Question

A survey is done among 200 people. 60% like tea and 72% like coffee. Let \(x\) be the number who like both. Let \(m\le x\le n\). Find the correct option.

• A. \(n-m=56\)
• B. \(n-m=28\)
• C. \(n-m=32\)
• D. \(n+m=92\)

Hint:

For two-set survey problems: \[ \text{Minimum intersection}=n(A)+n(B)-N \] and \[ \text{Maximum intersection}=\min(n(A),n(B)). \]

Answer 1

The Correct Option is A

Step 1: Find how many like each drink.
Total people $= 200$.
Tea lovers $= 60\%$ of $200 = 120$.
Coffee lovers $= 72\%$ of $200 = 144$.

Step 2: Recall the union formula.
\[ n(A \cup B) = n(A) + n(B) - n(A \cap B) \]
Here $x = n(A \cap B)$ is the number who like both.

Step 3: Find the smallest possible $x$.
The union can be at most the whole group, $200$. So $120 + 144 - x \le 200$, which gives $x \ge 64$. So the minimum is $m = 64$.

Step 4: Find the largest possible $x$.
The both-group cannot be bigger than the smaller single group. The smaller group is tea $= 120$. So $x \le 120$, giving $n = 120$.

Step 5: So $x$ lies between these.
\[ 64 \le x \le 120 \]
Step 6: Compute $n - m$.
\[ n - m = 120 - 64 = 56 \]
This matches option (A).
\[ \boxed{n - m = 56} \]