Please keep at least one language enabled. || కనీసం ఒక భాషను ఎంచుకోండి.
Question 1
ప్రశ్న 1
For a building with walls of uniform cross-section and a consistent structural layout, what is the central estimating principle of the centreline method?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The centreline method organizes quantities around the axes of wall runs. For a uniform strip, its cross-sectional area is multiplied by an effective centreline length. That effective length must account for any overlap at junctions. The breadth and height or depth correspond to the particular measured work layer. A centreline calculation does not replace accurate item descriptions and drawing checks.
వివరణ:
Question 2
ప్రశ్న 2
A rectangular building has EXTERNAL plan dimensions 12.40 m by 8.40 m and uniform external wall thickness 0.20 m. There are no internal walls. What is the closed-wall centreline perimeter?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The wall centreline lies halfway through the thickness of each opposite wall. The centreline length dimension is 12.40 - 0.20 = 12.20 m. The centreline width dimension is 8.40 - 0.20 = 8.20 m. Their closed perimeter is 2(12.20 + 8.20) = 40.80 m. The four ordinary rectangular corners need no additional T-junction deductions.
వివరణ:
Question 3
ప్రశ్న 3
A rectangular closed external wall circuit measures 10.30 m by 7.30 m OUTSIDE, with uniform thickness 0.30 m and height 3.00 m. There are no internal walls, openings or other deductions. What masonry volume follows from the centreline method?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The centreline dimensions are 10.30 - 0.30 = 10.00 m and 7.30 - 0.30 = 7.00 m. The closed centreline perimeter is 2(10.00 + 7.00) = 34.00 m. The uniform wall cross-sectional area is 0.30 x 3.00 = 0.90 m2. The wall volume is 34.00 x 0.90 = 30.60 m3. No additional adjustment is needed for the four standard corners of this uniform rectangular circuit.
వివరణ:
Question 4
ప్రశ్న 4
A uniform-thickness masonry wall follows a simple closed rectangular centreline without partitions or intersecting branches. What adjustment is normally required solely for its four right-angled L-corners?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The centreline of a closed rectangular wall circuit provides the correct area basis for uniform-width strips. At an ordinary right-angled corner, the apparent external excess is balanced by the internal shortage. Consequently, no independent deduction is generally needed for those L-corners. This differs from the treatment of branch walls meeting a continuous wall at T-junctions. The geometrical assumptions and uniform width still need to be satisfied.
వివరణ:
Question 5
ప్రశ్న 5
For a uniform-width work layer, the sum of all booked wall-centreline lengths is 30.60 m BEFORE corrections. Exactly one branch meets a continuous wall at a T-junction, and the work-layer breadth is 0.40 m throughout. What is the effective centreline length?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A T-junction can cause the branch measurement to overlap the continuous wall. Under the stated uniform-width convention, half the work-layer breadth is deducted per T-junction. Half of 0.40 m is 0.20 m. The corrected length is therefore 30.60 - 0.20 = 30.40 m. The resulting effective length is used with the cross-sectional dimensions of this same layer.
వివరణ:
Question 6
ప్రశ్న 6
A quantity take-off lists 42.00 m as the uncorrected sum of centreline lengths for a uniform 0.50 m wide masonry layer. There are exactly TWO T-junctions and no other intersections. What is the effective centreline length?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Each T-junction requires a correction equal to half the layer breadth under the stated convention. Half of 0.50 m is 0.25 m. Two junctions therefore require a combined deduction of 0.50 m. The effective length is 42.00 - 0.50 = 41.50 m. This deduction prevents overlapping material from being included twice.
వివరణ:
Question 7
ప్రశ్న 7
The uncorrected sum of centrelines for two uniform-width strips crossing at ONE full plus-shaped intersection is 38.00 m. Both strips are 0.30 m wide and there are no other junctions. What effective centreline length represents their combined plan area?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A full crossing differs from a T-junction because both strips continue through the intersection. Their overlap is a square of side equal to the strip breadth. Dividing that overlapping area by the common breadth gives a length correction of 0.30 m. Therefore the effective centreline length is 38.00 - 0.30 = 37.70 m. The deduction ensures the common intersection is counted only once.
వివరణ:
Question 8
ప్రశ్న 8
A 0.12 m thick internal wall meets a continuous 0.30 m thick main wall at a perpendicular T-junction. Its centreline is booked up to the MAIN wall centreline, and the main wall is measured continuously. Both walls are 3.00 m high. What overlap VOLUME must be excluded from the sum of their unreconciled volumes at that junction?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The internal wall extends within the measured main-wall region for half the main-wall thickness. That overlap length is 0.30/2 = 0.15 m. The overlapping plan area is 0.15 x 0.12 = 0.018 m2. At the common 3.00 m height, the duplicate volume is 0.018 x 3.00 = 0.054 m3. The correction depends on the actual intersection geometry rather than assuming both walls have the same thickness.
వివరణ:
Question 9
ప్రశ్న 9
An outer brickwork circuit is 0.30 m thick, while several partitions are 0.12 m thick and have different footing arrangements. Which take-off strategy is safest?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Walls with different thicknesses cannot always share one uniform cross-section. Their foundations may also have different widths and levels. Grouping all centreline lengths under one breadth can misstate the measured volume. A hybrid method retains the convenient closed-circuit calculation for the main walls while treating the partitions appropriately. Junction geometry should still be checked to avoid double counting.
వివరణ:
Question 10
ప్రశ్న 10
For a uniform-width trench layer, the uncorrected total centreline length is 31.20 m, and there are TWO T-junctions. The trench breadth is 0.90 m and depth is 0.30 m. What in-situ excavation volume follows from the corrected centreline length?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
At each T-junction, the effective-length correction is half the trench breadth. The total correction is 2 x (0.90/2) = 0.90 m. The effective centreline length becomes 31.20 - 0.90 = 30.30 m. The excavation volume is 30.30 x 0.90 x 0.30 = 8.181 m3. The trench breadth rather than a different masonry thickness controls this layer's junction correction.
వివరణ:
Question 11
ప్రశ్న 11
At a separate masonry layer on the same layout, the UNCORRECTED centreline sum is 31.20 m and there are TWO T-junctions. All wall runs have breadth 0.23 m and height 3.00 m. What is the masonry volume BEFORE ROUNDING?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The masonry breadth at the level being measured is 0.23 m. Two T-junctions require a combined effective-length deduction of 2 x (0.23/2) = 0.23 m. The corrected masonry centreline length is 31.20 - 0.23 = 30.97 m. Multiplying 30.97 x 0.23 x 3.00 gives 21.3693 m3. Using the excavation-breadth deduction for this narrower wall would produce an incorrect quantity.
వివరణ:
Question 12
ప్రశ్న 12
A wall and its continuous strip footing share the same geometrical axes, but the excavation breadth is 1.00 m, PCC footing breadth is 0.80 m, and superstructure wall thickness is 0.23 m. The plan contains several T-junctions. Which procedure is technically sound?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A symmetrical strip footing may share its centreline axis with the wall above it. However, the width of excavated and constructed strips changes from layer to layer. Their overlapping areas at T-junctions therefore change as well. The junction correction must correspond to the layer that is being measured. An invariant axis does not imply an invariant effective length after corrections.
వివరణ:
Question 13
ప్రశ్న 13
A uniform 0.30 m thick and 2.50 m high wall layout consists of an outer closed circuit with centreline perimeter 44.00 m and one internal cross wall booked for 6.00 m BETWEEN the two opposite outer-wall centrelines. The cross wall meets the circuit at TWO T-junctions. What is the net masonry volume, assuming no openings?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The booked uncorrected centreline sum is 44.00 + 6.00 = 50.00 m. The two T-junctions each require a half-breadth correction of 0.15 m. Their combined correction is 0.30 m, giving an effective length of 49.70 m. The wall cross-sectional area is 0.30 x 2.50 = 0.75 m2. The net masonry volume is therefore 49.70 x 0.75 = 37.275 m3.
వివరణ:
Question 14
ప్రశ్న 14
A uniform wall is supported on centred and symmetrically stepped strip footings. What happens to the geometrical wall-axis centreline through the successive footing levels, assuming the layout does not change?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A symmetrically widened footing extends an equal distance to either side of the supporting wall axis. Its geometric centreline therefore remains aligned with the wall centreline. The footing breadth can nevertheless differ substantially from the masonry breadth. Different breadths produce different quantities even over the same geometrical axis. Junction corrections must be calculated for each layer where intersections occur.
వివరణ:
Question 15
ప్రశ్న 15
A wall centreline is shown on a plan, but a proposed foundation widening is ECCENTRIC, extending 0.15 m beyond one wall face and 0.55 m beyond the other. What should an estimator do before applying a centreline formula to the foundation?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
An eccentric foundation is not centred on the wall above it. Unequal projections can shift the centre of the foundation footprint. Therefore the superstructure centreline may not represent the foundation strip axis. The estimator should use the actual foundation plan and cross-section to define the geometry. A blind multiplication using an unchanged wall centreline can misstate the quantity.
వివరణ:
Question 16
ప్రశ్న 16
A building has ONE closed rectangular wall circuit with an effective centreline perimeter of 40.00 m and NO internal junctions. Its continuous PCC strip is 0.70 m wide and 0.25 m deep at Rs 6,000/m3. The masonry above is 0.23 m thick and 3.00 m high at Rs 6,500/m3. No openings or other costs apply. What is the combined estimated amount?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The PCC quantity is 40.00 x 0.70 x 0.25 = 7.00 m3. Its cost is 7.00 x Rs 6,000 = Rs 42,000. The masonry volume is 40.00 x 0.23 x 3.00 = 27.60 m3. Its cost is 27.60 x Rs 6,500 = Rs 1,79,400. Combining both item amounts gives Rs 2,21,400 without any unstated allowance.
వివరణ:
Question 17
ప్రశ్న 17
A circular masonry wall has a CENTRELINE radius of 4.00 m, uniform thickness 0.25 m, and height 2.00 m. Assume no openings and use pi = 3.1416. What wall volume follows from its centreline circumference?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The centreline circumference of the circular wall is 2 times pi times radius. With a radius of 4.00 m, it is 2 x 3.1416 x 4.00 = 25.1328 m. The cross-sectional area of the wall strip is 0.25 x 2.00 = 0.50 m2. Their product gives 12.5664 m3. Rounding to two decimal places produces 12.57 m3.
వివరణ:
Question 18
ప్రశ్న 18
For uniformly 0.45 m wide strips, the sum of the uncorrected centreline runs is 50.00 m. The layout includes TWO independent T-junctions and ONE independent full plus-shaped crossing, with no overlaps among the junction regions. What effective centreline length should be used?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Two T-junctions require 2 x (0.45/2) = 0.45 m of length deduction. A full crossing requires a further deduction of one complete strip breadth, or 0.45 m. The junction regions are independent under the stated assumptions. Thus the total deduction is 0.90 m. The effective centreline length is 50.00 - 0.90 = 49.10 m.
వివరణ:
Question 19
ప్రశ్న 19
A uniform outer wall circuit has EXTERNAL dimensions 11.40 m by 8.40 m and wall thickness 0.30 m. Its symmetrically centred excavation strip is 1.00 m wide and 1.20 m deep, and no inner walls, overlaps or additions occur. What is the excavation volume by the centreline method?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The wall centreline dimensions are 11.40 - 0.30 = 11.10 m and 8.40 - 0.30 = 8.10 m. The closed centreline perimeter is 2(11.10 + 8.10) = 38.40 m. Symmetrical excavation keeps the strip centred on the same geometric axes. Multiplication by 1.00 m breadth and 1.20 m depth gives 46.08 m3. No T-junction corrections are needed because the circuit has no internal walls.
వివరణ:
Question 20
ప్రశ్న 20
A plan labels the external dimensions of a building but not the centre-to-centre dimensions of its external walls. The wall thicknesses and face alignments are available on coordinated drawings. What should the quantity surveyor do?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Centreline calculations must be based on the wall axes rather than unadjusted external boundaries. For a uniform rectangle, one wall thickness is removed from each external overall dimension to locate opposite centrelines. Irregular wall thicknesses or offsets require their own geometric interpretation. The resulting centreline plan should be checked against sections and junction details. This prevents systematic errors in all quantities derived from the wall axes.
వివరణ:
Question 21
ప్రశ్న 21
A 0.20 m thick partition wall meets a continuous 0.40 m thick main wall at ONE right-angled T-junction. The partition centreline is booked for 4.00 m up to the main-wall centreline; the main wall has already been measured in full. Both walls are 3.00 m high. What ADDITIONAL NET partition masonry volume is required?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The partition overlaps the already measured main wall for half the main-wall thickness. That overlap length is 0.40/2 = 0.20 m. Its net additional length is 4.00 - 0.20 = 3.80 m. Multiplying by partition thickness and height gives 3.80 x 0.20 x 3.00 = 2.28 m3. The correction must use the main-wall intersection geometry and not half the narrower partition thickness.
వివరణ:
Question 22
ప్రశ్న 22
A 0.20 m thick internal wall spans between TWO parallel continuous walls, each 0.50 m thick. Its booked centreline length is 6.00 m from one MAIN-wall centreline to the other; both main walls have already been measured through the junctions. The internal wall is 2.50 m high and there are no other overlaps. What net additional internal-wall masonry volume is required?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The internal wall penetrates the measured width of each main wall by half of 0.50 m. The two overlap lengths together are 0.25 + 0.25 = 0.50 m. Hence the additional internal-wall length is 6.00 - 0.50 = 5.50 m. Its cross-sectional area is 0.20 x 2.50 = 0.50 m2. The additional masonry volume is 5.50 x 0.50 = 2.75 m3.
వివరణ:
Question 23
ప్రశ్న 23
At a T-junction, an internal foundation strip joins a continuous external strip. At masonry level, the continuous external wall is 0.30 m thick; at the relevant footing level, the continuous external strip is 0.90 m wide. The joining strip is measured to the continuous strip centreline in both cases. Which branch-length corrections are geometrically appropriate?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Where a branch ends at the centreline of a previously measured continuous strip, it overlaps half the continuous strip breadth. At masonry level, half of 0.30 m is 0.15 m. At the specified footing level, half of 0.90 m is 0.45 m. The branch-width correction therefore differs between the two layers. Both calculations rely on the stated alignment and measurement convention for the continuous strip.
వివరణ:
Question 24
ప్రశ్న 24
For a rectangular building of constant wall thickness, both the centreline method and the long-wall-short-wall method are correctly applied to the same measured item without openings. How should their net quantities compare?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Both methods aim to measure the same physical volume of construction. The centreline method uses effective axis lengths and the relevant cross-sectional area. The long-wall-short-wall method uses appropriate out-to-out and in-to-in lengths. If all boundaries and overlaps are handled consistently, they describe the same occupied space. Differences between the results usually indicate inconsistent assumptions or a take-off error.
వివరణ:
Question 25
ప్రశ్న 25
An irregular pump-house combines 300 mm external walls, 115 mm partitions, a stepped foundation with asymmetric offsets, and several T- and cross-junctions. Which estimating decision is most defensible?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Different wall thicknesses and foundation offsets prevent the entire building from being represented by one uniform strip. Several junctions introduce opportunities for double counting. Geometrically consistent groups can still be measured efficiently with centreline calculations. Other sections may require individual wall measurements or a hybrid procedure. A final reconciliation against the drawings and measurement sheets ensures that each volume is accounted for only once.
వివరణ:
Answer Key సమాధానాల పట్టిక
-
Question 1 ప్రశ్న 1Answer: B. Determine the effective total wall-centreline length for each work layer and multiply it by that layer's breadth and height or depth. సమాధానం: B.
-
Question 2 ప్రశ్న 2Answer: D. 40.80 m. సమాధానం: D.
-
Question 3 ప్రశ్న 3Answer: A. 30.60 m3. సమాధానం: A.
-
Question 4 ప్రశ్న 4Answer: C. No separate corner deduction is normally needed when the closed centreline circuit is used correctly. సమాధానం: C.
-
Question 5 ప్రశ్న 5Answer: B. 30.40 m. సమాధానం: B.
-
Question 6 ప్రశ్న 6Answer: D. 41.50 m. సమాధానం: D.
-
Question 7 ప్రశ్న 7Answer: A. 37.70 m. సమాధానం: A.
-
Question 8 ప్రశ్న 8Answer: C. 0.054 m3. సమాధానం: C.
-
Question 9 ప్రశ్న 9Answer: A. Measure the outer circuit using its centreline and handle the partitions separately with appropriate junction corrections or a hybrid long-wall-short-wall calculation. సమాధానం: A.
-
Question 10 ప్రశ్న 10Answer: B. 8.181 m3. సమాధానం: B.
-
Question 11 ప్రశ్న 11Answer: C. 21.3693 m3. సమాధానం: C.
-
Question 12 ప్రశ్న 12Answer: D. Use the proper breadth and corresponding overlap adjustment for each layer while keeping common axes where the sections remain symmetrically aligned. సమాధానం: D.
-
Question 13 ప్రశ్న 13Answer: A. 37.275 m3. సమాధానం: A.
-
Question 14 ప్రశ్న 14Answer: C. It remains in the same geometrical position, although breadths and junction corrections for quantities may change by layer. సమాధానం: C.
-
Question 15 ప్రశ్న 15Answer: D. Determine the actual foundation strip geometry and shifted axis, then measure that layer with appropriate corner and junction treatment. సమాధానం: D.
-
Question 16 ప్రశ్న 16Answer: B. Rs 2,21,400. సమాధానం: B.
-
Question 17 ప్రశ్న 17Answer: C. 12.57 m3. సమాధానం: C.
-
Question 18 ప్రశ్న 18Answer: D. 49.10 m. సమాధానం: D.
-
Question 19 ప్రశ్న 19Answer: A. 46.08 m3. సమాధానం: A.
-
Question 20 ప్రశ్న 20Answer: B. Derive the actual wall-centreline runs from the stated dimensions and thicknesses, checking whether the plan includes offsets or thickness changes. సమాధానం: B.
-
Question 21 ప్రశ్న 21Answer: C. 2.28 m3. సమాధానం: C.
-
Question 22 ప్రశ్న 22Answer: D. 2.75 m3. సమాధానం: D.
-
Question 23 ప్రశ్న 23Answer: A. 0.15 m at masonry level and 0.45 m at footing level. సమాధానం: A.
-
Question 24 ప్రశ్న 24Answer: B. They should agree when geometric boundaries and junction corrections are consistently handled. సమాధానం: B.
-
Question 25 ప్రశ్న 25Answer: C. Divide the work into geometrically consistent sections, combine centreline and wall-by-wall methods where useful, and reconcile intersections for each measured layer. సమాధానం: C.