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$1000$ gardeners wants to plant two types of flower. $900$ of them plants Roses and $500$ of them plants Dandelions. let $X$ be the minimum amount of gardeners which plants both flowers, and let $Y$ be the maximum amount of gardener which plants both flowers. Then $X+Y$ is
a. $400$
b. $500$
c. $600$
d. $900$
e. $1000$

It is roughly translated from a question in Bahasa Indonesia, you may translate it yourself. I feel like the question is missing a few key words.

What I have done:

  • Use venn: only creates one value, cannot find Min or Max
  • Asked others: they told me that the maximum value is $500$ (from $400$ gardeners planting roses, and $500$ gardeners planting both roses and dandelions) and the minimum value is acquired from using venn. But if you check the maximum value, there will only be $900$ gardeners, not $1000$.

Sorry if my English is a bit rusty

user804886
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LucasWykes
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    You need to solve two LPs. Let $a,b,c,d$ be the only roses, both, only dandelions and neither counts. All must be non negative. $a+b+c+d = 1000$, $a+b = 900$, $b+c = 500$. Now find the $\max,\min$ of $b$ subject to these constraints. It is straightforward to solve by hand. – copper.hat Jul 07 '20 at 05:55

1 Answers1

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The question is ambiguous, must the $1000$ gardeners plants at least one of the flowers?

If you assume that everyone has to plant something, that is $|R \cup D|=1000$, yes, you get a single value, of which you will realized that it is not one of the option of the question.

Suppose they don't have to.

$|R \cup D| \le 1000$, $|R|=900$, $|D|=|500|$

$|R \cup D|=|R|+|D|-|R\cap D|=1400-|R \cap D|$

$1400-|R \cap D| \le 1000$

$|R \cap D| \ge 400$, also, we have $|R \cap D| \le \min(|R|, |D|)$. Leaving the final steps to you.

Siong Thye Goh
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