# How many children have two colored flags?

The puzzle is as follows:

A classroom in kindergarten has 35 children. Each of them carry a flag of one, two or three colors. The number of only two-colored flags is double that of the monochrome ones, and the number of flags that have red color is equal to the number of flags that have blue color and equal to the number of flags that have the green color. If eight children have a tricolored flag and two children have a yellow flag only, how many children carry flags of only two colors?

The alternatives given in my book are as follows:

1. 19 children
2. 18 children
3. 15 children
4. 16 children

For reference, this problem comes from my Reason and logic book from 2000s. It seems to be an adaptation from a reprinted copy of Martin Gardner's Puzzle carnival from 1970s.

I am confused on how to approach this without much fuss.

I'm not very savvy with these kinds of puzzles, but I could spot here that there are four colors mentioned. Those are: red, green, blue and yellow.

Since it is mentioned that two children carry only a yellow flag. The answer is to find how many flags there are.

But the thing is that the problem mentions bicolored, tricolored and monochrome flags. This makes it confusing.

Can someone help me solve this puzzle relying on Venn-Euler diagrams? I'm not sure how to do this. Since it mentions three sets I believe.

I don't know if a better way to do this exists. But I believe this is the form which would let me better understand.

Can someone help me here please? Because I'm lost.

• By the statement "the number of flags that have red color is equal to the number of flags that have blue color and equal to the number of flags that have the green color.", are you referring to flags that have ONLY these colors...are monochrome flags...or that include that color, possibly along with others? I don't see how one can be sure which in meant by the statement alone. I assume, as others have, that you are talking about monochrome flags. I think it is best, however, to not have to make that assumption but to have it spelled out more clearly in the puzzle's definition. Jan 11, 2021 at 22:09

The key to this puzzle is finding the relevant information and ignoring the irrelevant information. There are several red herrings in this question that make it seem more complicated.

If you focus on what the question asks for, and write down expressions for how many one, two and three colored flags there are, the solution should become clear.

First, let's start:

Let $$N_i$$ ($$i=1,2,3$$) be the number of children carrying a flag of $$i$$ colors. We know that $$N_1+N_2+N_3=35$$, $$N_3=8$$ (so $$N_1+N_2=27$$), and $$N_2=2N_1$$ (so $$N_1+2N_1=27$$ or $$3N_1=27$$). So, we have that $$N_1=27/3=9$$ and $$N_2=27-9=18$$.

It seems that we have a solution, but wait!

There are many additional (unused by us) clauses which seems to be irrelevant, but we should still check them if they contain any contradictions (in that case our solution will be incorrect, in fact there would be no solution at all). To resolve this problem, we construct a solution that satisfies all those clauses.
Let's assume that all 8 children with tricolored flags carry the flag of Germany (black-red-yellow (or gold)). 3 children carry the flag of Poland (white-red), 11 children carry the flag of Ukraine (blue-yellow), 4 children carry the flag of Pakistan (white-green), 7 children carry the pre-2011 flag of Libya (monochrome green), and 2 children carry the monochrome yellow flag (as specified). Now there are equal number of children (11 each) carrying the flags which contain red, blue and green colors.

Why the second part is needed:

The additional clauses may seem completely unrelated to the problem but nevertheless can change or invalidate the solution. For example, consider the following (much simpler) problem: There are cars and trucks (and no other vehicles) at the parking. There are two cars and twice more trucks. How many vehicles are at the parking? The answer is straightforward ($$2+2\times2=6$$). But let's modify the question adding a statement that is irrelevant to the problem at the first glance ("colors don't matter"): How many vehicles are at the parking if eight of them are black? The problem is now unsolvable, since we cannot have 8 black vehicles out of only 6 of them.

• I don't think there's any implied requirement that these flags be actual national flags. They could just be any old random colors. That said, I'm curious if there's enough information to tell exactly what colors are on all the flags - I suspect there is not, and this information was only provided as a distraction. (Especially if you consider there may be additional colors to the 4 that were listed.) Jan 11, 2021 at 17:53
• @DarrelHoffman Of course no, I just gave an example. Jan 11, 2021 at 19:55

Just suppose there are 2 yellow flags. And all other flags are painted with black, white and gray. So there are 0 green's, 0 red's, and, 0 blue's. Have a double check that above situation just fit all requirements. And you may finally find out that, what colors painted has nothing to do with this question.

Then what left:

• mono + two + three = 35
• two = 2 × mono
• three = 8
• mono >= 2

Then you simply have:

mono = 9; two = 18; three = 8

• Note that there are several different color distributions that only use the three mentioned colors, but this answer as wel as Jays one are entirely correct: Colors do not matter Jan 11, 2021 at 9:38
• @Retudin But there are 4 colors mentioned, not 3. Also, adding extra restrictions make the question unnecessarily complex. When one try to solve a question, they should reduce it as simple as possible.
– tsh
Jan 11, 2021 at 10:18
• You are right about the 4th color, my bad. I was only trying to give an argument that the problem was not badly formulated (as could be the case if it had 1 'solution' with 1 extra implicit assumption), but intentionally laced with disinformation. That 4 colors were mentioned only makes that argument stronger though. Jan 11, 2021 at 10:32
• @Retudin I wouldn't call it "disinformation". Just "irrelevant information". "Disinformation" implies that the statement is not true. In this case, there's no reason to believe that the provided color information is false, just that it does not affect the results. Jan 11, 2021 at 16:00
• @DarrelHoffman Disinformation indeed is not the right word. But I would say it is relevant to the puzzle, in the sense that part of the skill needed to get to the answer (easily) is filtering correctly. Maybe obfuscating information is what I meant?! Jan 11, 2021 at 19:50

The number of only two-colored flags is double that of the monochrome ones

So the answer must be even as the duplicate of any whole number is even

If eight children have a tricolored flag

35 - 8 = 27. So 27 Children have a mono or duo flag.

With that left:

27 / 3 = 9. 9 wil be mono and 18 (2*9) will be two colored

What the colors are is not important

Option 2 (18) Children.

There are just a lot of confusing details which we can ignore. Here is just the needed explanation to understand solution:

Let, no of children with one color flag = x Then, no of children with two color flag = 2*x (According to the above question) Given, by question: no of children with three color flag = 8

Now, since each child has a flag,

Total no of flags =35, so

x + 2x+ 8 =35, 3x = 35-8, 3x =27, x=27/3, x=9

So, the children with only two color flag = 2x = (2) (9) =18

A classroom in kindergarten has 35 children. Each of them carry a flag of one, two or three colors. The number of only two-colored flags is double that of the monochrome ones, and the number of flags that have red color is equal to the number of flags that have blue color and equal to the number of flags that have the green color. If eight children have a tricolored flag and two children have a yellow flag only, how many children carry flags of only two colors?

Let $$x$$ be the number of children with monochrome flags, $$y$$ with two-colored flags, and $$z$$ with tricolored flags. Now, just translate the remaining parts of word problem to math:

$$x + y + z = 35\\y = 2 x\\z = 8\\\text{Solve for }y$$

And solving that should be a piece of cake.