BROTHER CAN YOU EXPLAIN THIS A BIT WHY THIS IS WRONG

Total students, $n(C \cup H \cup B) = 280$

$n(C) = 160$

$n(H) = 130$

$n(B) = 110$

Let the number of students who study all three subjects be $x$.

$n(C \cap H \cap B) = x$

The problem states that students choosing more than one subject is $40\%$ more than $x$.

"More than one" means (Exactly 2 subjects) + (Exactly 3 subjects).

• $\text{More than one} = x + \frac{40}{100}x = 1.4x$
• Therefore, students who study exactly two subjects $= 1.4x - x = 0.4x$

The standard formula for three sets is:

$n(C \cup H \cup B) = n(C) + n(H) + n(B) - [n(C \cap H) + n(H \cap B) + n(C \cap B)] + n(C \cap H \cap B)$

In a Venn diagram, the sum of the dual intersections $[n(C \cap H) + n(H \cap B) + n(C \cap B)]$ actually counts the "exactly two" region once and the "exactly three" region three times.

$\text{Sum of intersections} = (\text{Exactly 2}) + 3 \times (\text{Exactly 3})$

$\text{Sum of intersections} = 0.4x + 3x = 3.4x$

Now, substitute these values into the main formula:

$280 = 160 + 130 + 110 - (3.4x) + x$

$280 = 400 - 3.4x + x$

$280 = 400 - 2.4x$

$2.4x = 400 - 280$

$2.4x = 120$

$x = \frac{120}{2.4} = \frac{1200}{24}$

$x = 50$