Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 7 · Chapter companion part
Count overlapping groups without double counting, apply set rules to factors and remainders, and recognise when a question lacks enough information.
Class 8 English Version: part C of the chapter 7 companion. Eight sections, 30 written tasks and a notebook challenge. Selected companion coverage, not complete textbook coverage; use all three parts. Counts are checked, while rosters, equivalent set-builder rules, diagrams and general proofs need guardian or teacher review.
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For a finite set A, n(A) denotes the number of its distinct elements, sometimes called its cardinality. In two finite sets, n(A∪B)=n(A)+n(B)−n(A∩B), because adding the separate counts counts shared elements twice. For A with 18 members, B with 15 and 6 shared, the union has 27. This identity counts the elements in these named groups; it does not measure their ability or quality.
If U has 40 members, n(A)=18,n(B)=15 and n(A∩B)=6, A-only has 12, B-only 9, both 6 and neither 13. These four disjoint regions total 40. Exactly one means A-only or B-only, giving 21. At least one means the union, 27. Neither means outside both, not merely outside one chosen set.
In an invented group of exactly 100 learners, 60% like activity A, 50% like B and 30% like both. That means 60,50 and 30 learners; 80 like at least one and 20 neither. In a group of 200 with those same proportions, the counts double. Source creative 6 gives percentages but no total: the actual student counts, and hence their factor sets, are not uniquely determined. Do not silently assume its population is 100.
The positive factors of 18 are {1,2,3,6,9,18}; its prime factors form {2,3}, with repeated prime occurrences still listed once. For 18 and 24, their common positive factors are {1,2,3,6}. Their largest common member is 6, the greatest common divisor. A union combines either number’s factors; an intersection selects factors dividing both.
For positive N divided by a positive integer d with remainder r, write N=qd+r with 0≤r<d. Thus d divides N−r and d>r. For N=74,r=4, the valid divisors are {5,7,10,14,35,70}; factors 1,2 are excluded by the remainder bound. For N=104,r=4, the valid divisors are {5,10,20,25,50,100}. The shared divisors are {5,10}. Divisibility of N−r alone is not sufficient.
For integer x with x²<17, the set is {−4,−3,−2,−1,0,1,2,3,4}. If the domain were positive naturals, only 1,2,3,4 would remain. Within U={1,...,8}, x²−5x+6=0 factors as (x−2)(x−3)=0, so its solution set is {2,3}. A condition and a domain are both needed to determine a roster.
Let P be the positive multiples of 6 and Q the positive multiples of 8. Their union begins {6,8,12,16,18,24,...} and continues indefinitely; specify the rule rather than giving a finite list as the whole set. Their intersection is the positive multiples of 24. A bounded task, such as x≤48, instead has a finite count. Source short 8(d) gives no finite upper bound, so its union is infinite under the stated positive-multiple interpretation.
If a question asks for the complement of (B∪C)ᶜ, it asks for ((B∪C)ᶜ)ᶜ=B∪C inside one fixed U. This is how the wording of source creative 7(c) reads; taking only one complement gives the opposite region. For B={1,3,5},C={2,3,5}, the union is {1,2,3,5}; one complement depends on U, while a second restores that union. When a task is ambiguous or missing a total, state the problem and any explicitly chosen illustration before calculating.
For a group of 40 with 18 in A,15 in B and 6 in both, the four disjoint regions are 12,9,6,13. Adding them gives the whole group, and adding 12+9+6 gives the union 27.
Use the numbers and membership labels. A larger coloured region does not mean more learners unless a chart explicitly establishes an area scale.
Listing multiples through 48 answers that bounded question only; it does not finish an unbounded union.
U has 40 members, A has 18, B has 15 and 6 are in both.
| Region | Calculation | Count |
|---|---|---|
| A only | 18−6 | 12 |
| Both | Given overlap | 6 |
| B only | 15−6 | 9 |
| Neither | 40−(12+6+9) | 13 |
The four disjoint region counts sum to 40. A-only plus both recovers A=18; B-only plus both recovers B=15.
For two finite sets, n(S) means the number of distinct members of S.
n(A∪B) = n(A)+n(B)−n(A∩B) = 18+15−6 = 27 Neither = 40−27 = 13
Exactly one is a different count: 12+9=21. Do not confuse it with at least one.
Require positive integer d and remainder 4.
| Number N | N−4 | Valid divisors d>4 |
|---|---|---|
| 74 | 70 | {5,7,10,14,35,70} |
| 104 | 100 | {5,10,20,25,50,100} |
| Shared valid divisors | Divide both differences; remain >4 | {5,10} |
A divisor of N−4 that is at most 4 fails the original remainder rule. Check the original division rather than only the difference.
Keep U={1,...,7},B={1,3,5},C={2,3,5}.
B∪C = {1,2,3,5} (B∪C)ᶜ = {4,6,7} ((B∪C)ᶜ)ᶜ = {1,2,3,5}
Do not silently give a single complement when the question names two. If wording is disputed, state both readings and their different results.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original membership model, not copied source artwork. Circle areas and distances do not encode member counts. Read the labels and element rosters.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original membership model, not copied source artwork. Circle areas and distances do not encode member counts. Read the labels and element rosters.
Show your working on paper. Check your answer, then choose Next.
27
18+15−6=27; shared members were counted twice before subtraction.
12
18−6=12 excludes the shared region.
9
15−6=9.
13
The union has 27, so 40−27=13 are outside both.
21
A-only 12 plus B-only 9 gives 21; both are excluded.
20
The union has 60+50−30=80 learners; 20 of the 100 remain.
60
Thirty out of every hundred gives 0.30×200=60 learners.
2
The set is {2,3}; repeated powers do not add members.
4
They are {1,2,3,6}.
6
The divisors of 70 greater than 4 are 5,7,10,14,35,70.
2
The shared valid divisors are 5 and 10.
9
They are the nine integers from −4 through 4, including zero.
2
The roots 2 and 3 are both in the stated U.
12
There are 8 multiples of 6 and 6 of 8, with 2 common multiples of 24; 8+6−2=12.
4
The second complement returns B∪C={1,2,3,5}.
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Adding n(A) and n(B) counts each shared member twice but every A-only or B-only member once. Subtract n(A∩B) once to count every union member exactly once.
Identify what is counted twice and why it is subtracted once.
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A-only 12, B-only 9, both 6, neither 13. The A circle totals 12+6=18, B totals 9+6=15 and all regions total 40.
Do not place the full set counts in the exclusive regions.
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The counts are 30,20,30,20 respectively. They total 100; at least one is 80 and exactly one is 50. This is an invented group with a stated total.
Explain that each percent becomes that many learners only because the total is 100.
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A missing population total leaves the counts unknown. For the source proportions 65%,55%,40% both, only-one groups are 25% and 15%. At total 100 their counts are 25,15 with common factors {1,5}; at total 200 they are 50,30 with common factors {1,2,5,10}. Both populations fit the proportions but give different factor sets.
State the missing total and label examples as illustrations, not the source’s actual headcount.
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A={1,2,3,6,9,18}; B={1,2,3,4,6,8,12,24}. Their intersection is {1,2,3,6}; union is {1,2,3,4,6,8,9,12,18,24}.
Check every member and count duplicates once.
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Its positive factors are {1,2,3,4,6,8,12,24}. Its distinct prime factors are {2,3}, since 24=2³×3. One is a factor but not prime.
Do not list repeated 2s as separate set elements.
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d divides 74−4=70 and d>4. From {1,2,5,7,10,14,35,70}, retain {5,7,10,14,35,70}. For instance 74=14×5+4.
Enforce the strict remainder bound as well as divisibility.
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The valid sets are {5,7,10,14,35,70} and {5,10,20,25,50,100}. Their intersection is {5,10}; its subsets are ∅,{5},{10},{5,10}.
Verify each common divisor in both original divisions.
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A remainder must be smaller than its positive divisor. Four is not smaller than 2, and the actual division 74÷2 has remainder 0. Divisibility of N−r without d>r does not suffice.
Use the original division as well as the bound.
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Integers give {−4,−3,−2,−1,0,1,2,3,4}; positive naturals give {1,2,3,4}. The inequality is the same but the domains differ.
Include negative values and zero when the domain is all integers.
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Factor (x−2)(x−3)=0 to get A={2,3}. Its proper subsets are ∅,{2},{3}; the whole A is not proper.
Check both roots belong to U and distinguish the two uses of braces.
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Union is {x∈N:6 divides x or 8 divides x}; it begins 6,8,12,16,18,24,... and is infinite. Intersection is {24,48,72,...}, the positive multiples of their least common multiple 24.
A few listed elements need an ellipsis and an explicit continuing rule.
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B∪C={1,2,3,5}. One complement gives {4,6,7}; complementing that set returns {1,2,3,5}. Source creative 7(c) asks for the complement of an already complemented expression, so read both operations.
Do not silently turn two complements into a single complement.
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The overlap must be a subset of each circle. Seven shared members cannot fit within a B set of only five. The B-only formula would give −2, showing that the supplied counts are inconsistent.
Explain the impossible membership relation rather than reporting a negative group size.
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Specify positive integers or another domain, the divisor rules, an inclusive or exclusive upper/lower bound, and the operation. Without a finite bound, a positive-multiple union is infinite and should not be marked by a finite total.
Give one well-defined bounded task and one clearly unbounded rule.
Build a four-region count model and a common-remainder divisor investigation.
Keep your work on paper or in a local file. This page does not upload it or automatically grade the project.
This original Kidemy practice follows the topic order of 2026 Class 8 Mathematics · English Version, chapter 7, printed pages 121–132. It does not replace the complete chapter or its practical teaching.
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