0

There are $36$ girls in a class. $19$ of the girls have dark hair and $22$ of them have blue eyes. How many girls have both dark hair and blue eyes?

My attempt was this: Let $d_w$ be girls with dark hair without blue eyes, $d_b$ be girls with dark hair with blue eyes, $nd_w$ be girls with non-dark hair without blue eyes and $nd_b$ be girls with non-dark hair with blue eyes.

Then $d_w + d_b + nd_w + nd_b =36 ; d_w + d_b=19 ; d_b+nd_b=22$

But I can't find $d_b$. What am I doing wrong in this method?

Brevan Ellefsen
  • 14,793
underop
  • 3
  • 1
    use n(A U B) = n(A) + n(B) - n(A $\cap$ B) – Sriharsha Madala Sep 18 '15 at 21:18
  • Thank you! I know that as i used it in the tag, but I don't understand what is wrong in this approach. – underop Sep 18 '15 at 21:18
  • 1
    This is unsolvable without information of $nd_w$. you assume it to be zero to calculate $d_b$. – Sriharsha Madala Sep 18 '15 at 21:22
  • Why should I assume $nd_w$ to be $0$? I don't understand. Isn't it logical to take it into account as it might exist in that class girls with non-dark hair without blue eyes? – underop Sep 18 '15 at 21:26
  • The point is that it is possible that there are 19 girls with dark hair and blue eyes. This happens if out of the 17 blondes only three have blue eyes. But it is also possible that all the 17 blondes have blue eyes, when only 5 girls would have both dark hair and blue eyes. Anything between 5 and 19 is also possible. You were not given enough information. – Jyrki Lahtonen Sep 18 '15 at 21:33
  • It is logical to take this into account. But as you can see you have 4 unknowns and three equations to solve. Its an underdetermined system. One would expect that $nd_w$ is given in the problem. If not you can assume it to be zero or state that problem has insufficient information. – Sriharsha Madala Sep 18 '15 at 21:35

1 Answers1

1

There is a range of possible answers.

It may be that all 19 dark-haired girls are among those who have blue eyes, giving us the answer 19. (We can't have more than this as we don't have enough dark-haired girls).

The minimum can be found by seeing how many girls we'd need in total if there were no dark-haired, blue-eyed girls. Then the 22 and the 19 must have no intersection and there would need to be 41 girls altogether. As in fact we have only 36, there must be at least 5 girls in both categories.

So the answer to the question is "between 5 and 19, inclusive". It is not possible to pin down the answer to a single number with the information given.

IanF1
  • 1,499