Estimating, Cost and Valuation MCQ Quiz in తెలుగు - Objective Question with Answer for Estimating, Cost and Valuation - ముఫ్త్ [PDF] డౌన్లోడ్ కరెన్
Last updated on Mar 8, 2025
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Estimating, Cost and Valuation Question 1:
Find the book value of the machine at the end of 6 years if a machine is purchased for Rs. 15,00000 and has estimated life of 15 years. Given salvage value (Rs) at the end of 15 years is 2,25,000. Use straight line method of evaluation of depreciation.
Answer (Detailed Solution Below)
Estimating, Cost and Valuation Question 1 Detailed Solution
Concept:
1. Straight line method: This assumes that the loss in the value of the property is same every year and at the end of its useful life it is equal to its scrap value.
\({\rm{Annual\;Depreciation}} = \frac{{{\rm{Purchasing\;cost}} - {\rm{salvage\;value}}}}{{{\rm{life\;of\;machine}}}}\)
2. Constant percentage method: This assumes that thew property loses it value by a constant percentage of its value at the begining of each year. (n = life of machine)
\(Annual{\rm{ }}\;Depreciation = 1 - {\left( {\frac{{Scrap{\rm{ }}value}}{{original{\rm{ }}cost}}} \right)^{\frac{1}{n}}}\)
Calculation:
Annual Depreciation is given by:
\(D = \;\frac{{C - V}}{n}\)
C = Cost Price of Machine = Rs. 15,00000
V = Salvage value = Rs. 2,25,000
n = estimated life of machine in years = 15
\(D = \frac{{1500000 - 225000}}{{15}} = 85000\)
Book value (B) is given by:
B = C - nD
n = No. of years after which Book value is to be computed
B = 1500000 - 6 × 85000
B = 990000Estimating, Cost and Valuation Question 2:
The expected out turn (square metre) of sawing of the soft wood per mason per day is _____.
Answer (Detailed Solution Below)
Estimating, Cost and Valuation Question 2 Detailed Solution
Explanation:
The quantity of work which can be done by an artisan for trade working of 8 hours is known as out - turn work.
The outturn for different types of work are as follows:
1. Sawing of softwood = 5.5 m2 per mason per day
2. Earthwork in the excavation in foundation trenches = 2.10 m3 per mason per day.
3. Cement concrete work = 5 m3 per mason per day.
4. Earthwork in the excavation in foundation = 2.75 m3 per mason per day.Estimating, Cost and Valuation Question 3:
Calculate the quantity of the earthworks in the cubic meter for a canal embankment of 100 m long having heights of 3 m and 5 m at the two extreme edges. Top widths of the embankment are 2 m and 4 m at the two extreme edges and the side slope is 2 : 1. Use the trapezoidal method.
Answer (Detailed Solution Below)
Estimating, Cost and Valuation Question 3 Detailed Solution
Concept:
The trapezoidal formula is given as,
\(V=\frac{L}{2}\left[ \left( {{A}_{1}}+{{A}_{n}} \right)+2\left( {{A}_{2}}+{{A}_{3}}+\ldots +{{A}_{n-1}} \right) \right]\)
Calculation:
The section across the embankment would be trapezoidal in shape.
For extreme edge (1): Top Width = 2 m, Heigth = 3, and
Bottom width = 2 + 2 × 3 + 2 × 3 = 14 m
\(⇒ A_1 = \frac{Top\;width+Bottom\;width}{2}× h=\frac{2+14}{2}× 3=24\;m^2\)
For extreme edge (2): Top Width = 4 m, Heigth = 5,
Bottom width = 4 + 2 × 5 + 2 × 5 = 24 m
\(⇒ A_2 = \frac{Top\;width+Bottom\;width}{2}× h=\frac{4+24}{2}× 5=70\;m^2\)
The volume of earthwork \(\Rightarrow V = \frac{L}{2}\left[ {\left( {{A_1} + {A_2}} \right)} \right]=\frac{100}{2}\left[ {\left( {{24} + {70}} \right)} \right]=4700\) m3.
Estimating, Cost and Valuation Question 4:
The expected out turn of the cement concrete 1 : 2 : 4 for per mason per day will be:
Answer (Detailed Solution Below)
Estimating, Cost and Valuation Question 4 Detailed Solution
The capacity of doing work by an artesian or skilled labour in the form of quantity of work per day is known as out turn of the labour.
|
Particulars of Item |
Expected out-turn (m3) |
|
Cement concrete (1 : 2 : 4) |
5.0 |
|
Lime concrete in foundation |
8.50 |
|
Reinforced Brick work |
1.00 |
|
R.C.C work |
3.00 |
|
Brickwork in lime or cement mortar in foundation and Plinth |
1.25 |
Estimating, Cost and Valuation Question 5:
Which of the following is the multiplying factor for the estimation of lead for sandy tracks?
Answer (Detailed Solution Below)
Estimating, Cost and Valuation Question 5 Detailed Solution
Explanation:
The distance between the source of availability of material and the construction site is known as Lead and is measured in km.
The cost of conveyance of material depends on lead.
The environment lead on the metalled roads are arrived by multiplying by a factor as follows:
a) For Sandy tracks – Lead × 1.4
b) For metal tracks – Lead × 1.0
c) For cartze tracks – Lead × 1.1
Estimating, Cost and Valuation Question 6:
A construction equipment has an initial cost of Rs. 2,00,000 and salvage value of Rs. 50,000 at the end of an economic life of 5 years. The rate of straight-line depreciation and total depreciation will be
Answer (Detailed Solution Below)
Estimating, Cost and Valuation Question 6 Detailed Solution
Concept:
1. Straight line method: This assumes that the loss in the value of the property is same every year and at the end of its useful life it is equal to its scrap value.
\({\rm{Annual\;Depreciation}} = \frac{{{\rm{Purchasing\;cost}} - {\rm{salvage\;value}}}}{{{\rm{life\;of\;machine}}}}\)
2. Constant percentage method: This assumes that thew property loses it value by a constant percentage of its value at the begining of each year. (n = life of machine)
\(Annual{\rm{ }}\;Depreciation = 1 - {\left( {\frac{{Scrap{\rm{ }}value}}{{original{\rm{ }}cost}}} \right)^{\frac{1}{n}}}\)
Calculation:
Given,
Fixed cost (FC) or purchasing Cost = Rs. 2,00,000
Scrap Value (SV) = Rs. 50,000
Life of Machine, n = 5 Years
Total depreciation = FC - SV
= 2,00,000 - 50,000
= Rs. 1,50,000/-
Depreciation per year = \(\frac{{{\rm{Total\;Depreciation}}}}{{{\rm{Deign\;life}}}}\)
= \(\frac{{{\rm{\;}}1,50,000}}{5}\)
= Rs. 30,000
For Straight line method, depreciation method annual depreciation rate (K) is given by the following formula:
\({\rm{K}} = \frac{{{\rm{Depreciation\;per\;year}}}}{{{\rm{total\;depreciation}}}}\)
K = \(\frac{{30,000}}{{150000}}\) = 0.2
Hence, depreciation and total depreciation will be 0.2 and Rs. 1,50,000 respectively.
Estimating, Cost and Valuation Question 7:
What is the lift of soil material for the canal shown in the figure?
Answer (Detailed Solution Below)
Estimating, Cost and Valuation Question 7 Detailed Solution
Concept:
Lift of material:
- It is the vertical distance calculated from the ground level up to which the laborers have to excavate the soil and remove it for the standard rate.
- It is the average height through which the earth has to be lifted from source to the place of spreading or disposing.
Calculation:
The lift of soil material for the canal = \((1.50\ +\ 1.20) \over 2\) = 1.35 m
Estimating, Cost and Valuation Question 8:
Whenever colour washing on Asbestos Cement corrugated sheets is done, in the estimation the plane area of the sheets is increased by ______.
Answer (Detailed Solution Below)
Estimating, Cost and Valuation Question 8 Detailed Solution
Concept:
- When color washing on asbestos cement sheets is done then the plane area of the sheets increased by 20 %.
- After colour washing, area = Area + (20/100) × Area.
- Corrugated surfaces shall be measured flat as fixed and not girthed.
Quantities so measured shall be increased by the following percentages and the results shall be included in general areas:
- Corrugated steel sheets ⇒ 14%
- Corrugated asbestos cement sheets ⇒ 20%
- Semi-corrugated asbestos cement sheets ⇒ 10%
Estimating, Cost and Valuation Question 9:
The approximate proportion of dry cement mortar required for brick work with bricks of nominal size 19 cm × 9 cm × 9 cm, with 15% extra for frog filling and wastage is: (assume 1 m3 of wet mortar = 1.25 m3 of dry mortar)
Answer (Detailed Solution Below)
Estimating, Cost and Valuation Question 9 Detailed Solution
For 1 m3 volume of brick work, the bricks required are 500.
Mortar needed = 1 – 500 × 0.19 × 0.09 × 0.09 = 0.23 m3
Add 15% extra for frog filling and wastage
Volume of wet mortar = 0.23 × 1.15 = 0.265 m3
1 m3 of wet mortar = 1.25 m3 of dry mortar
∴ 0.265 m3 of wet mortar = 0.33 m3 of dry mortar
∴ Dry mortar as a percentage of brick work = 33%
Materials for different item work:
|
1 |
Bricks used for brickwork |
500 Nos. for 1 cu.m |
|
2 |
Dry mortar for brickwork |
33% of the volume of brickwork |
|
3 |
Dry mortar for rubble stone masonry |
42% of the volume of stone masonry |
|
4 |
Volume of dry concrete |
1.54 times the volume of wet concrete |
|
5 |
Volume of dry mortar |
1.25 times the volume of wet mortar |
|
6 |
Dry mortar for 12 mm plastering |
2 cu.m for 100 sq.m |
Estimating, Cost and Valuation Question 10:
The total area of floor in-between walls and consists of floor of all rooms, verandahs passages, corridors, stair case, entrance halls, kitchen, stores, bath and latrines is known as ________.
Answer (Detailed Solution Below)
Estimating, Cost and Valuation Question 10 Detailed Solution
Different types of area are as follows:
Floor area:
The floor area of the building is the total area of the floor in between walls and consists of the floor of all rooms verandas, corridors, etc. The floor area is equal to the plinth area minus area occupied by walls.
Following are not included in the floor area of the building:
1. Sills of doors and openings are not included in the floor area of the building.
2. Area occupied by walls, pillars, and any other intermediate support.
Plinth area:
It is the covered built-up area measured at the floor level of any storey or at the floor level of the basement.
The plinth area is also called a built-up area and is the entire area occupied by the building including internal and external walls. It is generally 10-20% more than the carpet area.
Carpet area:
It is a useful area or liveable area or lettable area.
It is the total floor area minus the circulation area, verandahs, corridors, passages, staircase, lifts, entrance hall, etc., and minus other non-useable areas as sanitary accommodations, air conditioning room, etc.
Circulation area:
1. Vertical circulation area of the building is the area or space occupied by staircases, lifts, and the entrance halls adjacent to them which are required for the vertical movement of the users of the building.
It may be taken as (4 to 5) % of the plinth area of the building.
2. Horizontal circulation area of a building in the area of the verandahs, passages, corridors, balconies, porches, etc. which is required for the horizontal movement of the users of the building.
It may be taken as (10 to 15)% of the plinth area of the building.