GELE Photogrammetry & Cartography — Philippine Plane Coordinate System and UTMExam Answer Templates

Marks

1

Topic

UTM Overview

Difficulty

easy

Template Id

T1

Examiner Tip

For a 1-mark definition, one complete technical sentence with at least two correct parameters is sufficient. Do not write a paragraph — it wastes time and earns the same mark.

Model Answer

The Universal Transverse Mercator (UTM) is a worldwide plane coordinate system based on the Transverse Mercator projection, dividing the globe into 60 zones each 6° of longitude wide, with a central-meridian scale factor k₀ = 0.9996 and a false easting of 500 000 m to keep all coordinates positive.

Question Type

very_short_answer

Answer Structure

• Single sentence: Name the projection type (Transverse Mercator), state the number of zones (60), zone width (6°), and at least one defining parameter (k₀ or false easting) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying 'Universal Mercator' without 'Transverse' — wrong projection type
• Stating 60 zones but wrong width (e.g., saying 5° or 8°)
• Omitting any defining parameter — definition is too vague to credit

Key Phrases To Include

• Transverse Mercator projection
• 60 zones
• 6° of longitude
• k₀ = 0.9996
• false easting 500 000 m

Marks

2

Topic

PPCS vs UTM Parameters

Difficulty

easy

Template Id

T2

Examiner Tip

Memorize: PPCS has the HIGHER scale factor (0.99995) because it covers narrower zones (≈2°) and needs less correction. UTM zones are wider (6°) so k₀ is lower (0.9996). The narrower the zone, the closer k₀ is to 1.

Model Answer

(a) UTM central-meridian scale factor: k₀ = 0.9996 (b) PPCS central-meridian scale factor: k₀ = 0.99995 Note: Both systems use a scale factor slightly less than 1 so that the projection cylinder intersects the ellipsoid along two secant lines, reducing overall distortion across the zone.

Question Type

very_short_answer

Answer Structure

• Part (a): State k₀ = 0.9996 for UTM [1 mark]
• Part (b): State k₀ = 0.99995 for PPCS [1 mark]

Scoring Breakdown

Common Mark Deductions

• Interchanging the two values — most common error in this chapter
• Writing '1.0' or '1' for either — scale factor at CM is always less than 1 in TM
• Rounding 0.99995 to 0.9999 or 1.0000 — must be exact to 5 decimal places

Key Phrases To Include

• k₀ = 0.9996
• k₀ = 0.99995
• central-meridian scale factor
• Transverse Mercator

Marks

2

Topic

PPCS Zones

Difficulty

easy

Template Id

T3

Examiner Tip

A tabular format earns the mark faster and eliminates ambiguity. The examiner can check all five values in 5 seconds. Write it as a table even for a 2-mark question.

Model Answer

The PPCS has five (5) zones. Their central meridians are: | Zone | Central Meridian | |------|-----------------| | I | 117°E | | II | 119°E | | III | 121°E | | IV | 123°E | | V | 125°E | The zones are spaced 2° apart in longitude, covering the Philippine Archipelago from western Palawan to the eastern Visayas and Mindanao.

Question Type

short_answer

Answer Structure

• State total number of PPCS zones = 5 [1 mark]
• List all five central meridians correctly (117°, 119°, 121°, 123°, 125°E) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating 6 or 4 zones instead of 5
• Starting at 115°E or 116°E instead of 117°E
• Using 3° spacing instead of 2° — results in wrong central meridians
• Omitting the '°E' designation

Key Phrases To Include

• five (5) zones
• 117°E, 119°E, 121°E, 123°E, 125°E
• 2° spacing
• PPCS

Marks

2

Topic

UTM Zone Determination

Difficulty

easy

Template Id

T4

Examiner Tip

Always write ⌊ ⌋ (floor brackets) explicitly. Examiners award a method mark for showing the floor operation. Without it, even a correct final answer may not earn the method mark.

Model Answer

Given: λ = 121°E Formula: UTM Zone = ⌊(λ + 180) / 6⌋ + 1 Solution: UTM Zone = ⌊(121 + 180) / 6⌋ + 1 = ⌊301 / 6⌋ + 1 = ⌊50.1667⌋ + 1 = 50 + 1 ∴ UTM Zone = 51 (Zone 51N, since the Philippines is in the Northern Hemisphere)

Question Type

numerical

Answer Structure

• Write the given longitude [stated]
• Write the UTM zone formula [½ mark]
• Substitute and compute: (121+180)/6 = 50.1667, floor = 50 [½ mark]
• Add 1: zone = 51, specify '51N' for Northern Hemisphere [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using ceiling function instead of floor — gives Zone 51 from rounding, which may be coincidentally correct, but method mark is lost
• Forgetting to add 1 after the floor operation — gives Zone 50 (wrong)
• Not specifying N hemisphere — loses the hemisphere designation mark
• Rounding 50.1667 to 50 without explicitly writing the floor symbol — method unclear

Key Phrases To Include

• UTM Zone = ⌊(λ + 180) / 6⌋ + 1
• floor function
• Zone 51N
• Northern Hemisphere

Marks

2

Topic

False Easting and False Origin

Difficulty

easy

Template Id

T5

Examiner Tip

Always interpret the sign of the displacement. An easting less than 500 000 m means WEST of CM; greater than 500 000 m means EAST of CM. State this explicitly — it is a separate mark.

Model Answer

Given: Grid Easting E = 497 500 m False Easting (at Central Meridian, CM) = 500 000 m Solution: Displacement from CM = E − False Easting = 497 500 − 500 000 = −2 500 m The negative sign indicates the point lies 2 500 m WEST of the central meridian. ∴ The point is 2 500 m west of the zone's central meridian.

Question Type

numerical

Answer Structure

• Identify false easting = 500 000 m at the central meridian [stated]
• Apply formula: displacement = E − 500 000 [1 mark for method]
• Compute: 497 500 − 500 000 = −2 500 m [arithmetic]
• Interpret sign: negative = west of CM [1 mark for correct direction]

Scoring Breakdown

Common Mark Deductions

• Computing 500 000 − 497 500 = 2 500 without stating direction — loses interpretation mark
• Saying 'east of CM' instead of 'west' — direction error, loses 1 mark
• Not identifying 500 000 m as the false easting at the CM — conceptual error

Key Phrases To Include

• false easting 500 000 m
• central meridian
• displacement = E − 500 000
• west of central meridian
• negative value indicates west

Marks

2

Topic

PPCS Zone Identification

Difficulty

medium

Template Id

T6

Examiner Tip

Show the distance calculation for at least two candidate zones. This demonstrates the selection method and earns the method mark even if you state the answer confidently.

Model Answer

Given: λ = 120.5°E PPCS Central Meridians: Zone I = 117°E, Zone II = 119°E, Zone III = 121°E, Zone IV = 123°E, Zone V = 125°E Procedure: Assign the point to the PPCS zone whose central meridian is closest to the point's longitude. Distances from λ = 120.5°E: Zone II (CM = 119°E): |120.5 − 119| = 1.5° Zone III (CM = 121°E): |120.5 − 121| = 0.5° The nearest central meridian is 121°E (Zone III) — difference of only 0.5°. ∴ The point falls in PPCS Zone III (CM = 121°E).

Question Type

short_answer

Answer Structure

• List relevant PPCS central meridians (at least Zone II and III) [stated]
• Compute angular distance from the point's longitude to the two nearest CMs [1 mark]
• Identify Zone III (CM = 121°E) as the nearest and therefore correct zone [1 mark]

Scoring Breakdown

Common Mark Deductions

• Choosing Zone II (CM = 119°E) — failing to compute both distances and compare
• Stating the zone without showing the distance comparison — method mark lost
• Confusing PPCS zone numbering with UTM zone numbering

Key Phrases To Include

• nearest central meridian
• Zone III
• CM = 121°E
• 0.5° from central meridian

Marks

2

Topic

False Easting and Grid Coordinates

Difficulty

easy

Template Id

T7

Examiner Tip

State the false easting value before applying the formula. This one extra line demonstrates conceptual understanding and protects you from losing the method mark.

Model Answer

Given: Displacement east of CM = +4 500 m False Easting (FE) = 500 000 m (standard UTM) Formula: Grid Easting = False Easting + Displacement Solution: E = 500 000 + 4 500 E = 504 500 m ∴ Grid Easting = 504 500 m

Question Type

numerical

Answer Structure

• State false easting = 500 000 m [stated]
• Apply formula E = FE + displacement [1 mark]
• Compute: 500 000 + 4 500 = 504 500 m [1 mark]

Scoring Breakdown

Common Mark Deductions

• Answering E = 4 500 m — ignoring the false easting entirely
• Subtracting instead of adding — sign error for eastward displacement
• Omitting units (metres) from the final answer

Key Phrases To Include

• false easting 500 000 m
• E = 500 000 + 4 500
• 504 500 m
• grid easting

Marks

3

Topic

UTM Zone Determination

Difficulty

medium

Template Id

T8

Examiner Tip

Write the formula once at the top and reference it for each part. This saves time, looks organized, and earns the formula mark regardless of which part the examiner checks first.

Model Answer

Formula: UTM Zone = ⌊(λ + 180) / 6⌋ + 1 (a) λ = 117°E Zone = ⌊(117 + 180) / 6⌋ + 1 = ⌊297 / 6⌋ + 1 = ⌊49.5⌋ + 1 = 49 + 1 ∴ Zone = 50 (Zone 50N) (b) λ = 124°E Zone = ⌊(124 + 180) / 6⌋ + 1 = ⌊304 / 6⌋ + 1 = ⌊50.6667⌋ + 1 = 50 + 1 ∴ Zone = 51 (Zone 51N) Summary: Both 117°E and 124°E fall within UTM Zones 50N and 51N respectively — confirming that the Philippines occupies UTM Zones 50 and 51 North.

Question Type

numerical

Answer Structure

• Write the UTM zone formula once at the top [1 mark]
• Part (a): Correct computation and result — Zone 50N [1 mark]
• Part (b): Correct computation and result — Zone 51N [1 mark]

Scoring Breakdown

Common Mark Deductions

• Computing only one part and running out of time — always budget time for multi-part questions
• Forgetting to add +1 in either part
• Not specifying 'N' (North) hemisphere for Philippine locations
• Arithmetic error: 117+180 = 297, not 296 or 298

Key Phrases To Include

• ⌊(λ + 180) / 6⌋ + 1
• Zone 50N
• Zone 51N
• Philippines in Zones 50 and 51
• floor function

Marks

3

Topic

False Easting and False Origin

Difficulty

medium

Template Id

T9

Examiner Tip

Structure as a numbered list. Examiners matching marks to your answer can clearly award 1 mark per numbered point. Unstructured prose paragraphs often miss marks even when the content is correct.

Model Answer

A false easting (FE) of 500 000 m is added to all computed grid eastings for the following reasons: 1. Elimination of negative coordinates: Points located west of the zone's central meridian have negative natural eastings (measured from the CM = 0). Adding FE = 500 000 m shifts the origin 500 000 m west of the CM, ensuring that all eastings within the zone remain positive. 2. Single-valued coordinates: Without the false easting, symmetric points on opposite sides of the CM would have equal-magnitude but opposite-sign eastings, causing ambiguity. The false origin makes every easting within the zone unique and positive. 3. Practical field use: Positive-only coordinates prevent sign errors in field computations, traverse closure calculations, and data entry — critical for cadastral and engineering surveys referenced under PD 1529 and CA 141. The false easting is applied at the false origin, which is located 500 000 m west of the central meridian on the equator (or its projection). The CM thus carries an easting value of exactly 500 000 m.

Question Type

short_answer

Answer Structure

• State the primary reason: elimination of negative coordinates west of CM [1 mark]
• Explain the false origin concept: origin shifted 500 000 m west of CM [1 mark]
• Give a practical justification: unique, positive coordinates for field use or legal surveys [1 mark]

Scoring Breakdown

Common Mark Deductions

• Simply saying 'to make coordinates positive' without explaining WHY they would be negative — loses the conceptual mark
• Not mentioning the position of the false origin relative to the CM
• Failing to distinguish between false easting (easting shift) and false northing (northing shift)

Key Phrases To Include

• false easting 500 000 m
• negative coordinates
• west of central meridian
• false origin
• CM easting = 500 000 m
• all coordinates remain positive

Marks

3

Topic

UTM and PPCS Zone Identification

Difficulty

medium

Template Id

T10

Examiner Tip

When a point is between two PPCS zones, always compute the distance to BOTH candidate central meridians and compare. Never guess by inspection — the examiner expects a numerical comparison.

Model Answer

Given: λ = 123.9°E (Cebu City) (a) UTM Zone: Formula: Zone = ⌊(λ + 180) / 6⌋ + 1 Zone = ⌊(123.9 + 180) / 6⌋ + 1 = ⌊303.9 / 6⌋ + 1 = ⌊50.65⌋ + 1 = 50 + 1 ∴ UTM Zone = 51N (b) PPCS Zone: PPCS Central Meridians: Zone III = 121°E, Zone IV = 123°E, Zone V = 125°E Distances from λ = 123.9°E: Zone IV (CM = 123°E): |123.9 − 123| = 0.9° Zone V (CM = 125°E): |123.9 − 125| = 1.1° Nearest CM = 123°E → Zone IV ∴ PPCS Zone = IV (CM = 123°E) Note: This illustrates that UTM Zone 51N and PPCS Zone IV overlap in the Cebu area — surveyors must specify which coordinate system is used in all legal documents under PD 1529.

Question Type

numerical

Answer Structure

• Part (a): Apply UTM zone formula and show floor computation [1 mark]
• Part (a): State Zone 51N [stated under part a]
• Part (b): List nearest PPCS CMs and compute angular distances [1 mark]
• Part (b): Select Zone IV (CM = 123°E) as nearest [1 mark]

Scoring Breakdown

Common Mark Deductions

• Choosing PPCS Zone V instead of Zone IV — failing to compare distances
• Forgetting to show the floor operation for the UTM computation
• Not checking both candidate PPCS zones (IV and V)

Key Phrases To Include

• Zone 51N
• PPCS Zone IV
• CM = 123°E
• nearest central meridian
• PD 1529

Marks

3

Topic

Central Meridian Scale Factor

Difficulty

hard

Template Id

T11

Examiner Tip

Draw a simple cross-section sketch showing the cylinder intersecting the ellipsoid at two lines. Label 'k < 1 at CM', 'k = 1 at secant lines', 'k > 1 at edges'. This diagram alone can earn 1 mark in a diagram-based question.

Model Answer

Reason for k₀ < 1.0: In a Transverse Mercator projection, setting k₀ < 1.0 at the central meridian means the projection surface (cylinder) is secant — it intersects the ellipsoid along two lines parallel to and equidistant from the central meridian, rather than being tangent at the CM. Distortion-control advantage: 1. At the central meridian (CM): scale = k₀ < 1.0 (slight compression) 2. At the two secant lines: scale = 1.0 exactly (true scale, zero distortion) 3. At the zone edges: scale > 1.0 (slight elongation) This 'balanced' distortion ensures the maximum scale error within the zone is minimized. If k₀ = 1.0 (tangent), scale error would grow continuously from zero at the CM to a large value at the zone edges. For UTM (k₀ = 0.9996, 6° zones): maximum scale error ≈ 1 part in 2 500 For PPCS (k₀ = 0.99995, ≈2° zones): maximum scale error ≈ 1 part in 20 000 The PPCS achieves higher accuracy because its narrower zones require less scale correction.

Question Type

short_answer

Answer Structure

• Explain secant vs tangent projection: k₀ < 1 creates two secant lines of exact scale [1 mark]
• Describe the distortion pattern: compression at CM, exact scale at secant lines, elongation at edges [1 mark]
• Compare PPCS vs UTM: narrower PPCS zones yield smaller maximum distortion [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying k₀ < 1 'to reduce scale' without explaining the secant geometry — vague, loses the conceptual mark
• Not mentioning the two secant lines where scale = exactly 1.0
• Confusing scale factor with area distortion — they are different concepts

Key Phrases To Include

• secant cylinder
• two lines of exact scale
• k₀ < 1 at central meridian
• scale > 1 at zone edges
• balanced distortion
• maximum scale error minimized

Marks

5

Topic

PPCS vs UTM Comparison

Difficulty

hard

Template Id

T12

Examiner Tip

For 5-mark comparison questions, ALWAYS include a summary table. It organizes your answer, helps the examiner award marks quickly, and ensures you cover all four comparison parameters. A well-drawn table often earns the 'structure' mark even if one detail is wrong.

Model Answer

Comparison of UTM and PPCS Coordinate Systems Both UTM and PPCS are Transverse Mercator (TM) plane coordinate systems that project ellipsoidal positions onto a flat grid using a secant cylindrical projection. They differ in scale, datum, and design purpose as follows: (a) Number of Zones and Zone Width: UTM: 60 zones worldwide, each 6° of longitude wide. Zones are numbered 1–60 eastward from 180°. The Philippines occupies Zones 50N and 51N. PPCS: 5 zones nationwide, each approximately 2° wide (±1° from CM), with central meridians at 117°, 119°, 121°, 123°, and 125°E. PPCS zones are PURPOSE-DESIGNED for the Philippine Archipelago. (b) Central-Meridian Scale Factor (k₀): UTM: k₀ = 0.9996 — lower value because 6° zones require greater scale correction to limit edge distortion. PPCS: k₀ = 0.99995 — higher value (closer to 1) because 2° zones are narrower and require less scale correction. PPCS achieves approximately 8× better accuracy than UTM. (c) Datum: UTM: Based on WGS84 (World Geodetic System 1984) — the global GPS datum. For international and GPS-referenced surveys. PPCS: Based on PRS92 (Philippine Reference System of 1992), which uses the Clarke 1866 ellipsoid. PRS92 is the nationally mandated datum for all geodetic surveys in the Philippines under RA 8560. (d) False Easting: UTM: False Easting = 500 000 m, applied at the false origin (500 000 m west of CM on the equator). PPCS: False Easting = 500 000 m (same value). Both systems use this convention to keep all eastings positive. Note: False Northing = 0 for Northern Hemisphere (both systems); UTM uses FN = 10 000 000 m for the Southern Hemisphere. Summary Table: | Parameter | UTM | PPCS | |-----------------|-------------------|--------------------------| | Projection | Transverse Mercator | Transverse Mercator | | No. of zones | 60 (worldwide) | 5 (Philippines only) | | Zone width | 6° | ≈2° | | k₀ | 0.9996 | 0.99995 | | Datum | WGS84 | PRS92 (Clarke 1866) | | False Easting | 500 000 m | 500 000 m | | PH Zones | 50N, 51N | I, II, III, IV, V | For legal cadastral surveys in the Philippines, PPCS is the preferred system because it references PRS92 and achieves higher accuracy across the archipelago.

Question Type

long_answer

Answer Structure

• Introduction: State both are Transverse Mercator systems [½ mark]
• Part (a): UTM = 60 zones × 6°; PPCS = 5 zones × ≈2°; PH in UTM 50N–51N [1 mark]
• Part (b): UTM k₀ = 0.9996; PPCS k₀ = 0.99995; explain accuracy difference [1 mark]
• Part (c): UTM on WGS84; PPCS on PRS92/Clarke 1866; reference RA 8560 [1 mark]
• Part (d): Both use FE = 500 000 m; correct FN for Southern Hemisphere [½ mark]
• Summary table or organized presentation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Interchanging k₀ values between UTM and PPCS — most critical error
• Saying UTM uses Clarke 1866 or PPCS uses WGS84 — datum confusion
• Omitting the Philippine UTM zone numbers (50N, 51N)
• Failing to include a summary table for a 5-mark comparison question — misses the structure mark
• Not citing RA 8560 or PRS92 — misses the legal/datum mark

Key Phrases To Include

• Transverse Mercator
• k₀ = 0.9996 (UTM)
• k₀ = 0.99995 (PPCS)
• WGS84
• PRS92
• Clarke 1866
• RA 8560
• false easting 500 000 m
• Zone 50N and 51N
• PPCS Zones I–V

Marks

5

Topic

PPCS Coordinate Interpretation and Legal Context

Difficulty

hard

Template Id

T13

Examiner Tip

Case study questions with cadastral survey context almost always have a legal-reference mark. Mention PD 1529 (land registration), RA 8560 (geodetic engineering practice), or CA 141 (public lands) in your answer to capture this mark reliably.

Model Answer

Given: PPCS Zone: III (CM = 121°E, PRS92 datum) Northing N = 1 245 000 m Easting E = 503 750 m False Easting (FE) = 500 000 m False Northing (FN) = 0 m Step 1 — Physical meaning of Northing: N = 1 245 000 m means the point is 1 245 000 m north of the false origin (equator projection in PPCS Zone III). This places the point approximately at 11.2° North latitude, consistent with locations in the Visayas–Mindanao region. Step 2 — Displacement east of Central Meridian: Displacement_E = E − FE = 503 750 − 500 000 = +3 750 m Positive value → point is 3 750 m EAST of the Zone III central meridian (CM = 121°E). Step 3 — Displacement north of False Origin: Displacement_N = N − FN = 1 245 000 − 0 = 1 245 000 m north of the false origin Step 4 — Summary of position: • East-West: 3 750 m east of CM (121°E) → approximate longitude ≈ 121°E + 3 750/(111 320 × cos11.2°) ≈ 121.034°E • North-South: 1 245 000 m north of equatorial false origin → approximately 11.2°N • The point lies within PPCS Zone III jurisdiction, appropriate for surveys in Eastern Visayas or northern Mindanao. Step 5 — Legal and practical context: Under PD 1529 (Property Registration Decree) and RA 8560, all geodetic control points and cadastral surveys must reference the national geodetic datum PRS92. The PPCS Zone III system is the mandated coordinate framework for this area, and these coordinates would be acceptable for TCT (Transfer Certificate of Title) boundary descriptions.

Question Type

case_study

Answer Structure

• Identify PPCS Zone III parameters: CM = 121°E, FE = 500 000 m, FN = 0 [1 mark]
• Compute east displacement: E − 500 000 = +3 750 m → east of CM [1 mark]
• Interpret northing: N = 1 245 000 m north of false origin [1 mark]
• Approximate geographic position (latitude/longitude estimate) [1 mark]
• Legal context: PD 1529, RA 8560, PRS92 mandate [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using UTM parameters instead of PPCS parameters for Zone III
• Stating point is west of CM when E > 500 000 m — direction error
• Not citing any legal reference — misses the law mark in a cadastral context question
• Computing northing displacement as N − 10 000 000 — wrong hemisphere (S hemisphere formula applied to N hemisphere)

Key Phrases To Include

• PPCS Zone III
• CM = 121°E
• PRS92 datum
• false easting 500 000 m
• 3 750 m east of CM
• 1 245 000 m north of false origin
• PD 1529
• RA 8560

Marks

1

Topic

PPCS Datum and Legal Basis

Difficulty

easy

Template Id

T14

Examiner Tip

For law-based questions, always cite both the law number AND its short title. 'RA 8560' alone or 'Philippine Geodetic Engineering Act' alone — either earns partial credit, but both together ensures full credit.

Model Answer

The PPCS uses the Philippine Reference System of 1992 (PRS92), based on the Clarke 1866 ellipsoid. Its mandatory use for all geodetic surveys in the Philippines is prescribed by Republic Act 8560 (The Philippine Geodetic Engineering Act of 1998).

Question Type

very_short_answer

Answer Structure

• Single answer: Name PRS92/Clarke 1866 AND cite RA 8560 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Citing WGS84 instead of PRS92 — wrong datum
• Naming PD 1529 instead of RA 8560 — PD 1529 covers registration, not datum mandate
• Naming the ellipsoid as GRS80 or WGS84 ellipsoid — Clarke 1866 is correct for PRS92

Key Phrases To Include

• PRS92
• Clarke 1866
• RA 8560
• Philippine Reference System of 1992

Marks

1

Topic

PPCS Zone Coverage

Difficulty

easy

Template Id

T15

Examiner Tip

Memorize the PPCS zone-to-region mapping: Zone I (117°E) = Palawan; Zone II (119°E) = western Visayas/western Mindanao; Zone III (121°E) = Luzon/central Visayas; Zone IV (123°E) = eastern Visayas/Samar/Leyte; Zone V (125°E) = eastern Mindanao/Caraga.

Model Answer

PPCS Zone V has a central meridian of 125°E and primarily covers the eastern portions of Mindanao, including the regions of Caraga (Region XIII) and parts of the BARMM (Bangsamoro Autonomous Region in Muslim Mindanao).

Question Type

very_short_answer

Answer Structure

• State CM = 125°E AND identify correct Philippine region coverage [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating CM = 123°E (that is Zone IV, not Zone V)
• Placing Zone V in Palawan or western Philippines — geographic error

Key Phrases To Include

• 125°E
• Zone V
• eastern Mindanao
• Caraga