Midsem Maths Study Guide
Math-I mid-sem · tonight's study guide

Midsem Maths Study Guide

Work top to bottom. For each question type: read the method, follow the worked example, then do every practice question on paper before opening its solution. Write Part B style answers (full working) for anything marked Part B.

0:00 – 0:10Skim the cheatsheet
0:10 – 0:45Sets & Probability
0:45 – 1:20Matrices & Determinants
1:20 – 2:30Logic (biggest Part B block)
2:30 – 3:00Counting & Binomial
3:00 – 3:35Calculus
3:35 – 4:05Stats, mental maths, word problems
4:05 – 4:15AP / GP quick pass
4:15 – 5:00Mock test, then check solutions
Topic 1 · Part A

Sets

You knew this once. Five question types bring it all back. Notation reminder: ∈ "belongs to", ⊆ subset, ∪ union (or), ∩ intersection (and), A′ complement, A − B difference.

1.1Roster ↔ set-builder

  1. Read the domain first: N = {1, 2, 3, …}, Z = all integers (negatives and 0 too), R = all reals.
  2. Read the condition after the bar | ("such that").
  3. Walk through the domain and keep only the values that pass. Watch < vs ≤ at the ends.
  4. Going the other way (roster → builder): spot the rule (squares, evens, divisors), then write {x | x = rule, n ∈ N, limits}.
Worked example

B = {x ∈ Z | −2 ≤ x < 3}. Domain Z, so negatives allowed. −2 is included (≤), 3 is excluded (<). B = {−2, −1, 0, 1, 2}, so |B| = 5.

Practice (7)
  1. {x ∈ N | x² < 30} in roster form.
  2. {x ∈ Z | x² = 9} in roster form.
  3. {x ∈ Z | −3 < x ≤ 2} in roster form, and its cardinality.
  4. Write {1, 4, 9, 16, 25, 36} in set-builder form.
  5. Write {2, 4, 6, 8, 10} in set-builder form.
  6. {x ∈ N | x divides 12} in roster form.
  7. {x ∈ R | x² + 1 = 0} in roster form.
Show solutions
  1. 1²=1, 2²=4, 3²=9, 4²=16, 5²=25 pass; 6²=36 fails. {1, 2, 3, 4, 5}
  2. x = 3 or x = −3 (Z allows negatives). {−3, 3}
  3. −3 excluded, 2 included: {−2, −1, 0, 1, 2}, cardinality 5.
  4. They are squares of 1 to 6: {x | x = n², n ∈ N, n ≤ 6}
  5. Evens up to 10: {x | x = 2n, n ∈ N, 1 ≤ n ≤ 5}
  6. {1, 2, 3, 4, 6, 12}
  7. No real number squares to −1. ∅ (empty set)

1.2Operations: ∪, ∩, −, ′, ⊕, ×, subsets

  1. Write U, A and B in a column so you can scan them.
  2. ∪: everything in either (no repeats). ∩: only what is in both.
  3. A − B: start with A, delete anything also in B. (B − A is different.)
  4. A′: everything in U that is not in A.
  5. A ⊕ B = (A − B) ∪ (B − A): in exactly one of them.
  6. A × B: every ordered pair (a, b); count = |A|·|B|. Subsets of an n-element set: 2ⁿ; proper subsets: 2ⁿ − 1.
Worked example

U = {1,…,10}, A = {1,2,3,4,5}, B = {4,5,6,7}.

A∪B = {1,…,7} · A∩B = {4,5} · A−B = {1,2,3} · B−A = {6,7} · A′ = {6,7,8,9,10} · A⊕B = {1,2,3,6,7}.

Practice (7)

Use U = {1, 2, …, 10}, P = {1, 3, 5, 7, 9}, Q = {2, 3, 5, 7}.

  1. P ∪ Q
  2. P ∩ Q
  3. P − Q and Q − P. Are they equal?
  4. (P ∪ Q)′, then check it equals P′ ∩ Q′ (De Morgan).
  5. P ⊕ Q
  6. X = {0, 1}, Y = {a, b, c}: list X × Y. If |A| = 4, |B| = 3, |C| = 5, find |A × B × C|.
  7. C = {a, b, c, d}: how many subsets, proper subsets, and subsets containing a?
Show solutions
  1. {1, 2, 3, 5, 7, 9}
  2. {3, 5, 7}
  3. P − Q = {1, 9}, Q − P = {2}. Not equal, so difference is not commutative.
  4. (P∪Q)′ = {4, 6, 8, 10}. P′ = {2,4,6,8,10}, Q′ = {1,4,6,8,9,10}, P′∩Q′ = {4,6,8,10}. Same. ✓
  5. (P−Q) ∪ (Q−P) = {1, 2, 9}
  6. {(0,a), (0,b), (0,c), (1,a), (1,b), (1,c)}: 6 pairs. |A×B×C| = 4·3·5 = 60.
  7. 2⁴ = 16 subsets, 15 proper. Containing a: decide b, c, d freely, 2³ = 8.

1.3Two-set Venn word problems

  1. Translate the sentence into notation: |U|, |A|, |B|, and "neither" = |(A∪B)′|.
  2. If "neither" is given: |A∪B| = |U| − neither.
  3. Both: |A∩B| = |A| + |B| − |A∪B|.
  4. Only A = |A| − |A∩B|; only B = |B| − |A∩B|.
  5. Check: only A + both + only B + neither = |U|.
Worked example

100 CS students: 65 know Python, 45 know Java, 10 know neither. |P∪J| = 100 − 10 = 90. |P∩J| = 65 + 45 − 90 = 20. Only Python = 45, only Java = 25. Check: 45 + 20 + 25 + 10 = 100 ✓.

Practice (7)
  1. 50 people: 30 drink tea, 25 drink coffee, 10 drink both. How many drink neither?
  2. 120 students: 70 play cricket, 50 football, 30 both. How many play only cricket? How many play neither?
  3. 200 people: 120 use Android, 90 use iOS, 30 use neither. How many use both?
  4. |A ∪ B| = 35, |A| = 20, |B| = 25. Find |A ∩ B| and |A − B|.
  5. Class of 40: 25 like Maths, 18 like Physics, 7 like neither. How many like exactly one subject?
  6. A and B are disjoint, |A| = 12, |B| = 9. Find |A ∪ B|.
  7. |U| = 60, |A| = 40, |B| = 30. What are the smallest and largest possible values of |A ∩ B|?
Show solutions
  1. |T∪C| = 30 + 25 − 10 = 45. Neither = 50 − 45 = 5.
  2. Only cricket = 70 − 30 = 40. Union = 70 + 50 − 30 = 90, neither = 30.
  3. Union = 200 − 30 = 170. Both = 120 + 90 − 170 = 40.
  4. |A∩B| = 20 + 25 − 35 = 10. |A − B| = 20 − 10 = 10.
  5. Union = 40 − 7 = 33. Both = 25 + 18 − 33 = 10. Exactly one = 33 − 10 = 23.
  6. Disjoint means |A∩B| = 0, so 21.
  7. Largest: B fits inside A, so 30. Smallest: |A∪B| can be at most 60, so |A∩B| ≥ 40 + 30 − 60 = 10.

1.4Three-set Venn word problems

  1. Draw three circles. Fill from the centre outwards: first |A∩B∩C|.
  2. "Exactly A and B" = |A∩B| − centre. Do all three pairs.
  3. "Only A" = |A| − (A∩B only) − (A∩C only) − centre. Do all three.
  4. Union = |A|+|B|+|C| − |A∩B| − |B∩C| − |A∩C| + |A∩B∩C|. Neither = |U| − union.
Worked example

100 students: Maths 50, Physics 40, Chem 30; M∩P 20, P∩C 10, M∩C 15; all three 5.
Union = 50+40+30 − 20−10−15 + 5 = 80, neither = 20. M∩P only = 15, M∩C only = 10, so only M = 50 − 15 − 10 − 5 = 20.

Practice (7)

150 people surveyed: Netflix N = 80, Prime A = 60, Hotstar H = 50; N∩A = 30, A∩H = 20, N∩H = 25; all three = 10.

  1. How many use at least one service?
  2. How many use none?
  3. How many use exactly Netflix and Prime (not Hotstar)?
  4. How many use only Netflix?
  5. How many use only Prime? Only Hotstar?
  6. How many use exactly one service?
  7. How many use exactly two services?
Show solutions
  1. 80 + 60 + 50 − 30 − 20 − 25 + 10 = 125
  2. 150 − 125 = 25
  3. 30 − 10 = 20
  4. Pair-onlys: N∩A only = 20, N∩H only = 15, A∩H only = 10. Only N = 80 − 20 − 15 − 10 = 35
  5. Only A = 60 − 20 − 10 − 10 = 20. Only H = 50 − 15 − 10 − 10 = 15
  6. 35 + 20 + 15 = 70
  7. 20 + 15 + 10 = 45. Check: 70 + 45 + 10 = 125 ✓

1.5Regions, shading and De Morgan

  1. Translate symbols to words: ∩ = "and", ∪ = "or", ′ = "not", − = "but not".
  2. Read the expression as a sentence and shade that region.
  3. For a complement of a bracket, use De Morgan: flip ∩↔∪ and complement each set.
Worked example

(A ∩ B)′ = A′ ∪ B′: "not in both" = everything except the lens where A and B overlap. This was mock Q38 (answer B).

Practice (7)
  1. Describe the region A ∩ B′.
  2. Describe (A ∪ B)′.
  3. A′ ∩ B′ equals which single expression?
  4. Three sets: describe A ∩ B ∩ C′.
  5. Describe A ∪ (B ∩ C).
  6. Simplify A ∩ (A ∪ B).
  7. Simplify (A′)′ ∪ ∅.
Show solutions
  1. In A and not in B: the only-A crescent (= A − B).
  2. Outside both circles (neither).
  3. (A ∪ B)′ by De Morgan.
  4. The overlap of A and B that lies outside C.
  5. All of A, plus the B∩C lens.
  6. A (absorption: anything in A is already in A ∪ B).
  7. A (double complement, then union with empty set changes nothing).
Topic 2 · Part A + Part B Q5

Probability with sets

Every probability formula is a set formula with |·| replaced by P(·). The mock asked 2 MCQs here and a full 15-mark question mixed with combinations (that one is in Topic 6).

2.1Classical probability and the complement

  1. List or count the sample space n(S). Two dice = 36, three coins = 8, a deck = 52.
  2. Count the favourable outcomes n(E).
  3. P(E) = n(E)/n(S); simplify the fraction.
  4. For "at least one", find P(none) and use 1 − P(none).
Worked example

Two dice, P(sum = 7): (1,6) (2,5) (3,4) (4,3) (5,2) (6,1) = 6 outcomes. 6/36 = 1/6.

Practice (7)
  1. A fair 20-sided die is rolled. P(multiple of 4)?
  2. A coin is tossed 3 times. Write Ω and find P(exactly 2 heads).
  3. Same coin: P(at least one head).
  4. Two dice: P(sum ≥ 10).
  5. Two dice: P(doubles).
  6. One card from a deck: P(king or heart).
  7. A classifier is correct with probability 0.92. P(it is wrong)?
Show solutions
  1. 4, 8, 12, 16, 20 = 5 outcomes. 5/20 = 1/4
  2. Ω = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. Exactly 2 heads: HHT, HTH, THH. 3/8
  3. P(no heads) = 1/8, so 7/8
  4. Sum 10: 3 ways, 11: 2, 12: 1 → 6/36 = 1/6
  5. 6/36 = 1/6
  6. 4/52 + 13/52 − 1/52 (king of hearts counted twice) = 16/52 = 4/13
  7. 1 − 0.92 = 0.08

2.2Union, intersection, difference (the mock Q39 type)

  1. Write the formula P(A∪B) = P(A) + P(B) − P(A∩B).
  2. Substitute the three values you know and solve for the fourth.
  3. Only A = P(A) − P(A∩B). Neither = 1 − P(A∪B). Exactly one = P(A∪B) − P(A∩B).
  4. Mutually exclusive means P(A∩B) = 0.
Worked example (mock Q39)

P(X) = 0.7, P(Y) = 0.4, P(X∪Y) = 0.9. Then 0.9 = 0.7 + 0.4 − P(X∩Y), so P(X∩Y) = 1.1 − 0.9 = 0.2 (option B).

Practice (7)
  1. P(A) = 0.5, P(B) = 0.4, P(A∩B) = 0.2. Find P(A∪B).
  2. Same events: P(neither)?
  3. Same events: P(A only) and P(B only)?
  4. P(A∪B) = 0.8, P(A) = 0.6, P(A∩B) = 0.25. Find P(B).
  5. A, B mutually exclusive, P(A) = 0.3, P(B) = 0.45. Find P(A∪B).
  6. P(CPU spike) = 0.3, P(memory leak) = 0.2, P(both) = 0.05. P(at least one)? P(exactly one)?
  7. P(A) = 0.6, P(B) = 0.5. Can A and B be mutually exclusive?
Show solutions
  1. 0.5 + 0.4 − 0.2 = 0.7
  2. 1 − 0.7 = 0.3
  3. A only = 0.5 − 0.2 = 0.3; B only = 0.4 − 0.2 = 0.2
  4. 0.8 = 0.6 + P(B) − 0.25 → P(B) = 0.45
  5. 0.3 + 0.45 = 0.75
  6. At least one = 0.3 + 0.2 − 0.05 = 0.45; exactly one = 0.45 − 0.05 = 0.40
  7. No. P(A∩B) ≥ 0.6 + 0.5 − 1 = 0.1 > 0.

2.3Conditional probability and independence

  1. Conditional: P(A|B) = P(A∩B)/P(B). B is the new "universe".
  2. Independence test: compute P(A)·P(B) and compare with P(A∩B). Equal ⇒ independent.
  3. From a table: divide the cell by the row or column total for a conditional, by the grand total for a joint probability.
Worked example

P(A) = 0.4, P(B) = 0.5, P(A∩B) = 0.2. P(A)·P(B) = 0.2 = P(A∩B), so independent. P(A|B) = 0.2/0.5 = 0.4 = P(A): knowing B changed nothing.

Practice (7)
  1. P(A∩B) = 0.12, P(B) = 0.3. Find P(A|B).
  2. P(A) = 0.3, P(B) = 0.6, P(A∩B) = 0.18. Independent?
  3. P(A) = 0.5, P(B) = 0.4, P(A∪B) = 0.7. Independent?
  4. Two dice. A = "first die shows 6", B = "sum is 7". Independent?
  5. 200 people: 120 men (90 drink coffee), 80 women (50 drink coffee). Find P(coffee), P(coffee | woman), P(woman ∩ coffee). Are "woman" and "coffee" independent?
  6. A, B independent with P(A) = 0.3, P(B) = 0.5. Find P(A∪B).
  7. Why can two mutually exclusive events with non-zero probability never be independent?
Show solutions
  1. 0.12/0.3 = 0.4
  2. 0.3 × 0.6 = 0.18 = P(A∩B). Independent.
  3. P(A∩B) = 0.5 + 0.4 − 0.7 = 0.2, and 0.5 × 0.4 = 0.2. Independent.
  4. P(A) = 1/6, P(B) = 6/36 = 1/6, P(A∩B) = P(6 then 1) = 1/36 = 1/6 × 1/6. Independent.
  5. P(C) = 140/200 = 0.7; P(C|W) = 50/80 = 0.625; P(W∩C) = 50/200 = 0.25. P(W)·P(C) = 0.4 × 0.7 = 0.28 ≠ 0.25, so not independent.
  6. P(A∩B) = 0.15, so P(A∪B) = 0.3 + 0.5 − 0.15 = 0.65
  7. Mutually exclusive gives P(A∩B) = 0, but P(A)·P(B) > 0. They can't be equal.
Topic 3 · Linear Algebra (syllabus, Part A)

Matrices

Not in your worksheets or the mock, but it is on the syllabus, so expect at most a few MCQs. Learn the mechanics; don't over-invest.

3.1Order, equality, addition, scalar, transpose

  1. Order = rows × columns. Add or subtract only matrices of the same order, entry by entry.
  2. Scalar kA: multiply every entry by k.
  3. Equal matrices: same order and every matching entry equal (gives equations).
  4. Transpose Aᵀ: row 1 becomes column 1, and so on. An m×n matrix becomes n×m.
  5. AB exists only if columns of A = rows of B; result is (rows of A) × (columns of B).
Worked example

A = 1 2; 3 4, B = 5 0; −1 2. 2A − B = 2−5 4−0; 6+1 8−2 = −3 4; 7 6

Practice (7)
  1. State the order of 1 2 3; 4 5 6 and write its transpose.
  2. With A, B from the example, find A + B.
  3. Find 3A − 2B.
  4. Find x and y if x+y 2; 3 x−y = 7 2; 3 1.
  5. Is 1 2; 2 5 symmetric? Why?
  6. Write the transpose of 2 −1 0; 3 4 7.
  7. A is 3×2 and B is 2×4. Is AB defined? Its order? Is BA defined?
Show solutions
  1. 2×3; transpose 1 4; 2 5; 3 6 (3×2).
  2. 6 2; 2 6
  3. 3A = 3 6; 9 12, 2B = 10 0; −2 4. Difference −7 6; 11 8
  4. x + y = 7 and x − y = 1. Add: 2x = 8 → x = 4, y = 3.
  5. Yes: Aᵀ = A (entry (1,2) = entry (2,1) = 2).
  6. 2 3; −1 4; 0 7
  7. AB: inner 2 = 2 ✓, order 3×4. BA: (2×4)(3×2), inner 4 ≠ 3, not defined.

3.2Matrix multiplication

  1. Check inner dimensions match. Write the result's order.
  2. Entry in row i, column j = (row i of A) · (column j of B): multiply pairs and add.
  3. Do it entry by entry, writing each sum. Never multiply entry-by-entry in place.
  4. Remember AB ≠ BA in general.
Worked example

1 2; 3 45 0; −1 2 = 1·5+2·(−1) 1·0+2·2; 3·5+4·(−1) 3·0+4·2 = 3 4; 11 8. Reversed order gives 5 10; 5 6, which is different.

Practice (7)
  1. 2 1; 0 31 4; 2 −1
  2. 1 2 34; 5; 6
  3. 4; 5; 61 2 3
  4. A = 1 1; 0 1. Find A² and A³.
  5. 1 0 2; 0 1 −11 2; 0 1; 3 0
  6. A = 2 0; 0 3. Show A² − 5A + 6I = O.
  7. A = 0 1; 0 0, B = 0 0; 1 0. Find AB and BA.
Show solutions
  1. 2+2 8−1; 0+6 0−3 = 4 7; 6 −3
  2. 1·4 + 2·5 + 3·6 = [32] (a 1×1 matrix)
  3. 3×3: 4 8 12; 5 10 15; 6 12 18
  4. A² = 1 2; 0 1, A³ = A²·A = 1 3; 0 1
  5. Row 1: [1+0+6, 2+0+0], row 2: [0+0−3, 0+1+0] → 7 2; −3 1
  6. A² = 4 0; 0 9, 5A = 10 0; 0 15, 6I = 6 0; 0 6. 4−10+6 = 0 and 9−15+6 = 0, so O. ✓
  7. AB = 1 0; 0 0, BA = 0 0; 0 1. Different.

3.3Inverse of a 2×2 and solving AX = B

  1. Find det = ad − bc. If det = 0 the matrix is singular: no inverse.
  2. Swap a and d; change the signs of b and c.
  3. Multiply by 1/det.
  4. For a system: write it as AX = B, then X = A⁻¹B.
Worked example

A = 4 7; 2 6. det = 24 − 14 = 10. A⁻¹ = (1/10)6 −7; −2 4.

Practice (7)
  1. Inverse of 3 1; 5 2.
  2. Inverse of 2 4; 1 2.
  3. Inverse of 1 2; 3 4.
  4. For what k is k 3; 4 6 singular?
  5. Solve 2x + y = 5, x + y = 3 using the inverse.
  6. Solve 3x + 2y = 12, x − y = −1 using the inverse.
  7. If A⁻¹ = 2 1; 1 1, find A.
Show solutions
  1. det = 6 − 5 = 1 → 2 −1; −5 3
  2. det = 4 − 4 = 0 → singular, no inverse
  3. det = 4 − 6 = −2 → (−1/2)4 −2; −3 1 = −2 1; 1.5 −0.5
  4. 6k − 12 = 0 → k = 2
  5. A = 2 1; 1 1, det 1, A⁻¹ = 1 −1; −1 2. X = A⁻¹5; 3 = 5−3; −5+6 → x = 2, y = 1
  6. A = 3 2; 1 −1, det = −3 − 2 = −5, A⁻¹ = (−1/5)−1 −2; −1 3. X = (−1/5)−12+2; −12−3 = (−1/5)−10; −15 → x = 2, y = 3. Check 6 + 6 = 12 ✓
  7. A = (A⁻¹)⁻¹. det = 1 → 1 −1; −1 2
Topic 4 · Linear Algebra (syllabus, Part A)

Determinants

The one skill that matters: expanding a 3×3 with the + − + signs without slipping a sign.

4.1Evaluating 2×2 and 3×3 determinants

  1. 2×2: a b; c d = ad − bc (down-right minus up-right).
  2. 3×3: pick row 1, or the row/column with the most zeros.
  3. Sign pattern + − +; − + −; + − +.
  4. For each entry: cover its row and column, find the 2×2 minor, multiply entry × sign × minor.
  5. Add the three terms. Triangular matrix shortcut: multiply the diagonal.
Worked example

2 1 3; 0 4 −1; 5 2 1 = 2(4·1 − (−1)·2) − 1(0·1 − (−1)·5) + 3(0·2 − 4·5)
= 2(6) − 1(5) + 3(−20) = 12 − 5 − 60 = −53

Practice (7)
  1. 5 3; 2 4
  2. −2 7; 3 −1
  3. 1 2 3; 4 5 6; 7 8 9
  4. 3 0 0; 2 4 0; 1 5 −2
  5. 1 0 2; 3 1 0; 0 2 1
  6. 2 −1 0; 1 3 4; 0 5 −2
  7. Solve x 2; 3 x = 10.
Show solutions
  1. 20 − 6 = 14
  2. 2 − 21 = −19
  3. 1(45−48) − 2(36−42) + 3(32−35) = −3 + 12 − 9 = 0 (rows are in AP, a known zero)
  4. Lower triangular: 3 × 4 × (−2) = −24
  5. 1(1·1 − 0·2) − 0(…) + 2(3·2 − 1·0) = 1 + 12 = 13
  6. 2(3·(−2) − 4·5) − (−1)(1·(−2) − 4·0) + 0 = 2(−26) + (−2) = −54
  7. x² − 6 = 10 → x² = 16 → x = ±4

4.2Properties (fast MCQ marks)

  1. Before computing anything, look for: a zero row, two equal or proportional rows, a triangular shape.
  2. |AB| = |A||B| · |Aᵀ| = |A| · |A⁻¹| = 1/|A| · |kA| = kⁿ|A| for n×n.
  3. Swapping two rows flips the sign. Adding a multiple of one row to another changes nothing.
Worked example

A is 3×3 with |A| = 5. |2A| = 2³ × 5 = 40, not 10. The k is raised to the order n.

Practice (7)
  1. A is 2×2 with |A| = 6. Find |3A|.
  2. |A| = 3, |B| = −2. Find |AB|.
  3. |A| = 7. Find |Aᵀ|.
  4. |A| = 4. Find |A⁻¹|.
  5. A matrix has determinant 9. Two rows are swapped. New determinant?
  6. Evaluate 1 2 3; 4 5 6; 1 2 3 without expanding.
  7. In a 3×3 matrix, row 3 = row 1 + row 2. What is the determinant?
Show solutions
  1. 3² × 6 = 54
  2. 3 × (−2) = −6
  3. 7
  4. 1/4
  5. −9
  6. Rows 1 and 3 are equal → 0
  7. Rows are dependent → 0

4.3Cramer's rule and triangle area

  1. D = determinant of the coefficients.
  2. Dₓ: replace the x-column with the constants. D_y: replace the y-column.
  3. x = Dₓ/D, y = D_y/D. If D = 0 there is no unique solution.
  4. Area of triangle = ½ |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|. Area 0 ⇒ collinear.
Worked example

2x + 3y = 8, x − y = −1. D = 2 3; 1 −1 = −5. Dₓ = 8 3; −1 −1 = −5 → x = 1. D_y = 2 8; 1 −1 = −10 → y = 2.

Practice (7)
  1. x + y = 5, x − y = 1
  2. 3x + 2y = 7, 2x − y = 0
  3. 4x − 3y = 6, 2x + y = 8
  4. Area of the triangle (0,0), (4,0), (0,3).
  5. Area of the triangle (1,2), (4,6), (7,2).
  6. Are (1,1), (2,2), (3,3) collinear?
  7. For what k does kx + 2y = 3, 3x + 6y = 9 have no unique solution?
Show solutions
  1. D = −1 − 1 = −2; Dₓ = −5 − 1 = −6 → x = 3; D_y = 1 − 5 = −4 → y = 2. (3, 2)
  2. D = −3 − 4 = −7; Dₓ = 7 2; 0 −1 = −7 → x = 1; D_y = 3 7; 2 0 = −14 → y = 2. (1, 2)
  3. D = 4 + 6 = 10; Dₓ = 6 −3; 8 1 = 30 → x = 3; D_y = 4 6; 2 8 = 20 → y = 2. (3, 2)
  4. ½ |0 + 4(3 − 0) + 0| = 6
  5. ½ |1(6−2) + 4(2−2) + 7(2−6)| = ½ |4 + 0 − 28| = 12
  6. ½ |1(2−3) + 2(3−1) + 3(1−2)| = ½ |−1 + 4 − 3| = 0 → collinear
  7. D = 6k − 6 = 0 → k = 1
Topic 5 · Part A + Part B Q1

Propositional logic

The mock gave this 4 MCQs and a full 15-mark question. Three facts carry most of the marks: P⇒Q is false only when T⇒F, P⇒Q ≡ ¬P∨Q, and ¬(P⇒Q) ≡ P∧¬Q.

5.1Evaluating connectives

PQ¬PP∧QP∨QP⊕QP⇒QP⇔Q
TTFTTFTT
TFFFTTFF
FTTFTTTF
FFTFFFTT
  1. Substitute the truth values (1 = T, 0 = F).
  2. Bracket by precedence: ¬ first, then ∧, then ∨, then ⇒ and ⇔ last.
  3. Evaluate from the innermost bracket outwards, writing each intermediate value.
Worked example

P = T, Q = F, R = T. Evaluate ¬P ∨ Q ⇒ R. Bracketed: ((¬P) ∨ Q) ⇒ R = (F ∨ F) ⇒ T = F ⇒ T = T.

Practice (7)
  1. P = 1, Q = 0: evaluate P ∧ ¬Q.
  2. P = T, Q = F: evaluate P ⇒ Q.
  3. "Pigs can fly" (F) ⇒ "The moon is cheese" (F). True or false?
  4. A = 1, B = 1: evaluate A ⊕ B.
  5. "5 + 5 = 10 if and only if 2 is odd." True or false?
  6. Add full brackets to ¬P ∨ Q ∧ R ⇒ S.
  7. P = T, Q = F, R = F: evaluate ((P ∨ Q) ⇒ (Q ∧ R)) ⇔ ¬P.
Show solutions
  1. 1 ∧ 1 = 1
  2. T ⇒ F = F
  3. F ⇒ F = T (vacuous truth: a false premise never breaks the promise)
  4. Same inputs → 0
  5. T ⇔ F = F
  6. ((¬P) ∨ (Q ∧ R)) ⇒ S
  7. P∨Q = T, Q∧R = F, so T ⇒ F = F. ¬P = F. F ⇔ F = T

5.2Truth tables and classification

  1. Count distinct variables n. Rows = 2ⁿ.
  2. Inputs: first column T for the top half, F for the bottom half; next column halves again; last column alternates T, F.
  3. Add one column per sub-expression, inside out.
  4. Final column all T = tautology; all F = contradiction; mixed = contingency.
  5. Two expressions are equivalent (≡) if their final columns match in every row.
Worked example: (P ∧ (P⇒Q)) ⇒ Q
PQP⇒QP∧(P⇒Q)(P∧(P⇒Q))⇒Q
TTTTT
TFFFT
FTTFT
FFTFT

All T: a tautology. This is why Modus Ponens is a valid rule.

Practice (7)
  1. How many rows for W = (P ∧ Q) ⇒ (¬R ∨ S)?
  2. How many rows for (A ∨ B) ∧ (B ⇒ C) ∧ ¬A ∧ D ∧ E?
  3. Classify P ∨ ¬P.
  4. Classify A ∧ ¬A, and explain without a table.
  5. Classify P ⇒ Q.
  6. Use a truth table to show ¬(P ∧ Q) ≡ ¬P ∨ ¬Q.
  7. Classify (P ⇒ Q) ∨ (Q ⇒ P).
Show solutions
  1. 4 variables (P, Q, R, S) → 2⁴ = 16
  2. 5 variables → 32
  3. Tautology
  4. Contradiction: A and ¬A always have opposite values, so their AND is never T.
  5. Contingency (F only in the T, F row)
  6. Rows (TT, TF, FT, FF): ¬(P∧Q) = F, T, T, T and ¬P∨¬Q = F, T, T, T. Columns match, so equivalent.
  7. Tautology: if Q is T, then P⇒Q is T; if Q is F, then Q⇒P is T. One side is always T.

5.3Converse, inverse, contrapositive and English

  1. Find the hypothesis P (the "if" part) and conclusion Q.
  2. Converse Q⇒P · Inverse ¬P⇒¬Q · Contrapositive ¬Q⇒¬P (swap AND negate both). Only the contrapositive is equivalent to the original.
  3. English: "P only if Q" = P⇒Q. "P if Q" = Q⇒P. "Q is necessary for P" = P⇒Q. "P is sufficient for Q" = P⇒Q. "P unless Q" = ¬Q⇒P.
  4. When negating a part that is already negated, cancel: ¬(¬A) = A.
Worked example (mock Q7)

Contrapositive of ¬A ⇒ B: swap to B ⇒ ¬A, then negate both: ¬B ⇒ ¬(¬A) = ¬B ⇒ A (option B).

Practice (7)
  1. Write the converse, inverse and contrapositive of "If a system is compromised (C), an alert is triggered (A)".
  2. Contrapositive of P ⇒ ¬Q.
  3. Inverse of ¬P ⇒ Q.
  4. Which two of converse, inverse, contrapositive are equivalent to each other?
  5. J = "you get the job", D = "you have a degree". Translate "You get the job only if you have a degree".
  6. E = "you have experience". Translate "You get the job if you have a degree or experience".
  7. R = "you ride", T = "you are taller than 4 ft". Translate "You cannot ride unless you are taller than 4 ft".
Show solutions
  1. Converse A⇒C; inverse ¬C⇒¬A; contrapositive ¬A⇒¬C.
  2. Swap: ¬Q ⇒ P; negate both: Q ⇒ ¬P
  3. Negate both, no swap: P ⇒ ¬Q
  4. Converse and inverse (each is the contrapositive of the other).
  5. J ⇒ D
  6. (D ∨ E) ⇒ J
  7. R ⇒ T (equivalently ¬T ⇒ ¬R)

5.4Simplifying with the laws

  1. Remove every ⇒ with P⇒Q ≡ ¬P∨Q.
  2. Push each ¬ inside brackets with De Morgan (flip ∧↔∨, negate each part).
  3. Cancel double negations.
  4. Finish with distributive, absorption, identity (P∧T = P, P∨F = P), domination (P∨T = T, P∧F = F), complement (P∨¬P = T, P∧¬P = F).
  5. Name the law on every line; in Part B that is where the marks are.
Worked example

¬(¬A ∨ B) ∨ A
≡ (¬¬A ∧ ¬B) ∨ A  (De Morgan)
≡ (A ∧ ¬B) ∨ A  (double negation)
≡ A  (absorption)

Practice (7)
  1. (X ∨ 1) ∧ (Y ∨ 0)
  2. ¬(¬(A ∨ A ∨ A))
  3. P ∧ (P ∨ Q)
  4. ¬(P ∧ ¬Q), and say which implication it equals.
  5. (P ∧ Q) ∨ (P ∧ ¬Q)
  6. ¬(P ⇒ Q)
  7. Show (P ⇒ Q) ∧ (P ⇒ R) ≡ P ⇒ (Q ∧ R).
Show solutions
  1. X∨1 = 1 (domination), Y∨0 = Y (identity), 1 ∧ Y = Y
  2. Double negation → A∨A∨A → A (idempotent)
  3. P (absorption)
  4. De Morgan: ¬P ∨ Q, which is P ⇒ Q
  5. P ∧ (Q ∨ ¬Q) (distributive) = P ∧ T (complement) = P (identity)
  6. ¬(¬P ∨ Q) = ¬¬P ∧ ¬Q = P ∧ ¬Q
  7. (¬P ∨ Q) ∧ (¬P ∨ R) = ¬P ∨ (Q ∧ R) (distributive) = P ⇒ (Q ∧ R)

5.5Part B Policy → OR/NOT form → violation → truth values

This is exactly the mock's Question 1 structure. Practise the four steps until they are automatic.

  1. (A) Define. One letter per simple statement, written in words. Compound "if … and …" goes in brackets: (E ∧ Q) ⇒ F.
  2. (B) OR/NOT only. X ⇒ Y ≡ ¬X ∨ Y. If X is an AND, apply De Morgan: ¬(E∧Q) ∨ F ≡ ¬E ∨ ¬Q ∨ F. If an AND survives elsewhere, rewrite it as ¬(¬L ∨ ¬R).
  3. (C) Violation = negation. ¬(X ⇒ Y) ≡ X ∧ ¬Y. Or negate your Part B answer with De Morgan. Push all ¬ onto single letters.
  4. (D) Read off values. An AND is true only when every part is true, so each letter's value is forced. State each one in words.
Worked example

"If the input is invalid (I), then an error is logged (L) and the request is rejected (R)."

(A) I ⇒ (L ∧ R). (B) ¬I ∨ (L ∧ R); to remove the ∧: ¬I ∨ ¬(¬L ∨ ¬R). (C) I ∧ ¬(L ∧ R) ≡ I ∧ (¬L ∨ ¬R). (D) I must be T, but L, R are only known to be "not both true", so the values are not unique here. Compare with the mock, where the AND sits on the left and every value is forced.

Practice (6)
  1. "If the battery is low and the charger is disconnected, the phone shuts down." Do all four parts.
  2. "If a password is reused, the account is locked." Do all four parts.
  3. "If the user is an admin or a manager, the report is visible." Do parts A to C. Are the values in D unique?
  4. Write the violation of "If the input is valid, the program runs" in plain English.
  5. Write the contrapositive of (E ∧ Q) ⇒ F with no negation outside brackets.
  6. Prove with a 4-row truth table that P ⇒ Q ≡ ¬P ∨ Q.
Show solutions
  1. B: battery low, C: charger disconnected, S: phone shuts down. (A) (B ∧ C) ⇒ S. (B) ¬(B∧C) ∨ S ≡ ¬B ∨ ¬C ∨ S. (C) ¬(¬B ∨ ¬C ∨ S) ≡ B ∧ C ∧ ¬S. (D) B = T, C = T, S = F: battery low, charger disconnected, phone did not shut down.
  2. R: password reused, L: account locked. (A) R ⇒ L. (B) ¬R ∨ L. (C) R ∧ ¬L. (D) R = T, L = F.
  3. A: admin, M: manager, V: visible. (A) (A ∨ M) ⇒ V. (B) ¬(A ∨ M) ∨ V (already only OR and NOT). (C) (A ∨ M) ∧ ¬V. (D) V = F is forced, but A and M are only "at least one true": not unique.
  4. "The input is valid and the program does not run."
  5. ¬F ⇒ ¬(E ∧ Q) ≡ ¬F ⇒ (¬E ∨ ¬Q)
  6. Rows TT, TF, FT, FF: P⇒Q = T, F, T, T. ¬P = F, F, T, T so ¬P∨Q = T, F, T, T. Identical columns.
Topic 6 · Part A + Part B Q2

Rules of inference and proofs

Learn the seven rules as patterns. In a proof, every new line is "result, rule (line numbers)".

6.1Name the rule, spot the fallacy

RulePatternMemory hook
Modus Ponens (MP)P⇒Q, P ∴ Qaffirm the front
Modus Tollens (MT)P⇒Q, ¬Q ∴ ¬Pdeny the back
Hypothetical Syllogism (HS)P⇒Q, Q⇒R ∴ P⇒Rchain
Disjunctive Syllogism (DS)P∨Q, ¬P ∴ Qelimination
ResolutionP∨Q, ¬P∨R ∴ Q∨Rcancel P with ¬P
Addition / Simplification / ConjunctionP ∴ P∨Q · P∧Q ∴ P · P, Q ∴ P∧Qbuild up / break down
✗ Affirming the consequentP⇒Q, Q ∴ Pinvalid
✗ Denying the antecedentP⇒Q, ¬P ∴ ¬Qinvalid

Valid = the conclusion follows from the form. Sound = valid and every premise is actually true.

Practice (7) — name the rule or fallacy
  1. If the API key is valid, access is granted. Access is not granted. So the key is not valid.
  2. If x > 10 then x² > 100. If x² > 100 then x² > 50. So if x > 10 then x² > 50.
  3. The database is local or in the cloud. It is not local. So it is in the cloud.
  4. If a shape is a square, it has four sides. This shape has four sides. So it is a square.
  5. If a process exceeds 2 GB, it is terminated. It did not exceed 2 GB. So it was not terminated.
  6. I will sleep or code. I will not sleep or I will drink coffee. So I will code or drink coffee.
  7. User 404 is an admin. So user 404 is an admin or a guest.
Show solutions
  1. Modus Tollens
  2. Hypothetical Syllogism
  3. Disjunctive Syllogism
  4. Fallacy: affirming the consequent (a rectangle also has four sides)
  5. Fallacy: denying the antecedent (it could be terminated for another reason)
  6. Resolution (S∨C, ¬S∨D ∴ C∨D)
  7. Addition

6.2Writing a formal proof

  1. Number every premise 1, 2, 3, …
  2. Look at the goal. Ask: which rule outputs this shape? (A single letter often comes from MP or DS; a negated letter from MT.)
  3. Work forward: each new line = statement + rule + the lines it uses.
  4. Stop when the goal appears; write "∴ goal" and box it.
Worked example

Premises 1) P⇒Q 2) Q⇒R 3) ¬R. Prove ¬P.
4) P⇒R  HS (1, 2)
5) ¬P  MT (4, 3) ∎

Practice (7)
  1. 1) P⇒Q 2) P 3) Q⇒R. Prove R.
  2. 1) A⇒B 2) C⇒D 3) A∨C 4) ¬B. Prove D.
  3. 1) P∧Q 2) P⇒(R∨S) 3) ¬R. Prove S.
  4. 1) ¬P∨Q 2) P. Prove Q.
  5. 1) (P∧Q)⇒R 2) P 3) Q. Prove R.
  6. 1) P⇒Q 2) ¬Q 3) P∨R. Prove R ∧ ¬P.
  7. 1) H⇒S 2) S⇒B 3) ¬B. Prove ¬H in two lines.
Show solutions
  1. 4) Q MP(1,2) · 5) R MP(3,4)
  2. 5) ¬A MT(1,4) · 6) C DS(3,5) · 7) D MP(2,6)
  3. 4) P Simplification(1) · 5) R∨S MP(2,4) · 6) S DS(5,3)
  4. P is true so ¬P is false. 3) Q DS(1,2)
  5. 4) P∧Q Conjunction(2,3) · 5) R MP(1,4)
  6. 4) ¬P MT(1,2) · 5) R DS(3,4) · 6) R ∧ ¬P Conjunction(5,4)
  7. 4) H⇒B HS(1,2) · 5) ¬H MT(4,3)

6.3Part B Proof by contradiction with disjunctions

This is the mock's Question 2 structure.

  1. (A) Define each letter in words, translate each premise.
  2. (B) State the conclusion (usually a chain ending in Modus Tollens gives a "not"). Write its negation: this is your assumption.
  3. (C) Rewrite each implication: X ⇒ Y ≡ ¬X ∨ Y.
  4. (D) Start from the assumption. Use DS on each disjunction to walk along the chain until you derive something that clashes with a premise. Write "X ∧ ¬X, contradiction", so the assumption is false and the conclusion holds.
Worked example

"If it rains (R), the ground is wet (W). If the ground is wet, the match is postponed (M). The match was not postponed."

(A) 1) R⇒W 2) W⇒M 3) ¬M. (B) Conclusion ¬R; assume R. (C) 1′) ¬R ∨ W, 2′) ¬W ∨ M.
(D) 4) R assumption · 5) W DS(1′, 4) · 6) M DS(2′, 5) · 7) M ∧ ¬M Conjunction(6, 3), a contradiction. So ¬R: it did not rain. ∎

Practice (6)
  1. 1) A⇒B 2) ¬B. Prove ¬A by contradiction.
  2. 1) P⇒Q 2) Q⇒R 3) R⇒S 4) ¬S. Prove ¬P by contradiction.
  3. 1) P∨Q 2) ¬P. Prove Q by contradiction.
  4. "If the server is overloaded (O), latency rises (L). If latency rises, users complain (C). No users complained." Do parts A to D.
  5. 1) (P∧Q)⇒R 2) P 3) ¬R. Prove ¬Q.
  6. 1) A∨B 2) A⇒C 3) B⇒C. Prove C by contradiction.
Show solutions
  1. 1′) ¬A∨B. 3) A assumption · 4) B DS(1′,3) · 5) B ∧ ¬B contradiction(4,2). So ¬A.
  2. Disjunctions ¬P∨Q, ¬Q∨R, ¬R∨S. 5) P assumption · 6) Q DS · 7) R DS · 8) S DS · 9) S ∧ ¬S with 4, contradiction. So ¬P.
  3. 3) ¬Q assumption · 4) P DS(1,3) · 5) P ∧ ¬P with 2, contradiction. So Q.
  4. (A) O⇒L, L⇒C, ¬C. (B) Conclusion ¬O (not overloaded); assume O. (C) ¬O∨L, ¬L∨C. (D) O assumption · L DS · C DS · C ∧ ¬C contradiction. So the server was not overloaded.
  5. 4) Q assumption · 5) P∧Q Conjunction(2,4) · 6) R MP(1,5) · 7) R ∧ ¬R contradiction with 3. So ¬Q.
  6. 4) ¬C assumption · 5) ¬A MT(2,4) · 6) ¬B MT(3,4) · 7) B DS(1,5) · 8) B ∧ ¬B contradiction(7,6). So C.
Topic 7 · Part A + Part B Q3

Predicate logic and quantifiers

Three mock MCQs plus the "hard" Part B question. The two rules that decide most answers: ∀ goes with ⇒, ∃ goes with ∧, and to negate, flip each quantifier and negate the inside.

7.1English → predicate logic

  1. Name the domain and each predicate in words.
  2. "All / every / each / any / a (generic)" → ∀x (Type(x) ⇒ Property(x)).
  3. "Some / there is / at least one" → ∃x (Type(x) ∧ Property(x)).
  4. "No / none" → ∀x (Type(x) ⇒ ¬Property(x)) or ¬∃x (Type(x) ∧ Property(x)).
  5. "Only A are P" → ∀x (P(x) ⇒ A(x)): the word after "only" goes on the right.
Worked example (mock Q11)

"Only premium members receive discounts." Anyone with a discount must be premium: ∀x (D(x) ⇒ P(x)), option B.

Practice (7)
  1. D(x): developer, C(x): writes code. "Every developer writes code."
  2. "Some developers write code."
  3. S(x): student, H(x): happy. "No student is happy." Give two forms.
  4. A(x): admin, P(x): has system privileges. "Only admins have system privileges."
  5. Domain all creatures, B(x): bird, F(x): flies. "Some birds cannot fly."
  6. P(x): passed. "Not all students passed." Give the form with no ¬ in front.
  7. C(y): class, T(x,y): x has taken y. "Every student has taken some class."
Show solutions
  1. ∀x (D(x) ⇒ C(x))
  2. ∃x (D(x) ∧ C(x)) (∃ with ⇒ would be true for anyone who isn't a developer, which is wrong)
  3. ∀x (S(x) ⇒ ¬H(x)) and ¬∃x (S(x) ∧ H(x))
  4. ∀x (P(x) ⇒ A(x))
  5. ∃x (B(x) ∧ ¬F(x))
  6. ¬∀x (S(x) ⇒ P(x)) ≡ ∃x (S(x) ∧ ¬P(x))
  7. ∀x (S(x) ⇒ ∃y (C(y) ∧ T(x, y)))

7.2Negating quantified statements

  1. Move ¬ rightwards one symbol at a time. Each ∀ it passes becomes ∃, each ∃ becomes ∀.
  2. When it reaches the inside: ¬(A⇒B) = A ∧ ¬B, ¬(A∧B) = ¬A ∨ ¬B, ¬(A∨B) = ¬A ∧ ¬B.
  3. Check in English: "every … is" becomes "some … is not".
Worked example (mock Q10)

¬∃x∀y Q(x,y) → ∀x ¬∀y Q(x,y) → ∀x∃y ¬Q(x,y), option A.

Practice (7)
  1. ¬∀x P(x)
  2. ¬∃x (A(x) ∧ B(x))
  3. ¬∀x (S(x) ⇒ L(x))
  4. ¬∀x∃y L(x, y)
  5. ¬∃x∃y (x + y = 0)
  6. Negate in English: "Every server is online."
  7. ¬∀x (P(x) ⇒ ∃y Q(x, y))
Show solutions
  1. ∃x ¬P(x)
  2. ∀x (¬A(x) ∨ ¬B(x)), i.e. ∀x (A(x) ⇒ ¬B(x))
  3. ∃x (S(x) ∧ ¬L(x))
  4. ∃x∀y ¬L(x, y)
  5. ∀x∀y (x + y ≠ 0)
  6. "Some server is not online." (not "No server is online")
  7. ∃x (P(x) ∧ ∀y ¬Q(x, y))

7.3Truth values, domains and nested quantifiers

  1. ∀ is false if you find one counterexample. ∃ is true if you find one example.
  2. The domain matters: the same statement can flip between integers and reals.
  3. Nested: read left to right. ∀x∃y = "for each x you may pick a different y". ∃y∀x = "one single y works for every x" (much stronger).
  4. A variable is bound if a quantifier names it, free otherwise.
Practice (7)
  1. Domain positive integers: is ∀x (x² ≥ x) true?
  2. Q(x): x² = 2. Is ∃x Q(x) true over the integers? Over the reals?
  3. Over the integers, is ∃!x (|x| = 5) true?
  4. L(x, y): x likes y. Translate ∀x∃y L(x, y) and ∃y∀x L(x, y).
  5. Domain integers: truth of ∀x∃y (x + y = 0)? Of ∃y∀x (x + y = 0)?
  6. Give a counterexample to ∀x ∈ ℝ (x² > x).
  7. In ∀x (P(x, y) ⇒ ∃z Q(x, y, z)), which variables are free?
Show solutions
  1. True (x² − x = x(x − 1) ≥ 0 for x ≥ 1)
  2. Integers false; reals true (x = √2)
  3. False: both 5 and −5 work, so it is not unique.
  4. "Everybody likes somebody." / "There is someone whom everybody likes."
  5. First true (y = −x). Second false (no single y cancels every x).
  6. x = 0.5 (0.25 > 0.5 is false). x = 0 or x = 1 also work.
  7. y is free; x and z are bound.

7.4Part B Quantifier proofs with instantiation

This is the mock's Question 3 structure (labelled HARD, but it is the same four moves every time).

  1. (A) Define predicates in words. Write each policy with ∀ and ⇒. "No X does Y" becomes ∀x (X(x) ⇒ ¬Y(x)).
  2. (B) Universal Instantiation (UI): drop ∀ and plug in the named object, e.g. a. Then chain with Hypothetical Syllogism (or assume P(a) and use MP twice).
  3. (C) Rewrite the result with X ⇒ Y ≡ ¬X ∨ Y. Watch double negations.
  4. (D) Plug in the given fact (it makes one side false), then Disjunctive Syllogism forces the other side. Answer the question in words.
Worked example

Domain: files. "Every encrypted file is safe. No safe file is flagged." Show: if report.pdf (r) is encrypted, it is not flagged; then r is flagged, so is it encrypted?

(A) E(x), S(x), F(x). 1) ∀x (E(x) ⇒ S(x)) 2) ∀x (S(x) ⇒ ¬F(x)).
(B) 3) E(r) ⇒ S(r) UI(1) · 4) S(r) ⇒ ¬F(r) UI(2) · 5) E(r) ⇒ ¬F(r) HS(3,4).
(C) ¬E(r) ∨ ¬F(r).
(D) F(r) = T so ¬F(r) = F. For the OR to be true, ¬E(r) = T. report.pdf is not encrypted.

Practice (6)
  1. 1) ∀x (P(x) ⇒ Q(x)) 2) P(k). Prove Q(k).
  2. "All humans are mortal. Socrates is human." Prove Socrates is mortal, naming each rule.
  3. 1) ∀x (A(x) ⇒ B(x)) 2) ∀x (B(x) ⇒ C(x)) 3) ¬C(m). Prove ¬A(m).
  4. "Every CS student takes Maths. Riya does not take Maths." Prove Riya is not a CS student.
  5. "Every verified user can post. Some user is verified." Prove ∃x Post(x).
  6. "No cat is a dog. Tom is a cat." Do parts A to D: formalise, prove Cat(Tom) ⇒ ¬Dog(Tom), rewrite with ∨ and ¬, and conclude.
Show solutions
  1. 3) P(k) ⇒ Q(k) UI(1) · 4) Q(k) MP(3,2)
  2. 1) ∀x (H(x) ⇒ M(x)) 2) H(s). 3) H(s) ⇒ M(s) UI(1) · 4) M(s) MP(3,2)
  3. 4) A(m)⇒B(m) UI(1) · 5) B(m)⇒C(m) UI(2) · 6) A(m)⇒C(m) HS(4,5) · 7) ¬A(m) MT(6,3)
  4. 1) ∀x (CS(x) ⇒ M(x)) 2) ¬M(r). 3) CS(r) ⇒ M(r) UI(1) · 4) ¬CS(r) MT(3,2)
  5. 1) ∀x (V(x) ⇒ Post(x)) 2) ∃x V(x). 3) V(c) for some c, Existential Instantiation(2) · 4) V(c) ⇒ Post(c) UI(1) · 5) Post(c) MP(4,3) · 6) ∃x Post(x) Existential Generalization(5)
  6. (A) ∀x (Cat(x) ⇒ ¬Dog(x)), Cat(t). (B) Cat(t) ⇒ ¬Dog(t) by UI. (C) ¬Cat(t) ∨ ¬Dog(t). (D) Cat(t) = T so ¬Cat(t) = F, so ¬Dog(t) = T: Tom is not a dog.
Topic 8 · Syllabus Part B (not in the mock)

Karnaugh maps and circuits

On the syllabus but with no worksheet and no mock question. One method, one practice set, then move on.

8.1Minimising with a K-map

  1. Write the minterms in binary (for A, B, C: m5 = 101 = A B′ C).
  2. Draw the map. 3 variables: rows A = 0, 1; columns BC in Gray order 00, 01, 11, 10.
  3. Put a 1 in each minterm cell.
  4. Circle groups of 1, 2, 4 or 8 adjacent 1s (rectangles only, edges wrap, overlaps allowed). Make groups as large as possible and as few as possible.
  5. For each group keep the variables that stay constant (1 → plain, 0 → primed). OR the groups together.
Worked example

F(A,B,C) = Σm(1, 3, 5, 7) = 001, 011, 101, 111. All four have C = 1 and A, B change, so one group of 4: F = C.

Practice (7)
  1. F(A,B) = Σm(1, 2, 3)
  2. F(A,B) = Σm(0, 1)
  3. F(A,B,C) = Σm(0, 2, 4, 6)
  4. F(A,B,C) = Σm(6, 7)
  5. F(A,B,C) = Σm(0, 1, 2, 3, 7)
  6. F(A,B,C) = Σm(1, 3, 4, 6)
  7. Simplify AB + AB′ algebraically, and list the gates needed for AB + C′.
Show solutions
  1. 01, 10, 11 → groups B (01, 11) and A (10, 11): A + B
  2. 00, 01 → A = 0 constant: A′
  3. 000, 010, 100, 110 → C = 0 constant: C′
  4. 110, 111 → A = 1, B = 1: AB
  5. m0 to m3 is the whole A = 0 row → A′; m3 + m7 (011, 111) → BC. A′ + BC
  6. 001, 011 → A′C; 100, 110 → AC′. A′C + AC′ = A ⊕ C
  7. A(B + B′) = A·1 = A. AB + C′ needs one AND, one NOT, one OR.
Topic 9 · Part A + Part B Q5

Counting and the binomial theorem

First question every time: does order matter? Yes (codes, rankings, roles) → permutation. No (teams, committees, handfuls) → combination.

9.1Arrangements and permutations

  1. Draw one slot per position. Fill each slot with "how many choices are left".
  2. Repetition allowed: every slot has n choices (nʳ). Not allowed: n, n−1, n−2, … = ⁿPᵣ.
  3. Identical items: divide by the factorial of each repeat count.
  4. Circle: fix one person, (n−1)!. Flippable ring/necklace: (n−1)!/2.
  5. "Must be together": tie them as one block, arrange, then × internal order. "Must be apart": arrange the others, then place them in the gaps.
Worked example (mock Q29)

4-letter codes from L, O, G, I, C, no repetition: 5 × 4 × 3 × 2 = 120 (option C).

Practice (7)
  1. How many 4-digit PINs (digits may repeat)?
  2. 15 teams compete for gold, silver and bronze. How many podium results?
  3. Arrangements of the letters of BANANA.
  4. 8 board members around a round table.
  5. 7 distinct LEDs on a ring that can be flipped over.
  6. 6 books on a shelf, with Volumes 1 and 2 next to each other.
  7. 4 distinct Type-A and 3 distinct Type-B requests in a queue, with no two Type-B adjacent.
Show solutions
  1. 10 × 10 × 10 × 10 = 10,000
  2. ¹⁵P₃ = 15 × 14 × 13 = 2,730
  3. 6 letters, A×3, N×2: 6!/(3!·2!) = 720/12 = 60
  4. (8 − 1)! = 5,040
  5. (7 − 1)!/2 = 720/2 = 360
  6. Block + 4 books = 5 items: 5! × 2! = 120 × 2 = 240
  7. Arrange A's: 4! = 24. They create 5 gaps; place the 3 B's in order: ⁵P₃ = 60. Total 1,440

9.2Part B Combinations, teams and probability

This is the mock's Question 5 structure: total → exact split → probability → "at least one" by complement.

  1. Total: C(all people, team size).
  2. Exact split: C(group 1, k₁) × C(group 2, k₂) (AND ⇒ multiply).
  3. Probability = favourable ÷ total; simplify the fraction fully.
  4. At least one from a group: 1 − P(none from that group), where "none" = choose everyone from the other group.
  5. "At least k" by cases: add the case counts (OR ⇒ add).
  6. Always write ⁿCᵣ = n!/(r!(n−r)!), then the numbers, then the value.
Worked example

Committee of 4 from 6 men and 4 women.
Total = C(10,4) = 210. Exactly 2 women = C(6,2) × C(4,2) = 15 × 6 = 90. P = 90/210 = 3/7.
At least one woman: none = C(6,4) = 15, so P = 1 − 15/210 = 195/210 = 13/14.

Practice (7)
  1. 8 teams each play every other team once. How many matches?
  2. A VC firm picks 4 startups from 12.
  3. Choose 3 developers from 8 and 2 designers from 5.
  4. A task force of 5 from 6 seniors and 4 juniors must have at least 3 seniors. How many ways?
  5. For question 4, if the 5 are picked at random, what is the probability the rule is met?
  6. 10 points, no three collinear. How many lines? How many triangles?
  7. A bag has 5 red and 3 blue balls; draw 2. P(both red)? P(at least one blue)?
Show solutions
  1. C(8,2) = 8·7/2 = 28
  2. C(12,4) = (12·11·10·9)/24 = 495
  3. C(8,3) × C(5,2) = 56 × 10 = 560
  4. 3S2J + 4S1J + 5S0J = C(6,3)C(4,2) + C(6,4)C(4,1) + C(6,5) = 120 + 60 + 6 = 186
  5. Total C(10,5) = 252. P = 186/252 = 31/42
  6. Lines C(10,2) = 45; triangles C(10,3) = 120
  7. P(both red) = C(5,2)/C(8,2) = 10/28 = 5/14; at least one blue = 1 − 5/14 = 9/14

9.3Binomial expansion and specific coefficients

  1. Write the general term Tₖ₊₁ = ⁿCₖ · (first)ⁿ⁻ᵏ · (second)ᵏ, keeping signs and constants inside the brackets.
  2. Collect the power of x in terms of k.
  3. Set it equal to the target power and solve for k (must be a whole number 0…n).
  4. Substitute k and compute the number, including signs.
  5. Sum of all coefficients: set every variable to 1.
Worked example

Coefficient of x³ in (2x + 1)⁵: Tₖ₊₁ = ⁵Cₖ (2x)⁵⁻ᵏ (1)ᵏ. Power 5 − k = 3 → k = 2. ⁵C₂ · 2³ = 10 · 8 = 80.

Practice (7)
  1. Fully expand (x + 2y)⁴.
  2. Coefficient of x³y⁵ in (x + y)⁸.
  3. Coefficient of x⁴ in (x − 3)⁶.
  4. Coefficient of x² in (2x − 1/x)⁴.
  5. Constant term of (x + 1/x)⁶.
  6. Evaluate ¹⁰C₀ + ¹⁰C₁ + … + ¹⁰C₁₀.
  7. Sum of coefficients in (x + y)⁵, and in (2x − 1)⁷.
Show solutions
  1. Coefficients 1, 4, 6, 4, 1 with powers of 2: x⁴ + 8x³y + 24x²y² + 32xy³ + 16y⁴
  2. ⁸C₅ = 56
  3. 6 − k = 4 → k = 2: ⁶C₂ (−3)² = 15 × 9 = 135
  4. Power of x: (4 − k) − k = 4 − 2k = 2 → k = 1: ⁴C₁ · 2³ · (−1)¹ = −32
  5. 6 − 2k = 0 → k = 3: ⁶C₃ = 20
  6. 2¹⁰ = 1,024
  7. (1 + 1)⁵ = 32 (mock Q32); (2 − 1)⁷ = 1
Topic 10 · Part A

Functions, logs, limits and derivatives

Mostly quick MCQs. Domain and the 0/0 limit were both in the mock.

10.1Domain, range and function types

  1. List every restriction: denominator ≠ 0; expression under √ ≥ 0 (strictly > 0 if the √ is in a denominator); inside log > 0.
  2. Solve each, then combine with "and".
  3. Write intervals: [ ] includes, ( ) excludes, ∪ joins pieces.
  4. Injective: different inputs never share an output. Surjective: every codomain value is hit. Inverse: swap x and y, solve for y.
Worked example (mock Q14)

g(x) = √(x − 3)/(x − 5). Need x − 3 ≥ 0 → x ≥ 3, and x ≠ 5. Domain [3, 5) ∪ (5, ∞), option C.

Practice (7)
  1. Domain of 1/(x − 5).
  2. Domain of √(2x − 8).
  3. Domain of ln(x + 2).
  4. Domain of 1/√(x − 1).
  5. f(x) = x² from ℝ to ℝ. Range? Injective? Surjective?
  6. Prove f(x) = 3x + 2 is injective.
  7. Find the inverse of f(x) = (2x + 3)/(x − 1).
Show solutions
  1. (−∞, 5) ∪ (5, ∞)
  2. 2x − 8 ≥ 0 → [4, ∞)
  3. x + 2 > 0 → (−2, ∞)
  4. x − 1 > 0 (√ in denominator) → (1, ∞)
  5. Range [0, ∞). Not injective (f(2) = f(−2) = 4). Not surjective (−1 is never hit).
  6. Suppose f(a) = f(b): 3a + 2 = 3b + 2 → 3a = 3b → a = b. ∎
  7. y(x − 1) = 2x + 3 → xy − y = 2x + 3 → x(y − 2) = y + 3 → f⁻¹(x) = (x + 3)/(x − 2)

10.2Exponentials and logarithms

  1. log_b(y) = x means bˣ = y. ln means base e.
  2. Laws: product → sum, quotient → difference, power → multiplier. e^(ln x) = x, ln(eˣ) = x.
  3. To solve aeᵏˣ = c: isolate the exponential, take ln of both sides, solve.
  4. Growth/decay: A = A₀·(factor)^(t/period); continuous: A = Pe^(rt).
Practice (7)
  1. Write log₂(1/16) = −4 in exponential form.
  2. Simplify e^(3 ln x).
  3. Expand ln(x²y/z³).
  4. Solve 5e^(2x − 1) = 40.
  5. A server load of 100 doubles every 4 hours. Load after 12 hours?
  6. P(t) = 500(0.8)ᵗ + 20. What value does P approach long-term?
  7. A = 1000e^(0.05 × 10). Evaluate.
Show solutions
  1. 2⁻⁴ = 1/16
  2. e^(ln x³) = x³
  3. 2 ln x + ln y − 3 ln z
  4. e^(2x−1) = 8 → 2x − 1 = ln 8 → x = (1 + ln 8)/2 ≈ 1.54
  5. 12/4 = 3 doublings: 100 × 2³ = 800
  6. (0.8)ᵗ → 0, so P → 20 (horizontal asymptote)
  7. 1000e^0.5 ≈ 1648.72

10.3Limits

  1. Substitute directly. A number? Done.
  2. 0/0? Factor top and bottom, cancel the common factor, substitute again.
  3. k/0 with k ≠ 0? A vertical asymptote: the limit is ±∞ or does not exist.
  4. x → ∞ for a fraction of polynomials: compare highest powers. Same degree → ratio of leading coefficients; top smaller → 0; top bigger → ∞.
  5. If left and right limits differ, the two-sided limit does not exist.
Worked example (mock Q15)

lim x→3 (x² − 9)/(x − 3): substitution gives 0/0. Factor: (x − 3)(x + 3)/(x − 3) = x + 3 → 6 (option C).

Practice (7)
  1. lim x→−2 (3x³ + 4x² − x + 7)
  2. lim x→1 (x² + 2x − 3)/(x − 1)
  3. lim x→2 (x³ − 8)/(x − 2)
  4. lim x→∞ (3x² + 1)/(5x² − x)
  5. lim x→∞ (2x + 1)/(x² + 3)
  6. Where are the vertical asymptotes of (x + 1)/(x² − 9)?
  7. f(x) = −1 for x < 0 and 1 for x ≥ 0. Find both one-sided limits at 0 and the limit.
Show solutions
  1. −24 + 16 + 2 + 7 = 1
  2. (x + 3)(x − 1)/(x − 1) → 1 + 3 = 4
  3. x³ − 8 = (x − 2)(x² + 2x + 4) → 4 + 4 + 4 = 12
  4. Same degree: 3/5
  5. Top degree smaller: 0
  6. x² − 9 = 0 and top ≠ 0: x = 3 and x = −3
  7. Left −1, right 1, so the limit does not exist.

10.4Derivatives and tangent lines

  1. Rewrite roots and fractions as powers: √x = x^(1/2), 3/x² = 3x⁻².
  2. Power rule term by term: bring the power down, subtract 1. Constants vanish.
  3. Tangent at x = a: point (a, f(a)), slope f′(a), then y − f(a) = f′(a)(x − a).
  4. Average rate of change on [a, b] = (f(b) − f(a))/(b − a).
Worked example

f(x) = 2x² − 3x + 1 at x = 2. f(2) = 3. f′(x) = 4x − 3, f′(2) = 5. Tangent: y − 3 = 5(x − 2) → y = 5x − 7.

Practice (7)
  1. d/dx of x⁷
  2. dy/dx for y = 5x³ − ½x² + 8x − 15
  3. d/dx of √x
  4. d/dx of 3/x²
  5. Average rate of change of x² from x = 1 to x = 4.
  6. f′(x) = 3x² − 4. Find f′(2).
  7. Tangent line to y = x³ at x = 1.
Show solutions
  1. 7x⁶
  2. 15x² − x + 8
  3. ½x^(−1/2) = 1/(2√x)
  4. 3x⁻² → −6x⁻³ = −6/x³
  5. (16 − 1)/3 = 5
  6. 12 − 4 = 8
  7. Point (1, 1), slope 3: y = 3x − 2
Topic 11 · Part A + Part B Q4

Integration and area

Five mock MCQs plus the full area-between-curves question. If you skip one Part B question, this is a strong one to keep because the method never changes.

11.1Indefinite and definite integrals

  1. Simplify first (expand brackets, split fractions, rewrite roots as powers).
  2. Power rule: add 1 to the power, divide by the new power. Special: ∫1/x = ln|x|, ∫eˣ = eˣ. Indefinite → + C.
  3. Definite: find F(x), then write F(b) − F(a) with both substitutions shown.
  4. Linearity: pull constants out, split sums. Swapping limits flips the sign.
  5. Initial value: integrate, then use the given point to find C.
Worked example (mock Q17)

∫₀³ 2x dx = [x²]₀³ = 9 − 0 = 9 (option C).

Practice (7)
  1. ∫(4x³ − 6x + 2) dx
  2. ∫(1/x³ + √x) dx
  3. ∫₁² 3x² dx
  4. ∫₁ᵉ (1/x) dx
  5. ∫₀¹ 2eˣ dx
  6. ∫₀⁶ f = 10 and ∫₀⁶ g = −2. Find ∫₀⁶ (2f − 3g).
  7. f′(x) = 2x − 3 and f(1) = 4. Find f(x).
Show solutions
  1. x⁴ − 3x² + 2x + C
  2. x⁻³ → x⁻²/(−2); x^(1/2) → x^(3/2)/(3/2). −1/(2x²) + (2/3)x^(3/2) + C
  3. [x³]₁² = 8 − 1 = 7
  4. [ln x]₁ᵉ = 1 − 0 = 1
  5. [2eˣ]₀¹ = 2e − 2
  6. 2(10) − 3(−2) = 26
  7. f = x² − 3x + C; 1 − 3 + C = 4 → C = 6. f(x) = x² − 3x + 6

11.2Part B Area between two curves

  1. (A) Set the two functions equal, move everything to one side, factor, solve. These x-values are the limits.
  2. (B) Pick a test x between the limits (0 is often easiest). The larger value is the top curve. Write Area = ∫ₐᵇ (top − bottom) dx and simplify the integrand.
  3. (C) Antiderivative of the simplified integrand, term by term.
  4. (D) Show F(b), show F(a) separately (careful with negative a), subtract. Area must be positive; units²
Worked example

y = x² and y = x. (A) x² = x → x(x − 1) = 0 → x = 0, 1. (B) At x = 0.5: line 0.5 > parabola 0.25, so Area = ∫₀¹ (x − x²) dx. (C) F(x) = x²/2 − x³/3. (D) F(1) = 1/2 − 1/3 = 1/6; F(0) = 0. Area = 1/6.

Practice (7)
  1. y = x² and y = 4.
  2. y = x² and y = 2x.
  3. y = x² and y = x + 2.
  4. y = 6x − x² and the x-axis.
  5. y = x³ and y = x for 0 ≤ x ≤ 1.
  6. Find the net signed area and the total area of y = x − 1 from 0 to 2.
  7. y = 4 − x² and y = x + 2.
Show solutions
  1. x = ±2; top is 4. ∫₋₂² (4 − x²) dx = [4x − x³/3]₋₂² = (8 − 8/3) − (−8 + 8/3) = 32/3
  2. x² − 2x = 0 → x = 0, 2; top 2x. [x² − x³/3]₀² = 4 − 8/3 = 4/3
  3. x² − x − 2 = 0 → x = −1, 2; at 0: 2 > 0, top is the line. F = x²/2 + 2x − x³/3. F(2) = 10/3, F(−1) = −7/6. Area = 10/3 + 7/6 = 9/2
  4. x = 0, 6. [3x² − x³/3]₀⁶ = 108 − 72 = 36
  5. [x²/2 − x⁴/4]₀¹ = 1/2 − 1/4 = 1/4
  6. Net: [x²/2 − x]₀² = 2 − 2 = 0. Total: two triangles of ½ each = 1
  7. x² + x − 2 = 0 → x = −2, 1; at 0: 4 > 2, top is the parabola. ∫₋₂¹ (2 − x − x²) dx; F = 2x − x²/2 − x³/3; F(1) = 7/6, F(−2) = −10/3. Area = 7/6 + 20/6 = 9/2

11.3u-substitution

  1. Choose u = the "inside" function whose derivative also appears (up to a constant).
  2. du = u′(x) dx. Balance any missing constant (e.g. x dx = du/2).
  3. Rewrite everything in u, integrate, substitute x back.
  4. Definite: change the limits to u-values and never return to x.
Worked example (mock Q18)

∫3x² cos(x³) dx: u = x³, du = 3x² dx (option B). Then ∫cos u du = sin u + C = sin(x³) + C.

Practice (7)
  1. ∫2x(x² + 1)⁵ dx
  2. ∫x(x² + 1)⁵ dx
  3. ∫2x e^(x²) dx
  4. ∫x²/(x³ + 1) dx
  5. ∫sin x cos x dx
  6. ∫₀¹ 2x(x² + 1)³ dx (change the limits)
  7. ∫₀^(π/2) cos x sin² x dx
Show solutions
  1. u = x² + 1, du = 2x dx: (x² + 1)⁶/6 + C
  2. x dx = du/2: (x² + 1)⁶/12 + C
  3. e^(x²) + C
  4. u = x³ + 1, x² dx = du/3: (1/3) ln|x³ + 1| + C
  5. u = sin x: sin²x/2 + C
  6. u: 1 → 2. ∫₁² u³ du = [u⁴/4]₁² = 4 − 1/4 = 15/4
  7. u = sin x: 0 → 1. ∫₀¹ u² du = 1/3
Topic 12 · Part A

Descriptive statistics

Four mock MCQs, all conceptual. Population formulas (divide by N) are what your notes use.

12.1Mean, median, mode and choosing between them

  1. Sort the data first. Always.
  2. Mean = sum ÷ count. Median = middle value (even count: average the two middles). Mode = most frequent.
  3. Outliers or skew → report the median. The mean is pulled towards the long tail: right skew ⇒ mean > median.
  4. Combined mean = (n₁m₁ + n₂m₂)/(n₁ + n₂), never the plain average of the two means.
Practice (7)
  1. Mean, median and mode of {3, 7, 7, 2, 9, 11, 4}.
  2. Median of {12, 5, 8, 20, 15, 3}.
  3. A class of N students has mean M. A new student scores X. New mean?
  4. Cluster A: 40 servers averaging 60%. Cluster B: 60 servers averaging 80%. Combined mean?
  5. 50 staff earn ₹7 lakh and the CEO earns ₹5 crore. Which average describes a typical salary?
  6. A histogram has a long tail to the right. Compare mean and median.
  7. Sorted data {2, 5, x, y, 14, 20} has median 9.5 and mean 10. If y − x = 3, find x and y.
Show solutions
  1. Sorted 2, 3, 4, 7, 7, 9, 11. Mean 43/7 ≈ 6.14, median 7, mode 7
  2. Sorted 3, 5, 8, 12, 15, 20 → (8 + 12)/2 = 10
  3. (NM + X)/(N + 1)
  4. (40·60 + 60·80)/100 = 7200/100 = 72%
  5. The median; the outlier drags the mean up.
  6. Mean > median
  7. Median (x + y)/2 = 9.5 → x + y = 19. Check mean: (2+5+14+20+19)/6 = 60/6 = 10 ✓. With y − x = 3: x = 8, y = 11

12.2Range, variance and standard deviation

  1. Find the mean μ.
  2. Make a column of deviations (x − μ). They always sum to 0, a good check.
  3. Square each deviation and add them.
  4. Variance σ² = that sum ÷ N. Standard deviation σ = √variance.
  5. Shortcut: σ² = (Σx²)/N − μ².
Worked example

{2, 4, 4, 4, 5, 5, 7, 9}: μ = 40/8 = 5. Squared deviations 9, 1, 1, 1, 0, 0, 4, 16 → 32. σ² = 32/8 = 4, σ = 2.

Practice (7)
  1. Variance and σ of {1, 2, 3, 4, 5}.
  2. σ and range of {10, 10, 10, 10}.
  3. Variance and σ of {2, 4, 6, 8}.
  4. σ = 4. What is the variance?
  5. Variance of {0, 10}.
  6. Use the shortcut formula on {1, 2, 3, 4, 5} to check question 1.
  7. Range of {4, 19, 7, 12}.
Show solutions
  1. μ = 3; squares 4, 1, 0, 1, 4 = 10; σ² = 2, σ = √2 ≈ 1.41
  2. No spread: σ = 0, range = 0
  3. μ = 5; 9, 1, 1, 9 = 20; σ² = 5, σ = √5 ≈ 2.24
  4. 4² = 16 (mock Q34)
  5. μ = 5; (25 + 25)/2 = 25
  6. Σx² = 55; 55/5 − 3² = 11 − 9 = 2 ✓
  7. 19 − 4 = 15

12.3Transforming data and the bell curve

  1. Add c to every value: mean + c, median + c, σ unchanged (spread doesn't move).
  2. Multiply every value by k: mean × k, σ × |k|, variance × k².
  3. New = aX + b: μ_new = aμ + b, σ_new = |a|σ.
  4. Normal curve: 68% within μ ± σ, 95% within μ ± 2σ, 99.7% within μ ± 3σ. Each half is symmetric, so split percentages in two.
Practice (7)
  1. Every score gets +10 marks. Old mean 65, σ 8. New mean and σ?
  2. Every value is multiplied by 3. What happens to σ and to the variance?
  3. Temperatures have mean 20°C, σ 5°C. Convert both to °F (F = 1.8C + 32).
  4. X has μ = 8, σ = 3. Y = −2X + 5. Find μ_Y and σ_Y.
  5. Normal, mean 70, σ 10. Percentage between 50 and 90?
  6. Same distribution: percentage above 80?
  7. Same distribution: percentage between 60 and 90?
Show solutions
  1. Mean 75, σ 8
  2. σ × 3, variance × 9 (mock Q36: B)
  3. Mean 1.8·20 + 32 = 68°F; σ = 1.8·5 = 9°F (the +32 doesn't affect spread)
  4. μ_Y = −16 + 5 = −11; σ_Y = |−2|·3 = 6
  5. μ ± 2σ: 95%
  6. 68% is inside 60 to 80, leaving 32% split over two tails: 16%
  7. 60 to 70 is 34%, 70 to 90 is 47.5%: 81.5%
Topic 13 · Part A

Mental maths and number theory

Four free marks in the mock if you know the tricks. No calculator thinking needed.

13.1Calculation tricks

  1. ×5: ×10 then halve. ×25: ×100 then ÷4. ÷5: ×2 then ÷10.
  2. ×11: write the outer digits, put their sum in the middle (carry if ≥ 10).
  3. N5²: N × (N + 1), then append 25.
  4. Near 100: cross-subtract the differences for the front, multiply the differences for the last two digits.
  5. Two numbers equally spaced around a round base: (base)² − (gap)².
Worked example (mock Q1)

75 × 45: midpoint 60, gap 15. 60² − 15² = 3600 − 225 = 3375 (option A).

Practice (7)
  1. 48 × 25
  2. 815 ÷ 5
  3. 45 × 11 and 78 × 11
  4. 85²
  5. 91 × 96
  6. 44 × 36
  7. 64 × 5
Show solutions
  1. 4800 ÷ 4 = 1200
  2. 1630 ÷ 10 = 163
  3. 4|9|5 = 495; 7|15|8 → carry → 858
  4. 8 × 9 = 72, append 25: 7225
  5. Differences −9, −4. 91 − 4 = 87; 9 × 4 = 36 → 8736
  6. 40² − 4² = 1600 − 16 = 1584
  7. 640 ÷ 2 = 320

13.2Fractions, decimals and repeating decimals

  1. Memorise: 1/2 = .5, 1/4 = .25, 3/4 = .75, 1/5 = .2, 1/8 = .125, 1/6 ≈ .167, 1/3 ≈ .333.
  2. k/9 = 0.kkk…; k/11 = 0.(9k)(9k)… e.g. 3/11 = 0.2727…; k/7 cycles through 142857.
  3. Repeating decimal → fraction: repeating block over as many 9s as its length (0.ababab… = ab/99), then simplify.
  4. With a whole part: keep it aside, convert the decimal part, then add.
Worked example (mock Q4)

0.363636… = 36/99. Divide by 9: 4/11 (option B).

Practice (7)
  1. 3/4 + 2/5 as a decimal.
  2. First three digits of 4/9.
  3. 5/11 as a decimal.
  4. 3/7 as a decimal (use the 142857 cycle).
  5. 0.454545… as a fraction.
  6. 2.135135… as a fraction.
  7. Estimate 7/8 + 1/6.
Show solutions
  1. 0.75 + 0.40 = 1.15 (mock Q2: A)
  2. 0.444 (mock Q3: C)
  3. 9 × 5 = 45 → 0.4545…
  4. Cycle starting at 4: 0.428571…
  5. 45/99 = 5/11
  6. 2 + 135/999 = 2 + 5/37 = 79/37
  7. 0.875 + 0.167 ≈ 1.04

13.3Divisibility, primes, factors, GCD and LCM

  1. Divisibility: 3 and 9 by digit sum; 4 by last two digits; 8 by last three; 11 by alternating digit sum; 7 by "drop the last digit, subtract twice it"; composite divisors by coprime parts (36 = 4 × 9).
  2. Prime test: divide only by primes up to √n.
  3. Prime factorise by ladder division. Number of factors of p^a·q^b·r^c = (a+1)(b+1)(c+1).
  4. GCD: common primes, smallest powers. LCM: all primes, largest powers. GCD × LCM = product of the two numbers.
  5. Word problems: "repeat together / same time again" = LCM; "largest equal pieces / groups" = GCD.
Practice (7)
  1. The 4-digit number 4x52 is divisible by 9. Find the digit x.
  2. Is 81,752 divisible by 11?
  3. Is 223 prime?
  4. How many factors does 360 have? How many are odd?
  5. A = 2⁴·3²·5 and B = 2²·3³·7. Find GCD and LCM.
  6. GCD of two numbers is 12, LCM is 360, one number is 60. Find the other.
  7. Backups run every 12, 18 and 30 minutes, all starting at noon. When do they next coincide?
Show solutions
  1. 4 + x + 5 + 2 = 11 + x must be 18 → x = 7
  2. From the right: 2 − 5 + 7 − 1 + 8 = 11, divisible by 11 → yes
  3. √223 ≈ 14.9. Not divisible by 2, 3, 5, 7, 11, 13 → prime
  4. 360 = 2³·3²·5 → 4·3·2 = 24 factors. Odd ones ignore the 2s: 3·2 = 6
  5. GCD = 2²·3² = 36; LCM = 2⁴·3³·5·7 = 15,120
  6. 12 × 360 / 60 = 72
  7. LCM(12, 18, 30) = 2²·3²·5 = 180 minutes → 3:00 pm
Topic 14 · Part A

Word problems

Four mock MCQs. Translate phrase by phrase; never solve in your head first.

14.1Translating and solving

  1. Let x = the unknown, written in words.
  2. Translate: "is" → =, "more than" → +, "times" → ×, "less than" → subtract in reverse order ("5 less than x" = x − 5).
  3. Consecutive integers x, x+1; consecutive even or odd x, x+2.
  4. Rate = amount ÷ time. Profit = Revenue − Total cost; Revenue = price × quantity; break-even when Revenue = Cost.
  5. Solve, then substitute back to check.
Worked example (mock Q21)

"Four more than three times a number is twenty-five" → three times x is 3x, four more is 3x + 4, "is" → 3x + 4 = 25, so x = 7.

Practice (7)
  1. "Five less than twice a number is 17." Find the number.
  2. Two consecutive even integers sum to 46. Find them.
  3. A machine processes 5 blocks in 2 hours. Its rate? Time for 20 blocks?
  4. Price ₹50 per unit; costs ₹2000 fixed + ₹30 per unit. Profit at 150 units?
  5. Same business: break-even quantity?
  6. 20% of a number is 36. Find it.
  7. A train runs at 60 km/h for 2.5 hours. Distance?
Show solutions
  1. 2x − 5 = 17 → x = 11
  2. x + (x + 2) = 46 → x = 22: 22 and 24
  3. 2.5 blocks per hour; 20 ÷ 2.5 = 8 hours
  4. Revenue 7500, cost 2000 + 4500 = 6500, profit ₹1000
  5. 50q = 2000 + 30q → q = 100
  6. 0.2x = 36 → 180
  7. 150 km
Topic 15 · Part A · quick revision

AP, GP and patterns

You've practised these, so this is a speed check. If you get all of a set right in a few minutes, move on.

15.1Arithmetic progressions

  1. aₙ = a + (n − 1)d; Sₙ = n/2 [2a + (n − 1)d] = n/2 (a + l).
  2. Two terms given: subtract the equations to get d, then a.
  3. Three unknown terms: a − d, a, a + d.
Practice (7)
  1. 8th term of 5, 9, 13, 17, …
  2. How many terms in 7, 11, …, 99?
  3. Sum of the first 20 terms of 3, 7, 11, …
  4. Sum of all odd numbers from 101 to 199.
  5. For 50, 47, 44, …, which n gives the largest sum, and what is it?
  6. a₃ = 7 and a₇ = 19. Find a₁₅.
  7. Three numbers in AP have sum 15 and product 80. Find them.
Show solutions
  1. 5 + 7·4 = 33 (mock Q25: B)
  2. (99 − 7)/4 + 1 = 24
  3. 10 × (6 + 76) = 820
  4. n = 50; 50/2 × (101 + 199) = 7500
  5. Terms stay positive while 50 − 3(n − 1) > 0 → n ≤ 17 (a₁₇ = 2). S₁₇ = 17/2 × (50 + 2) = 442
  6. 4d = 12 → d = 3, a = 1. a₁₅ = 1 + 42 = 43
  7. 3a = 15 → a = 5; 5(25 − d²) = 80 → d = 3: 2, 5, 8

15.2Geometric progressions

  1. aₙ = arⁿ⁻¹; Sₙ = a(rⁿ − 1)/(r − 1); S∞ = a/(1 − r) only when |r| < 1.
  2. r = second term ÷ first term. Check it with the third.
  3. Inserting k means between a and b: b = a·r^(k+1).
  4. Two numbers from AM and GM: a + b = 2·AM, ab = GM².
Practice (7)
  1. Sum to infinity of 12, 4, 4/3, 4/9, …
  2. 6th term of 3, 6, 12, …
  3. Insert four geometric means between 3 and 96.
  4. Sum of the first 6 terms of 2, 6, 18, …
  5. For what x does 1 + x/2 + x²/4 + … converge?
  6. A ball dropped from 20 m rebounds to 3/4 of its height each time. Total distance travelled?
  7. Two positive numbers have AM 10 and GM 8. Find them.
Show solutions
  1. r = 1/3: 12/(2/3) = 18 (mock Q26: B)
  2. 3 × 2⁵ = 96
  3. 96 = 3r⁵ → r = 2: 6, 12, 24, 48
  4. 2(3⁶ − 1)/(3 − 1) = 728
  5. |x/2| < 1 → −2 < x < 2
  6. 20 + 2 × 15/(1 − 3/4) = 20 + 120 = 140 m
  7. a + b = 20, ab = 64 → 16 and 4

15.3Number patterns

  1. Take differences until a row is constant. Constant at row k ⇒ polynomial of degree k.
  2. Quadratic: an² + bn + c with 2a = second difference. Subtract an² from each term and find the linear rest.
  3. Check for alternating signs, factorials, interleaved sequences (odd and even positions follow separate rules).
Practice (7)
  1. What kind of sequence are the first differences of 1, 8, 27, 64, …?
  2. nth term of 4, 11, 22, 37, 56, …
  3. The 10th triangular number.
  4. How many squares of all sizes are on an 8 × 8 chessboard?
  5. 12 people all shake hands once. How many handshakes?
  6. Next term: 2, 1/2, 4, 1/4, 8, 1/8, …
  7. Verify Cassini's identity F₄F₆ − F₅² = (−1)⁵ for Fibonacci.
Show solutions
  1. 7, 19, 37, 61 have constant second difference → quadratic (mock Q27: C)
  2. Second difference 4 → 2n². Leftover 2, 3, 4, 5, 6 = n + 1. 2n² + n + 1
  3. 10 × 11/2 = 55
  4. 1² + 2² + … + 8² = 8·9·17/6 = 204
  5. C(12, 2) = 66
  6. Odd positions double, even positions halve: 16
  7. F₄ = 3, F₅ = 5, F₆ = 8: 24 − 25 = −1 ✓