Set Notation and Venn Diagrams: past paper questions & worked solutions

\(A=\{1,2,3,4,6,12\}\). Remove these factors from the integers 1 to 12.

The complement contains precisely the universal-set elements not in A.

Within the universal set, \(A=\{20,21,22,23,24,25,26\}\) and \(B=\{25,26,27,28,29,30\}\). Their common members are \(25\) and \(26\).

Intersection consists of elements in both sets.

\(A\cup B=\{20,21,\ldots,30\}\). The members of \(\mathcal E=\{20,21,\ldots,32\}\) outside this union are \(31\) and \(32\).

Take the complement relative to the stated universal set, not all integers.

The multiples of \(3\) are

The numbers in both sets are

The numbers in either set are

\(A=\{1,2,4,5,10\}\) and \(B=\{5,10,15\}\), so \(A\cup B=\{1,2,4,5,10,15\}\).

Include every member of either set once; 20 lies outside the universal set.

\(A\cap B^{\prime}=\{1,2,4\}\).

Retain the factors of 20 in the universal set which are not multiples of 5.

Both 2 and 5 are prime factors of 20, so \(A\cap C=\{2,5\}\ne\varnothing\).

A shared member is enough to disprove the claim that the intersection is empty.

Regions in \(L\) or \(S\): \(5+x+2x+4x=5+2+4+8=19\).

The members of \(B\) are all the numbers inside circle \(B\), including the overlap:

The members of \(A\cap B\) are the numbers in both circles:

\(A'\) means all the numbers in the universal set that are not in \(A\):

The members of \(A\) are all the numbers inside circle \(A\), including the overlap:

The members of \(A\cap B\) are the numbers in both circles:

\((A\cup B)'\) means outside both circles:

The numbers inside \(A\) are the numbers in the \(A\)-only region and the overlap with \(B\):

\(B\cup C\) means every number inside \(B\) or \(C\):

Dominic writes \(9\notin C\). This is correct because \(9\) is outside circle \(C\) in the Venn diagram.

The regions inside \(A\cup B\) are

So

For \(A'\), add every region not inside \(A\):

For \(B'\cap C\), add the regions that are in \(C\) but not in \(B\):

Also,

so

And

so

Therefore the Venn diagram regions are:

- \(A\cap B:\ \{5,10,15\}\) - \(A\) only: \(\{9,11,12,13\}\) - \(B\) only: \(\{4,6\}\) - outside both sets: \(\{7,8,14\}\)

The union contains every letter that is in \(B\) or \(I\):

\(B\cap I'\) means letters in \(B\), but not in \(I\):

For \(K=\{k,e,n,y,a\}\), Cody is not correct because the letter \(a\) is in both \(B\) and \(K\).

From the Venn diagram:

- \(A\) only is \(12\) - \(A\cap B\) only is \(5\) - \(B\) only is \(9\) - \(A\cap C\) only is \(10\) - all three is \(6\) - \(B\cap C\) only is \(3\) - \(C\) only is \(4\) - outside is \(8\)

For \(A\cup B\), add every region inside \(A\) or \(B\):

For \(A'\cap B'\), add the regions outside both \(A\) and \(B\):

For \((A\cap B)\cup C\), add \(A\cap B\) and all of \(C\):

The complement of \(A\), using the given universal set, is \[ A'=\{21,23,25,27,29\} \]

is everything not in \(A\) and not in \(B\), so \[ C=A'\cap B' \]

Equivalently, \[ C=(A\cup B)' \]

\(B \cup G\) means all the letters that are in \(B\) or in \(G\).

\(W \cap G'\) means letters that are in \(W\), but not in \(G\). Since \(e\) is in \(G\), it is not included.

Serena is not correct because \(e\) is in all three sets.

\(B'=\{3,4,6,7,9,11\}\), so \(B=\{1,2,5,8,10,12\}\).

\(A\cap B=\{2,5\}\); \(A\) only \(=\{3,7,11\}\); \(B\) only \(=\{1,8,10,12\}\); outside both \(=\{4,6,9\}\).

\(15\notin B\cap C\) because \(15\) is not prime.

Since \(D\cup C=A\) and \(D\) has two members,

Therefore

and

No number is both even and odd, so

Also \(13\notin \mathcal{E}\) because \(13\) is not in the universal set.

Therefore

and

No number is both even and odd, so

Also \(13\notin \mathcal{E}\) because \(13\) is not in the universal set.

The complement of \(A \cup B\) contains the regions outside both \(A\) and \(B\). These are \(5x\) and \(3x+2\).

So \(5x+(3x+2)=26\), hence \(8x+2=26\) and \(x=3\).

Now \(A' \cap C\) means inside \(C\), but not inside \(A\). These are the regions \(5x\) and \(2x\).

So \(n(A' \cap C)=5x+2x=7x=7(3)=21\).

From the Venn diagram:

There is no region in both \(A\) and \(C\), and there is no triple intersection.

For \(A'\cup B'\cup C'\), use

Since \(A\cap B\cap C\) is empty, this includes everyone in the universal set.

\(A\cap B=\{2,4,6,12\}\)

Everything outside circle \(B\): \(B'=\{5,7,8,9,10,11,13,14\}\)

\(\{8,10,14\}\): \(3\) of the \(14\) numbers.

The French-and-German-only region is \(16-x\), and the German-and-Italian-only region is \(15-x\).

The totals 16 and 15 each include the triple intersection x.

The four disjoint regions in the German set must total 24.

Solve for the common triple-intersection count.

Use the triple-intersection value from part (a).

The French-and-Italian-only count is the given 4.

Sum all seven mutually exclusive interior regions.

Subtract the union from the total number of students.

So

Sophie is wrong because \(30\) is both a multiple of \(3\) and a multiple of \(5\), so \(30\in B\cap C\).

For \(D\), include \(18\) and \(26\), but no other even numbers. One possible set is

So

and

Harpreet is not correct because \(2\) and \(3\) are in both \(A\) and \(B\).

For \(C\), it has \(4\) members, \(n(A\cap C)=2\), \(n(B\cap C)=2\), and no member is in both of those overlaps. So choose the two elements of \(A\) only and the two elements of \(B\) only:

So

Also \(A'\) is the set of odd numbers in \(\mathcal{E}\), so

The statement \(B\cap C=\emptyset\) is true because none of \(23,29,31\) is a multiple of \(3\).

For \(D\), we need \(D\cap(A\cup C)=\emptyset\). So \(D\) must contain the four numbers outside \(A\cup C\):

The multiples of \(3\) in \(\mathcal{E}\) are

The odd numbers in \(\mathcal{E}\) are

So

and

The statement \(24\in A\) is false because \(24\) is not in the universal set \(\mathcal{E}\).

For \(C\), we need \(C\cap B'=\{10,18\}\) and \(C\) has \(4\) members. One valid choice is

Maths only \(=3\), Maths and English only \(=18-x\), Maths and French only \(=11-x\), all three \(=x\).

\(M\cap E\) only \(=11\), \(M\cap F\) only \(=4\), centre \(=7\), \(E\cap F\) only \(=2\).

Outside: \(4\). (Check: \(3+11+4+7+0+2+1+4=32\).)

Maths only and Maths-and-English only: \(3+11=14\) of the \(25\).

\(A\) only \(=62-24=38\), \(B\) only \(=30-24=6\), outside \(=80-68=12\).

\(A\cup B'\) is everything except "\(B\) only": \(80-6=74\).

There are 120 gardeners in total and 29 are outside all three sets, so

So the pair-only regions are:

Since 43 grow carrots,

Let the number who grow only potatoes be \(y\). The number who grow only tomatoes is also \(y\).

Now add the regions in the union:

So the Venn regions are:

For part (b),

For part (c), one gardener who grows carrots is chosen.

The number studying all three languages is \(4\).

So the two-language-only regions are

The German-only region is

The French-only region is \(x\) and the Russian-only region is \(2x\). There are \(11\) students outside all three sets.

Using the total \(80\),

Let the sets be \(D\) for dogs, \(C\) for cats, and \(R\) for rabbits.

The number in all three sets is \(6\).

So the remaining one-set regions are

There are \(5\) students outside all three sets. The total number of students is \(50\), so

For \(C' \cap D'\), we need students who are not in \(C\) and not in \(D\). These are the rabbit-only students and the students outside all three sets.

All three: \(9\).

\(58\) students are inside the circles (\(60-2\)):

Outside: \(2\).

\(28\) chose photography; \(8+9=17\) of them also chose knitting.

\(P\) only and \(P\cap E\) only: \(4+7=11\).

\(8+9+11=28\)

All three: \(6\). Then the "two only" regions:

\(80\) farmers are inside the circles (\(100-20\)), so

Outside the circles: \(20\).

Only "\(C\) only" and the outside region: \(24+20=44\).

"\(C\) only" and "\(S\cap C\) only": \(24+11=35\).

\(53\) farmers have chickens; \(12+6=18\) of them also have goats.

The number with goats, sheep and chickens is \(6\).

Since \(20\) have none, the total inside the circles is \(80\).

Out of the \(53\) farmers with chickens, \(18\) also have goats.