VI. Quantitative Ability / Reasoning

CSS GSA — Paper 7, Part-II, Section VI (Quantitative Ability/Reasoning)

Overlaps heavily with the dedicated CSS Math / General Ability — Index page (under MPT — Quantitative / Math), which holds the full formulas, worked examples, and drills. This page indexes the GSA-Paper-7-specific sub-topics from the KIPS material below.

This is the index for Quantitative Ability/Reasoning. Each sub-topic below has its own sub-page with rules, worked/practice questions and answers, MCQ traps, and self-test questions.

Past Paper Weightage

• Part-II (General Ability) = 40 marks; Quantitative Ability is its single biggest contributor. Solid Geometry, Plane Geometry, Algebra, Ratios/Proportion/Average, Work-Rate/Pipes, and Profit & Loss are the most consistently recurring in 2021–2025 solved papers and the MPT General Ability section — prioritize these first.

Sub-topics

Exam Angle

• CSS quant questions in GSA are typically applied word problems, not abstract proofs — practice translating a sentence into an equation quickly
• Time management matters more than depth here: most questions are solvable in under a minute once the underlying formula/technique is recognized

Inheritance, Zakat, Ushr, and Tax Calculations

These are a recurring CSS GSA word-problem category (Inheritance Distribution already appeared in 2020 Q8d above). Each topic below gives the formula/rule plus one worked example in numbered-steps format.

Zakat Calculation

Rule: Zakat = 2.5% per annum on wealth held above the Nisab threshold for one full lunar (Hijri) year — applies to savings, cash, gold, silver, and business/trade assets (not on personal-use items like a house or car you live in/use).

Worked Example: A person has Rs. 800,000 in savings held for over a year. Calculate the Zakat due.

Steps to write in exam:

1. Zakat rate = 2.5% = 0.025.
2. Zakat due = 800,000 × 0.025.
3. Zakat due = Rs. 20,000.

Quick Notes:

• Zakat = 2.5% of qualifying wealth held for 1 lunar year, above Nisab
• Applies to: savings, gold/silver, business stock — not personal-use assets
• Formula: Zakat = Wealth × 0.025

Ushr Calculation

Rule: Ushr is a levy on agricultural produce. It is 10% of produce from land irrigated by natural means (rainfall, rivers, springs — no extra cost to the farmer), and 5% of produce from land irrigated by artificial/paid means (tube wells, canals requiring payment, tractors etc.), since less net benefit is assumed when irrigation costs money.

Worked Example: A farmer harvests crop worth Rs. 200,000 from rain-fed land, and separately Rs. 150,000 from land irrigated by tube well. Calculate total Ushr payable.

Steps to write in exam:

1. Rain-fed land → rate = 10%: Ushr = 200,000 × 0.10 = Rs. 20,000.
2. Tube-well irrigated land → rate = 5%: Ushr = 150,000 × 0.05 = Rs. 7,500.
3. Total Ushr payable = 20,000 + 7,500 = Rs. 27,500.

Quick Notes:

• Natural irrigation (rain/river) → Ushr = 10% of produce value
• Artificial irrigation (tube well/canal, paid) → Ushr = 5% of produce value
• Total = sum of each portion's Ushr separately

Islamic Inheritance (Fara'idh) — Basic Shares

Rule: Islamic inheritance law assigns fixed fractional shares to specific heirs before the remainder is divided; the most commonly tested shares are: Wife = 1/8 (if children exist) or 1/4 (if no children); Husband = 1/4 (if children exist) or 1/2 (if no children); Son : Daughter ratio = 2:1 (a son gets double a daughter's share) when both inherit together; Mother = 1/6 (if children exist) or 1/3 (if not); Father = 1/6 (if children exist), plus residue if no children. Debts and funeral expenses are deducted from the estate FIRST, before any share is calculated.

Worked Example: A man dies leaving an estate of Rs. 2,400,000 and debts of Rs. 400,000. He is survived by his wife, one son, and one daughter (no other heirs). Distribute the estate.

Steps to write in exam:

1. Deduct debts first: net estate = 2,400,000 − 400,000 = 2,000,000.
2. Wife's share (children exist) = 1/8 of net estate = 2,000,000 × 1/8 = Rs. 250,000.
3. Remaining estate after wife's share = 2,000,000 − 250,000 = 1,750,000.
4. Son : Daughter = 2:1, so son gets 2 parts and daughter gets 1 part → 3 parts total.
5. Each part = 1,750,000/3 ≈ Rs. 583,333.
6. Son's share = 2 × 583,333 ≈ Rs. 1,166,667; Daughter's share = 1 × 583,333 ≈ Rs. 583,333.

Quick Notes:

• Order of operations: pay debts/funeral costs FIRST → then fixed shares (wife/husband/parents) → then residue split among children
• Wife: 1/8 (with children) or 1/4 (without); Husband: 1/4 (with children) or 1/2 (without)
• Son : Daughter = 2:1 when splitting the residue
• Mother: 1/6 (with children) or 1/3 (without); Father: 1/6 (with children) + possible residue

Tax Calculation (Income Tax / GST word problems)

Rule: Two common CSS-style tax problems: (1) Slab-based income tax — a portion of income is tax-free, and tax is applied only to the amount above each slab boundary; (2) Flat-rate tax/GST — a fixed percentage applied to a price or bill (as already seen in the 2021 Q6c restaurant-bill example above, which stacked a discount, service charge, and GST).

Worked Example (GST on a bill): An item's marked price is Rs. 50,000. A 10% discount is applied, followed by 17% GST on the discounted price. Find the final amount payable.

Steps to write in exam:

1. Apply discount first: 50,000 × (1 − 0.10) = 50,000 × 0.90 = Rs. 45,000.
2. Apply GST on the discounted price: 45,000 × 0.17 = Rs. 7,650.
3. Final amount payable = 45,000 + 7,650 = Rs. 52,650.

Worked Example (slab-based income tax): A person's annual income is Rs. 1,200,000. The tax rule is: no tax on the first Rs. 600,000; 5% tax on the amount from Rs. 600,001 to Rs. 1,200,000. Find the tax payable.

Steps to write in exam:

1. Tax-free portion = first Rs. 600,000 → tax = 0.
2. Taxable portion = 1,200,000 − 600,000 = Rs. 600,000.
3. Tax on taxable portion = 600,000 × 5% = 600,000 × 0.05 = Rs. 30,000.
4. Total tax payable = Rs. 30,000 (only the amount above the first slab is taxed, not the whole income).

Quick Notes:

• GST/flat tax: apply percentage directly to the (already-discounted, if any) amount — always apply discount before GST unless the question states otherwise
• Slab-based tax: only the portion of income ABOVE each slab boundary is taxed at that slab's rate — never apply the rate to the whole income
• Common trap: candidates mistakenly tax the full amount instead of only the amount exceeding the tax-free threshold

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Past paper practice

• Number series, ages/ratio word problems, geometry (area/volume), percentage, and probability/sampling are the most consistently recurring Quant types — master these first. Some questions duplicate the KIPS Probability sub-page exactly (noted below).

Past Paper Questions (FPSC GSA Paper 7, 2020–2024)

Each question is solved as numbered steps (write these in the exam) plus Quick Notes for fast revision — no dense single-paragraph answers.

2024 — Q6(a) [5] Four Numbers Puzzle

Steps to write in exam:

1. Let the four numbers be a, b, c, d.
2. Set up the equal-value chain: a+3 = 2b = 5c = d = k.
3. Express each in terms of k: a = k−3, b = k/2, c = k/5, d = k.
4. Substitute into the sum a+b+c+d = 105: (k−3) + k/2 + k/5 + k = 105.
5. Combine: 2.7k − 3 = 105 → 2.7k = 108 → k = 40.
6. So a = 37, b = 20, c = 8, d = 40 (check: 37+20+8+40 = 105 ✓).

Answer: Ascending order: 8, 20, 37, 40

Quick Notes:

• Set a+3=2b=5c=d=k → solve 2.7k=108 → k=40
• Numbers (ascending): 8, 20, 37, 40

2024 — Q6(c) [5] Missing Numbers in Series

Steps to write in exam:

1. (i) 121, 11, 81, 9, ?, 7 → pairs of (square, root): 121=11², 81=9², so missing = 7² = 49.
2. (ii) 100, 50, 25, ?, 6.25 → each term is halved: 25→12.5→6.25, so missing = 12.5.
3. (iii) 4, 9, 64, 125, 1296, ? → pattern is 2², 3², 4³, 5³, 6⁴, so missing = 7⁴ = 2401.
4. (iv) 2, 5, 12, 24, 48, ? → last three terms double each time (12→24→48), so missing = 96.
5. (v) 44, 22, 66, 33, 132, ? → pattern alternates ÷2, ×3, ÷2, ×4, ÷2: 132÷2 = 66.

Quick Notes:

• (i) 49 (ii) 12.5 (iii) 2401 (iv) 96 (v) 66
• Always test: differences, ratios, alternating operations, and square/cube patterns first

2024 — Q6(d) [5] Three-Digit Number Puzzle

Steps to write in exam:

1. Let digits be h (hundreds), t (tens), u (units).
2. Given: h+t+u = 15 and t+u = 12 → h = 15−12 = 3.
3. Given: t−u = 2. Combine with t+u = 12: adding the two equations gives 2t = 14 → t = 7.
4. Then u = 12−7 = 5.
5. Number = 375.

Answer: 375

Quick Notes:

• h=3 (from 15−12), then t+u=12 and t−u=2 → t=7, u=5
• Number = 375

2024 — Q7(a) [5] Right-Triangle Path — Total Distance and Displacement

Steps to write in exam:

1. Base=4km, hypotenuse=5km → by the 3-4-5 Pythagorean triple, the third side = 3km.
2. One complete round of the triangle = 4+5+3 = 12 km.
3. After the loop, he continues 6 km, then turns 90° and continues 8 km.
4. Total distance travelled = 12 + 6 + 8 = 26 km.
5. The final 6 km and 8 km legs are perpendicular, so displacement from start = √(6²+8²) = √100 = 10 km.

Quick Notes:

• Triangle sides: 4-5-3 (Pythagorean triple) → perimeter = 12 km
• Total distance travelled = 12+6+8 = 26 km
• Distance from start = √(6²+8²) = 10 km (6-8-10 triple)

2024 — Q7(b) [5] Pocket Money Ratio Problem

Steps to write in exam:

1. Let Nasir's money = n.
2. Akbar has thrice Nasir's amount → Akbar = 3n.
3. Ali has five times Akbar's amount → Ali = 15n.
4. Hassan has one-third of Ali's amount → Hassan = 5n.
5. Shahbaz gets Nasir + Ali combined → Shahbaz = n+15n = 16n.
6. Total: 5n+15n+3n+n+16n = 40n = 8000 → n = 200.
7. So Nasir=200, Akbar=600, Ali=3000, Hassan=1000, Shahbaz=3200 (check: sum=8000 ✓).

Quick Notes:

• Nasir=200, Akbar=600, Ali=3000, Hassan=1000, Shahbaz=3200

2024 — Q7(c) [5] Surface Area and Volume of a Sphere (r = 7m)

Steps to write in exam:

1. Use π = 22/7 since r=7 makes the arithmetic clean.
2. Surface area = 4πr² = 4×(22/7)×49 = 4×22×7 = 616 m².
3. Volume = (4/3)πr³ = (4/3)×(22/7)×343 = (4×22×49)/3 = 4312/3 ≈ 1437.33 m³.

Quick Notes:

• Surface area formula: 4πr² → 616 m²
• Volume formula: (4/3)πr³ → ≈ 1437.33 m³
• Use π=22/7 when radius is a multiple of 7 — makes arithmetic clean

2024 — Q7(d) [5] Distribute Rs. 4320 in Ratio 2:3:7

Steps to write in exam:

1. Total parts = 2+3+7 = 12.
2. Each part = 4320/12 = Rs. 360.
3. Zain (2 parts) = 720, Aslam (3 parts) = 1080, Ashraf (7 parts) = 2520 (check: 720+1080+2520=4320 ✓).

Quick Notes:

• Total parts = 12, each part = Rs. 360
• Zain=720, Aslam=1080, Ashraf=2520

2024 — Q8(a) [5] Original Price and Profit

Steps to write in exam:

1. Let original price = x.
2. Price paid includes one-fourth extra profit: x + x/4 = 2,400,000 → (5/4)x = 2,400,000.
3. x = Rs. 1,920,000 (original price).
4. Profit = 2,400,000 − 1,920,000 = Rs. 480,000.

Quick Notes:

• (5/4)x = 2,400,000 → original price = Rs. 1,920,000, profit = Rs. 480,000

2024 — Q8(b) [5] Work Completion with Workers Leaving

Steps to write in exam:

1. Total work = 12 men × 24 days = 288 man-days.
2. Work done in first 4 days = 12×4 = 48 man-days.
3. Remaining work = 288−48 = 240 man-days.
4. After 4 men quit, 8 men remain.
5. Days needed = 240/8 = 30 more days.

Quick Notes:

• Total work = 288 man-days; first 4 days done = 48 man-days; remaining = 240 man-days
• 8 men remain → 240/8 = 30 more days needed

2024 — Q8(c) [5] Shadow on a Building — Similar Triangles

Steps to write in exam:

1. Let D = distance from light source to wall; tree (10m) stands 20m from wall, so it is (D−20) from the source.
2. Similar triangles give: 10/(D−20) = 100/D.
3. Cross-multiply: 10D = 100(D−20) = 100D − 2000.
4. 90D = 2000 → D = 2000/90 ≈ 22.22 m.

Quick Notes:

• Similar triangles: 10/(D−20) = 100/D → D ≈ 22.22 m

2024 — Q8(d) [5] Relative Distances of Three Moving Cars

Steps to write in exam:

1. A and B move opposite at 60 and 100 km/h → relative speed = 160 km/h.
2. In 15 min (0.25 hr): A-B distance = 160×0.25 = 40 km.
3. C moves perpendicular to A/B at 80 km/h → C covers 20 km; A covers 15 km; B covers 25 km in 15 min.
4. A-C distance = √(15²+20²) = √625 = 25 km.
5. B-C distance = √(25²+20²) = √1025 ≈ 32.02 km.

Quick Notes:

• (i) A-B = 40 km (relative speed 160 km/h × 0.25hr)
• (ii) A-C = 25 km (√(15²+20²))
• (iii) B-C ≈ 32.02 km (√(25²+20²))

2023 — Q6(a) [5] Telephone Cost Expression

Steps to write in exam:

1. Fixed initial charge = Rs. 0.50.
2. Per-minute charge = Rs. 0.11, for N minutes → 0.11N.
3. Total cost expression = 0.50 + 0.11N.

Quick Notes:

• Cost = 0.50 + 0.11N (fixed charge + per-minute rate)

2023 — Q6(b) [5] Missing Numbers in Series

Steps to write in exam:

1. (i) 1, 8, 4, 27, 9, ? → pattern alternates cubes/squares of increasing integers (1³,2³,2²,3³,3²,...) → next = 4³ = 64.
2. (ii) 3, 6, 8, 16, 18, ? → alternates ×2, +2: 18×2=36.
3. (iii) 2, 8, 512, ? → each term is the previous cubed (2³=8, 8³=512) → missing = 512³ = 134,217,728.
4. (iv) 81, 9, 64, 8, ?, 12 → (square, root) pairs: 81=9², 64=8², so missing = 12² = 144.
5. (v) 6, 11, 21, 36, 56, ? → differences increase by 5 (5,10,15,20,25) → missing = 56+25 = 81.

Quick Notes:

• (i) 64 (ii) 36 (iii) 134,217,728 (iv) 144 (v) 81

2023 — Q6(d) [5] Shadow on a Wall — Similar Triangles

Steps to write in exam:

1. Ahmad (1.6m) stands 2m from spotlight; wall is 10.5m from him, so 12.5m from the light.
2. Similar triangles: BC/12.5 = 1.6/2.
3. BC = (1.6×12.5)/2 = 10 m.

Quick Notes:

• Similar triangles: BC/12.5 = 1.6/2 → BC = 10 m

2023 — Q7(a) [5] Height of a Tree — Line of Sight

Steps to write in exam:

1. Ali stands 10m from tree, eyes at 1.5m height, line of sight to treetop = 15m (hypotenuse).
2. Right triangle: vertical rise = √(15²−10²) = √125 ≈ 11.18 m.
3. Tree height = rise + eye height = 11.18 + 1.5 ≈ 12.68 m.

Quick Notes:

• Vertical rise = √(15²−10²) = √125 ≈ 11.18 m
• Tree height = 11.18 + 1.5 ≈ 12.68 m

2023 — Q7(c) [5] Lines of Symmetry — Hexagon, Octagon, Circle

Steps to write in exam:

1. Rule: a regular n-sided polygon has exactly n lines of symmetry.
2. Regular hexagon (n=6) → 6 lines of symmetry.
3. Regular octagon (n=8) → 8 lines of symmetry.
4. Circle → infinite lines of symmetry (any line through the center works).

Quick Notes:

• Regular hexagon: 6 lines of symmetry
• Regular octagon: 8 lines of symmetry
• Circle: infinite lines of symmetry
• Rule: a regular n-sided polygon has exactly n lines of symmetry

2023 — Q7(d) [5] Volume of the Egyptian Pyramid

Steps to write in exam:

1. Base side = 230.6 m, height = 146.6 m.
2. Volume formula = (1/3) × base area × height.
3. Base area = 230.6² = 53,176.36 m².
4. Volume = (1/3) × 53,176.36 × 146.6 ≈ 2,598,551 m³.

Quick Notes:

• Volume of pyramid = (1/3) × base area × height
• ≈ 2,598,551 m³ (~2.6 million m³)

2023 — Q8(a) [5] Selling Price with Percentage Gain

Steps to write in exam:

1. Cost price = Rs. 36,800; gain = 8.5%.
2. Selling price = Cost × (1 + gain%) = 36,800 × 1.085.
3. Selling price = Rs. 39,928.

Quick Notes:

• SP = Cost × (1 + gain%) = 36,800 × 1.085 = Rs. 39,928

2023 — Q8(b) [5] Probability — Card Drawn from 12 Cards ⚠️ Duplicates KIPS Probability Sub-page

Steps to write in exam:

1. Total cards = 12 (numbered 1–12).
2. (i) P('8') = 1/12 (only one card shows 8).
3. (ii) P(even) = 6/12 = 1/2 (evens: 2,4,6,8,10,12).
4. (iii) P(perfect square) = 3/12 = 1/4 (squares: 1,4,9).
5. (iv) P(negative) = 0/12 = 0 (no negative cards).
6. (v) P(<13) = 12/12 = 1 (all cards qualify — certain event).

Quick Notes:

• P(8) = 1/12; P(even) = 1/2; P(perfect square) = 1/4; P(negative) = 0; P(<13) = 1 (certain event)
• This exact question type appears on the KIPS Probability sub-page — cross-reference that page's worked examples for more practice

2023 — Q8(d) [5] Direct Proportion — y ∝ x²

Steps to write in exam:

1. Since y ∝ x², write y = kx² for constant k.
2. At value x, y = m, so m = kx².
3. Double x to 2x: y_new = k(2x)² = 4kx² = 4m.
4. Doubling x makes y become 4m (four times the original).

Quick Notes:

• y ∝ x² → doubling x quadruples y → y_new = 4m

2022 — Q6(b) [5] Successive Percentage Decrease

Steps to write in exam:

1. First-year decrease of 10% → retained fraction = 0.90.
2. Second-year decrease of 15% → retained fraction = 0.85.
3. Multiply retained fractions (never add percentages): 0.90×0.85 = 0.765.
4. Overall decrease = 1 − 0.765 = 23.5%.

Quick Notes:

• Successive decrease: multiply retained fractions (0.90 × 0.85 = 0.765) — never simply add the two percentages
• Overall decrease = 23.5%

2022 — Q7(a) [5] Sum of Three Consecutive Odd Numbers

Steps to write in exam:

1. Let numbers be (n−2), n, (n+2).
2. Sum = 3n = 273 → n = 91.
3. Numbers = 89, 91, 93.

Quick Notes:

• Middle number = 273/3 = 91 → numbers are 89, 91, 93

2022 — Q7(b) [5] Missing Numbers in Series

Steps to write in exam:

1. (i) 4,16,36,64,?,144 → squares of even numbers (2²,4²,6²,8²,10²,12²) → missing = 10² = 100.
2. (ii) 30,29,27,?,20,15 → subtract increasing amounts (1,2,3,4,5): 27−3=24.
3. (iii) 1,7,15,25,?,51 → differences increase by 2 (6,8,10,12,14): 25+12=37.
4. (iv) 0,2,6,12,20,30,? → differences 2,4,6,8,10,12 (n(n+1) pattern): 30+12=42.
5. (v) 48,24,72,36,108,? → alternates ÷2, ×3: 108÷2=54.

Quick Notes:

• (i) 100 (ii) 24 (iii) 37 (iv) 42 (v) 54

2022 — Q7(d) [5] Ages Problem — Sara, Ali, Mother

Steps to write in exam:

1. Let Sara's age = s.
2. Mother = 6s, Ali = 2s.
3. In 3 years: (s+3)+(6s+3)+(2s+3) = 72 → 9s+9 = 72.
4. 9s = 63 → s = 7.
5. Sara = 7, Ali = 14, Mother = 42.

Quick Notes:

• 9s + 9 = 72 → s = 7 → Sara=7, Ali=14, Mother=42

2022 — Q8(b) [5] Tiling Cost for a Room

Steps to write in exam:

1. Tile area = (1/2)×12×4 = 24 cm² (right triangle, legs 12cm & 4cm).
2. Room area = 8m×6m = 800cm×600cm = 480,000 cm².
3. Number of tiles = 480,000/24 = 20,000 tiles.
4. Total cost = 20,000×15 = Rs. 300,000.

Quick Notes:

• Tile area = 24 cm²; room area = 480,000 cm² → 20,000 tiles needed
• Total cost = Rs. 300,000

2022 — Q8(c) [5] Original Price from a Pricing Mistake

Steps to write in exam:

1. Let original price = P.
2. Mistaken price (+20%) = 1.2P; correct price (−20%) = 0.8P.
3. Difference = 1.2P − 0.8P = 0.4P = 80.
4. P = Rs. 200.

Quick Notes:

• Difference between +20% and −20% of P is 0.4P = 80 → P = Rs. 200

2021 — Q6(a) [5] Man and Son's Ages

Steps to write in exam:

1. Let son's age = s, so man's age = 3s.
2. In 10 years: (s+10)+(3s+10) = 76 → 4s+20 = 76.
3. 4s = 56 → s = 14; man = 3×14 = 42.
4. Man's age when son was born = 42−14 = 28.

Quick Notes:

• 4s + 20 = 76 → son = 14, man = 42 → man was 28 when son was born

2021 — Q6(b) [5] Tiles for a Footpath Around a Plot

Steps to write in exam:

1. Outer rectangle (plot+footpath) = (24+2)×(14+2) = 26×16 = 416 m².
2. Inner plot area = 24×14 = 336 m².
3. Footpath area = 416−336 = 80 m² = 800,000 cm².
4. Tiles needed = 800,000/20 = 40,000 tiles.

Quick Notes:

• Outer area 416 m² − inner area 336 m² = 80 m² footpath → 40,000 tiles needed

2021 — Q6(c) [5] Restaurant Bill with Discount, Service Charge, and GST

Steps to write in exam:

1. Marked price = Rs. 15,000; 10% discount → 15,000×0.90 = 13,500.
2. Add 10% service charge on discounted price: 13,500×0.10 = 1,350 → subtotal = 14,850.
3. Add 17% GST on subtotal: 14,850×0.17 = 2,524.50.
4. Total = 14,850+2,524.50 = Rs. 17,374.50.

Quick Notes:

• Apply discount first, then service charge, then GST cumulatively: 15,000→13,500→14,850→17,374.50
• Total ≈ Rs. 17,374.50

2021 — Q6(d) [5] Speed of Running After Walking

Steps to write in exam:

1. Walking distance = 3 km/h × 0.75 hr = 2.25 km.
2. Remaining distance to cover by running = 6 − 2.25 = 3.75 km.
3. Running time = 0.5 hr, so speed = 3.75/0.5 = 7.5 km/h.

Quick Notes:

• Walk distance = 2.25 km → remaining 3.75 km covered in 0.5 hr running → speed = 7.5 km/h

2021 — Q7(b) [5] Finding a Number from Group Averages

Steps to write in exam:

1. Total of 11 numbers = 11×63 = 693.
2. Total of first 6 = 6×60 = 360; total of last 6 = 6×65 = 390.
3. The 6th number is counted in both groups, so: 6th number = (360+390) − 693.
4. 6th number = 750 − 693 = 57.

Quick Notes:

• 6th number = (sum of first 6 + sum of last 6) − total of all 11 = 750−693 = 57

2021 — Q7(c) [5] Completing a Table for y = x³ − 3x − 10

Steps to write in exam:

1. Substitute each x: x=−2→y=−12, x=−1→y=−8, x=1→y=−12, x=2→y=−8 (fills the table).
2. For x=1.8: y = 1.8³−3(1.8)−10 = 5.832−5.4−10 = −9.568.
3. For y=10: solve x³−3x−20=0 numerically → x ≈ 3.08.

Quick Notes:

• Complete table: x=−2→−12, x=−1→−8, x=1→−12, x=2→−8
• y at x=1.8 ≈ −9.57; x at y=10 ≈ 3.08

2021 — Q7(d) [5] Simple Interest Rate

Steps to write in exam:

1. Principal = Rs. 60,000; final amount after 5 years = Rs. 85,000 → interest earned = 25,000.
2. Use I = (P×r×t)/100: 25,000 = (60,000×r×5)/100 = 3,000r.
3. r = 25,000/3,000 ≈ 8.33% per year.

Quick Notes:

• Interest earned = 25,000 → I=Prt/100 → rate ≈ 8.33% per annum

2021 — Q8(d) [5] Missing Numbers in Series

Steps to write in exam:

1. (i) 4,18,?,100,180,294 → 2nd differences increase by 6, so missing difference=30 → missing term = 18+30 = 48.
2. (ii) 15,31,63,127,? → each term = previous×2+1 → 127×2+1 = 255.
3. (iii) 1,8,27,64,125,? → consecutive cubes (1³–6³) → missing = 6³ = 216.
4. (iv) 132,156,?,210,240 → differences increase by 2 (24,26,28,30) → missing = 156+26 = 182.
5. (v) 8,24,12,36,18,54,? → alternates ×3,÷2 → 54÷2 = 27.

Quick Notes:

• (i) 48 (ii) 255 (iii) 216 (iv) 182 (v) 27

2020 — Q6(a) [5] Work-Rate Problem — Three Workers

Steps to write in exam:

1. Rates: Tariq=1/6/hr, Sajid=1/4/hr, Irfan=1/8/hr.
2. First 2 hrs (Tariq+Sajid): (1/6+1/4)×2 = (5/12)×2 = 5/6 of job done; 1/6 remains.
3. Sajid leaves, Irfan joins: combined rate = 1/6+1/8 = 7/24/hr.
4. Time for remaining 1/6: (1/6)/(7/24) = 4/7 hr ≈ 34 min.
5. Total time = 2 hrs + 4/7 hr ≈ 2 hrs 34 min.

Quick Notes:

• First 2 hrs (Tariq+Sajid): complete 5/6 of job, 1/6 remains
• Remaining 1/6 done by Tariq+Irfan (rate 7/24/hr) → takes 4/7 hr (~34 min)
• Total time ≈ 2 hrs 34 min

2020 — Q6(b) [5] Missing Numbers to Complete Equations

Steps to write in exam:

1. (a) 9+8−5=12=2×? → ? = 6.
2. (b) 3×9−14=13=24−? → ? = 11.
3. (c) 15÷3×12=60=41+? → ? = 19.
4. (d) 24÷4+5=11=66÷? → ? = 6.
5. (e) 8×6−13+3=38=7×6−? → ? = 4.

Quick Notes:

• (a) 6 (b) 11 (c) 19 (d) 6 (e) 4
• Always calculate both sides of the equation first, then isolate the unknown

2020 — Q6(c) [5] Mean, Median, Mode, Range of Ages

Steps to write in exam:

1. Ages: 17,17,18,18,18,19,19 (7 values, sum=126).
2. Mean = 126/7 = 18.
3. Median = middle (4th) value when sorted = 18.
4. Mode = most frequent value = 18 (appears 3×).
5. Range = 19−17 = 2.

Quick Notes:

• Mean = 18, Median = 18, Mode = 18, Range = 2
• Mean = sum÷count; Median = middle value (sorted); Mode = most frequent; Range = max−min

2020 — Q7(b) [5] Wall Area Around a Farm

Steps to write in exam:

1. Farm area = 484 m² (square) → side = √484 = 22 m.
2. Perimeter = 4×22 = 88 m (3 sides × 22m + 1 side × 22m).
3. Wall area = (3 sides×22m×2m) + (1 side×22m×3m) = 132+66 = 198 m².

Quick Notes:

• Square farm: side = √484 = 22 m
• Wall area = (3×22×2) + (22×3) = 132+66 = 198 m²

2020 — Q8(b) [5] Probability — Blood Groups ⚠️ Duplicates KIPS Probability Sub-page

Steps to write in exam:

1. Total people = 600+650+550+200 = 2,000.
2. P(B) = 650/2000 = 32.5%.
3. P(O) = 200/2000 = 10%.

Quick Notes:

• Total = 2000; P(B) = 650/2000 = 32.5%; P(O) = 200/2000 = 10%
• Same calculation type as the KIPS Probability sub-page's blood-group example — cross-reference it

2020 — Q8(c) [5] Road Construction — Men and Days

Steps to write in exam:

1. Total work = 50 men × 40 days = 2,000 man-days.
2. For 70 men: days = 2,000/70 ≈ 28.6 days.

Quick Notes:

• Man-days constant: 50×40 = 2000 = 70×days → days ≈ 28.6

2020 — Q8(d) [5] Inheritance Distribution

Steps to write in exam:

1. Net estate after debt = 1,750,000 − 150,000 = 1,600,000.
2. Let daughter's share = x, son's share = 2x → x+2x = 1,600,000.
3. 3x = 1,600,000 → x = Rs. 533,333 (daughter), 2x = Rs. 1,066,667 (son).

Quick Notes:

• Net estate = 1,600,000; split 1:2 (daughter:son) → daughter ≈ Rs. 533,333, son ≈ Rs. 1,066,667

2024 — Part-I MCQs (Quantitative items)

Steps to write in exam:

1. Combinations of 9 things taken 3 at a time: C(9,3) = 9!/(3!×6!) = (9×8×7)/(3×2×1) = 504/6 = 84.
2. Circle radius tripled: Area = πr², so tripling r gives π(3r)² = 9πr² → area becomes nine times the original. (Same reasoning also appears in the Physical Sciences page's 2024 MCQ set for this question.)

Quick Notes:

• C(9,3) = 9!/(3!6!) = 84
• Radius ×3 → Area ×9 (since area depends on r²)