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# Flowers

## 53 posts in this topic

While the first two answers are absolutely valid, there IS a single answer that is more precisely correct: "There is insufficient data to determine exactly."

Either of the first two answers has a 50/50 chance of being wrong, but this one is absolutely correct because there could be 2 or 3.

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First off, if two flowers doesn't work, you all seem to prove it by saying he must have one rose/daisy/tulip. Where does it say that? All means all in the answer, it never says he has any of the above.

Two: Mathematically. All = x, but 2 = -2, and a = the number of that flower.

If there is only 2 flowers, then all = 2. This mean 2 = x. If we subtitute for roses, then (2)-2=a. Therefore a, roses, is equal to 0. As long as it works when you substitute all three, then the statement is true.

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Brain teasers mathmatical or not are not generally presented in ways that people who have not taken advanced mathmatics can't solve. Even if it is possible for the answer to be null, that is an extreem leap from the question posed.

i think there are 3 flowers. one of each kind.

and i think it is obnoxious to think otherwise.

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Three Flowers. XD

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couldnt there be six

see with smilies:

all are happy except 2 all are mean except 2 and all are embarrsed except 2

see!

No all are happy ecept 4 (2 evil and 2 embarrased 2+2=4) and same with the rest, u just have 2 that are happy, ect. duh

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in total there are Three flowers.

one rose

one tulip

one daisy

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Flowers - Back to the River Crossing Puzzles

How many flowers do I have if all of them are roses except two, all of them are tulips except two, and all of them are daisies except two?

Flowers - solution

There are 2 solutions:

Three flowers: rose, tulip, daisy.

Two flowers: carnation, geranium.

3

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Flowers

How many flowers do I have if all of them are roses except two, all of them are tulips except two, and all of them are daisies except two?

Flowers - solution

There are 2 solutions:

Three flowers: rose, tulip, daisy.

Two flowers: carnation, geranium.

Easy - 3.

I liked the twisted solution of 2 though.

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The right answer is three. let R be roses, T be tulips and D be daisies.

R T D All are roses except 2 so R and two others (T and D) all are tulips except 2 (R and D) and all are daisies except 2 (R and T)

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THen again, there could be no flowers at all. all are roses/tulips/daisies, except two as in the number two.

3

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Here is why I don't like the two answer:

If I have no cars and I say "All of my cars are Fords" am I telling the truth? I don't see how that can be true

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I'm new...but I say 3...one of each...then two are not the other one!!

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r= t= d=

x flower y flower - x and y can't be r, t, or d

1st answer - 3 flowers - - 3 minus 2 is one - one of each

2nd answer - 2 flowers - - 2 minus 2 is zero - zero of each

just some basic algebra guys

Three flowers

i rule

3

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two....you don't have daisies, roses or tulips...u have some fifth variety

6

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Three flowrs are there

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"How many flowers do I have if all of them are roses except two, all of them are tulips except two, and all of them are daisies except two?"

There is three flowers

Rose, Tulip, and Daisies

It says (and i quote) "all of themare roses except two, all of them are tulips except two, and all of them are daisies except two

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Flowers - Back to the River Crossing Puzzles

How many flowers do I have if all of them are roses except two, all of them are tulips except two, and all of them are daisies except two?

Flowers - solution

There are 2 solutions:

Three flowers: rose, tulip, daisy.

Two flowers: carnation, geranium.

You have 3, one of each. 1 rose, one daisy and 1 whatever the other 1 was. That's why they are all except for 2.

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The answer is 3. Regardless of how many flowers you attempt to include in this situation... the max result is 1 per if 3 types are included but you only have the option to inlude each with a -2... this is a logic question, not mathematics. This is not a debate for there is truth stated in the rule of the question.

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3; 1 rose, 1 tulip and 1 daisy coz 1r except 1t 1d, 1t except 1r 1d and 1d except 1r 1t

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