SETS AND VENN DIAGRAMS

$$35-7=28 \text{ students}$$

$n(F\cup S)=n(F)+n(S)-n(F\cap S) \Rightarrow 28=22+18-n(F\cap S) \Rightarrow n(F\cap S)=40-28=12$

$$22-12=10 \text{ students}$$

$$60-11=49$$

$n(C\cap T)=34+29-49=14$. Tea only $=29-14=15$.

$$25+20+18-10-7-6+8 = 48$$

$$50-48=2 \text{ students}$$

Students in all three clubs get counted three times in $n(A)+n(D)+n(R)$, then subtracted three times (once in each pairwise overlap) — leaving them counted zero times. Adding back $n(A\cap D\cap R)$ once corrects this, so they end up counted exactly once, as they should be.

Classmate's formula: $22+18+7=47$, which does not equal the actual total of 35. It's wrong because students who study BOTH languages get counted twice — once in the French count, once in the Spanish count — so the overlap must be subtracted to correct for this.

$$\text{total} = n(F) + n(S) - n(F\cap S) + \text{neither}$$

$n(T)=2x$. Using $\text{total}=n(L)+n(T)-n(\text{both})+\text{neither}$: $$200 = x + 2x - 30 + 20$$

$200=3x-10 \Rightarrow 3x=210 \Rightarrow x=70$. So $n(L)=70$ and $n(T)=2\times70=140$. Check: $70+140-30+20=200$ ?

Both $=90-25-30-8=27$.

French total $=25+27=52$.

$n(T\cup W)=150-20=130$. $n(T\cap W)=95+70-130=35$.

Exactly one $=130-35=95$ (i.e. the union minus the overlap).

$$110+95+120-45-50-40+20=210$$ This does NOT match the stated total of 200 — suggesting an inconsistency in the given data (the numbers as stated are not mutually consistent with 200 total attendees).

Possible explanations: measurement/counting errors in the original tallying, some attendees being double-counted across categories, or the sub-totals being independently (and imperfectly) estimated rather than precisely cross-tabulated. This demonstrates why VERIFYING that detailed breakdown numbers actually reconcile with a known total is an essential check — inconsistent data can lead to flawed conclusions if used without this kind of validation.

$n(T\cup C)=60-10=50$. $n(T\cap C)=40+35-50=25$ — sensible, since 25 is less than both 40 and 35 (the both-group can't exceed either individual group).

$n(T\cup C)=60-5=55$. $n(T\cap C)=40+35-55=20$ — actually still achievable numerically (20 is still less than both 40 and 35), so this particular change does NOT cause a problem. This shows that not every change in given data leads to an impossible result — the key danger zone is when the calculated intersection would need to EXCEED one of the individual set sizes, which requires more careful checking of the SPECIFIC numbers involved, not just assuming any change causes trouble.

Every region of a Venn diagram represents a COUNT of real people/objects, which can never be negative (the smallest possible count for any group is 0, representing an empty group). A negative result like $-3$ is mathematically impossible for a real-world count, so it must indicate either an arithmetic error in the working, or that the ORIGINAL problem's given numbers are inherently inconsistent (e.g. the stated totals don't actually fit together logically).

Check 1: re-verify all arithmetic steps carefully, since simple calculation errors are the most common cause. Check 2: verify the ORIGINAL given data is internally consistent — e.g. that no individual set's size is smaller than a claimed overlap involving that set, and that all given totals genuinely add up correctly across all four regions of a 2-set Venn diagram (or eight regions for 3 sets).

Let $X=\{\text{respondents who like Brand X}\}$ and $Y=\{\text{respondents who like Brand Y}\}$, with $n(X)=55$, $n(Y)=48$, and $n(X\cup Y)=60$. Using the formula $n(X\cup Y)=n(X)+n(Y)-n(X\cap Y)$: $$60=55+48-n(X\cap Y) \Rightarrow n(X\cap Y)=103-60=43$$ Therefore, 43 respondents like both Brand X and Brand Y.

Clearly defining sets and notation at the start makes the ENTIRE solution understandable to someone reading it independently — it removes ambiguity about what each symbol represents, allows the reader to follow the logical structure of the argument, and demonstrates that the solver understands the underlying mathematical framework being used, not just performing calculations by rote.

Without labels or explanation, a reader cannot tell WHAT each number represents, WHY each operation was performed, or WHICH Venn diagram region or set relationship is being calculated at each step — good mathematical communication requires the WORKING to be understandable on its own, not just the final answer to be correct. A grader or reader has no way to verify the REASONING was sound, only that a number happened to be right.

Example: 'Total surveyed: 80. Number who like neither product: 15, so number who like at least one: $80-15=65$. Given $n(A)=50$: $65+50=115$ represents an intermediate sum before removing double-counted overlap. Given $n(A\cup B)=95$ was actually the true union... [continuing to derive] $n(A\cap B)=115-95=20$, so 20 people like both products.' (Any well-labeled, logically explained version demonstrating clear communication is acceptable.)

$n(X\cup B)=130+95-45=180$.

X-ray capacity needed: 130 patients. 2 technicians $\times8\text{h}\times8\text{ patients/h}=128$ patients — INSUFFICIENT (130 needed, only 128 capacity, short by 2). Blood test capacity needed: 95 patients. 1 phlebotomist $\times8\text{h}\times12\text{ patients/h}=96$ patients — sufficient (barely, with 1 to spare). The BOTTLENECK is X-ray staffing, which falls just short of meeting demand within a single 8-hour shift; the hospital should consider adding staff or extending X-ray hours specifically, while blood test staffing is adequate.

$n(M\cup S)=400-60=340$. $n(M\cap S)=220+180-340=60$.

Dual-enrolled: 60 students. Room capacity: 50. Percentage accommodated: $\frac{50}{60}\times100\approx83.3\%$. Since not all 60 can fit, a fair selection method could be a random lottery/draw among the 60 dual-enrolled students, or a first-come-first-served registration system — either approach avoids favoritism and gives every eligible student an equal, transparent chance at one of the 50 spots.

Maximum $n(A\cap B)=18$ — this occurs if EVERY element of $A$ is also in $B$ (i.e. $A$ is entirely a subset of $B$); the overlap can never exceed the size of the smaller set.

Minimum $n(A\cap B)=0$ — this occurs if $A$ and $B$ share no elements at all (disjoint sets); an intersection can never be negative, so 0 is always achievable and is the theoretical floor.

Without shown working, there is no way to verify HOW the answer was reached, whether the correct method was used, or whether the student genuinely understands the underlying relationship (versus guessing or recalling a memorized number) — clear working demonstrates the LOGICAL PATH to a solution, which is central to communicating mathematics effectively, not just stating a result.

$84-38-29=17$ donors.

Cost for all 17 dual donors: $17\times12=\$204$, which EXCEEDS the \$150 budget by $204-150=\$54$. Maximum donors affordable: $150\div12=12.5\to12$ donors. So $17-12=5$ donors would need to be excluded from receiving the special gift to stay within budget.

Both $=120-45=75$ employees.

Total training time: $75\times30\text{ min}=2250$ minutes $=37.5$ hours. With 1 trainer and a 40-hour week, this is JUST barely feasible ($37.5<40$), leaving only 2.5 hours of slack — this is a tight but technically achievable schedule, though it leaves little room for other trainer duties, breaks, or unexpected delays, suggesting the company might benefit from a second trainer or spreading the training across more than one week for a more comfortable margin.