Question

In a class of 40 students, who study mathematics, physics and chemistry, the number of students studying mathematics is 2 more than 40% of those studying chemistry, while 20 of the students study physics. Three less than one fifth of the total students in the class study all the three subjects. The number of students studying only physics is 2 less than the number of students studying mathematics. The number of students studying only mathematics and only chemistry is 3 and 15 respectively. The number of students studying mathematics as well as physics is the same as the number of students studying mathematics as well as chemistry. How many students are studying both mathematics and chemistry but not all the three subjects?

• A. 25
• B. 15
• C. 10
• D. 3
• E. 2

Hint:

Let x = only M and P, y = only M and C, z = only P and C, with the "all three" region fixed at 5. The condition n(M∩P) = n(M∩C) forces x = y; combine the physics-total and grand-total equations to solve for x directly.

Answer 1

Answer: (E)

Step 1: Set up the Venn diagram regions.
Let only-M = 3, only-C = 15 (given). Let "all three" = \(\frac{1}{5}(40)-3 = 8-3 = 5\).
Let \(x\) = students studying M and P only (not C), \(y\) = students studying M and C only (not P), \(z\) = students studying P and C only (not M).
Given: "M as well as P" = "M as well as C", i.e. \(x+5 = y+5\), so \(x=y\).

Step 2: Express the total and physics-total equations.
Total: only-M + only-P + only-C + \(x+y+z\) + all-three = 40
\(3+\text{only-P}+15+x+y+z+5=40 \Rightarrow \text{only-P}+x+y+z=17\)
Since \(x=y\): \(\text{only-P}+2x+z=17\) ... (i)

Physics total = 20: only-P + \(x+z\) + 5 = 20 \(\Rightarrow\) only-P + \(x+z\) = 15 ... (ii)

Step 3: Solve for x.
Subtracting (ii) from (i): \(2x - x = 17-15 \Rightarrow x = 2\), so \(y=2\) as well.

Step 4: Cross-check using the mathematics-total condition (not required for the final answer, but confirms consistency).
Mathematics total \(=\) only-M \(+x+y+\)all-three \(=3+2+2+5=12\)
only-P \(=\) M-total \(-2=12-2=10\)
From (ii): \(10+2+z=15\wedge... \) actually \(10+x+z=15 \Rightarrow 10+2+z=15\Rightarrow z=3\)
Chemistry total \(=\) only-C \(+y+z+\)all-three \(=15+2+3+5=25\)
Check: \(0.4\times25+2=10+2=12\), which matches M-total \(=12\), confirming all values are consistent.
Overall total check: \(3+10+15+2+2+3+5=40\ ✓\)

Step 5: Answer the question.
"Both mathematics and chemistry but not all three" is exactly the region \(y\), which was found to be 2.
So the answer is 2, matching option (e).