GELE Surveying (Geomatics) — Measurements and Theory of ErrorsExam Answer Templates

Marks

1

Topic

Most Probable Value

Difficulty

easy

Template Id

T1

Examiner Tip

Examiners reward the formula x̄ = Σx/n alongside the word 'arithmetic mean.' A definition in words alone, without the formula, often gets only partial credit in numerical-subject boards.

Model Answer

The most probable value (MPV) is the arithmetic mean of a series of repeated measurements of the same quantity under the same conditions. It is the value that has the highest probability of being closest to the true value and is computed as x̄ = Σx / n.

Question Type

very_short_answer

Answer Structure

• Line 1: State the definition using the term 'arithmetic mean' and 'repeated measurements' [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing 'average value' without linking it to the formula or the concept of probability (-0 to -1 depending on strictness)
• Omitting that measurements must be of the same quantity under the same conditions
• Confusing MPV with the true value (they are not the same)

Key Phrases To Include

• arithmetic mean
• repeated measurements
• x̄ = Σx / n
• highest probability
• same conditions

Marks

2

Topic

Types of Errors

Difficulty

easy

Template Id

T2

Examiner Tip

Frame your answer as a contrast — examiners for distinction/comparison questions look for the parallel structure: systematic (law-governed, correctable) vs. random (probabilistic, not individually correctable).

Model Answer

Systematic errors follow a definite physical law or cause, always biasing the measurement in the same direction (either consistently positive or consistently negative). They can be corrected by applying computed corrections (e.g., temperature correction to a steel tape). Random (accidental) errors, in contrast, are small residual errors that remain after all systematic corrections are applied; they are equally likely to be positive or negative and are treated statistically using the theory of probability.

Question Type

short_answer

Answer Structure

• Part A: Define systematic error — follows a law, consistent direction, correctable [1 mark]
• Part B: Define random error — residual, bidirectional, treated statistically [1 mark]

Scoring Breakdown

Common Mark Deductions

• Mentioning only one type and ignoring the other
• Calling random errors 'mistakes' or 'blunders' (those are a third distinct category)
• Failing to state that systematic errors are correctable while random errors are not individually correctable

Key Phrases To Include

• definite physical law
• same direction (consistently biased)
• correctable
• residual errors
• equally likely positive or negative
• theory of probability

Marks

2

Topic

Tape Corrections — Temperature

Difficulty

easy

Template Id

T3

Examiner Tip

The sign of C_t is conceptually important: if T > T_s, the steel tape expands and is physically longer than its nominal length, so measured distances read short and the correction is positive (add to get true length).

Model Answer

Given: L = 300 m, α = 11.6 × 10⁻⁶/°C, T = 40°C, T_s = 20°C Formula: C_t = α L (T − T_s) Solution: C_t = (11.6 × 10⁻⁶)(300)(40 − 20) C_t = (11.6 × 10⁻⁶)(300)(20) C_t = +0.0696 m The temperature correction is +0.0696 m (positive because field temperature exceeds standard; the tape is longer, so measured distances read short — the true length is longer).

Question Type

numerical

Answer Structure

• Step 1: List all given values with symbols and units [0.5 mark]
• Step 2: Write the formula C_t = αL(T − T_s) [0.5 mark]
• Step 3: Substitute and compute the numerical answer [0.5 mark]
• Step 4: State the sign and physical interpretation [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Reversing (T_s − T) instead of (T − T_s), giving wrong sign
• Omitting units on the final answer
• Not stating whether the correction is added or subtracted

Key Phrases To Include

• C_t = αL(T − T_s)
• α = 11.6 × 10⁻⁶/°C
• +0.0696 m
• field temperature exceeds standard
• tape is longer

Marks

3

Topic

Most Probable Value and Probable Error

Difficulty

medium

Template Id

T4

Examiner Tip

Always write out the residual table in full columnar form (x | v | v²). Board-exam rubrics allocate a dedicated mark for this table. A correct final answer without the table may still lose 1 mark.

Model Answer

Given measurements: x₁ = 100.12 m, x₂ = 100.08 m, x₃ = 100.10 m; n = 3 (a) Most Probable Value: x̄ = (100.12 + 100.08 + 100.10) / 3 x̄ = 300.30 / 3 = 100.10 m (b) Probable Error of a Single Measurement: Residuals: v₁ = 100.12 − 100.10 = +0.02 m v₂ = 100.08 − 100.10 = −0.02 m v₃ = 100.10 − 100.10 = 0.00 m Σv² = (0.02)² + (0.02)² + (0.00)² = 0.0004 + 0.0004 + 0 = 0.0008 m² E = 0.6745 √(Σv² / (n − 1)) E = 0.6745 √(0.0008 / 2) E = 0.6745 √(0.0004) E = 0.6745 × 0.02 E = ±0.0135 m Final Answers: MPV = 100.10 m Probable error (single observation) = ±0.0135 m

Question Type

numerical

Answer Structure

• Step 1: Compute the arithmetic mean x̄ [1 mark]
• Step 2: Tabulate residuals v = x − x̄ and compute Σv² [1 mark]
• Step 3: Apply E = 0.6745√(Σv²/(n−1)) and compute final answer with sign [1 mark]

Scoring Breakdown

Common Mark Deductions

• Dividing by n instead of n−1 in the formula (uses population standard deviation instead of sample)
• Forgetting the ± sign on the probable error
• Not showing the residual table — step marks require the working to be visible
• Using E = 0.6745√(Σv²/n) — the n−1 denominator is mandatory for sample observations

Key Phrases To Include

• x̄ = Σx / n
• residuals v = x − x̄
• Σv²
• E = 0.6745√(Σv²/(n−1))
• n − 1 (degrees of freedom)
• ±0.0135 m

Marks

1

Topic

Types of Errors

Difficulty

easy

Template Id

T5

Examiner Tip

The key distinguishing phrase is 'gross error' and 'must be eliminated.' Examiners will not award the mark if you say blunders are reduced by taking the mean.

Model Answer

A blunder (mistake) is a gross human error caused by carelessness, misreading, or poor judgment (e.g., misreading a rod, recording a wrong digit). Blunders are not treated statistically because they do not follow any probability law — they must be detected and eliminated through field checks, repetition, and verification before statistical analysis begins.

Question Type

very_short_answer

Answer Structure

• Line 1: Define blunder as a gross human error with at least one example [0.5 mark]
• Line 2: Explain that it does not follow a probability law and must be eliminated [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing blunders with systematic errors
• Saying blunders are 'minimised' statistically — they are eliminated, not minimised

Key Phrases To Include

• gross human error
• does not follow probability law
• must be eliminated
• carelessness or misreading

Marks

3

Topic

Slope Correction

Difficulty

medium

Template Id

T6

Examiner Tip

State explicitly that slope correction C_h is always negative — this is a high-frequency board-exam pitfall and examiners specifically look for this statement as an indicator of conceptual understanding.

Model Answer

Given: L = 250 m (slope distance), h = 10 m (elevation difference) (a) Exact Formula: H = √(L² − h²) H = √(250² − 10²) H = √(62500 − 100) H = √62400 H = 249.80 m (b) Approximate Slope Correction: C_h = −h² / (2L) C_h = −(10²) / (2 × 250) C_h = −100 / 500 C_h = −0.20 m H ≈ L + C_h = 250 + (−0.20) = 249.80 m Both methods give H = 249.80 m (the approximate formula is accurate for gentle slopes where h/L < 0.10).

Question Type

numerical

Answer Structure

• Step 1: State given data (L and h) [0.5 mark]
• Step 2: Apply exact formula H = √(L² − h²) and compute [1 mark]
• Step 3: Apply approximate formula C_h = −h²/2L and compute H [1 mark]
• Step 4: State that slope correction is always negative (horizontal < slope distance) [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using h²/2L without the negative sign
• Adding C_h instead of subtracting (wrong sign application)
• Not showing the formula before substituting
• Rounding intermediate values leading to a different final answer

Key Phrases To Include

• H = √(L² − h²)
• C_h = −h²/2L
• slope correction is always negative
• horizontal distance is always less than slope distance
• 249.80 m

Marks

2

Topic

Tape Corrections — Tape Too Long / Too Short

Difficulty

medium

Template Id

T7

Examiner Tip

Always state the physical logic first: 'tape too long → measured reads short → true is longer.' This guards against formula inversion, which is the most common error on this type of board question.

Model Answer

Given: Nominal tape length = 30 m Actual tape length = 30.02 m Measured distance = 450 m Since the tape is too long (actual > nominal), each 30 m interval laid down by the tape actually covers 30.02 m on the ground — the true distance is greater than the measured reading. Formula: True distance = Measured distance × (Actual length / Nominal length) Solution: True distance = 450 × (30.02 / 30) True distance = 450 × 1.000667 True distance = 450.30 m The correct distance is 450.30 m.

Question Type

numerical

Answer Structure

• Step 1: Identify whether the tape is too long or too short and state the direction of error [0.5 mark]
• Step 2: Write the formula: True = Measured × (Actual/Nominal) [0.5 mark]
• Step 3: Substitute values and compute [0.5 mark]
• Step 4: State the final answer with units [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Inverting the ratio: using (nominal/actual) instead of (actual/nominal)
• Incorrectly concluding that tape too long → measured reads long (it reads SHORT — each interval is longer, so fewer intervals cover the same ground)
• Not explaining the physical reasoning before the formula

Key Phrases To Include

• tape is too long (actual > nominal)
• true distance = measured × (actual/nominal)
• true distance is greater than measured distance
• 450.30 m

Marks

1

Topic

Probable Error of the Mean

Difficulty

easy

Template Id

T8

Examiner Tip

Relate the formula conceptually: increasing n reduces E_m, demonstrating why surveyors repeat measurements. This statement, if added, shows depth of understanding and can earn bonus credit from lenient examiners.

Model Answer

Given: E = ±0.0135 m (probable error of a single observation), n = 3 E_m = E / √n E_m = 0.0135 / √3 E_m = 0.0135 / 1.732 E_m = ±0.0078 m The probable error of the mean is ±0.0078 m.

Question Type

very_short_answer

Answer Structure

• Line 1: Write formula E_m = E/√n [0.5 mark]
• Line 2: Substitute and compute final answer with sign and units [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Multiplying by √n instead of dividing
• Forgetting the ± sign
• Omitting units

Key Phrases To Include

• E_m = E / √n
• ±0.0078 m
• more observations → smaller error of the mean

Marks

2

Topic

Error Propagation

Difficulty

medium

Template Id

T9

Examiner Tip

The board exam frequently tests whether a student knows that errors of a sum combine as √(ΣEᵢ²) — NOT as a simple arithmetic sum. Write this formula explicitly before computing to earn the formula mark even if arithmetic is off.

Model Answer

Given: E₁ = ±0.015 m, E₂ = ±0.020 m, E₃ = ±0.012 m, E₄ = ±0.018 m For the sum of independently measured quantities, errors propagate as: E_total = √(E₁² + E₂² + E₃² + E₄²) E_total = √[(0.015)² + (0.020)² + (0.012)² + (0.018)²] E_total = √[0.000225 + 0.000400 + 0.000144 + 0.000324] E_total = √0.001093 E_total = ±0.0331 m The probable error of the total perimeter is ±0.0331 m.

Question Type

numerical

Answer Structure

• Step 1: State the error propagation formula for sums: E = √(ΣEᵢ²) [1 mark]
• Step 2: Compute each Eᵢ², sum them, take the square root [0.5 mark]
• Step 3: State the final answer with sign and units [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Adding errors directly (E₁ + E₂ + E₃ + E₄) instead of combining in quadrature
• Forgetting to square the individual errors before summing
• Not including the ± sign on the final answer

Key Phrases To Include

• E_total = √(ΣEᵢ²)
• independent measurements
• errors propagate as square root of sum of squares
• ±0.0331 m

Marks

3

Topic

Tape Corrections — Sag

Difficulty

hard

Template Id

T10

Examiner Tip

The sag correction is ALWAYS negative — write this statement explicitly. Examiners consistently note that students who drop the negative sign on C_sag lose the interpretation mark even when arithmetic is correct.

Model Answer

Given: W = 0.80 kg → W = 0.80 × 9.81 = 7.848 N (total weight of tape) L = 30 m (length of tape) P = 80 N (applied pull) Formula (using total weight W): C_sag = −W²L / (24P²) Solution: C_sag = −(7.848)² × 30 / (24 × 80²) C_sag = −(61.59) × 30 / (24 × 6400) C_sag = −1847.7 / 153600 C_sag = −0.01203 m The sag correction per tape length is −0.0120 m (negative because sag always shortens the horizontal distance between tape ends).

Question Type

numerical

Answer Structure

• Step 1: Convert W from kg to Newtons using W = mg [0.5 mark]
• Step 2: State the sag correction formula C_sag = −W²L/(24P²) [1 mark]
• Step 3: Substitute values and compute [1 mark]
• Step 4: State the sign and physical meaning (sag always negative) [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using W in kg without converting to Newtons
• Omitting the negative sign on C_sag
• Using the per-unit-length formula (−w²L³/24P²) incorrectly when total weight W is given
• Confusing P (pull) with W (weight)

Key Phrases To Include

• C_sag = −W²L/(24P²)
• W in Newtons (convert from kg)
• sag correction is always negative
• supported at ends only
• −0.0120 m

Marks

3

Topic

Tape Corrections — Tension (Pull)

Difficulty

hard

Template Id

T11

Examiner Tip

Unit conversion is the No. 1 source of wrong answers for C_p. Always write out 'A = 6.0 mm² = 6.0×10⁻⁶ m²' and 'E = 200 GPa = 200×10⁹ N/m²' explicitly — this also earns the conversion marks.

Model Answer

Given: P = 75 N (applied pull) P_s = 50 N (standard pull) L = 300 m (measured distance) A = 6.0 mm² = 6.0 × 10⁻⁶ m² E = 200 GPa = 200 × 10⁹ Pa = 200 × 10⁹ N/m² Formula: C_p = (P − P_s) × L / (A × E) Solution: C_p = (75 − 50) × 300 / (6.0 × 10⁻⁶ × 200 × 10⁹) C_p = (25 × 300) / (1200) C_p = 7500 / 1200 C_p = +0.00625 m Note: A × E = 6.0 × 10⁻⁶ m² × 200 × 10⁹ N/m² = 1.20 × 10⁶ N The tension correction is +0.00625 m (positive because applied pull > standard pull, elongating the tape).

Question Type

numerical

Answer Structure

• Step 1: List all given values with correct unit conversions (A in m², E in N/m²) [1 mark]
• Step 2: Write formula C_p = (P − P_s)L / (AE) [0.5 mark]
• Step 3: Substitute and compute AE, then the full expression [1 mark]
• Step 4: State final answer with sign and physical interpretation [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Failing to convert A from mm² to m² (off by factor of 10⁶)
• Failing to convert E from GPa to Pa/N/m²
• Using wrong formula (e.g., mixing up tension and temperature formulas)
• Dropping the sign of C_p

Key Phrases To Include

• C_p = (P − P_s)L / (AE)
• A in m² (convert from mm²)
• E in N/m² (convert from GPa)
• +0.00625 m
• P > P_s → tape elongates → positive correction

Marks

1

Topic

Error Propagation

Difficulty

easy

Template Id

T12

Examiner Tip

Know the two square-root formulas: E_sum = E√n (error grows with more measurements in a series) vs. E_m = E/√n (error of the mean shrinks with more measurements). Mixing these up is a perennial board-exam trap.

Model Answer

Given: n = 5 equal measurements, each with E = ±0.03 m For n measurements of equal probable error E, the probable error of the sum is: E_sum = E × √n E_sum = 0.03 × √5 E_sum = 0.03 × 2.236 E_sum = ±0.0671 m The probable error of the sum is ±0.0671 m.

Question Type

numerical

Answer Structure

• Line 1: Write formula E_sum = E√n [0.5 mark]
• Line 2: Substitute and compute final answer [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Multiplying directly: E × n = 0.15 m (ignoring the square root)
• Confusing this with the error of the mean formula (E/√n)

Key Phrases To Include

• E_sum = E√n
• equal probable errors
• ±0.0671 m

Marks

2

Topic

Tape Corrections — Tape Too Long / Too Short

Difficulty

medium

Template Id

T13

Examiner Tip

The conceptual logic is: a short tape requires more intervals to cover the same distance → the tape reading (number of intervals × nominal length) exceeds the true distance → correction is negative. Memorise this logic, not just the formula.

Model Answer

Given: Nominal length = 30 m Actual length = 29.97 m (tape is too short) Measured distance = 600 m Physical reasoning: The tape is shorter than nominal. Each 30 m reading actually covers only 29.97 m on the ground — the true distance is less than the measured distance. Formula: True distance = Measured × (Actual / Nominal) True distance = 600 × (29.97 / 30) True distance = 600 × 0.999 True distance = 599.40 m The correction is −0.60 m (negative; tape too short → subtract from measured distance).

Question Type

numerical

Answer Structure

• Step 1: Identify tape is too short and state direction (true < measured) [0.5 mark]
• Step 2: Apply formula True = Measured × (Actual/Nominal) [0.5 mark]
• Step 3: Compute True = 599.40 m [0.5 mark]
• Step 4: State correction = −0.60 m with correct sign and direction [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using nominal/actual instead of actual/nominal
• Concluding that tape too short → measured reads short (wrong — measured reads LONG when tape is short)
• Adding the correction instead of subtracting

Key Phrases To Include

• tape too short (actual < nominal)
• true distance is less than measured distance
• True = Measured × (Actual/Nominal)
• 599.40 m
• correction = −0.60 m

Marks

5

Topic

Most Probable Value and Probable Error

Difficulty

hard

Template Id

T14

Examiner Tip

For 5-mark long-answer questions on statistics, the residual table with the Σv = 0 verification is mandatory. Examiners specifically allocate 1 mark for the table setup. The interpretation in part (d) is what separates 4-mark from 5-mark answers — never skip it.

Model Answer

Given measurements (n = 4): x₁ = 250.12 m, x₂ = 250.08 m, x₃ = 250.15 m, x₄ = 250.09 m (a) Most Probable Value (MPV): x̄ = (250.12 + 250.08 + 250.15 + 250.09) / 4 x̄ = 1000.44 / 4 x̄ = 250.11 m (b) Probable Error of a Single Observation: Residual Table: ┌────────┬───────────────────┬───────────┐ │ x (m) │ v = x − x̄ (m) │ v² (m²) │ ├────────┼───────────────────┼───────────┤ │ 250.12 │ +0.01 │ 0.0001 │ │ 250.08 │ −0.03 │ 0.0009 │ │ 250.15 │ +0.04 │ 0.0016 │ │ 250.09 │ −0.02 │ 0.0004 │ │ Σ │ 0.00 ✓ │ 0.0030 │ └────────┴───────────────────┴───────────┘ Check: Σv = +0.01 − 0.03 + 0.04 − 0.02 = 0.00 ✓ (verifies x̄) E = 0.6745 √(Σv² / (n−1)) E = 0.6745 √(0.0030 / 3) E = 0.6745 √(0.0010) E = 0.6745 × 0.031623 E = ±0.0213 m (c) Probable Error of the Mean: E_m = E / √n E_m = 0.0213 / √4 E_m = 0.0213 / 2 E_m = ±0.0107 m (d) Physical Interpretation: E = ±0.0213 m means there is a 50% probability that any single measurement of this line differs from the true value by no more than 0.0213 m. E_m = ±0.0107 m means there is a 50% probability that the computed mean (MPV = 250.11 m) differs from the true value by no more than 0.0107 m — the mean is a more reliable estimate of the true distance than any individual measurement. Final Answers: MPV = 250.11 m E (single observation) = ±0.0213 m E_m (mean) = ±0.0107 m

Question Type

long_answer

Answer Structure

• Part (a): Compute x̄ = Σx/n [1 mark]
• Part (b): Set up residual table with v = x − x̄ and v²; verify Σv = 0; apply E = 0.6745√(Σv²/(n−1)) [2 marks]
• Part (c): Apply E_m = E/√n [1 mark]
• Part (d): State physical meaning of E and E_m in terms of 50% probability [1 mark]

Scoring Breakdown

Common Mark Deductions

• Omitting the residual table and jumping straight to Σv² — loses the table mark
• Not verifying Σv = 0 (this is an important self-check that examiners look for)
• Using n instead of n−1 in the denominator
• Omitting the physical interpretation in part (d)
• Leaving out the ± sign on both E and E_m

Key Phrases To Include

• x̄ = Σx/n
• residual v = x − x̄
• Σv = 0 (verification check)
• E = 0.6745√(Σv²/(n−1))
• E_m = E/√n
• 50% probability
• mean is more reliable than a single observation
• ±0.0213 m
• ±0.0107 m

Marks

5

Topic

Combined Tape Corrections

Difficulty

hard

Template Id

T15

Examiner Tip

For multi-correction tape problems, always solve in sequence: C_t → C_p → C_sag → C_total → True distance. The structured format earns step marks even when the final answer is slightly off due to rounding. Examiners follow the same sequence in their rubric.

Model Answer

Given: L (measured) = 500 m, nominal tape = 50 m α = 11.6 × 10⁻⁶/°C, T = 35°C, T_s = 20°C A = 6.45 mm² = 6.45 × 10⁻⁶ m² E_steel = 200 GPa = 200 × 10⁹ N/m² P_s = 50 N, P = 80 N No sag (tape fully supported) Step 1 — Temperature Correction: C_t = α × L × (T − T_s) C_t = (11.6 × 10⁻⁶)(500)(35 − 20) C_t = (11.6 × 10⁻⁶)(500)(15) C_t = +0.0870 m (Positive: field temp > standard → tape longer → measured reads short → add) Step 2 — Tension (Pull) Correction: C_p = (P − P_s) × L / (A × E) A × E = 6.45 × 10⁻⁶ × 200 × 10⁹ = 1.29 × 10⁶ N C_p = (80 − 50) × 500 / (1.29 × 10⁶) C_p = (30 × 500) / 1290000 C_p = 15000 / 1290000 C_p = +0.01163 m (Positive: applied pull > standard → tape stretched → measured reads short → add) Step 3 — Sag Correction: C_sag = 0 (tape fully supported throughout — no sag) Step 4 — Total Correction: C_total = C_t + C_p + C_sag C_total = +0.0870 + 0.01163 + 0 C_total = +0.09863 m Step 5 — True (Corrected) Distance: True distance = Measured + C_total True distance = 500 + 0.09863 True distance = 500.099 m ≈ 500.10 m Final Answer: The corrected horizontal distance is 500.10 m.

Question Type

long_answer

Answer Structure

• Step 1: Temperature correction C_t = αL(T−T_s) with sign [1 mark]
• Step 2: Tension correction C_p = (P−P_s)L/(AE) with unit conversion and sign [1.5 marks]
• Step 3: Sag correction = 0, with reason stated [0.5 mark]
• Step 4: Sum all corrections [0.5 mark]
• Step 5: Add C_total to measured distance for final answer [0.5 mark]
• Correct final answer with units [0.5 mark if not already credited above; combined with Step 5]

Scoring Breakdown

Common Mark Deductions

• Forgetting to state that C_sag = 0 when tape is fully supported
• Unit conversion error for A (keeping it in mm²) or E (keeping it in GPa without converting)
• Adding all corrections without checking individual signs
• Not writing the formula for each correction before substituting — formula marks are allocated separately

Key Phrases To Include

• C_t = αL(T−T_s) = +0.0870 m
• C_p = (P−P_s)L/(AE) = +0.01163 m
• tape fully supported — C_sag = 0
• A in m², E in N/m²
• C_total = +0.09863 m
• True distance = Measured + C_total
• 500.10 m