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Re: The town rules in Kid-Town require each house to have at [#permalink]
I would say E

20 + 35 - 5 = 50
20 + 40 - 10 = 50

Please correct me if I am wrong.

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Alex
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Re: The town rules in Kid-Town require each house to have at [#permalink]
I was surprised everyone picked C. It seems to me that this is an impossible question. If 20 have "only" ping-pong tables and 40 soccer tables, this means that, with no overlapping, the population will be at least 60 houses. Even with 60 houses, there will be no houses having both tables
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Re: The town rules in Kid-Town require each house to have at [#permalink]
Yes, this question was still in my mind, because my previous post is absolutely wrong.

I figured my problem by using an venn diagram and calculation. They each seraparately give an different value for both ping pong and football.

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Re: The town rules in Kid-Town require each house to have at [#permalink]
This Q is not correct
If X be num of fam having both pingpng+soccer

Then
Tot num of fam = num ping + num soc - ping and soc
= (20+x) + 40 - x = 60 <> 50!!!
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Re: The town rules in Kid-Town require each house to have at [#permalink]
I choose E.
We know that there are exactly 20 ping pong tables alone, maybe more including double equipment.
So that there are at least and no more than 30 soccer tables, 40 in the stem...
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Re: The town rules in Kid-Town require each house to have at [#permalink]
Hi all
Here is the answer. can soemone explain this to me or this question is incorrect.

The best answer is E.
Pay attention to the statements, (1) gives you the exact number of houses who have a ping-pong table only as opposed to the other statement, which tells you that 40 houses have a soccer-table.
There are two possible answers to the question: there are (40 + 20 – 50 = 10) houses with both tables in them or there are 20 houses (the size of the small group).
The answer is not distinct and therefore more sufficient data is required.



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