrong but I think this is an error in the OP)

The Niece is Miss Acute

The Aunt is Mrs Acute (nee Obtuse)

This would mean that the travelling part would consist of 7 people with 2 extra seats for the amply bottomed aunt, would leave a requirement for 9 seats.

So Mr Cheap would need to pay $3430.58 (assuming the Clerk counted 14 people and added the 2 extra seats for Mrs Acute). I Dont quite know why Mr cheap was buying the tickets, It seems he does not live up to his name!

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Posted · Report post

or the answer is none at all.... Mr. Cheap wasnt paying, Mr. Obtuse was! lol

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or the answer is none at all.... Mr. Cheap wasnt paying, Mr. Obtuse was! <!-- s:mrgreen: --><!-- s:mrgreen: --> lol

It seems from the way the question was phrased, Mr. Cheap was paying for the tickets. Mr. Obtuse was merely dealing with the ticket attendant.

I believe it can be as done as cheap as:

$2246.93

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Posted · Report post

It seems from the way the question was phrased, Mr. Cheap was paying for the tickets. Mr. Obtuse was merely dealing with the ticket attendant.

I must apologise to everyone. It seems that making up one of these problems is harder than I thought. The math/logic is easy - properly editing the text is mindblowingly difficult!

The reference to Mr. Cheap is an error - a missed edit when I changed the name of the family. He is really Mr Obtuse. It doesn't really change the problem or the solution - but it was a mistake none-the-less.

I can only imagine the eye-rolling and muttered swearing that you will be doing when you read this.

I stand here with my butt dutifully and ideally positioned - kick away.

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There Are 2 nephews Mr Obtuse III <!-- s:P --><!-- s:P --> and Mr Acute <!-- s:) --><!-- s:) --> (the OP says 1 but I can not see how you can have 3 cousins and only one each nephew and niece. I could be proved wrong but I think this is an error in the OP)

Despite all my other errors ( <!-- s:oops: --><!-- s:oops: --> ), I am proud to say that this was not an error. There are 3 cousins and only one nephew and one niece. Sorry normdeplume.

So Mr Cheap would need to pay $3430.58 (assuming the Clerk counted 14 people and added the 2 extra seats for Mrs Acute).

Again, sorry to say that this is not correct.

he paid less than that

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Posted · Report post

I believe it can be as done as cheap as:

$2246.93

Maybe so, but I can see it being done for less

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Posted · Report post

I have added the answer to the original post.

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Posted · Report post

Note: It is also possible ...

... to have D be the daughter of A, if E becomes the sister of C.

Who are the three cousins in that case?

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Note: It is also possible ...

... to have D be the daughter of A, if E becomes the sister of C.

Who are the three cousins in that case?

ahh...ummm... well, the answer is... i goofed. AGAIN.

The correct variation is "if D is the niece of B". Basically, move D and E down one generation of the chart. D, E, and C should be cousins. The space occupied currently by D would be another x (D's an orphan)

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