Unit 3: Mensuration, Calendar and Clocks

PEA308 — Advanced Analytical Skills-Ii 9 min read

I. Foundations — Measurement, Time and Modular Reasoning

Mensuration measures geometric boundaries, surfaces and occupied space, while calendar and clock problems locate events within repeating cycles of days, dates and hours.

A. Governing Principles and Conventions

These principles provide the common framework for solving all problems in the unit.

Units: Perimeter uses linear units such as cm; area uses square units such as cm²; volume uses cubic units such as cm³.
Dimensions: Length, breadth, height and radius must be converted to the same unit before applying a formula.
Exact and approximate values: Keep answers in terms of π when exact values are required; otherwise use π ≈ 22/7 or 3.1416.
Composite measurement: Divide an irregular figure or solid into standard shapes, calculate each required measure, and add or subtract appropriately.
Cyclic arithmetic: Calendar calculations operate modulo 7, while clock-hand positions operate modulo 12 hours or 360°.
Reference convention: In calendars, seven complete days produce no change in weekday; in clocks, one complete revolution returns a hand to its initial direction.

II. Mensuration — Measurement of Plane Figures and Solids

Mensuration applies standard formulae to determine boundaries, enclosed regions, exposed surfaces and capacities.

A. Perimeter and area of 2-D figures

Perimeter is the total boundary length of a plane figure, whereas area is the region enclosed by that boundary.

Rectangle: For length l and breadth b:
TEXT
```
  Perimeter = 2(l + b)
  Area = lb
```
Square: For side a, perimeter is 4a, area is a², and diagonal is a√2.
Triangle: For sides a, b, c, base b₀ and corresponding height h:
TEXT
```
  Perimeter = a + b + c
  Area = 1/2 × b₀h
```
Heron’s formula: If only three sides are known:
TEXT
```
  s = (a + b + c)/2
  Area = √[s(s-a)(s-b)(s-c)]
```
Here, s is the semiperimeter.
Equilateral triangle: For side a, area is (√3/4)a² and height is (√3/2)a.
Parallelogram and rhombus: A parallelogram has area bh; a rhombus with diagonals d₁ and d₂ has area d₁d₂/2.
Trapezium: For parallel sides a and b separated by height h, area is (a+b)h/2.
Circle: For radius r, circumference is 2πr and area is πr²; a semicircle has area πr²/2.

Sector: For central angle θ in degrees:

TEXT

  Arc length = (θ/360)2πr
  Sector area = (θ/360)πr²

Worked example: A rectangular field of 20 m × 12 m has perimeter 64 m and area 240 m².

B. Problems on surface area and volume of cube and cuboid

Cube and cuboid problems distinguish the total enclosing surface from lateral surfaces and internal capacity.

Cube:
- Measures: For edge a, lateral surface area is 4a², total surface area is 6a², and volume is a³.
- Diagonal: A face diagonal is a√2, while the space diagonal is a√3.
Cuboid:
- Measures: For length l, breadth b and height h:
  TEXT
```
     LSA = 2h(l + b)
     TSA = 2(lb + bh + hl)
     Volume = lbh
```
  LSA denotes lateral surface area and TSA denotes total surface area.
- Diagonal: The space diagonal is √(l²+b²+h²).
- Open surfaces: Subtract the omitted face from total surface area; an open cuboid without a top has area lb + 2lh + 2bh.
- Recasting principle: When a solid is melted and reshaped without wastage, its volume remains constant.
- Worked example: A cube of edge 5 cm has total surface area 150 cm² and volume 125 cm³.

C. Problems on surface area and volume of sphere and hemisphere

A sphere has one continuous curved surface, while a hemisphere includes a curved surface and, when closed, a circular base.

Sphere:
- Formulae: For radius r, surface area is 4πr² and volume is 4πr³/3.
- Scaling: If the radius is multiplied by k, surface area becomes k² times and volume becomes k³ times.
Hemisphere:
- Curved surface area: 2πr².
- Total surface area: Including the base, 2πr² + πr² = 3πr².
- Volume: 2πr³/3.
- Displacement: A fully immersed spherical object displaces liquid equal to its own volume.
- Worked example: A sphere of radius 3 cm has surface area 36π cm² and volume 36π cm³.

D. Problems on surface area and volume of cone and cylinder

Cones and cylinders share circular bases, but a cone narrows to a vertex and requires its slant height for curved-surface calculations.

Right circular cylinder:
- Formulae: For radius r and height h:
  TEXT
```
     CSA = 2πrh
     TSA = 2πr(h + r)
     Volume = πr²h
```
  CSA is curved surface area; total surface area includes two circular bases.
Right circular cone:
- Slant height: l = √(r²+h²), where l is slant height.
- Formulae:
  TEXT
```
     CSA = πrl
     TSA = πr(l + r)
     Volume = 1/3 πr²h
```
- Capacity and material: Capacity is determined by volume; sheet material is determined by the exposed surface area.
- Worked example: A cylinder with r = 3 cm and h = 10 cm has volume 90π cm³ and curved surface area 60π cm².

III. Calendar — Weekday Cycles and Date Calculation

A calendar arranges days into weeks, months and years; weekday calculations depend on counting completed seven-day cycles and their remainders.

A. Basic concept of calendar

Calendar problems use the number of “odd days,” meaning days remaining after complete weeks are removed.

Ordinary year: 365 = 52 × 7 + 1, so an ordinary year contributes 1 odd day.
Leap year: 366 = 52 × 7 + 2, so a leap year contributes 2 odd days.
Gregorian leap rule: A year divisible by 4 is normally leap; a century year must also be divisible by 400. Thus 2000 was leap, but 1900 was not.
Month lengths: April, June, September and November have 30 days; February has 28 or 29; the remaining months have 31.
Weekday cycle:
TEXT
```
  Weekday shift = Total days mod 7
```
A remainder of 0 means no shift; 1 means one weekday forward.

B. Date and days

The relationship between dates and days is obtained by counting the interval and reducing it modulo seven.

Inclusive counting: The number of dates from date A through date B includes both endpoints, but the weekday shift counts elapsed days after A.
Same month: The shift from the pth to the qth date is (q-p) mod 7.
Same weekday dates: Dates differing by 7, such as the 5th, 12th, 19th and 26th, fall on the same weekday.
Month transition: A 31-day month shifts the next month’s first day by 3; a 30-day month by 2; February shifts it by 0 in an ordinary year and 1 in a leap year.
Worked example: If 6 March is Monday, 27 March is also Monday because 27-6 = 21, a multiple of 7.

C. Finding the exact day

An exact weekday can be found by adding odd days from completed years, completed months and elapsed dates to a known reference weekday.

Reference date: Under the Gregorian calendar, 1 January 2000 was Saturday.
Procedure:
1. Count complete years between the reference year and target year.
2. Add 1 for each ordinary year and 2 for each leap year.
3. Add days in completed months of the target year.
4. Add date - 1, because no day has elapsed on the first date itself.
5. Reduce the total modulo 7 and move forward from the reference weekday.
Worked example: From 1 January 2000 to 1 January 2001, the leap year 2000 contributes 2 odd days; therefore 1 January 2001 was Monday.

D. Advanced concept of calendar

Advanced calendar analysis uses long-cycle repetition, century effects and structural properties of months.

Century odd days: In the Gregorian calendar, 100, 200, 300 and 400 years contribute 5, 3, 1 and 0 odd days respectively.
Four-hundred-year cycle: There are 303 ordinary and 97 leap years:
TEXT
```
  Total days = 303(365) + 97(366) = 146097
  146097 mod 7 = 0
```
Therefore, the Gregorian weekday pattern repeats every 400 years.
Identical calendars: Two years share a calendar when they have the same leap status and their first days fall on the same weekday.
Five-weekday occurrence: In a 31-day month, the first three weekdays occur five times; in a 30-day month, the first two do.
Counting Sundays: A 28-day February has exactly four of every weekday; extra days determine which weekdays occur five times.

IV. Clocks — Relative Motion of Hands

Clock problems model the hour and minute hands as rotating objects with constant angular speeds on a 360° circular scale.

A. Concept of clock

The positions of clock hands are determined by their angular movement from the 12 o’clock direction.

Minute hand: It covers 360° in 60 minutes, so its speed is 6° per minute.
Hour hand: It covers 360° in 12 hours, so its speed is 0.5° per minute.
Relative speed: The minute hand gains on the hour hand at 6-0.5 = 5.5° per minute.
Position after m minutes past h:
TEXT
```
  Hour-hand angle = 30h + 0.5m
  Minute-hand angle = 6m
```
Here, h is the hour number and m is minutes past that hour.
Coincidence: Consecutive coincidences occur every 360/5.5 = 720/11 minutes, approximately 65 5/11 minutes.

B. Gain and loss of time

A faulty clock’s displayed interval differs from the real interval in a constant ratio when its rate of gain or loss is uniform.

Gaining clock:
- Relation: If a clock gains g minutes in T real minutes, it shows T+g minutes.
- Conversion:
  TEXT
```
     Real time = Shown time × T/(T + g)
```
Losing clock:
- Relation: If it loses l minutes in T real minutes, it shows T-l minutes.
- Conversion:
  TEXT
```
     Real time = Shown time × T/(T - l)
```
- Error accumulation: Rate errors are proportional to elapsed real time; a clock gaining 5 minutes daily gains 20 minutes in 4 days.
- Correct-again condition: On a 12-hour dial, a continuously faulty clock appears correct after gaining or losing a full 12 hours.

C. Angle based problems

Angle problems use the absolute difference between the angular positions of the two hands.

General formula:
TEXT
```
  θ = |30h - 5.5m|
  Smaller angle = min(θ, 360° - θ)
```
Here, θ is the raw angular separation, h is the hour, and m is the number of minutes past it.
Required-angle equation: For an angle α, solve |30h-5.5m| = α; a second position may arise from separation 360°-α.
Right angles: Set the smaller angle to 90°; the hands are perpendicular 22 times in 12 hours.
Opposite hands: Set the separation to 180°; this occurs 11 times in 12 hours.
Coincident hands: Set the separation to 0°; the hands coincide 11 times in 12 hours.
Worked example: At 3:20, the hour hand is at 100° and the minute hand at 120°, so the smaller angle is 20°.