Question:hard

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?

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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.
Updated On: Jul 20, 2026
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The Correct Option is

Solution and Explanation

Here is an alternative route using the inclusion-exclusion formula directly instead of building the Venn diagram region by region.

Let $M$, $P$, $C$ be the sets of students studying mathematics, physics and chemistry, with $n(M\cap P\cap C)=5$ (from "three less than one-fifth of 40").

We are told $n(M\cap P) = n(M\cap C)$; call this common value $k$. We want $n(M\cap C) - n(M\cap P\cap C) = k - 5$, so it is enough to find $k$.

Since only-M $=3$ and all-three $=5$:
$n(M) = \text{only-M} + [n(M\cap P)-5] + [n(M\cap C)-5] + 5 = 3 + (k-5)+(k-5)+5 = 2k-2$

Physics total is 20, and only-P $= n(M)-2$:
$n(P) = \text{only-P} + [n(M\cap P)-5] + [n(P\cap C)-5] + 5 = 20$
Also, using the grand total of 40 with only-M$=3$, only-C$=15$:
$40 = 3+\text{only-P}+15+(k-5)+(k-5)+n(P\cap C \text{ only})+5$
$40 = 18+\text{only-P}+2k-10+z+5 \Rightarrow \text{only-P} = 27-2k-z$ ... where $z=n(P\cap C)-5$
From physics total: $\text{only-P}+(k-5)+z+5=20 \Rightarrow \text{only-P}=15-k-z$
Equating: $27-2k-z=15-k-z \Rightarrow 27-2k=15-k \Rightarrow k=12$? This does not match, so instead solve the two-variable system directly:
only-P$+k-5+z+5=20$ and $\text{only-P}+2(k-5)+z=17$ (from grand total, since $x=y=k-5$). Subtracting: $(k-5)-0 = 3 \Rightarrow k-5=2 \Rightarrow k=7$.

So $n(M\cap C) = k = 7$, and the region "M and C but not all three" $= k-5 = 7-5=2$.
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