Question

In a class of 60, along with English as a common subject, students can opt to major in Mathematics, Physics, Biology or a combination of any two. 6 students major in both Mathematics and Physics, 15 major in both Physics and Biology, but no one majors in both Mathematics and Biology. In an English test, the average mark of \emph{Mathematics majors is 45 and that of \emph{Biology majors} is 60. The combined average of students of these two majors together is 50. What is the maximum possible number of students who major ONLY in Physics?}

• A. 30
• B. 25
• C. 20
• D. 15
• E. None of the above

Hint:

When a combined average of two groups is given, convert it to a weight relation: here $\;50=\frac{45m+60b}{m+b}\Rightarrow m=2b$. Then translate set information into variables and optimize under nonnegativity.

Answer 1

The Correct Option is D

Step 1: Notation.
Let $x=$ only Mathematics, $y=$ only Physics, $z=$ only Biology.
Given: $|M\cap P|=6$, $|P\cap B|=15$, and $|M\cap B|=0$. No triple majors.
Total students: \[ x+y+z+6+15=60 \;\Rightarrow\; x+y+z=39. \tag{1} \]

Step 2: Use the average condition.
Let $m=$ number of Mathematics majors (only $M$ or $M\cap P$) and $b=$ number of Biology majors (only $B$ or $P\cap B$). Then \[ m=x+6,\qquad b=z+15. \] Combined average of these two groups is $50$: \[ \frac{45m+60b}{m+b}=50 \;\Rightarrow\; 45m+60b=50(m+b) \;\Rightarrow\; 10b=5m \;\Rightarrow\; m=2b. \tag{2} \] Using $m=x+6$ and $b=z+15$ in (2): \[ x+6=2(z+15)\;\Rightarrow\; x=2z+24. \tag{3} \]

Step 3: Express $y$ and maximize.
From (1): $y=39-x-z$. Substitute (3): \[ y=39-(2z+24)-z=15-3z. \] To maximize $y$, minimize $z$ subject to $z\ge 0$ (and $x\ge 0$ holds automatically).
The minimum feasible $z$ is $0$, giving \[ y_{\max}=15-3\cdot 0=15. \] (Then $x=2\cdot 0+24=24$, and the counts are nonnegative and consistent.)

Final Answer: \[ \boxed{\text{(D) 15}} \]