Operations on Sets · 4.1.3
Solving problems with intersection of sets
Students apply their understanding of set intersection to solve real-life or numerical problems, such as finding common elements between groups, using given information to determine unknown sets, or working with the number of elements in overlapping sets.
The official learning standard (4.1.3)
“Solve problems involving the intersection of sets.”
What it means
Students apply their understanding of set intersection to solve real-life or numerical problems, such as finding common elements between groups, using given information to determine unknown sets, or working with the number of elements in overlapping sets.
How it is examined
This standard is examined mainly in Paper 2 as structured or problem-solving questions set in real-life contexts, such as surveys about student preferences, often requiring the number of elements, n(A ∩ B), to be found using a Venn diagram before answering a related question.
Worked example
In a class of 40 students, 25 like Mathematics and 20 like Science. If 10 students like both subjects, find the number of students who like neither Mathematics nor Science.
- Let M be the set of students who like Mathematics and S the set who like Science. n(M) = 25, n(S) = 20, n(M ∩ S) = 10.
- Number who like at least one subject: n(M ∪ S) = n(M) + n(S) − n(M ∩ S) = 25 + 20 − 10 = 35.
- Total students = 40, so students who like neither = 40 − n(M ∪ S) = 40 − 35 = 5.
Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)
Frequently asked questions
Why do we subtract n(A ∩ B) when finding the total number in A or B?
Because elements in the intersection belong to both sets, simply adding n(A) and n(B) counts them twice. Subtracting n(A ∩ B) removes this double-counting, so n(A ∪ B) = n(A) + n(B) − n(A ∩ B) gives the correct total of distinct elements.
How do I set up a Venn diagram when given only totals, not element lists?
Draw two overlapping circles inside a rectangle for the universal set. Fill in n(A ∩ B) in the overlap first, then subtract this from each given total to find the 'A only' and 'B only' regions, and finally use the grand total to find anyone outside both circles.
What if a problem gives three overlapping sets instead of two?
The same logic extends, but you must work from the centre outward: start with the innermost triple intersection, then find each pairwise overlap minus that centre value, and finally each set's remaining 'only' region, checking that all regions sum to the given total.