SETS AND VENN DIAGRAMS

9 students (the region inside Chess only).

5 students (the overlapping region).

$$9+5+6+4 = 24 \text{ students}$$

$X\cap Y=\{6\}$. $X\cup Y=\{2,3,4,6,8,9,10\}$.

Elements of $U$ not in $X\cup Y$: $\{1,5,7\}$ — 3 elements.

$n(A\cup B) = n(A)+n(B)-n(A\cap B) = 10+10-0=20$.

$n(A\cup B)=10+10-10=10$.

As the overlap size increases, the union size decreases (for fixed $n(A)$ and $n(B)$) — more shared elements means fewer NEW elements are added by the second set, right down to a minimum union size equal to the larger individual set when full overlap occurs.

Add the 'Dogs only' region and the overlap region together: $3+7=10$ families own dogs (some of whom also own cats).

Add all three visible regions ($3+7+5=15$) and subtract from the total surveyed: $20-15=5$ families own neither.

$$80-15=65 \text{ members}$$

$n(W\cup C)=n(W)+n(C)-n(W\cap C) \Rightarrow 65=52+38-n(W\cap C) \Rightarrow n(W\cap C)=90-65=25$. So: weights only $=52-25=27$, cardio only $=38-25=13$, both $=25$.

15 students.

$$15+22+8=45 \text{ students}$$

$$15+22+8+5=50$$ ? — matches.

$$45+30=75 \text{ students}$$

$$\frac{7}{100}\times100=7\%$$

$A$ only $=30$, $B$ only $=25$, $A\cup B=30+25-0=55$.

$A$ only $=30-15=15$. $B$ only $=25-15=10$. $A\cup B=30+25-15=40$.

As overlap increases, BOTH 'only' regions shrink (since more of each set's elements are being 'shared' rather than exclusive), and the total union shrinks too. Visually, this makes sense because a bigger overlapping middle region means the two circles are covering MORE of the same space, so their combined (union) area covers LESS total ground than if they barely overlapped.

$n(A\text{ only})=20-25=-5$ — a NEGATIVE value, which is impossible for a real region of a Venn diagram (you can't have $-5$ elements).

This reveals the original claim is IMPOSSIBLE: the intersection $A\cap B$ is a SUBSET of $A$ itself, so it can never contain MORE elements than $A$ does. Since $n(A)=20$, we must have $n(A\cap B)\le20$ — the claimed value of 25 violates this basic requirement, which is exactly why the calculation produced a nonsensical negative result.

Total accounted for by known regions plus 'neither': $12+20+8=40$. Since the grand total is 60, the missing 'Tea only' region must be $60-40=20$.

The classmate's method missed subtracting the 8 people who drink neither — a Venn diagram for 2 sets within a universal set always has FOUR distinct regions (not three), and the total must account for every person, including those OUTSIDE both circles. Forgetting the 'neither' region leads to an answer (28) that's too large by exactly the size of that missed region (8), since $28-8=20$, the correct answer.

The number 9 represents $n(\text{Guitar only})$ — students who play guitar but NOT piano. The number 14 represents $n(\text{Guitar}\cap\text{Piano})$ — students who play BOTH instruments. The number 6 represents $n(\text{Piano only})$ — students who play piano but NOT guitar. Together, $n(\text{Guitar})=9+14=23$ and $n(\text{Piano})=14+6=20$.

24 volunteers (the 'Project A only' region).

New recruits needed: $12\times3=36$. Total people involved: $12\text{ leaders}+36\text{ new}=48$. The 8 people who help with 'neither' project currently should NOT be considered for 'new recruit' status if the goal is genuinely NEW volunteers from outside the current 60 — although they aren't in A or B, they ARE still part of the existing 60 registered volunteers, so recruiting them wouldn't bring in truly NEW people to the organization.

Households with neither: $120-38-45-22=15$. Budget needed: $15\times35=\$525$.

Shortfall: $525-400=\$125$. One option: reduce the number of kits — with \$400 available, $400\div35\approx11.4$, so the council could afford kits for only 11 of the 15 qualifying households (prioritizing perhaps by need or by first-come-first-served), leaving 4 households without a kit under the current budget.