GELE Cadastral Surveys & Land Laws — Property/Boundary Surveys and Lot ComputationExam Answer Templates

Examiner tip

The single mark is for the key action word 'establishes or re-marks' paired with the legal reference. Students who write a vague definition like 'a survey of boundaries' earn zero.

Model answer

A boundary survey is a survey that establishes or re-marks the corners and lines of a parcel of land based on its title, plan, and legal description, resolving overlaps and gaps with adjoining lots.

Answer structure

• One complete sentence: state what a boundary survey does (establishes/re-marks corners and lines) + legal basis (title and plan) [1 mark]

Scoring breakdown

Common mark deductions

• Saying only 'measuring the area of land' — this describes area computation, not boundary survey.
• Omitting the legal basis (title/plan); boundary surveys must be referenced to existing records.
• Confusing boundary survey with topographic survey (topographic surveys map terrain, not parcel corners).

Key phrases to include

• establishes or re-marks corners
• title and plan
• adjoining lots
• legal description

Examiner tip

Examiners check specifically for the ½, the absolute value, and the cross-product structure Eᵢ Nᵢ₊₁ − Eᵢ₊₁ Nᵢ. All three must be present.

Model answer

A = ½ |Σ(Eᵢ Nᵢ₊₁ − Eᵢ₊₁ Nᵢ)|, where the index wraps so that E_{n+1} = E₁ and N_{n+1} = N₁.

Answer structure

• Write the formula with correct notation [½ mark]
• State the wrap-around condition for the last corner [½ mark]

Scoring breakdown

Common mark deductions

• Forgetting the ½ factor — gives double the correct area.
• Omitting the absolute value bars — sign of the sum depends on traversal direction.
• Writing Nᵢ Eᵢ₊₁ instead of Eᵢ Nᵢ₊₁ — reversed subscript order is a common slip.

Key phrases to include

• ½ |Σ(Eᵢ Nᵢ₊₁ − Eᵢ₊₁ Nᵢ)|
• indices wrap
• E_{n+1} = E₁

Examiner tip

The key differentiator examiners look for: boundary survey deals with legal disputes/establishment; relocation survey deals with physical recovery of existing corners. State this contrast explicitly.

Model answer

A boundary survey establishes or re-marks the corners and lines of a parcel using its title and plan, often settling overlaps or gaps with adjacent lots. A relocation survey, on the other hand, re-establishes corners that have been lost or obliterated by locating them from the recorded technical description and existing control ties, without altering the original boundary positions.

Answer structure

• Line 1: Define boundary survey — emphasis on 'establishes or re-marks' using title/plan [1 mark]
• Line 2: Define relocation survey — emphasis on 're-establishes lost corners' from recorded technical description and control ties [1 mark]

Scoring breakdown

Common mark deductions

• Treating them as synonyms — one mark is immediately lost.
• Not mentioning that relocation does NOT move the boundary, only re-marks the existing position.
• Omitting 'lost or obliterated' for the relocation survey definition.

Key phrases to include

• establishes or re-marks
• title and plan
• re-establishes lost or obliterated corners
• recorded technical description
• control ties
• without altering original boundary

Examiner tip

Always write d = √(ΔN² + ΔE²) explicitly. Some students write the answer only and lose the method mark. Examiners in PRC board exams reward the process.

Model answer

Given: P(E=20, N=15), Q(E=80, N=55)

Step 1 — Compute differences:
ΔE = 80 − 20 = +60 m (East)
ΔN = 55 − 15 = +40 m (North)

Step 2 — Distance:
d = √(ΔN² + ΔE²) = √(40² + 60²) = √(1600 + 3600) = √5200 = 72.11 m

Step 3 — Bearing angle:
θ = arctan(|ΔE| / |ΔN|) = arctan(60/40) = arctan(1.5) = 56°18'36"

Step 4 — Quadrant: ΔN > 0 (North), ΔE > 0 (East) → N–E quadrant

Bearing of PQ = N 56°19' E

Distance = 72.11 m

Answer structure

• Step 1: Compute ΔE and ΔN with sign analysis [½ mark]
• Step 2: Apply distance formula d = √(ΔN² + ΔE²) with correct substitution [½ mark]
• Step 3: Compute bearing angle using arctan(|ΔE|/|ΔN|) [½ mark]
• Step 4: State quadrant from signs of ΔN and ΔE; write bearing in N/S–degrees–E/W format [½ mark]

Scoring breakdown

Common mark deductions

• Using arctan(ΔN/ΔE) instead of arctan(|ΔE|/|ΔN|) — gives the complement angle and wrong bearing.
• Writing bearing as a decimal degree azimuth (e.g., 56.31°) instead of the PH technical-description format.
• Omitting the distance formula — loses the method mark even if the numerical answer is correct.

Key phrases to include

• ΔE = +60 m (East)
• ΔN = +40 m (North)
• d = √(ΔN² + ΔE²)
• arctan(|ΔE|/|ΔN|)
• N 56°19' E

Examiner tip

Connect the geometric concept (closed loop) to the mathematical consequence (zero net displacement). Examiners want cause-and-effect reasoning, not just the formula.

Model answer

A property boundary is a closed polygon: starting at any corner and traversing all sides in order, you return to the same starting point. Each side's latitude (ΔN) is the northing increment and each side's departure (ΔE) is the easting increment. Because the start and end points are identical, the net northing change and net easting change over the complete loop must both equal zero, giving ΣΔN = 0 and ΣΔE = 0. Any non-zero sum is the misclosure, which indicates measurement error.

Answer structure

• Line 1: State that a lot boundary is a closed loop returning to the starting point [1 mark]
• Line 2: Therefore net northing and net easting changes = 0; define misclosure as non-zero sum [1 mark]

Scoring breakdown

Common mark deductions

• Saying 'because the area is balanced' — vague and incorrect reasoning.
• Not defining misclosure as the non-zero sum.
• Explaining only one of the two conditions (ΣΔN or ΣΔE) and omitting the other.

Key phrases to include

• closed polygon
• returns to the starting point
• net northing change = 0
• net easting change = 0
• misclosure
• measurement error

Examiner tip

Write each cross-product term on a separate line. Examiners follow your algebra term by term. Cramming everything on one line makes it easy to miss an error and hard to award partial credit.

Model answer

Given corners (E, N): A(0,0), B(60,0), C(30,50) — taken in order.

Shoelace Formula: A = ½|Σ(Eᵢ Nᵢ₊₁ − Eᵢ₊₁ Nᵢ)|

Term-by-term computation:
Term 1 (A→B): E_A·N_B − E_B·N_A = (0)(0) − (60)(0) = 0
Term 2 (B→C): E_B·N_C − E_C·N_B = (60)(50) − (30)(0) = 3000 − 0 = 3000
Term 3 (C→A): E_C·N_A − E_A·N_C = (30)(0) − (0)(50) = 0 − 0 = 0

Σ = 0 + 3000 + 0 = 3000

A = ½|3000| = 1500 m²

Conversion: 1 ha = 10,000 m²
A = 1500 / 10,000 = 0.15 ha

Area = 1,500 m² = 0.15 ha

Answer structure

• State the shoelace formula [½ mark]
• Set up and evaluate all three cross-product terms with wrap-around [1 mark]
• Compute Σ and apply ½|Σ| to get area in m² [1 mark]
• Convert correctly to hectares using 1 ha = 10,000 m² [½ mark]

Scoring breakdown

Common mark deductions

• Missing the C→A wrap-around term — the sum is wrong and both later marks are at risk.
• Forgetting to multiply by ½ — gives 3000 m² instead of 1500 m².
• Incorrect hectare conversion (e.g., dividing by 1000 or 100,000).
• No formula written — loses the method mark even with correct answer.

Key phrases to include

• Shoelace Formula
• A = ½|Σ(Eᵢ Nᵢ₊₁ − Eᵢ₊₁ Nᵢ)|
• index wrap
• 1500 m²
• 1 ha = 10,000 m²
• 0.15 ha

Examiner tip

When the question says 'technical-description format,' examiners expect the full sentence 'Thence [bearing], [distance] to corner [label].' This phrasing mirrors DENR and LRA document language and signals professional knowledge.

Model answer

Given: C(E=60, N=40), D(E=10, N=30)

Step 1 — Increments:
ΔE = 10 − 60 = −50 m → West
ΔN = 30 − 40 = −10 m → South

Step 2 — Distance:
d = √(ΔN² + ΔE²) = √(10² + 50²) = √(100 + 2500) = √2600 = 50.99 m ≈ 51.0 m

Step 3 — Bearing angle:
θ = arctan(|ΔE| / |ΔN|) = arctan(50/10) = arctan(5) = 78°41'24"

Step 4 — Quadrant determination:
ΔN < 0 → South; ΔE < 0 → West → S–W quadrant

Bearing = S 78°41' W

Technical Description Format:
"Thence S 78°41' W, 51.0 m to corner D"

Answer structure

• Step 1: Compute ΔE and ΔN with direction labels (East/West, North/South) [½ mark]
• Step 2: Apply d = √(ΔN² + ΔE²) and compute correctly [1 mark]
• Step 3: Compute bearing angle = arctan(|ΔE|/|ΔN|) [½ mark]
• Step 4: Determine correct quadrant from signs; write in 'Thence S ___° W, ___ m to corner ___' format [1 mark]

Scoring breakdown

Common mark deductions

• Writing the bearing as S 11°19' W — using arctan(ΔN/ΔE) instead of arctan(|ΔE|/|ΔN|) swaps the angle.
• Writing bearing as a numeric azimuth (e.g., 258.7°) instead of the Philippine quadrant format.
• Not writing the full technical-description sentence — the question specifically asks for the TD format.

Key phrases to include

• ΔE = −50 m (West)
• ΔN = −10 m (South)
• d = √(ΔN² + ΔE²)
• arctan(|ΔE|/|ΔN|)
• S 78°41' W
• Thence S 78°41' W, 51.0 m to corner D

Examiner tip

Philippine board exam questions frequently test the precision ratios for different survey classes. Memorize: cadastral = 1:10,000; subdivision = 1:10,000; isolated = 1:5,000. These numbers appear directly in DENR survey regulations.

Model answer

The linear misclosure is the resultant closing error of a traverse, computed as:
closure = √(ΣΔN)² + (ΣΔE)²
where ΣΔN and ΣΔE are the sums of latitudes and departures of all traverse sides. It represents the distance between the computed closing position and the starting point.

Under DENR survey regulations for subdivision surveys in the Philippines, the standard tolerance is a precision ratio of 1:10,000 (linear misclosure ≤ 1 part in 10,000 of the total traverse length).

Answer structure

• Line 1-2: Define linear misclosure with formula: √(ΣΔN)² + (ΣΔE)² [1 mark]
• Line 3: State DENR tolerance for subdivision surveys: 1:10,000 [1 mark]

Scoring breakdown

Common mark deductions

• Stating tolerance as 1:5,000 (cadastral) or 1:3,000 (reconnaissance) — wrong survey class.
• Defining misclosure as just 'the error in area' instead of the position-closure distance.
• Omitting the formula or geometric meaning of misclosure.

Key phrases to include

• linear misclosure
• √(ΣΔN)² + (ΣΔE)²
• closing error
• 1:10,000
• DENR subdivision surveys

Examiner tip

Use a two-column table. Separate the Eᵢ Nᵢ₊₁ column from the Eᵢ₊₁ Nᵢ column. Sum each column independently before subtracting. This organized approach earns partial marks at every step and eliminates sign errors.

Model answer

Given corners (E, N) in order:
A(0, 0), B(80, 0), C(90, 50), D(20, 40)

Formula: A = ½|Σ(Eᵢ Nᵢ₊₁ − Eᵢ₊₁ Nᵢ)|

Shoelace Table:

Σ(Eᵢ Nᵢ₊₁) = 0 + 4000 + 3600 + 0 = 7600
Σ(Eᵢ₊₁ Nᵢ) = 0 + 0 + 1000 + 0 = 1000

Σ(Eᵢ Nᵢ₊₁ − Eᵢ₊₁ Nᵢ) = 7600 − 1000 = 6600

A = ½|6600| = 3300 m²

Conversion to hectares:
A = 3300 / 10,000 = 0.33 ha

Area = 3,300 m² = 0.33 ha

Answer structure

• State the shoelace formula [½ mark]
• Set up the four cross-product terms correctly, including the D→A wrap-around [1.5 marks]
• Compute ΣEᵢNᵢ₊₁ and ΣEᵢ₊₁Nᵢ correctly [1 mark]
• Apply ½|Σ| to get area in m² [1 mark]
• Convert to hectares correctly [1 mark]

Scoring breakdown

Common mark deductions

• Missing the D→A term (wrap-around) — the most common blunder; removes at least 1 mark.
• Swapping Eᵢ Nᵢ₊₁ and Eᵢ₊₁ Nᵢ columns — gives wrong sign for the sum.
• Not applying the ½ factor — doubles the area.
• Incorrect hectare conversion — using 1000 or 100 instead of 10,000 m² per ha.

Key phrases to include

• A = ½|Σ(Eᵢ Nᵢ₊₁ − Eᵢ₊₁ Nᵢ)|
• wrap-around D→A
• ΣEᵢNᵢ₊₁ = 7600
• ΣEᵢ₊₁Nᵢ = 1000
• 3300 m²
• 0.33 ha
• 1 ha = 10,000 m²

Examiner tip

For a 5-mark TD question, examiners apply a point per component: increments, distances, bearings, TD sentences, and completeness. Even if one bearing is wrong, you can still earn 4/5. Work each line independently.

Model answer

Given (E, N): A(0,0), B(50,0), C(60,40), D(10,30)

Line AB: A→B
ΔE=+50 (E), ΔN=0; d=√(0²+50²)=50.00 m
Bearing: Due East → N 90°00' E (or 'East')
Technical desc: Thence N 90°00' E, 50.00 m to corner B

Line BC: B→C
ΔE=+10 (E), ΔN=+40 (N)
d=√(40²+10²)=√1700=41.23 m
θ=arctan(10/40)=arctan(0.25)=14°02'
ΔN>0, ΔE>0 → N–E
Bearing: N 14°02' E
Technical desc: Thence N 14°02' E, 41.23 m to corner C

Line CD: C→D
ΔE=−50 (W), ΔN=−10 (S)
d=√(10²+50²)=√2600=50.99 m
θ=arctan(50/10)=arctan(5)=78°41'
ΔN<0, ΔE<0 → S–W
Bearing: S 78°41' W
Technical desc: Thence S 78°41' W, 50.99 m to corner D

Line DA: D→A
ΔE=−10 (W), ΔN=−30 (S)
d=√(30²+10²)=√1000=31.62 m
θ=arctan(10/30)=arctan(0.333)=18°26'
ΔN<0, ΔE<0 → S–W
Bearing: S 18°26' W
Technical desc: Thence S 18°26' W, 31.62 m to corner A (Point of Beginning)

Complete Technical Description:
Beginning at corner A, coordinates E=0, N=0;
Thence N 90°00' E, 50.00 m to corner B;
Thence N 14°02' E, 41.23 m to corner C;
Thence S 78°41' W, 50.99 m to corner D;
Thence S 18°26' W, 31.62 m to corner A, the Point of Beginning.
Containing 1,700 m² more or less.

Answer structure

• Compute ΔE and ΔN for each of the 4 lines [1 mark]
• Compute distance for each of the 4 lines [1 mark]
• Determine bearing with correct quadrant for each line [1 mark]
• Write bearing-distance in TD sentence format for all 4 lines [1 mark]
• Write complete TD starting with 'Beginning at...' and closing with 'Point of Beginning' and area [1 mark]

Scoring breakdown

Common mark deductions

• Writing bearings as azimuths (0°–360°) instead of quadrant bearings — loses bearing marks.
• Missing the closing clause 'Point of Beginning' — incomplete TD.
• Not including the area statement at the end of the TD.
• Reversed arctan argument (arctan(ΔN/ΔE)) for any one line — wrong angle and wrong bearing.

Key phrases to include

• Beginning at corner A
• Point of Beginning
• Thence [bearing], [distance] m to corner
• Containing ___ m² more or less
• sign analysis for quadrant
• arctan(|ΔE|/|ΔN|)

Examiner tip

Board exam questions often test your ability to distinguish Bowditch (proportional to length) from Transit Rule (proportional to latitude/departure). State the distinguishing factor immediately in your first sentence.

Model answer

The Bowditch (Compass Rule) method distributes the traverse misclosure proportionally to each side's length relative to the total perimeter. For each course:

Correction to ΔN = −(ΣΔN) × (length of course / total perimeter)
Correction to ΔE = −(ΣΔE) × (length of course / total perimeter)

The correction is subtracted from each course's latitude and departure. Longer sides receive larger corrections; shorter sides receive smaller corrections.

The Bowditch rule is appropriate when the precision of angular and linear measurements is roughly equal, which is the condition assumed for typical property surveys using theodolites and tapes or total stations. Under DENR regulations, it is the standard adjustment method for cadastral and subdivision traverses where misclosure is within the allowable tolerance (e.g., 1:10,000).

Answer structure

• State the rule: corrections proportional to side length / total perimeter [1 mark]
• Write the correction formulas for ΔN and ΔE [1 mark]
• State when it is appropriate: equal angular/linear precision; within tolerance; DENR cadastral/subdivision surveys [1 mark]

Scoring breakdown

Common mark deductions

• Describing Transit Rule instead (corrections proportional to the magnitude of the latitude/departure, not the course length).
• Not writing the correction formula explicitly.
• Omitting the condition 'within tolerance' — Bowditch cannot be applied if misclosure exceeds the limit; the traverse must be re-measured.

Key phrases to include

• proportional to course length
• total perimeter
• correction to ΔN = −ΣΔN × (L/P)
• equal angular and linear precision
• DENR cadastral and subdivision surveys
• within allowable tolerance

Examiner tip

Philippine law citations earn marks only when the law number AND its subject are both written. 'RA 8560' alone may earn partial credit; 'RA 8560, the Geodetic Engineering Act of 1998' earns full credit.

Model answer

Republic Act No. 8560 (Geodetic Engineering Act of 1998) authorizes only duly licensed Geodetic Engineers registered with the Professional Regulation Commission to perform cadastral surveys, subdivision surveys, and other land surveys in the Philippines.

Consequences of non-compliance:

1. The unlicensed practitioner commits a criminal offense under RA 8560 and may be fined or imprisoned.
2. Any survey plan or technical description prepared by an unlicensed person has no legal validity and will not be accepted by DENR-LMB, the Register of Deeds, or courts for title registration under PD 1529.

Answer structure

• Line 1: Cite RA 8560 (Geodetic Engineering Act of 1998) and state it requires licensed GEs for land surveys [1 mark]
• Lines 2–3: State two distinct consequences: criminal liability under RA 8560 AND invalidity of survey product under PD 1529 [1 mark]

Scoring breakdown

Common mark deductions

• Citing RA 4374 alone (older GE law, superseded by RA 8560) without noting RA 8560.
• Listing consequences that are the same idea rephrased (e.g., 'fined' and 'penalized') — only one distinct consequence.
• Not citing any law — loses the first mark entirely.

Key phrases to include

• Republic Act No. 8560
• Geodetic Engineering Act of 1998
• licensed Geodetic Engineers
• PRC
• criminal offense
• no legal validity
• PD 1529
• Register of Deeds

Examiner tip

After computing the precision ratio, always state explicitly whether it meets the applicable DENR tolerance and what action follows. This decision-making step is what separates a complete professional answer from a student calculation.

Model answer

Given:
ΣΔN = +0.05 m, ΣΔE = −0.04 m
Perimeter = 2(100 + 60) = 320 m

Step 1 — Linear Misclosure:
LM = √(ΣΔN)² + (ΣΔE)² = √(0.05)² + (0.04)²
= √(0.0025 + 0.0016) = √0.0041 = 0.064 m

Step 2 — Precision Ratio:
PR = LM / Perimeter = 0.064 / 320 = 1/5000

Precision Ratio = 1 : 5,000

Step 3 — Tolerance check:
For DENR subdivision surveys, tolerance = 1:10,000.
Since 1:5,000 < 1:10,000 (larger error), the traverse DOES NOT meet the tolerance. It must be re-measured or re-computed before adjustment.

Answer structure

• Compute perimeter = 320 m [½ mark]
• Apply LM = √(ΣΔN² + ΣΔE²) correctly → 0.064 m [1 mark]
• Compute PR = LM/perimeter = 1:5000 [1 mark]
• State whether it meets DENR tolerance and conclusion [½ mark]

Scoring breakdown

Common mark deductions

• Using LM = |ΣΔN| + |ΣΔE| (Manhattan distance) instead of Pythagorean formula — wrong value.
• Expressing precision ratio as 0.0002 (decimal) instead of 1:5000 (ratio format).
• Not making a tolerance decision — the question implies it; leaving out the conclusion loses ½ mark.

Key phrases to include

• LM = √(ΣΔN)² + (ΣΔE)²
• 0.064 m
• perimeter = 320 m
• 1:5,000
• 1:10,000 tolerance
• does not meet tolerance
• must be re-measured

Examiner tip

Connect the law to the survey product. PD 1529 questions are not purely legal — they always connect to the GE's role in producing the survey plan that makes title registration possible.

Model answer

Presidential Decree No. 1529 (Property Registration Decree of 1978) governs the registration of land titles in the Philippines under the Torrens system. It requires that any application for original registration or subsequent subdivision/consolidation of titled land must be supported by a duly approved survey plan bearing the seal and signature of a licensed Geodetic Engineer. The survey plan — showing the parcel's boundaries, area, corner coordinates, and technical description — is the primary document that defines what property is registered. The Register of Deeds will not accept any title application without a DENR-LMB approved plan, ensuring that the physical parcel corresponds exactly to the title's legal description.

Answer structure

• Line 1: Identify PD 1529 and its subject (Torrens-system title registration, 1978) [½ mark]
• Line 2: State the survey plan requirement — must be DENR-LMB approved and signed by licensed GE [1 mark]
• Line 3: Explain the link — survey plan defines the property registered; Register of Deeds requires it [½ mark]

Scoring breakdown

Common mark deductions

• Confusing PD 1529 with CA 141 (Public Land Act) — different laws with different scopes.
• Not mentioning the survey plan requirement specifically.
• Saying 'any licensed surveyor' instead of 'licensed Geodetic Engineer' — the Philippine specific requirement.

Key phrases to include

• PD 1529
• Property Registration Decree
• Torrens system
• DENR-LMB approved survey plan
• licensed Geodetic Engineer
• Register of Deeds
• technical description
• original registration

Examiner tip

The sign of the Bowditch correction is OPPOSITE to the sign of ΣΔN/ΣΔE. If ΣΔN = +0.008 (too far north), then correction = −0.002 (pull south). Always check: after applying corrections, the new ΣΔN must equal zero.

Model answer

Given:
ΣΔN = +0.008 m, ΣΔE = −0.006 m
Perimeter P = 250 m, Length of AB = 60 m

Part (a) — Misclosure and Precision Ratio:
LM = √(ΣΔN)² + (ΣΔE)²
= √(0.008)² + (0.006)²
= √(0.000064 + 0.000036)
= √0.0001 = 0.01 m

Precision Ratio = LM / P = 0.01 / 250 = 1/25,000 = 1:25,000

Tolerance check: 1:25,000 > 1:10,000 → PASSES DENR subdivision tolerance.

Part (b) — Bowditch Applicability and Correction for AB:
Bowditch rule is applicable because:
(1) The precision ratio (1:25,000) is within the allowable tolerance of 1:10,000;
(2) Angular and linear precision are assumed approximately equal for a total-station survey.

Bowditch Correction Formulas:
corr ΔN_AB = −ΣΔN × (L_AB / P) = −(+0.008) × (60/250) = −0.008 × 0.24 = −0.00192 m ≈ −0.002 m
corr ΔE_AB = −ΣΔE × (L_AB / P) = −(−0.006) × (60/250) = +0.006 × 0.24 = +0.00144 m ≈ +0.001 m

These corrections are added to the measured ΔN and ΔE of side AB to obtain adjusted values.

Summary:
LM = 0.01 m; PR = 1:25,000; Bowditch applicable.
Correction to AB: −0.002 m in ΔN, +0.001 m in ΔE.

Answer structure

• Part (a): LM = √(ΣΔN² + ΣΔE²) = 0.01 m [1 mark]
• Part (a): PR = 1:25,000 and tolerance check (passes 1:10,000) [1 mark]
• Part (b): State two conditions for Bowditch applicability [1 mark]
• Part (b): Write Bowditch correction formulas [½ mark]
• Part (b): Compute corrections for AB: −0.002 m in ΔN, +0.001 m in ΔE [1.5 marks]

Scoring breakdown

Common mark deductions

• Applying corrections with the wrong sign — Bowditch corrections OPPOSE the misclosure; forgetting the negative sign is a common error.
• Computing PR as LM alone (0.01 m) without dividing by perimeter.
• Stating Bowditch is applicable without giving any condition — loses the condition mark.
• Using L = 250 m (full perimeter) instead of L = 60 m (AB only) in the correction formula.

Key phrases to include

• LM = √(ΣΔN² + ΣΔE²) = 0.01 m
• 1:25,000
• passes 1:10,000 DENR tolerance
• Bowditch applicable
• corr ΔN = −ΣΔN × (L/P)
• corr ΔE = −ΣΔE × (L/P)
• −0.002 m in ΔN
• +0.001 m in ΔE

Do

• Use exact technical vocabulary: 'establishes or re-marks corners', 'shoelace formula', 'misclosure'
• Write the formula with all symbols correctly placed (½, |·|, subscripts)
• If a definition is asked, include both the action and the reference (what the survey does AND what document it is based on)
• Answer in bold or underline key terms if writing on paper to draw examiner's eye

Don't

• Do not write a paragraph — 1 mark = 1 sentence or 1 formula
• Do not use colloquial or informal language
• Do not give an example unless the question asks for one
• Do not confuse similar terms (e.g., boundary vs relocation survey)

Do

• Label your two parts clearly: '(1) ...' and '(2) ...' or 'First ..., Second ...'
• For numerical: write the formula as Line 1, compute as Line 2
• For law citations: write the law number AND what it says — both are needed
• Check that both marks are addressed before moving on

Don't

• Do not write only one idea hoping partial marks cover both — they won't
• Do not write a long paragraph that hides the two key points
• Do not skip the formula step in numerical problems
• Do not omit units from numerical answers

Do

• For area problems: set up a 3-column table (i, Eᵢ Nᵢ₊₁, Eᵢ₊₁ Nᵢ) and show all rows
• For bearing problems: show ΔE/ΔN, distance, angle, and quadrant determination as separate lines
• For conceptual questions: state definition, formula, and application condition — one per line
• Reference Philippine regulations (DENR tolerance) when relevant — earns the application mark

Don't

• Do not jump straight to the final answer without showing computation steps
• Do not neglect the wrap-around term in shoelace calculations
• Do not write bearings in azimuth format for Philippine TD questions
• Do not omit the tolerance/applicability discussion in adjustment questions

Do

• Write 'Given:' and 'Required:' at the start to define the problem scope
• Use a table for shoelace or TD computations — one row per side
• Show every step with its intermediate result; do not skip arithmetic
• State the final answer clearly: box it or write 'ANSWER:' to highlight it
• For TD questions: write the full paragraph starting with 'Beginning at corner...' and ending with 'Point of Beginning' and area
• Check your answer: does ΣΔN ≈ 0 after adjustment? Does area make geometric sense?

Don't

• Do not attempt a 5-mark problem in 5 minutes — allocate proper time
• Do not omit the 'Given' and 'Required' framing — professional presentation earns marks
• Do not leave the area in m² only if the question asks for hectares too
• Do not forget the 'more or less' clause and area statement at the end of a TD
• Do not write bearings as decimal degrees in a TD question