Elements

**liptonpc** · January 23rd 2010, 01:21 AM

There are 3 sets A, B and C, such that B is a proper subset of A (BcA),intersection of A and B #0, intersection of C and B #0, whereby
n(U) = 95. There are altogether 22 elements in the intersection of the complements of these three sets. Set B contains 31 elements whereas the intersection of A and C has 34 elements. The number of elements in A and C but not in B is one more than half the number in A and C. The number of elements in C, not in A and not in B is 4 more than the number of elements in A, not in C and not in B.
Work out the numbers in the various regions and state them on the Venn diagram that depicts the relationship of A, B and C as given above.

pls help me solve this.
Thanks.

**Grandad** · January 23rd 2010, 07:48 AM

Hello liptonpc

Welcome to Math Help Forum!

Originally Posted by liptonpc

There are 3 sets A, B and C, such that B is a proper subset of A (BcA),intersection of A and B #0, intersection of C and B #0, whereby
n(U) = 95.

Look at the attached diagram.

There are altogether 22 elements in the intersection of the complements of these three sets.

This statement gives us the 22 outside the three set loops.

Set B contains 31 elements whereas the intersection of A and C has 34 elements.

Initially I wrote $x$ in the intersection of B and C (where the 16 is now), and therefore $(31 - x)$ in the area representing $B \cap C'$ (where the 15 is) and $34-x$ in $A\cap C\cap B'$ (where the 18 is).

The number of elements in A and C but not in B is one more than half the number in A and C.

This is potentially ambiguous. What does "The number of elements in A and C but not in B" mean? I am assuming it means "The number of elements that are in A and in C but not in B"; in other words the number of elements in $A \cap C \cap B'$ . And I'm also assuming that "the number in A and C" means the number that are in A and in C; in other words the number in $A \cap C$ , which (we've been told) is 34. So:

$n(A \cap C \cap B') = \tfrac12\times34+1 = 18$

So this gives us $x = 34-18 = 16$ , and the 15 in $B\cap C'$ .

The number of elements in C, not in A and not in B is 4 more than the number of elements in A, not in C and not in B.

Initially I wrote $y$ in $A\cap B' \cap C'$ (where the 10 is now); and therefore I wrote $y+4$ in $C\cap A'$ (where the 14 is). So I was then able to total up the numbers, and put the result equal to 95. This gave:

$y + 31 + 18 + y+4+22 = 95$

$\Rightarrow 2y + 75=95$

$\Rightarrow y = 10$

So those are my answers, which I believe to be correct based on the assumption that I have interpreted the phrase "the number in A and C but not in B" correctly.

Grandad

**liptonpc** · January 24th 2010, 04:07 AM

Hi Grandad,
i have something dont understand.

There are altogether 22 elements in the intersection of the complements of these three sets.

I think this statement "A∩B∩C =22" gives us the 22 inside the three set?

and

The number of elements in A and C but not in B is one more than half the number in A and C.

one more than half <-- is it mean 1 1/2 (1.5) ??

i have att exercise.
thanks for reply.
Lipton

**Grandad** · January 24th 2010, 06:00 AM

Hello liptonpc

There are altogether 22 elements in the intersection of the complements of these three sets.

This is $n(A'\cap B'\cap C')$ - not $n(A\cap B\cap C)$ .

The number of elements in A and C but not in B is one more than half the number in A and C.

one more than half <-- is it mean 1 1/2 (1.5) ??

No: this would be "one-and-a-half times as many".

"One more than half of" $x$ is $\tfrac12x+1$ .

$1.5x$ is "one-and-a-half times as many" as $x$ , or "half as many again" as $x$ .

English can be very confusing sometimes, can't it?

Grandad

**liptonpc** · January 24th 2010, 08:31 AM

yeah ^^ got it ^^ thanks Grandad ^^

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