LARE Practice Questions (2026) — Free Sample Questions with Answers

Answer: C

1. Point 2 to the near shore: from (station 190, elev. 80) to (station 510, elev. 20), the line drops 60 over 320 units.
2. At the ridge (station 370): 80 − 60 × 180 ÷ 320 = 46.3 < 70, so the view is blocked.
3. Point 3 to the shore: from (370, 70) to (510, 20), the line is at 45 at station 440 (ground 40) and 27.1 at station 490 (ground 24), so it is clear.
4. Point 4 (elev. 24) is below the 28 flood elevation.

Circle at Point 1: At elevation 50, the view east is blocked by the 80 hilltop and the 70 ridge.

Circle at Point 2: It is the highest point, but the sight line to the lake shore falls to about 46 at the ridge (station 370), which is below the ridge top at 70, so the ridge hides the lake.

Circle at Point 3: Correct. From the 70 ridge the ground falls in a concave slope to the shore, so the sight line clears the terrain to the whole lake, and the point is well above flood elevation 28.

Circle at Point 4: It sees the lake, but at elevation 24 it lies below flood elevation 28, inside the floodplain, which breaks the stated constraint.

In a visual resource analysis, check sight lines on a section: an intervening high point blocks the view whenever the straight line from viewer to target passes below it. The site must also meet every non-visual constraint.

Answer: C

1. Givens: Point A is on the 104 contour, Point B on the 96 contour; horizontal distance 40 units.
2. Rise = 104 − 96 = 8 units (four 2-unit intervals).
3. Slope % = rise ÷ run × 100.
4. Slope = 8 ÷ 40 × 100 = 20%.

5%: Uses only one 2-unit contour interval as the rise (2 ÷ 40 × 100); A and B are four intervals apart.

10%: Treats each of the four intervals as 1 unit (4 ÷ 40 × 100); the labels show a 2-unit interval.

20%: Correct. The rise is 104 − 96 = 8 units over a 40-unit run: 8 ÷ 40 × 100 = 20%.

25%: Counts five contour lines × 2 units = 10 units of rise; five lines bound only four intervals, so the rise is 8.

Slope % = rise ÷ horizontal run × 100; take the rise from the contour labels (number of intervals × interval), not from the number of contour lines.

Answer: A

North-west zone: Correct. It is outside the floodplain, next to the school, and farthest from the future six-lane arterial.

North-east zone: Quiet today, but it borders the county road that will become a six-lane arterial.

South-west zone: Near the school, but it lies within the 1-percent-annual-chance floodplain.

South-east zone: Within the floodplain and next to the future arterial, so it fails both criteria.

Contextual analysis looks beyond current conditions: adopted plans for future roads and development change which locations remain suitable.

Answer: D

Location A: Lies in the regulatory floodway, which is the most hazardous part of Zone AE.

Location B: Lies in Zone AE, the 1-percent-annual-chance flood hazard area.

Location C: Lies in shaded Zone X, the 0.2-percent-annual-chance flood hazard area.

Location D: Correct. It lies in unshaded Zone X (minimal flood hazard), outside both mapped flood hazard areas.

On a flood map, Zone AE (with its floodway) is the 1-percent-annual-chance hazard area and shaded Zone X is the 0.2-percent-annual-chance area; only unshaded Zone X is outside both.

Answer: Service yard: Zone 3; Quiet garden: Zone 4; Entry plaza: Zone 1; Visitor parking lot: Not placed; Café terrace: Zone 2

Turn a functional relationship diagram into a site layout by placing the most constrained uses first (street frontage, service access), then fitting the uses that must touch them. CLARB scores the placement all-or-nothing.

Answer: B

1. Rise = 110.0 − 100.0 = 10.0 units for every plan.
2. Slope % = rise ÷ length × 100.
3. Plan A: 10 ÷ 120 × 100 = 8.3% (fails, over 5%).
4. Plan B: 10 ÷ 230 × 100 = 4.3% (passes; no stream crossing).
5. Plan C: 10 ÷ 260 × 100 = 3.8% (slope passes, but it crosses the stream).

Plan A: Plan A: 10 ÷ 120 × 100 = 8.3%. Steeper than 5%, so it would have to be designed as a ramp.

Plan B: Correct. Plan B: 10 ÷ 230 × 100 = 4.3%, within the 5% maximum, and the switchbacks stay west of the stream.

Plan C: Plan C: 10 ÷ 260 × 100 = 3.8% meets the slope limit, but the loop crosses the stream twice, which the criteria prohibit.

An accessible route steeper than 1:20 (5%) becomes a ramp; lengthening the alignment with switchbacks lowers the grade, but every other stated constraint still applies.

Answer: A

1. Weights: slope 3, soil drainage 2, road access 1.
2. A = 3(3) + 2(2) + 1(1) = 9 + 4 + 1 = 14.
3. B = 2(3) + 2(2) + 3(1) = 6 + 4 + 3 = 13.
4. C = 1(3) + 3(2) + 3(1) = 3 + 6 + 3 = 12.
5. D = 2(3) + 3(2) + 1(1) = 6 + 6 + 1 = 13.
6. Highest weighted total: Area A (14). (Unweighted sums would favor B and C at 7, which is why weights matter.)

Area A: Correct. Weighted totals: A = 3×3 + 2×2 + 1×1 = 14; B = 13; C = 12; D = 13. Area A scores highest because it is best on slope, the most heavily weighted factor.

Area B: B totals 6 + 4 + 3 = 13. It ties for the highest unweighted sum (7), which is the error of ignoring the weights.

Area C: C totals 3 + 6 + 3 = 12, the lowest. Its strong soil and access scores cannot offset the poorest slope, which carries the most weight.

Area D: D totals 6 + 6 + 1 = 13, one point below Area A.

A weighted overlay multiplies each factor score by its weight before summing, so the factor the program cares most about (here slope for playing fields) drives the result.

Answer: B

Approve it, because strength and absorption exceed the requirements: Better strength and absorption do not offset the thickness shortfall; the paver still fails a stated requirement for a vehicular surface.

Return it marked revise and resubmit for the noncompliant thickness: Correct. The submitted paver is 60 mm thick where 80 mm is specified, so the submittal does not conform and must be corrected and resubmitted.

Approve it with a note that the pavers be installed 80 mm thick: A review note cannot change the thickness of a manufactured product; approving it would accept nonconforming material.

Forward it to the owner to decide whether thinner pavers are acceptable: Submittal review is the landscape architect's role; a deliberate deviation would need a substitution request or change, not owner review of a noncompliant submittal.

A submittal is reviewed against every specified requirement; if any required property is not met, it is returned for revision, however well it meets the others.

Answer: D

An addendum to the contract documents: Addenda modify the bidding documents before bid opening; this contract has already been awarded.

A change order signed by the owner and contractor: A change order records agreement on the change in price and time; that agreement does not exist yet, so a change order cannot be executed now.

A request for information from the contractor: An RFI asks for clarification of the documents; it does not authorize added work or adjust the contract.

A construction change directive: Correct. A construction change directive, issued by the owner through the landscape architect, directs the contractor to proceed with changed work before price and time are agreed; the adjustment is settled later and recorded by change order.

Use a change order when owner and contractor agree on price and time; use a construction change directive when the owner needs the change to proceed before that agreement.

Answer: B

West row, south end: Closest to the stepped main entrance, which is not accessible; the route to the accessible door is longer.

West row, north end: Correct. The access aisle opens straight onto the walk beside the accessible entrance: the shortest accessible route, with no drive aisle to cross.

East row, north end: Opposite the entrance, but the route crosses the vehicular drive aisle.

East row, south end: Farthest from the accessible entrance, and the route crosses the drive aisle.

Accessible parking goes on the shortest accessible route to an accessible entrance, with the access aisle adjoining that route and, where possible, no crossing of vehicular traffic.

Answer: C

1. Given: clear span 12'-0", joists at 16 in [400 mm] o.c.
2. Use only the 16 in o.c. column of Table 1.
3. Read down: 2x6 = 8'-6" (too short); 2x8 = 11'-0" (too short); 2x10 = 14'-0" (OK).
4. Smallest member whose maximum span ≥ 12'-0": 2x10.

2x6 [50x150]: At 16 in o.c. a 2x6 spans only 8'-6", well short of 12'-0".

2x8 [50x200]: At 16 in o.c. a 2x8 spans 11'-0", which is short of 12'-0". The 12'-6" value belongs to the 12 in o.c. column.

2x10 [50x250]: Correct. In the 16 in o.c. column a 2x10 spans 14'-0", the first value that equals or exceeds 12'-0".

2x12 [50x300]: A 2x12 (16'-6") works but is larger than needed; the question asks for the smallest member.

For a span table, pick the spacing column first, then move down to the first member whose maximum span equals or exceeds the required span.

Answer: (1) 94.00, (2) 93.90, (3) 92.70

1. Fall = slope × length.
2. Pipe 1 fall = 0.010 × 120 = 1.20; MH-1 invert in = 95.20 − 1.20 = 94.00.
3. MH-1 invert out = 94.00 − 0.10 = 93.90.
4. Pipe 2 fall = 0.015 × 80 = 1.20; outfall invert = 93.90 − 1.20 = 92.70.

Downstream invert = upstream invert − (slope × length); the outlet invert of a structure is set below its inlet invert by the stated drop.

Answer: B

Upper-left plan: The V's point toward higher contours, so it is a swale, but elevations rise toward the inlet; it drains north, away from the inlet.

Upper-right plan: Correct. The V's point uphill (north, toward the higher numbers), forming a swale, and elevations fall to 100 at the inlet.

Lower-left plan: The V's point downhill toward lower numbers, which forms a ridge that sheds water away from the centerline and the inlet.

Lower-right plan: The 101 contour crosses the 100 contour; contours of different elevations can never cross or touch except at a vertical face.

Contour V's that point uphill (toward higher numbers) mark a swale; V's pointing downhill mark a ridge. Check the numbering to confirm the flow direction.

Answer: C

1. Givens: C = 0.60, A = 5.0 ac, Tc = 15 min, 10-year storm.
2. Read i where duration = Tc = 15 min in the 10-year column: i = 5.0 in/hr.
3. q = CiA = 0.60 × 5.0 in/hr × 5.0 ac.
4. q = 15.0 cfs.

7.2 cfs: Uses the 60-minute intensity (2.4 in/hr); the storm duration must equal the 15-minute time of concentration.

11.4 cfs: Uses the 2-year column (3.8 in/hr); the design storm is the 10-year event.

15.0 cfs: Correct. i for a 15-minute, 10-year storm is 5.0 in/hr; q = 0.60 × 5.0 × 5.0 = 15.0 cfs.

18.0 cfs: Uses the 25-year column (6.0 in/hr), which over-designs the inlet for the stated 10-year storm.

In the Rational Method, read the intensity for the design return period at a duration equal to the time of concentration, then multiply by C and A (acres) to get cfs.