GELE Surveying (Geomatics) — LevelingExam Answer Templates

Marks

1

Topic

Differential Leveling — HI Method

Difficulty

easy

Template Id

T1

Examiner Tip

A 1-mark definition question is answered in one sharp sentence that contains the formula. Board examiners scan for the phrase 'line of sight' and the equation HI = elev + BS.

Model Answer

The Height of Instrument (HI) is the elevation of the line of sight of the level above the adopted datum, obtained by adding the backsight (BS) rod reading to the known elevation of the occupied station: HI = elev + BS.

Question Type

very_short_answer

Answer Structure

• Line 1: State the definition using 'elevation of the line of sight' and include the formula HI = elev + BS [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing 'height of the instrument above the ground' — the HI is above the datum, not the ground surface.
• Omitting the formula and giving only a vague narrative description.
• Confusing HI with the height of the instrument tripod.

Key Phrases To Include

• elevation of the line of sight
• datum
• HI = elev + BS
• backsight

Marks

1

Topic

Curvature and Refraction

Difficulty

easy

Template Id

T2

Examiner Tip

Examiners test whether you know the constant (0.0675) AND the sign. Write both explicitly even in a 1-mark question.

Model Answer

h_cr = 0.0675 K² (metres), where K is the sight distance in kilometres. The correction is SUBTRACTED from the observed rod reading because the Earth's curvature makes the distant rod appear to read higher than its true value.

Question Type

very_short_answer

Answer Structure

• Line 1: State the formula with correct constant and units [0.5 mark]
• Line 2: State the sign convention — subtracted from rod reading — with brief reason [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using K in metres instead of kilometres in the formula.
• Stating the correction is added — reversing the sign convention.
• Omitting the unit statement for h_cr (metres).

Key Phrases To Include

• h_cr = 0.0675K²
• K in kilometres
• subtracted from the rod reading
• curvature dominates

Marks

2

Topic

Profile Leveling — Intermediate Foresight vs. Turning Point

Difficulty

medium

Template Id

T3

Examiner Tip

The key discriminator is the new HI — say it explicitly for TP and explicitly deny it for IF. That single phrase earns the mark for each.

Model Answer

A turning point (TP) is a temporary, stable point where BOTH a foresight (FS) and a subsequent backsight (BS) are taken, thereby transferring the level to a new instrument position and establishing a new Height of Instrument. An intermediate foresight (IF) is a rod reading taken on a station along the profile whose elevation is computed from the current HI only; no backsight is taken on it, so it does NOT establish a new HI. Errors in an IF affect only that station's elevation, whereas errors at a TP propagate through all subsequent elevations.

Question Type

short_answer

Answer Structure

• Sentence 1: Define TP — both FS and BS taken, transfers instrument, establishes new HI [1 mark]
• Sentence 2: Define IF — rod reading only, no new HI, error is localised [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating that an IF establishes a new HI — this is the most critical conceptual error.
• Omitting the fact that a TP receives BOTH a FS and a subsequent BS.
• Confusing IF with a check shot on a benchmark.

Key Phrases To Include

• turning point
• backsight and foresight
• new HI
• intermediate foresight
• does not establish a new HI
• error propagation

Marks

2

Topic

Differential Leveling — HI Method

Difficulty

easy

Template Id

T4

Examiner Tip

Write both formulas explicitly before substituting. Even if you get the arithmetic wrong, you earn the formula mark — that is often the difference between passing and failing a numerical question.

Model Answer

Given: Elevation of BM-A = 100.000 m BS on BM-A = 1.525 m FS on TP-1 = 2.310 m Step 1 — Height of Instrument: HI = elev(BM-A) + BS HI = 100.000 + 1.525 HI = 101.525 m Step 2 — Elevation of TP-1: elev(TP-1) = HI − FS elev(TP-1) = 101.525 − 2.310 elev(TP-1) = 99.215 m

Question Type

numerical

Answer Structure

• Step 1: Write formula HI = elev + BS and substitute — answer 101.525 m [1 mark]
• Step 2: Write formula elev = HI − FS and substitute — answer 99.215 m [1 mark]

Scoring Breakdown

Common Mark Deductions

• Subtracting BS instead of adding it to get HI.
• Adding FS to HI instead of subtracting it.
• Missing units (m) on the final answers.
• Rounding to fewer than 3 decimal places when the problem data has 3 decimal places.

Key Phrases To Include

• HI = elev + BS
• elev = HI − FS
• 101.525 m
• 99.215 m

Marks

2

Topic

Arithmetic Check and Loop Closure

Difficulty

medium

Template Id

T5

Examiner Tip

For a closed loop, always state 'If the loop closed perfectly, Δelev = 0.' Then give the actual value and call it the misclosure. This two-step reasoning earns both marks.

Model Answer

Given: elev(BM-1) = 50.000 m (starting elevation) ΣBS = 8.540 m, ΣFS = 6.210 m (a) Net elevation change and computed final elevation: Δelev = ΣBS − ΣFS = 8.540 − 6.210 = +2.330 m elev_final = 50.000 + 2.330 = 52.330 m (b) Arithmetic check (loop returns to BM-1, true Δelev = 0): Misclosure = elev_final − elev_initial = 52.330 − 50.000 = +2.330 m Since misclosure ≠ 0, the loop does NOT close — there is a 2.330-m error in the run.

Question Type

numerical

Answer Structure

• Part (a): Apply Δelev = ΣBS − ΣFS and add to starting elevation [1 mark]
• Part (b): Compare computed final elevation with the true elevation of BM-1; state misclosure with correct sign [1 mark]

Scoring Breakdown

Common Mark Deductions

• Computing ΣFS − ΣBS (wrong sign) and getting a negative Δelev.
• Failing to state whether the loop closes — just computing a number without interpreting it.
• Forgetting that a closed loop must have Δelev = 0 for perfect closure.

Key Phrases To Include

• ΣBS − ΣFS
• arithmetic check
• misclosure
• Δelev
• loop closure

Marks

3

Topic

Curvature and Refraction

Difficulty

medium

Template Id

T6

Examiner Tip

Board examiners always look for the unit declaration for K. Write 'K = 3.5 km' boldly before substituting — it shows you know the formula's domain and prevents the most common error.

Model Answer

Given: K = 3.5 km Formula: h_cr = 0.0675 K² (h_cr in metres, K in kilometres) Substitution: h_cr = 0.0675 × (3.5)² h_cr = 0.0675 × 12.25 h_cr = 0.827 m (rounded to 3 decimal places) Sign convention: The correction is SUBTRACTED from the observed rod reading. Reason: The Earth's curvature causes the far rod to appear higher than its true position (the horizontal line of sight rises above the curved Earth surface). Atmospheric refraction partially offsets this effect, but curvature dominates. Therefore, the net apparent rod reading is too large by h_cr, and the correction is subtracted to obtain the true reading.

Question Type

numerical

Answer Structure

• Step 1: State the formula h_cr = 0.0675K² with units [1 mark]
• Step 2: Substitute K = 3.5 km and compute h_cr = 0.827 m [1 mark]
• Step 3: State 'subtracted' and explain why — curvature makes rod read too high, refraction partially offsets [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using K = 3500 m (metres) instead of K = 3.5 km — gives a wildly incorrect answer.
• Stating the correction is added instead of subtracted.
• Giving the sign convention without any physical justification — examiners require the 'why'.
• Rounding to only 1 decimal place when input data precision warrants 3.

Key Phrases To Include

• h_cr = 0.0675K²
• K in kilometres
• 0.827 m
• subtracted
• curvature
• atmospheric refraction
• rod reads too high

Marks

3

Topic

Profile and Cross-Section Leveling

Difficulty

medium

Template Id

T7

Examiner Tip

Use the words 'longitudinal' for profile and 'transverse/perpendicular' for cross-section — these contrasting terms immediately signal conceptual clarity to the examiner.

Model Answer

Profile Leveling Profile leveling determines ground elevations at regular intervals (stations) along a route's centerline. The results are plotted as a longitudinal section (profile) showing the existing ground line. Engineers use this profile to design the vertical alignment — computing cut and fill depths at each station, setting grade elevations, and ensuring proper drainage slopes. Cross-Section Leveling Cross-section leveling measures ground elevations at points perpendicular to the centerline at each station, typically at fixed offsets (e.g., 5 m, 10 m left and right). The resulting cross-sectional shapes are used to compute earthwork volumes (cut or fill) by the end-area method or prismoidal formula, which directly determines the cost of grading operations. Design Contribution Together, the profile provides the vertical control for alignment design, while cross-sections provide the data for earthwork quantity estimates — both essential inputs to road, railway, and canal design.

Question Type

short_answer

Answer Structure

• Paragraph 1: Define profile leveling, state it is along the centerline, and link to vertical alignment / longitudinal section design [1 mark]
• Paragraph 2: Define cross-section leveling, state it is perpendicular to centerline at offsets, and link to earthwork volume computation [1 mark]
• Paragraph 3: Synthesise both — how they complement each other in route engineering design [1 mark]

Scoring Breakdown

Common Mark Deductions

• Describing profile and cross-section leveling as identical — missing the perpendicular vs. longitudinal distinction.
• Failing to connect each method to its engineering application (design or earthwork).
• Treating profile leveling as a plan-view survey rather than an elevation survey along the centerline.

Key Phrases To Include

• centerline
• stations
• longitudinal section
• vertical alignment
• perpendicular offsets
• earthwork volumes
• cut and fill
• end-area method

Marks

5

Topic

Differential Leveling — Multi-Setup Run with Arithmetic Check

Difficulty

hard

Template Id

T8

Examiner Tip

Draw a vertical rod-and-level sketch or at minimum a ruled table before calculating. Board examiners for this type of question explicitly allocate 1 mark for a correct, complete table. A messy or unorganised solution loses that mark regardless of numerical accuracy.

Model Answer

Given starting elevation: elev(BM-1) = 215.400 m ───────────────────────────────────────────────────────── STATION | BS (m) | HI (m) | FS (m) | ELEVATION (m) ───────────────────────────────────────────────────────── BM-1 | 1.345 | 216.745 | — | 215.400 TP-1 | 2.120 | 218.110 | 0.875 | 215.870 TP-2 | 1.560 | 217.685 | 1.985 | 216.125 ← wait — recalculate BM-2 | — | — | 0.660 | 217.025 ───────────────────────────────────────────────────────── Step-by-step calculations: Setup 1 (instrument over BM-1): HI₁ = 215.400 + 1.345 = 216.745 m elev(TP-1) = HI₁ − FS(TP-1) = 216.745 − 0.875 = 215.870 m Setup 2 (instrument over TP-1): HI₂ = 215.870 + 2.120 = 217.990 m elev(TP-2) = HI₂ − FS(TP-2) = 217.990 − 1.985 = 216.005 m Setup 3 (instrument over TP-2): HI₃ = 216.005 + 1.560 = 217.565 m elev(BM-2) = HI₃ − FS(BM-2) = 217.565 − 0.660 = 216.905 m (Corrected table:) ───────────────────────────────────────────────────────── STATION | BS (m) | HI (m) | FS (m) | ELEVATION (m) ───────────────────────────────────────────────────────── BM-1 | 1.345 | 216.745 | — | 215.400 TP-1 | 2.120 | 217.990 | 0.875 | 215.870 TP-2 | 1.560 | 217.565 | 1.985 | 216.005 BM-2 | — | — | 0.660 | 216.905 ───────────────────────────────────────────────────────── (c) Arithmetic check: ΣBS = 1.345 + 2.120 + 1.560 = 5.025 m ΣFS = 0.875 + 1.985 + 0.660 = 3.520 m Δelev = ΣBS − ΣFS = 5.025 − 3.520 = +1.505 m Independent check: elev(BM-2) − elev(BM-1) = 216.905 − 215.400 = +1.505 m ✓ The arithmetic check closes — the computations are consistent.

Question Type

numerical

Answer Structure

• Setup layout: Neatly tabulated BS, HI, FS, Elevation columns [1 mark]
• HI₁ = 216.745 m; elev(TP-1) = 215.870 m [1 mark]
• HI₂ = 217.990 m; elev(TP-2) = 216.005 m [1 mark]
• HI₃ = 217.565 m; elev(BM-2) = 216.905 m [1 mark]
• Arithmetic check: ΣBS − ΣFS = +1.505 m = Δelev ✓ [1 mark]

Scoring Breakdown

Common Mark Deductions

• Reusing HI₁ for all subsequent setups instead of recomputing HI after each TP.
• Adding FS instead of subtracting it when computing an elevation.
• Omitting the arithmetic check entirely — this costs a guaranteed mark.
• Carrying 3 or more errors silently with only a final answer — examiners cannot award method marks without shown work.
• Confusing which BS belongs to which HI (a TP's BS is used for the NEXT setup, not the current one).

Key Phrases To Include

• HI = elev + BS
• elev = HI − FS
• ΣBS − ΣFS = Δelev
• arithmetic check
• turning point
• 216.905 m
• 1.505 m

Marks

1

Topic

Arithmetic Check

Difficulty

easy

Template Id

T9

Examiner Tip

Memorise the equation in the form shown — 'BS sums first, FS sums second, then elevation difference.' That order matches the mnemonic 'BS adds, FS subtracts.'

Model Answer

ΣBS − ΣFS = elev_last − elev_first ΣBS = sum of all backsight readings; ΣFS = sum of all foresight readings; elev_last and elev_first are the elevations of the final and initial benchmark, respectively. The equation is satisfied when all HI and elevation computations are arithmetically consistent.

Question Type

very_short_answer

Answer Structure

• Write the equation ΣBS − ΣFS = Δelev [0.5 mark]
• Define each term correctly [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Writing ΣFS − ΣBS (reversed sign).
• Stating the check as ΣBS = ΣFS — valid only for a closed loop returning to the same benchmark, not a general open-ended run.

Key Phrases To Include

• ΣBS − ΣFS
• elev_last − elev_first
• arithmetic check
• backsight
• foresight

Marks

2

Topic

Curvature and Refraction

Difficulty

medium

Template Id

T10

Examiner Tip

Always write 'K = 4.0 km' in your solution header — it signals unit awareness. Then the computation is mechanical and full marks follow.

Model Answer

Given: Observed rod reading = 3.750 m Distance K = 4.0 km Step 1 — Curvature-and-refraction correction: h_cr = 0.0675 K² h_cr = 0.0675 × (4.0)² h_cr = 0.0675 × 16 = 1.080 m Step 2 — Apply correction (subtract, since curvature makes rod appear too high): Corrected rod reading = observed reading − h_cr Corrected rod reading = 3.750 − 1.080 = 2.670 m

Question Type

numerical

Answer Structure

• Step 1: Compute h_cr = 0.0675(4.0)² = 1.080 m [1 mark]
• Step 2: Subtract correction — corrected reading = 2.670 m [1 mark]

Scoring Breakdown

Common Mark Deductions

• Adding 1.080 m instead of subtracting.
• Using K = 4000 m — gives h_cr = 1,080,000 m, an obviously absurd answer not flagged by the student.
• Computing the correction but failing to apply it to the rod reading (stopping after Step 1).

Key Phrases To Include

• h_cr = 0.0675K²
• 1.080 m
• subtract
• corrected rod reading = 2.670 m

Marks

3

Topic

Profile Leveling — Intermediate Foresight

Difficulty

medium

Template Id

T11

Examiner Tip

Part 2 of this question is worth 2 marks — the 'No' answer earns only 1. The second mark comes from the reason. Always write: 'Because no backsight is taken on the IF, the HI is NOT changed.'

Model Answer

Given: HI = 215.400 m IF on Sta. 1+060 = 1.850 m Step 1 — Ground elevation at Sta. 1+060: elev = HI − IF elev = 215.400 − 1.850 elev = 213.550 m Step 2 — Does this IF establish a new HI? No. An intermediate foresight is a rod reading taken on a profile station under the CURRENT instrument setup. Because no backsight is subsequently taken on Station 1+060, the level is not moved, and no new HI is computed. The HI remains 215.400 m for all subsequent IF and FS readings in this setup. Rule: Only turning points (TP) — where both a foresight AND a subsequent backsight are taken — establish a new Height of Instrument.

Question Type

numerical

Answer Structure

• Step 1: Apply elev = HI − IF; compute 213.550 m [1 mark]
• Step 2: State 'No' and explain — no BS taken, HI unchanged, only TPs establish new HI [2 marks]

Scoring Breakdown

Common Mark Deductions

• Adding IF to HI instead of subtracting — a direct formula error.
• Saying the IF does establish a new HI — fundamental conceptual error.
• Failing to state the general rule about turning points in Part 2 — loses the third mark.

Key Phrases To Include

• elev = HI − IF
• 213.550 m
• does not establish a new HI
• no backsight taken
• turning point
• HI remains unchanged

Marks

3

Topic

Collimation Error — Two-Peg Test

Difficulty

hard

Template Id

T12

Examiner Tip

Board examiners for the two-peg test question want to see: (1) midpoint = true difference, (2) near-peg = apparent difference, (3) the correction formula. Write all three components clearly, one per paragraph.

Model Answer

Purpose of the Two-Peg Test The two-peg test determines whether the line of collimation of a level is truly horizontal when the bubble is centred — i.e., it checks for collimation (line-of-sight) error. Procedure 1. Set up the level at the midpoint M between two pegs A and B, placed approximately 60–80 m apart. Read rods at A (r₁) and B (r₂). The difference in rod readings (r₁ − r₂) gives the TRUE difference in elevation Δh, because equal sight lengths cancel the collimation error: Δh_true = r₁ − r₂ (from midpoint setup) 2. Move the level close to peg A (within 1–3 m). Read the rod at A again (r₃, short sight — nearly free of error) and at B again (r₄, long sight — full collimation error ε present). 3. The APPARENT elevation difference from the near-A setup: Δh_apparent = r₃ − r₄ Computing the Collimation Error Collimation error per unit length: e = (Δh_apparent − Δh_true) / D_AB where D_AB is the full peg-to-peg distance. If e ≠ 0, the level needs adjustment. The correct rod reading at B (from the near-A position) should be: r₄_correct = r₃ − Δh_true

Question Type

short_answer

Answer Structure

• Para 1: State purpose — check if collimation is horizontal [0.5 mark]
• Para 2 Step 1: Level at midpoint — reads cancel collimation error, gives true Δh [1 mark]
• Para 2 Step 2: Level near one peg — long sight carries full error [0.5 mark]
• Para 3: Formula for collimation error e = (Δh_apparent − Δh_true) / D_AB [1 mark]

Scoring Breakdown

Common Mark Deductions

• Describing only one setup position — the two-peg test requires TWO specific instrument positions.
• Failing to explain WHY equal sight lengths eliminate the error (the mathematical reasoning earns marks).
• Not providing the correction formula — the 'compute' part of the question is left unanswered.

Key Phrases To Include

• collimation error
• midpoint setup
• equal sight lengths cancel error
• true elevation difference
• near-peg setup
• long sight
• r₄_correct = r₃ − Δh_true

Marks

5

Topic

Differential Leveling — Complete Run with Accuracy Assessment

Difficulty

hard

Template Id

T13

Examiner Tip

Five-mark questions in Surveying almost always have exactly five scorable elements. Identify them from the question parts and ensure each has a clearly labelled, complete response. Never merge parts — write (a), (b), (c) as separate sections.

Model Answer

Given: elev(BM-A) = 75.000 m Field notes as tabulated. ────────────────────────────────────────────────────────────────── STATION | BS (m) | HI (m) | FS (m) | ELEVATION (m) ────────────────────────────────────────────────────────────────── BM-A | 2.135 | 77.135 | — | 75.000 TP-1 | 1.820 | 75.940 | 3.015 | 74.120 TP-2 | 2.560 | 77.060 | 1.440 | 74.500 ← wait, recalc BM-B | — | — | 2.085 | 74.975 ────────────────────────────────────────────────────────────────── (a) Step-by-step: Setup 1: HI₁ = 75.000 + 2.135 = 77.135 m elev(TP-1) = 77.135 − 3.015 = 74.120 m Setup 2: HI₂ = 74.120 + 1.820 = 75.940 m elev(TP-2) = 75.940 − 1.440 = 74.500 m Setup 3: HI₃ = 74.500 + 2.560 = 77.060 m elev(BM-B) = 77.060 − 2.085 = 74.975 m (b) Arithmetic check: ΣBS = 2.135 + 1.820 + 2.560 = 6.515 m ΣFS = 3.015 + 1.440 + 2.085 = 6.540 m Δelev = ΣBS − ΣFS = 6.515 − 6.540 = −0.025 m Check: elev(BM-B) − elev(BM-A) = 74.975 − 75.000 = −0.025 m ✓ Arithmetic check closes. (c) Misclosure vs. published elevation: Misclosure = computed elev(BM-B) − published elev(BM-B) Misclosure = 74.975 − 74.800 = +0.175 m = +175 mm Total leveled distance: K = (number of setups × 2 sides × avg sight) / 1000 K = (3 setups × 2 × 60 m) / 1000 = 360 m / 1000 = 0.36 km (Note: 3 instrument positions cover 3 forward sights + 2 backsights forward = ~360 m total run) Allowable misclosure (third-order): Misclosure_allow = ±12√K mm = ±12√0.36 = ±12 × 0.6 = ±7.2 mm Since |misclosure| = 175 mm >> 7.2 mm, the leveling does NOT meet third-order accuracy requirements. The survey must be repeated or checked for blunders.

Question Type

numerical

Answer Structure

• Table: Correct HI and elevation columns [1 mark]
• Part (a): elev(BM-B) = 74.975 m with all intermediate values [1 mark]
• Part (b): ΣBS = 6.515, ΣFS = 6.540, Δelev = −0.025 m, arithmetic check confirms ✓ [1 mark]
• Part (c): Misclosure = +175 mm computed and compared with allowable ±7.2 mm [1 mark]
• Conclusion: Does not meet third-order accuracy, survey must be redone [1 mark]

Scoring Breakdown

Common Mark Deductions

• Failing to include a conclusion about whether the survey meets accuracy requirements — the question explicitly asks for this.
• Using K in metres instead of kilometres in the misclosure formula.
• Mixing up misclosure (vs. published BM) with the internal arithmetic check — these are two separate checks.
• Not showing the arithmetic check (Part b) separately from the elevation computation (Part a).

Key Phrases To Include

• HI = elev + BS
• elev = HI − FS
• ΣBS − ΣFS
• arithmetic check
• misclosure
• ±12√K mm
• third-order accuracy
• 74.975 m
• 175 mm

Marks

3

Topic

Errors in Leveling

Difficulty

medium

Template Id

T14

Examiner Tip

For a 3-mark enumeration question, give exactly 3 points, each with: name, one-sentence description, and classification. Do not write a paragraph — bullet or numbered format is fastest and clearest.

Model Answer

1. Collimation error (line-of-sight not horizontal when bubble is centred) — SYSTEMATIC. The error accumulates in the same direction with each setup; minimised by balancing BS and FS distances. 2. Rod not held vertically (rod lean) — ACCIDENTAL. The rod may lean in any direction, causing the observed reading to be larger than the true vertical reading; error magnitude and direction vary randomly. 3. Atmospheric refraction variation — SYSTEMATIC (for a single survey session) / ACCIDENTAL (session to session). Refraction bends the line of sight by an amount dependent on atmospheric conditions, generally overestimated by long sights; partially accounted for in the curvature-refraction correction h_cr = 0.0675K².

Question Type

short_answer

Answer Structure

• Error 1: Name + brief description + classification with minimisation note [1 mark]
• Error 2: Name + brief description + classification with minimisation note [1 mark]
• Error 3: Name + brief description + classification with reason [1 mark]

Scoring Breakdown

Common Mark Deductions

• Listing only instrument errors and no human or environmental errors — examiners expect variety across error categories.
• Confusing systematic with accidental — a blunder (e.g., misread rod) is neither; it is a gross error.
• Describing an error without classifying it — the question explicitly asks for classification.

Key Phrases To Include

• collimation error
• systematic
• accidental
• rod lean
• refraction
• balancing backsight and foresight distances

Marks

2

Topic

Benchmarks and Datum Reference

Difficulty

easy

Template Id

T15

Examiner Tip

Mention NAMRIA or mean sea level as the datum reference for Philippine surveys — this contextual detail signals awareness of local practice and earns the definition mark.

Model Answer

A benchmark (BM) is a permanent, stable physical object of known elevation above an adopted datum (in the Philippines, mean sea level as established by the National Mapping and Resource Information Authority, NAMRIA) from which leveling surveys originate or are referenced. Two characteristics required of a suitable benchmark: 1. Permanence — the structure must be stable and resistant to movement, settlement, or destruction over the long term (e.g., concrete monument, brass disc set in bedrock, or a stable building foundation). 2. Accessibility and identifiability — the BM must be easily found and precisely described so that any surveyor can re-occupy the exact point; an ambiguous or hidden mark is not useful as a reference.

Question Type

short_answer

Answer Structure

• Sentence 1: Define benchmark — permanent, stable, known elevation, referenced to datum [1 mark]
• Points 1 and 2: Two correct characteristics with brief justification [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining BM as merely 'a known point' without mentioning permanence and elevation above datum.
• Listing 'high accuracy' as a characteristic — this is a property of the elevation value, not of the physical structure.
• Mentioning the same characteristic twice in different words (e.g., 'stable' and 'unmovable').

Key Phrases To Include

• permanent
• stable
• known elevation
• datum
• mean sea level
• NAMRIA
• permanence
• accessibility