Unit 3: Mensuration, Calendar & Clocks

Opening Hook — Numbers Shape the Real World

Learning Outcomes — Bloom's Taxonomy Mapped

Concept Explanation — Mensuration, Calendar & Clocks from Scratch

PART I — MENSURATION (Measurement of Areas & Volumes)

Mensuration is the branch of mathematics that deals with measuring geometric figures — their perimeters, areas, surface areas, and volumes. Every engineering, architecture, and competitive exam problem ultimately comes down to knowing the right formula and applying it correctly.

1. Cube — The Perfect 3D Square

          +--------+
         /|       /|
        / |      / |          All edges = a
       +--------+  |          All faces = squares
       |  |     |  |
       |  +-----|--+
       | /      | /
       |/       |/
       +--------+
         side a

2. Cuboid — The Box Shape

          +------------+
         /|           /|
        / |     h    / |        l = length
       +------------+  |        b = breadth
       |  |         |  |        h = height
       |  +---------|--+
       | /     b    | /
       |/           |/
       +------------+
            l

3. Sphere & Hemisphere

     SPHERE                    HEMISPHERE
       _____                      _____
     /       \                  /       \
    |         |                |         |
    |    ·r   |                |    ·r   |
    |         |                |_________|
     \_______/                  flat base
   all points at                half sphere
   distance r                  + circular base

4. Cone — The Ice Cream Shape

           /\
          /  \
         / h  \ l (slant height)
        /      \
       /________\
      |----r----|
     
     l = √(r² + h²)

5. Cylinder — The Pipe/Tank Shape

       ___________
      /           \
     |  ○ r        |  ← top circle
      \___________/
      |           |
      |     h     |
      |           |
      |___________|
     /             \
    |  ○ r          |  ← bottom circle
     \_____________/

6. Combined Shapes — Real-World Structures

Real objects are rarely just cubes or spheres. A water tank is a cylinder + hemisphere. A rocket is a cone on a cylinder. A capsule tablet is a cylinder with two hemispheres. You need to add or subtract volumes and areas of basic shapes.

  WATER TANK               ROCKET               CAPSULE
  (cylinder +              (cone on              (cylinder +
   hemisphere)              cylinder)              2 hemispheres)
                             /\
       _____                /  \                 ___     ___
     /       \             / h₁ \              /   |   |   \
    |   hemi  |           /______\            | r  |   | r  |
    |_________|          |        |            |   |   |   |
    |         |          |   h₂   |             \__|   |__/
    |   cyl   |          |        |
    |    h    |          |________|
    |_________|

📊 Master Formula Comparison Table — ALL Shapes

PART II — CALENDAR (Odd Days & Day Finding)

7. Calendar Basics — Odd Days Concept

The calendar is a mathematical system based on the Earth's revolution around the Sun. To find the day of any date — past or future — we use the Odd Days method.

8. Finding the Day of Any Date — Step-by-Step Method

9. Complete Worked Solution — What day was 15 August 1947?

PART III — CLOCKS (Angles & Positions)

10. Clock Basics — Hand Speeds

           12
        11    1
      10   |    2
            |  ↗ minute hand
     9  ----+---- 3          Hour hand:   360° in 12 hrs = 0.5°/min
            |                Minute hand: 360° in 60 min = 6°/min
      8    |    4            Relative speed = 6 - 0.5 = 5.5°/min
        7     5
           6

11. The Clock Angle Formula

12. Clock Practice Problems

Learn by Doing — 3-Tier Practice Structure

Problem Set — 19 Questions

Formula-Based (5 Questions)

Word Problems (8 Questions)

Q6. A room is 15m long, 12m broad, and 4m high. Find the cost of painting all 4 walls at ₹12/m².

Solution: LSA = 2×4×(15+12) = 8×27 = 216 m². Cost = 216 × 12 = ₹2,592

Q7. A hemispherical dome of a temple has inner radius 10.5m. Find the cost of whitewashing at ₹8/m².

Solution: CSA = 2πr² = 2×(22/7)×110.25 = 693 m². Cost = 693 × 8 = ₹5,544

Q8. A tent is in the shape of a cylinder (r=14m, h=3m) surmounted by a cone (slant height 13m). Find canvas needed.

Solution: Canvas = CSA_cyl + CSA_cone = 2πrh + πrl = 2×(22/7)×14×3 + (22/7)×14×13 = 264 + 572 = 836 m²

Q9. A metallic sphere of radius 21 cm is melted and recast into a cone of radius 21 cm. Find the cone's height.

Solution: V_sphere = V_cone → (4/3)πr³ = (1/3)πR²h → (4/3)×21³ = (1/3)×21²×h → h = 4×21 = 84 cm

Q10. The calendar for the year 2003 will be the same as for which upcoming year?

Solution: 2003 starts on Wednesday (1 odd day). We need the next year starting on Wednesday with same leap pattern. Count: 2003(1) → 2004(2, leap) → 2005 → ... → 2014. The calendar repeats in 2014.

Q11. At what time between 4 and 5 o'clock will the minute hand and hour hand be at 180°?

Solution: At 4:00, angle = 120°. Need 180°. Using formula: 30×4 − 5.5M = −180 → 120 − 5.5M = −180 → 5.5M = 300 → M = 600/11 = 54(6/11) min. Answer: 4:54:33

Q12. A well of diameter 3m is dug 14m deep. The earth taken out is spread all around to form an embankment of width 4m. Find the height of the embankment.

Solution: Volume of earth = π×(1.5)²×14 = 99 m³. Embankment area = π(5.5² − 1.5²) = π×28 = 88 m². Height = 99/88 ≈ 1.125 m

Q13. How many days are there between 2 February and 15 April of the same (non-leap) year?

Solution: Feb remaining = 26 days + March 31 + April 15 = 72 days

SSC/Bank Previous Year (3 Questions)

Q14. [SSC CGL 2022] If the radius of a sphere is increased by 50%, by what percent does the volume increase?

Solution: New radius = 1.5r. New volume = (4/3)π(1.5r)³ = (4/3)π×3.375r³ = 3.375 × original. Increase = 237.5%. Answer: 237.5%

Q15. [IBPS PO 2021] What was the day on 17 June 1998?

Solution: 1900y→1 odd. 97y: 73ord+24leap = 73+48 = 121 → 121÷7 = 17w+2 → 2 odd. Jan(3)+Feb(0)+Mar(3)+Apr(2)+May(3)+Jun(0)=11 → 4 odd. 17 days → 3 odd. Total = 1+2+4+3 = 10 → 10÷7 = 1w+3 → 3 = Wednesday ✅

Q16. [SSC CHSL 2023] At what angle are the clock hands at 4:40?

Solution: Angle = |30×4 − 5.5×40| = |120 − 220| = 100°. Answer: 100°

Interview / Placement (3 Questions)

Q17. [TCS NQT] A cylindrical bucket (r=14cm, h=30cm) is full of water. The water is poured into rectangular tank (l=28cm, b=22cm). Find the rise in water level.

Solution: πr²h = l×b×rise → (22/7)×196×30 = 28×22×rise → 18480 = 616×rise → rise = 30 cm

Q18. [Infosys] How many times in a day do the minute and hour hands of a clock overlap?

Solution: 22 times in 24 hours (11 times in 12 hours). They overlap approximately every 65 5/11 minutes, not every 60 minutes.

Q19. [Wipro] The last day of a century cannot be which of these days: Tuesday, Thursday, Friday, or Saturday?

Solution: Century odd days cycle: 5, 3, 1, 0 → maps to Friday, Wednesday, Monday, Sunday. So last day of a century can ONLY be Sunday, Monday, Wednesday, or Friday. Cannot be Tuesday, Thursday, or Saturday.

MCQ Assessment Bank — 30 Questions (Bloom's Mapped)

Remember / Recall (Q1–Q6)

Understand / Explain (Q7–Q12)

Apply / Calculate (Q13–Q18)

Analyze / Compare (Q19–Q24)

Evaluate / Judge (Q25–Q27)

Create / Design (Q28–Q30)

Short Answer Questions (8)

Q1. Define mensuration and state the formulas for TSA and volume of a cuboid.

Answer: Mensuration is the branch of mathematics that deals with the measurement of geometric figures — their lengths, areas, and volumes. For a cuboid with dimensions l × b × h: TSA = 2(lb + bh + lh) and Volume = lbh.

Q2. Explain the "Odd Days" concept in calendar problems with an example.

Answer: Odd days are the extra days beyond complete weeks in a given period. For example, 365 days = 52 complete weeks + 1 extra day, so an ordinary year has 1 odd day. A leap year (366 days) has 2 odd days. We use odd days to determine the day of the week for any given date by computing total odd days modulo 7.

Q3. Differentiate between CSA and TSA of a hemisphere.

Answer: CSA (Curved Surface Area) of a hemisphere = 2πr² — this includes only the curved dome surface. TSA (Total Surface Area) = 3πr² — this adds the flat circular base (πr²) to the CSA. The difference is πr², the area of the base circle.

Q4. State the clock angle formula and find the angle at 6:20.

Answer: The angle between clock hands = |30H − 5.5M|. At 6:20: |30×6 − 5.5×20| = |180 − 110| = 70°.

Q5. Why is the volume of a cone exactly one-third the volume of a cylinder with the same base and height?

Answer: This can be demonstrated experimentally by filling a cone with water and pouring it into a cylinder of the same radius and height — it takes exactly 3 full cones to fill the cylinder. Mathematically, V_cone = (1/3)πr²h, while V_cyl = πr²h. This relationship arises from integral calculus — the cone tapers linearly from base to apex, reducing the cross-sectional area.

Q6. List the leap year rules and determine whether 1700, 1800, 1900, and 2000 are leap years.

Answer: Rules: (1) Divisible by 4 → leap year candidate. (2) If divisible by 100 → NOT a leap year. (3) If divisible by 400 → IS a leap year.
• 1700: div by 100, NOT by 400 → Not leap
• 1800: div by 100, NOT by 400 → Not leap
• 1900: div by 100, NOT by 400 → Not leap
• 2000: div by 400 → Leap year ✅

Q7. A solid sphere is melted and recast into 8 equal smaller spheres. What is the ratio of the surface area of the original sphere to the total surface area of all 8 smaller spheres?

Answer: Let original radius = R. Volume conserved: (4/3)πR³ = 8 × (4/3)πr³ → R³ = 8r³ → R = 2r.
SA_original = 4πR² = 4π(2r)² = 16πr²
SA_8_spheres = 8 × 4πr² = 32πr²
Ratio = 16πr² : 32πr² = 1 : 2
💡 Total surface area INCREASES when a solid is divided — that's why sugar dissolves faster when crushed!

Q8. How many times do clock hands overlap in 12 hours? Explain why it's not 12.

Answer: The hands overlap 11 times in 12 hours (22 times in 24 hours). It's not 12 because the minute hand takes more than 60 minutes to catch up to the hour hand each time (exactly 65 5/11 minutes). Between 11 and 1 (crossing 12), there's only ONE overlap (at 12:00), not two. So we lose one overlap per 12-hour cycle.

Long Answer Questions (3)

Q1. A toy is made by mounting a cone on a hemisphere. The radius of both is 7 cm. The total height of the toy is 31 cm. Find (i) CSA of the cone, (ii) TSA of the toy, (iii) Volume of the toy. [10 marks]

Q2. Describe the step-by-step method to find the day of the week for any given date. Use this method to find the day on 2 October 1869 (Mahatma Gandhi's birth) and verify for 30 January 1948 (his death). [10 marks]

Q3. A solid iron cylinder (r = 14 cm, h = 42 cm) is melted and recast into cones (r = 7 cm, h = 6 cm). Find the number of cones formed. Also find the clock angle at the time 7:42 and determine what day was 14 November 1889 (Nehru's birth). [10 marks]

Industry Spotlight — SSC CGL Topper

Earn With It — Coaching Centre Aptitude Faculty

Chapter Summary & Master Formula Sheet

📊 COMPLETE Formula Reference Table

📅 Calendar Quick Reference

⏰ Clock Quick Reference

Earning Checkpoint — Self-Assessment

✅ Unit 3 complete. Mensuration, Calendar & Clocks mastered!

[QR: Link to EduArtha video tutorial — Mensuration, Calendar & Clocks]