CELE Construction Management & Methods — Construction Materials and TestingExam Answer Templates

Marks

1

Topic

Concrete — Water-Cement Ratio

Difficulty

easy

Template Id

T1

Examiner Tip

Examiners look for the keyword 'mass' (or 'weight') and the fraction W_water/W_cement. A bare definition without the formula form earns borderline credit — include both.

Model Answer

The water-cement ratio (w/c) is the ratio of the mass of free water to the mass of cement in a concrete mix. It is expressed as: w/c = W_water / W_cement (both in kg).

Question Type

very_short_answer

Answer Structure

• Line 1: State the definition using the phrase 'ratio of mass of water to mass of cement' [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing 'volume of water to volume of cement' — w/c is by mass, not volume.
• Omitting the formula or writing the ratio inverted (cement/water).
• Using 'weight' loosely without specifying it means mass — though commonly accepted, add 'by weight' for clarity.

Key Phrases To Include

• mass of water
• mass of cement
• w/c = W_water / W_cement
• by weight / by mass

Marks

1

Topic

Concrete — Slump Test

Difficulty

easy

Template Id

T2

Examiner Tip

The single most common error is equating slump with strength. Examiners specifically test this distinction. Write 'workability' prominently.

Model Answer

The slump test measures the workability (consistency) of fresh concrete. It indicates how easily the concrete can be mixed, placed, and compacted without segregation.

Question Type

very_short_answer

Answer Structure

• Line 1: State that slump test measures workability/consistency of fresh concrete [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating that the slump test measures concrete strength — slump measures workability, NOT strength.
• Describing the procedure instead of the purpose when the question asks 'what is the purpose'.
• Confusing slump with compressive strength test.

Key Phrases To Include

• workability
• consistency
• fresh concrete
• vertical drop

Marks

2

Topic

Concrete — Compressive Strength Testing

Difficulty

easy

Template Id

T3

Examiner Tip

Always convert kN to N at the start — write '530 kN = 530,000 N' explicitly. Examiners deduct marks for unit inconsistency even when the numerical answer is correct.

Model Answer

Given: d = 150 mm, P = 530 kN = 530,000 N Step 1 — Compute cross-sectional area: A = (π/4) × d² = (π/4) × (150)² = 17,671 mm² Step 2 — Compute compressive strength: f'c = P / A = 530,000 / 17,671 = 30.0 MPa Answer: f'c = 30.0 MPa

Question Type

numerical

Answer Structure

• Line 1: State given data and convert P to Newtons [setup]
• Line 2: Write formula A = (π/4)d² and compute A [1 mark]
• Line 3: Write formula f'c = P/A, substitute, and state answer with unit MPa [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to convert kN to N before dividing (using 530 instead of 530,000 in the formula).
• Using d = 0.15 m and A in m² but then not converting result to MPa correctly.
• Skipping the area calculation step and jumping directly to the final answer without showing work.
• Omitting units in the final answer.

Key Phrases To Include

• A = (π/4)d²
• f'c = P/A
• 17,671 mm²
• MPa
• 530,000 N

Marks

2

Topic

Concrete — Water-Cement Ratio

Difficulty

easy

Template Id

T4

Examiner Tip

For part (b), the justification mentioning 'capillary pores' or 'voids' formed by excess water elevates the answer from partial to full marks. One engineering-specific reason is all that is needed.

Model Answer

(a) w/c = W_water / W_cement = 180 / 360 = 0.50 (b) A w/c of 0.50 produces moderate-to-high strength concrete. According to Abrams' strength law, a lower w/c yields higher compressive strength because there is less excess water to form voids (capillary pores) in the hardened cement paste. w/c = 0.50 is considered acceptable for general structural concrete (target f'c ≈ 28–35 MPa).

Question Type

numerical

Answer Structure

• Part (a): Write formula w/c = W_water/W_cement, substitute 180/360, state answer 0.50 [1 mark]
• Part (b): State relationship (lower w/c → higher strength), cite Abrams' law or capillary porosity concept [1 mark]

Scoring Breakdown

Common Mark Deductions

• Inverting the ratio (writing 360/180 = 2.0 — this is cement-water ratio, not w/c).
• Stating higher w/c gives higher strength — this is the opposite of the correct relationship.
• Giving part (b) answer without any justification — 'because of less water' alone earns only partial credit.

Key Phrases To Include

• w/c = W_water / W_cement
• 0.50
• lower w/c → higher strength
• capillary pores / voids
• Abrams' law

Marks

3

Topic

Concrete — w/c Ratio, Strength, Workability

Difficulty

medium

Template Id

T5

Examiner Tip

Use a small comparison table or two bullet points showing the opposing effects of w/c on strength and workability — this structure makes the trade-off instantly clear to the examiner and is very likely to earn full marks.

Model Answer

Compressive strength (f'c) is the maximum axial compressive load per unit area that hardened concrete can resist, measured at 28 days on 150×300 mm cylinders (unit: MPa). Workability is the ease with which fresh concrete can be mixed, transported, placed, and compacted without segregation. It is measured by the slump test (unit: mm of slump). Effect of water-cement ratio (w/c): - Decreasing w/c INCREASES compressive strength (fewer capillary pores in hardened paste, per Abrams' law). - Decreasing w/c DECREASES workability (stiffer mix, lower slump). Therefore, mix design involves a trade-off: the w/c is selected to meet the minimum required f'c while maintaining adequate workability for the placement method used.

Question Type

short_answer

Answer Structure

• Sentence 1–2: Define compressive strength with unit (MPa) and test method (28-day cylinder) [1 mark]
• Sentence 3–4: Define workability with unit (mm slump) and measurement method (slump test) [1 mark]
• Sentence 5–7: Explain inverse relationship — lower w/c → higher f'c but lower workability; state it is a design trade-off [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating slump measures strength — this common error eliminates the workability mark.
• Not mentioning the opposing effects on strength and workability — answering only one side.
• Omitting units for both strength (MPa) and workability (mm slump).
• Vague answer like 'w/c affects quality' without specifying direction of effect.

Key Phrases To Include

• f'c
• 28-day cylinders
• MPa
• workability
• slump test
• lower w/c → higher strength
• lower w/c → lower workability
• trade-off
• Abrams' law

Marks

3

Topic

Concrete — Compressive Strength and Acceptance

Difficulty

medium

Template Id

T6

Examiner Tip

Part (b) is where students lose the mark — they simply say 'yes it passes' without stating why. Write the ACI criterion (average of 3 consecutive ≥ f'c) even briefly, to show code knowledge.

Model Answer

(a) Compute compressive strength: Given: d = 150 mm, P = 620 kN = 620,000 N A = (π/4)(150)² = 17,671 mm² f'c = P/A = 620,000 / 17,671 = 35.1 MPa (b) Acceptance check: ACI 318 Section 26.12 requires that the average of any three consecutive strength tests must be ≥ f'c AND no individual test may fall below f'c by more than 3.5 MPa (for f'c ≤ 35 MPa). For this single cylinder: 35.1 MPa > 28 MPa ✓ This cylinder exceeds the specified strength. It passes the individual test criterion. (Note: final acceptance requires evaluation of consecutive cylinder sets, not a single result alone.)

Question Type

numerical

Answer Structure

• Part (a): Compute A = 17,671 mm² and f'c = 35.1 MPa with units [2 marks]
• Part (b): State ACI acceptance criteria and compare 35.1 MPa vs 28 MPa, state pass/fail with justification [1 mark]

Scoring Breakdown

Common Mark Deductions

• Converting kN incorrectly — writing P = 6,200 N instead of 620,000 N.
• Forgetting that a single cylinder result is not the final acceptance basis under ACI 318.
• Not citing any code reference for acceptance criteria in part (b).

Key Phrases To Include

• A = (π/4)d²
• f'c = P/A
• 35.1 MPa
• ACI 318
• acceptance criteria
• exceeds specified strength

Marks

3

Topic

Aggregates — Properties

Difficulty

medium

Template Id

T7

Examiner Tip

Structure each point as: Property name → mechanism → consequence on concrete. Examiners follow this chain. One-word answers ('gradation is important') earn zero for explanation.

Model Answer

Three key aggregate properties that affect concrete quality: 1. Gradation (Fineness Modulus): A well-graded aggregate fills voids efficiently, reducing the cement paste required. Poorly graded aggregate increases paste demand and cost, and may produce weaker, more permeable concrete. 2. Specific Gravity and Absorption: High water absorption means aggregate particles absorb mix water, effectively increasing the w/c ratio and reducing strength. Knowing specific gravity allows accurate proportioning of the mix by mass. 3. Cleanliness (Freedom from Deleterious Materials): Silt, clay, organic matter, and salts coat aggregate surfaces and weaken the paste-aggregate bond. Excessive fines (clay/silt) also increase water demand. ASTM C 33 sets limits on deleterious materials.

Question Type

short_answer

Answer Structure

• Point 1: Name property (Gradation/Fineness Modulus) + explain effect on concrete (void filling, paste demand) [1 mark]
• Point 2: Name property (Absorption) + explain effect (changes effective w/c, proportioning accuracy) [1 mark]
• Point 3: Name property (Cleanliness) + explain effect (bond weakening, increased water demand) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Listing properties without explaining WHY they affect concrete quality — the explanation earns the mark, not just the name.
• Naming the same property twice (e.g., 'gradation' and 'particle size' — these are the same concept).
• Vague explanations like 'affects concrete' without specifying the mechanism.

Key Phrases To Include

• gradation
• fineness modulus
• specific gravity
• absorption
• cleanliness
• deleterious materials
• paste demand
• w/c ratio
• paste-aggregate bond

Marks

2

Topic

Steel — Quality Control Tests

Difficulty

easy

Template Id

T8

Examiner Tip

Mention 'mill certificates' as the documentary verification — this extra detail signals practical knowledge and impresses examiners even though it is not worth a separate mark.

Model Answer

Reinforcing steel (rebars) are verified through the following tests: 1. Tension Test: Measures yield strength (fy), ultimate tensile strength (fu), and percent elongation (ductility). The steel must meet minimum fy (e.g., Grade 40: fy = 275 MPa; Grade 60: fy = 415 MPa per ASTM A615/A706). 2. Bend Test: The bar is bent around a specified pin diameter without cracking, verifying ductility and toughness. Mill certificates (test reports from the manufacturer) are also required and must be verified against delivered samples.

Question Type

short_answer

Answer Structure

• Test 1: Tension test — state what it measures (fy, fu, elongation) with at least one grade example [1 mark]
• Test 2: Bend test — state what it verifies (ductility/toughness, no cracking) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Naming only one test when two are expected for 2 marks.
• Confusing the bend test with the flexural test for concrete beams.
• Not specifying what the tension test measures — 'it checks the steel' is too vague.

Key Phrases To Include

• tension test
• yield strength (fy)
• ultimate tensile strength (fu)
• elongation
• ductility
• bend test
• mill certificate

Marks

3

Topic

Concrete — Required Average Strength / Quality Control

Difficulty

medium

Template Id

T9

Examiner Tip

Write both formulas explicitly, compute both results, then write 'Governing: f'cr = max(__, __) = __ MPa'. This three-step structure earns all three marks reliably even if there is a minor arithmetic error, because the method is demonstrably correct.

Model Answer

Given: f'c = 28 MPa, s = 3.5 MPa (f'c ≤ 35 MPa, so the two-formula set applies) ACI 318 requires the larger of: Formula 1: f'cr = f'c + 1.34s = 28 + 1.34(3.5) = 28 + 4.69 = 32.69 MPa Formula 2: f'cr = f'c + 2.33s − 3.5 = 28 + 2.33(3.5) − 3.5 = 28 + 8.155 − 3.5 = 32.66 MPa Governing: f'cr = max(32.69, 32.66) = 32.69 MPa ≈ 32.7 MPa Answer: The required average compressive strength is f'cr = 32.7 MPa. The mix must be designed to produce a mean cylinder strength of 32.7 MPa so that the probability of individual results falling below f'c = 28 MPa is acceptably low.

Question Type

numerical

Answer Structure

• Line 1: State applicability condition (f'c ≤ 35 MPa, use two-formula set) [setup]
• Lines 2–3: Compute Formula 1: f'cr = f'c + 1.34s = 32.69 MPa [1 mark]
• Lines 4–5: Compute Formula 2: f'cr = f'c + 2.33s − 3.5 = 32.66 MPa [1 mark]
• Line 6: State governing value = max of the two = 32.7 MPa [1 mark]

Scoring Breakdown

Common Mark Deductions

• Computing only one formula and not checking both — ACI requires the larger of the two.
• Subtracting 3.5 from the wrong term in Formula 2 (e.g., writing f'c + 2.33s + 3.5 instead of minus).
• Not stating which formula governs — just writing two numbers without identifying the maximum.
• Using the wrong formula set (e.g., applying formulas for f'c > 35 MPa when f'c = 28 MPa).

Key Phrases To Include

• f'cr = f'c + 1.34s
• f'cr = f'c + 2.33s − 3.5
• f'c ≤ 35 MPa
• governing = max
• 32.7 MPa
• ACI 318

Marks

2

Topic

Concrete — Water-Cement Ratio

Difficulty

easy

Template Id

T10

Examiner Tip

Always write the formula in its original form first, then rearrange algebraically, then substitute. Two-line numerical answers without algebra steps earn only partial credit in most boards.

Model Answer

Given: w/c = 0.45, W_cement = 400 kg From the definition of water-cement ratio: w/c = W_water / W_cement Solving for water: W_water = w/c × W_cement = 0.45 × 400 = 180 kg Answer: The mass of mixing water required is 180 kg.

Question Type

numerical

Answer Structure

• Line 1: Write the w/c formula [setup]
• Line 2: Rearrange to W_water = w/c × W_cement [1 mark]
• Line 3: Substitute and state answer with unit kg [1 mark]

Scoring Breakdown

Common Mark Deductions

• Dividing W_cement by w/c instead of multiplying (W_water = 400 / 0.45 = 889 kg — incorrect).
• Omitting the unit 'kg' in the final answer.
• Not showing the rearrangement step — going directly from w/c = 0.45 to W_water = 180 without showing algebra.

Key Phrases To Include

• w/c = W_water / W_cement
• W_water = w/c × W_cement
• 180 kg

Marks

3

Topic

Concrete — Required Average Strength / Quality Control

Difficulty

hard

Template Id

T11

Examiner Tip

This problem is designed to trap students who memorize 'Formula 1 governs' from the Example 3 in the reference. When s is larger (≥ ~4 MPa for f'c = 35), Formula 2 can govern. Always compute both — this is worth 1 full mark.

Model Answer

Given: f'c = 35 MPa, s = 4 MPa Note: f'c = 35 MPa satisfies the condition f'c ≤ 35 MPa; use the standard two-formula set. ACI 318 (Section 26.12.3.1): Formula 1: f'cr = f'c + 1.34s = 35 + 1.34(4) = 35 + 5.36 = 40.36 MPa Formula 2: f'cr = f'c + 2.33s − 3.5 = 35 + 2.33(4) − 3.5 = 35 + 9.32 − 3.5 = 40.82 MPa Governing: f'cr = max(40.36, 40.82) = 40.82 MPa ≈ 40.8 MPa Answer: f'cr = 40.8 MPa Note: Here Formula 2 governs because the higher s = 4 MPa makes the coefficient 2.33s dominant relative to the −3.5 offset — always check both.

Question Type

numerical

Answer Structure

• State applicable condition (f'c ≤ 35 MPa) and cite ACI 318 section [setup]
• Compute Formula 1 = 40.36 MPa [1 mark]
• Compute Formula 2 = 40.82 MPa [1 mark]
• Identify governing formula (Formula 2 governs) and state f'cr = 40.8 MPa [1 mark]

Scoring Breakdown

Common Mark Deductions

• Assuming Formula 1 always governs — at higher s values, Formula 2 governs (as in this problem).
• Arithmetic errors in Formula 2: miscalculating 2.33 × 4 (= 9.32, not 9.30).
• Not stating which formula governs — examiners require this conclusion explicitly.

Key Phrases To Include

• Formula 1: f'cr = f'c + 1.34s
• Formula 2: f'cr = f'c + 2.33s − 3.5
• 40.36 MPa
• 40.82 MPa
• Formula 2 governs
• f'cr = 40.8 MPa

Marks

3

Topic

Concrete — Slump Test / Workability

Difficulty

medium

Template Id

T12

Examiner Tip

Include at least one cone dimension (height = 300 mm or any of the three) to show you know the test equipment. Examiners look for this detail to distinguish students who know the standard from those guessing.

Model Answer

Slump Test Procedure (per ASTM C143 / AASHTO T119): 1. Moisten the slump cone (base 200 mm, top 100 mm, height 300 mm) and place on a flat, non-absorbent surface. 2. Fill the cone in three equal layers; rod each layer 25 times with a 16-mm tamping rod. 3. Strike off the top flush; lift the cone vertically in 5–10 seconds. 4. Measure the vertical drop (slump) from the top of the cone to the displaced top of the concrete specimen. Significance of Very High Slump: A very high slump (e.g., > 150–200 mm) indicates an excessively wet mix (high w/c). While workability is high, this causes: - Reduced compressive strength (weaker hardened concrete per Abrams' law). - Segregation and bleeding — coarse aggregate sinks, water rises to the surface. - Increased permeability and reduced durability of the finished structure. For structural pours, NSCP 2015 / ACI 318 specify maximum slump limits (typically 100 mm for normal reinforced concrete) to prevent these effects.

Question Type

short_answer

Answer Structure

• Steps 1–4: Describe the slump test procedure with key dimensions and rodding requirement [1 mark]
• Significance part 1: State that high slump = high w/c → reduced strength (mention Abrams' law) [1 mark]
• Significance part 2: State segregation/bleeding risk and cite slump limit per NSCP/ACI [1 mark]

Scoring Breakdown

Common Mark Deductions

• Describing the slump test as measuring strength — major conceptual error.
• Not giving cone dimensions or rodding detail — procedure marks require specifics.
• Stating only 'it is too wet' for the significance without mentioning strength reduction OR segregation.

Key Phrases To Include

• slump cone
• three layers, 25 rods each
• vertical drop
• workability
• high w/c → lower strength
• segregation
• bleeding
• maximum slump limit
• NSCP 2015 / ACI 318

Marks

2

Topic

Aggregates — Gradation / Fineness Modulus

Difficulty

medium

Template Id

T13

Examiner Tip

Even if you cannot recall all 7 standard sieves, listing at least 3 (#4, #8, #16) shows you know the concept. The formula structure (cumulative % retained / 100) is the critical part.

Model Answer

Fineness Modulus (FM): The fineness modulus of fine aggregate is a single index number representing the average particle size of the aggregate. It is computed as: FM = (Sum of cumulative percent retained on standard sieves #100, #50, #30, #16, #8, #4, 3/8") / 100 Typical FM for fine aggregate: 2.3 to 3.1 (coarser sand = higher FM). Influence on Mix Design: A higher FM (coarser sand) generally reduces the water demand for a given workability, allowing a lower w/c and potentially higher strength. A lower FM (finer sand) increases surface area, requires more paste/water, and can make the mix stickier. The FM is used in ACI mix design tables to estimate the optimum fine-to-coarse aggregate ratio.

Question Type

short_answer

Answer Structure

• Definition and formula for FM (cumulative % retained on standard sieves / 100) [1 mark]
• Influence: higher FM = coarser sand = lower water demand; used in ACI mix design [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing fineness modulus with particle size distribution curve — FM is a single number, not a curve.
• Not specifying the standard sieves used in the FM calculation.
• Not connecting FM to its effect on water demand or mix design.

Key Phrases To Include

• fineness modulus
• cumulative percent retained
• standard sieves
• 2.3 to 3.1
• coarser sand
• water demand
• ACI mix design

Marks

5

Topic

Concrete — Statistical Quality Control / Acceptance

Difficulty

hard

Template Id

T14

Examiner Tip

In a 5-mark problem, the examiner expects complete, organized work for each sub-part. Box your intermediate results (x̄, s, each triplet average). Show all squared deviations in a table format — this prevents arithmetic errors and makes checking easy for the examiner.

Model Answer

Given: Data set = {30.1, 28.5, 31.2, 29.8, 32.0} MPa, n = 5, f'c = 28 MPa (a) Sample Mean: x̄ = (30.1 + 28.5 + 31.2 + 29.8 + 32.0) / 5 = 151.6 / 5 = 30.32 MPa (b) Sample Standard Deviation: s = √[Σ(xi − x̄)² / (n−1)] Deviations (xi − x̄): 30.1 − 30.32 = −0.22 → (−0.22)² = 0.0484 28.5 − 30.32 = −1.82 → (−1.82)² = 3.3124 31.2 − 30.32 = 0.88 → (0.88)² = 0.7744 29.8 − 30.32 = −0.52 → (−0.52)² = 0.2704 32.0 − 30.32 = 1.68 → (1.68)² = 2.8224 Σ(xi − x̄)² = 0.0484 + 3.3124 + 0.7744 + 0.2704 + 2.8224 = 7.2280 s = √(7.2280 / 4) = √1.8070 = 1.344 MPa ≈ 1.34 MPa (c) ACI 318 Mean Criterion — Average of any 3 consecutive results ≥ f'c = 28 MPa: Set 1: (30.1 + 28.5 + 31.2)/3 = 89.8/3 = 29.93 MPa ≥ 28 MPa ✓ Set 2: (28.5 + 31.2 + 29.8)/3 = 89.5/3 = 29.83 MPa ≥ 28 MPa ✓ Set 3: (31.2 + 29.8 + 32.0)/3 = 93.0/3 = 31.00 MPa ≥ 28 MPa ✓ All three consecutive averages exceed f'c = 28 MPa. The strength record satisfies the ACI mean criterion and is ACCEPTABLE.

Question Type

numerical

Answer Structure

• Part (a): Sum all 5 values, divide by 5, state x̄ = 30.32 MPa [1 mark]
• Part (b): Compute deviations and their squares for all 5 data points [1 mark]
• Part (b): Sum squared deviations, divide by n−1 = 4, take square root, state s ≈ 1.34 MPa [1 mark]
• Part (c): Compute averages of all three consecutive triplets [1 mark]
• Part (c): Compare each triplet average to f'c = 28 MPa and state acceptance conclusion [1 mark]

Scoring Breakdown

Common Mark Deductions

• Dividing by n instead of n−1 in the standard deviation formula (population vs sample formula error).
• Computing only one consecutive triplet average instead of all three.
• Arithmetic errors in deviation calculation — show all work to earn partial marks.
• Concluding 'acceptable' without checking all triplets — if even one triplet average < f'c, it fails.

Key Phrases To Include

• sample mean x̄ = 30.32 MPa
• Σ(xi − x̄)²
• n − 1 = 4
• s = 1.34 MPa
• average of 3 consecutive
• ≥ f'c = 28 MPa
• ACCEPTABLE
• ACI 318

Marks

5

Topic

Concrete — Mix Design / Batch Proportions / Acceptance

Difficulty

hard

Template Id

T15

Examiner Tip

Part (c) earns 2 marks: one for the numerical comparison and one for the engineering conclusion with code reference. Many students write the numbers but forget to write 'ADEQUATE' clearly and cite ACI 318 — this single omission costs 1 mark.

Model Answer

Given: W_cement = 350 kg, W_water = 168 kg, W_FA = 680 kg, W_CA = 1,050 kg f'cr = 31.5 MPa (required average), actual mean from trials = 33.0 MPa (a) Water-Cement Ratio: w/c = W_water / W_cement = 168 / 350 = 0.48 (b) Total Batch Mass per cubic metre: Note: For a 1 m³ batch (unit weight basis), total mass is the sum of all ingredients. W_total = 350 + 168 + 680 + 1,050 = 2,248 kg/m³ This is the unit weight (density) of fresh concrete = 2,248 kg/m³ ≈ 2,250 kg/m³ (reasonable for normal-weight concrete, typically 2,300–2,400 kg/m³). (c) Adequacy Check: The mix design is adequate if the trial batch mean strength ≥ f'cr. Actual mean = 33.0 MPa Required average f'cr = 31.5 MPa 33.0 MPa > 31.5 MPa ✓ Conclusion: The mix design is ADEQUATE. The trial batch mean strength of 33.0 MPa exceeds the required average compressive strength of 31.5 MPa by 1.5 MPa, providing an acceptable margin. Per ACI 318, a mix design is approved when the trial batch mean meets or exceeds f'cr.

Question Type

numerical

Answer Structure

• Part (a): w/c = 168/350 = 0.48 [1 mark]
• Part (b): Sum all four batch masses = 2,248 kg/m³ and identify as unit weight of fresh concrete [2 marks]
• Part (c): Compare 33.0 MPa vs 31.5 MPa, state adequate with reference to ACI 318 criterion [2 marks]

Scoring Breakdown

Common Mark Deductions

• Including only cement and water in the total mass, omitting aggregates.
• Concluding 'adequate' without comparing numbers — state both values and the comparison.
• Not interpreting the total mass as unit weight per m³ — just summing without explaining significance.
• Confusing f'cr (required average) with f'c (specified minimum) in the adequacy check.

Key Phrases To Include

• w/c = 168/350 = 0.48
• 2,248 kg/m³
• unit weight of fresh concrete
• normal-weight concrete
• 33.0 MPa > 31.5 MPa
• ADEQUATE
• ACI 318
• f'cr = required average