Venn Diagrams Worked ExamplesIGCSE Maths 0580

Always fill the overlap first. Put $7$ in the centre region, where the two circles meet.

(a) The $22$ footballers include those $7$, so subtract to get football only.

Do the same for tennis.

Add the three regions inside the circles to get the union.

(b) Everyone else is outside both circles. Subtract from the group total.

(a) Use the three-set inclusion–exclusion formula. Since everyone studies at least one, the union is $50$.

Substitute everything you know, writing $x$ for the unknown centre.

Simplify the arithmetic on the right.

(b) Now fill the diagram from the centre outwards. Each pairwise figure includes the centre, so subtract $4$ from each.

For Physics only, take $n(P)$ and remove all three regions that overlap another subject.

Repeat for Chemistry and Biology.

Add the three "only" regions, then check all seven regions total $50$.

(a) "Exactly two" means the three pairwise regions, excluding the centre.

Write as a probability over the total, then simplify by dividing by $2$.

(b) "All three" is the centre region only.

(c) "Studies Chemistry" means anywhere inside the $C$ circle — all four of its regions.

This matches the given $n(C) = 30$, confirming the diagram. Write the probability.

(a) The four regions of the diagram — car only, both, bicycle only, and neither — must account for everybody.

Collect the $x$ terms.

Solve.

Fill in the regions: car only $= 24$, both $= 8$, bicycle only $= 16$, neither $= 12$.

(b) $n(C)$ is the whole car circle — car only plus the overlap.

(c) The union is everyone inside either circle, i.e. everyone except the $12$.

(a) Apply inclusion–exclusion to find how many bought at least one drink.

Work through it in stages: add the singles, subtract the pairs.

Then add back the triple overlap.

Those buying none are the rest of the $120$.

(b) Fill the diagram from the centre. Subtract the $12$ from each pairwise figure.

Now each "only" region: take the circle total and remove the other three regions.

Repeat for tea and juice.

Add the three "only" regions and write the probability. Check all regions plus the $4$ total $120$.

(a) Read the expression from the inside out: take everything in $C$ or $T$, then keep only the parts outside $J$.

Those are exactly three regions: coffee only, tea only, and coffee-and-tea only.

Add them.

(b) For the full union, use inclusion–exclusion on the two sets. The $30$ already covers everyone in both, including the triple overlap.

Check against the regions: the $74$ from part (a) plus the juice-containing parts of $C \cup T$, namely $C \cap J$ only $(6)$, $T \cap J$ only $(10)$ and the centre $(12)$.

First write out both sets from the universal set.

(a) The intersection: even and greater than $6$.

(b) Remove the shared elements from each set in turn.

(c) Anything left over from $\xi$ lies outside both circles.

Count the union, and check all regions sum to $10$.

(a)(i) Circle $O$ lies completely inside circle $M$: every orchestra member is also in the music club. $O$ is a subset of $M$.

(ii) Circles $O$ and $D$ do not touch, so no student is in both — their intersection is the empty set.

(b) $n(M)$ counts everything inside the large circle, including the $9$ inside $O$ and the $6$ shared with $D$.

All the regions together make $n(\xi) = 60$.

(c) $M' \cup D$ is everything outside $M$ (the $11$ and the $20$) together with all of $D$ (which adds the $6$). The only students left out are those in $M$ but not $D$: $9 + 14 = 23$.

(d) The student is chosen from the drama club, so the denominator is $n(D) = 6 + 11 = 17$. Of these, $6$ are in the music club.

(a) Add the four regions inside circle $A$.

Add the four regions inside circle $B$.

Set them equal and solve — this is the method mark.

(b) Substitute $x = 5$ into every region: $A$ only $10$, $B$ only $9$, $C$ only $14$, $A \cap B$ only $5$, $B \cap C$ only $5$, $A \cap C$ only $4$, centre $3$, outside $4$.

(c) "Exactly two" means the three regions where just two circles overlap — the centre is not included.

(d) Restrict to circle $C$: $n(C) = 14 + 5 + 4 + 3 = 26$. The part of $C$ that is also in $A$ is $4 + 3 = 7$.

Write out each set in full before checking anything.

(a) Test each number Sam has placed against all three definitions. $12$ is in $B$ only on his diagram, but $36 = 3 \times 12$, so $12 \in A$. It belongs in $A \cap B$ (not $C$).

$9$ is in $C$ only, but $36 = 4 \times 9$, so $9 \in A$. It is odd, so it belongs in $A \cap C$ (not $B$).

$16$ is in $C$ only, but $16$ is even, so $16 \in B$. It does not divide $36$, so it belongs in $B \cap C$ (not $A$). Every other number is correct — for example $4$ is a factor of $36$, even and square, so it is rightly in the centre.

(b) $A \cup C = \{1, 2, 3, 4, 6, 9, 12, 16\}$. Intersecting with $B'$ removes the even numbers.

(c) The square numbers in $\xi$ are $1, 4, 9, 16$. The odd ones, $1$ and $9$, are factors of $36$, so they are in $A$; the even ones, $4$ and $16$, are in $B$. Every square number is in $A$ or $B$, so none is in $C$ only.

(a) Read from the inside out. $P \cup Q$ is every region inside $P$ or $Q$. Intersecting with $R'$ keeps only the parts of those that lie outside circle $R$.

That leaves three regions to shade: $P$ only, $Q$ only, and the part where $P$ and $Q$ overlap outside $R$. Nothing inside circle $R$ is shaded — not even the centre.

(b) Add the numbers in the three shaded regions.

(c) First find how many students were asked — add all eight regions, including the $8$ outside the circles.

The probability is the shaded count out of the total. Simplify.

(d) $P' \cap Q'$ is everything outside both $P$ and $Q$: the $R$ only region and the region outside all three circles.