# Elements

• Jan 23rd 2010, 02:21 AM
liptonpc
Elements
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.
• Jan 23rd 2010, 08:48 AM
Hello liptonpc

Welcome to Math Help Forum!
Quote:

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.
Quote:

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.
Quote:

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).
Quote:

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 $B\cap C'$.
Quote:

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.

• Jan 24th 2010, 05:07 AM
liptonpc
i have something dont understand.

Quote:

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
Quote:

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.
Lipton
• Jan 24th 2010, 07:00 AM
Hello liptonpc
Quote:

There are altogether 22 elements in the intersection of the complements of these three sets.
This is $n(A'\cap B'\cap C')$ - not $n(A\cap B\cap C)$.

Quote:

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?