A class of 44 students was surveyed about the subjects they took. Let M be the set of students who took Maths and E be the set of students who took English. Given that the number of students who took

Let $M$ be the set of candidates who took Maths and $E$ be the set of candidates who took English.
Given:
Total candidates $= 44$
Number of candidates who took Maths, $|M| = x$
Number of candidates who took English, $|E| = 3x$
Number of candidates who took both, $|M \cap E| = \frac{1}{3}|M| = \frac{1}{3}x$

The formula for the union of two sets is:

Assuming all 44 candidates took at least one subject:

To combine the terms on the right side, find a common denominator:

Multiply both sides by 3:

Divide by 11:

Number of candidates who took Maths, $|M| = 12$
Number of candidates who took English, $|E| = 3 \times 12 = 36$
Number of candidates who took both, $|M \cap E| = \frac{1}{3} \times 12 = 4$

Number of candidates who took Maths only: $12 - 4 = 8$
Number of candidates who took English only: $36 - 4 = 32$

• Draw two intersecting circles. Label one circle 'M' for Maths and the other 'E' for English.
• In the intersection region (M $\cap$ E), write the number of candidates who took both subjects: 4.
• In the part of circle M that does not overlap with E (Maths only), write the number of candidates who took Maths only: 8.
• In the part of circle E that does not overlap with M (English only), write the number of candidates who took English only: 32.