CONSTRUCTION | Estimates | 10 Questions (MODERATE-CHALLENGING)

CONSTRUCTION ESTIMATES

(10 Questions, Difficulty Level: MODERATE to CHALLENGING)

by Raison John J. Bassig

Q#1: You are to install one (1) layer of 300mm x 300mm ceramic tiles as accent around the topmost interior wall finish of a circular pool. If the area of the pool is 98.12sqm, how many accent tiles will you need?

a. 117 pcs
b. 105 pcs
c. 149 pcs
d. 94 pcs

We first need to determine the circumference (or "length" of the circle), recall that,

Circumference (C) = 2 π r

where: r is the radius of the circle

However, in my given problem, the measurement of the radius (r) of the pool was not given. As there are two unknowns, (C) and (r), we must need another equation. Recall that,

Area (A) of a circle = π r²

From such area equation,

r = √ (A / π)

So, substituting (r) in the circumference equation using the (r) from the area equation, we now have,

C = 2 π [√ (A / π)]

Now we can solve the circumference of the pool in the given problem,

C = 2 x π x √ ([98.12sqm] / π)
C = 35.1m

Determining the tile quantity around the circular wall in 1 layer,

Number of Tiles in 1 Layer = [Circumference] / [Tile Dimension]
Number of Tiles in 1 Layer = [35.1m] / [0.3m]
Number of Tiles in 1 Layer = 117 pcs

Therefore, the correct answer is a. 117 pcs.

Q#2: You need 2 pints of tinting color mixed with 6-1/2 gallons of paint to achieve the desired hue. If it takes 23 gallons of paint to cover a 360-sqm surface for the 1st coat, and it takes 15% less paint for the 2nd coat of the same surface, how many pints of tinting color must you order to paint with the desired hue a total surface area of 185 square meters?
a. 7 pints
b. 8 pints
c. 9 pints
d. 10 pints

Since painting rate for the FIRST COAT is [ 23 gallons : 360 sqm ] and the amount of first coat paint (x) covers 185 sqm, i.e., [ x : 185 sqm ], then, by proportion,

[ 23 gallons / 360 sqm ] = [ x / 185 sqm ]
x = 185 sqm x [ 23 gallons / 360 sqm ]
x = 11.82 gallons (the amount of paint for FIRST COAT)

Since painting rate for the SECOND COAT is 15% less paint than the first coat on the same surface area, then, the amount of second coat paint (y) would be,

[ (23 gallons - 15% of 23 gallons) / 360 sqm ] = [ y / 185 sqm ]
y = 185 sqm x [ (23 gallons - 0.15 x 23 gallons) / 360 sqm ]
y = 185 sqm x [ 19.55 gallons / 360 sqm ]
y = 10.05 gallons (the amount of paint for SECOND COAT)

So, the total amount of paint is,

Total Required Paint = [ x ] + [ y ]
Total Required Paint = [11.82 gallons] + [10.05 gallons]
Total Required Paint = 21.87 gallons

For the TINTING COLOR, since the tinting rate is [ 2 pints of tint : 6.5 gallons of paint ] and the total required number of paint was computed at [21.87 gallons], then,

Total Required Tinting Color = 21.87 gallons x [ 2 pints / 6.5 gallons ]
Total Required Tinting Color = 6.73 pints

Since you cannot normally order "partial cans" of painting materials, we round 6.73 pints of tinting color to 7 pints. Likewise, for the required paint, we also round 21.87 gallons to 22 gallons.

Therefore, the correct answer is a. 7 pints (i.e., order 7 pints of tinting color and 22 gallons of paint).

Q#3: Your client required you as his architect to design a triangular-shaped personal gaming room. In your plan, the exact lengths of the walls measured inside the room came out to be 5.22m, 4.35m, and 8.75m. If you are to specify 40cm x 40cm ceramic tiles for the floor finish, how many pieces would be required considering 5% wastage?

a. 19 pcs
b. 56 pcs
c. 75 pcs
d. 91 pcs

To determine the Area (A) of a scalene triangle (a triangle with no equal sides), recall Heron's formula,

A = √ [ p (p - a) (p - b) (p - c) ]

where:
p is the semi-perimeter (i.e., half the perimeter)
a, b, and c are the lengths of the three sides, respectively

Let's first solve for the semi-perimeter (p),

p = (a + b + c) / 2
p = (5.22m + 4.35m + 8.75m) / 2
p = 9.16m

So, solving for the area (A) of the triangular gaming room,

A = √ [ 9.16m x (9.16m - 5.22m) x (9.16m - 4.35m) x (9.16m - 8.75m) ]
A = 8.435sqm

Now we can determine the quantity of the floor tile finish,

Number of Tiles = [Area of Room] / [Area of 1 Tile]
Number of Tiles = [8.435sqm] / [0.4m x 0.4m]
Number of Tiles = [8.435sqm] / [0.16m]
Number of Tiles = 52.728 pcs or 53 pcs

Adding 5% wastage as required in the question,

53 pcs x 1.05 = 55.65 pcs or 56 pcs
(Note: Using "52.728 pcs x 1.05" will yield 55.3644 pcs or 56 pcs as well)

Therefore, the correct answer is b. 56 pcs.

Q#4: Painter C can finish the job in 28 days with the help of Painter A. Painter B can finish the job in 34 days with the help of Painter C. Painter A can finish the job in 56 days with the help of Painter B. If you hire Painter C alone, how long will he finish the painting job?
a. Exactly 56 days
b. A little less than 30 days
c. About 42 days
d. Approximately 2 months (63 days)

Let:

1 = The Painting Job
1/t = Total Days to Finish The Painting Job
1/A = Work Rate of Painter A
1/B = Work Rate of Painter B
1/C = Work Rate of Painter C

Since we have three (3) unknowns, we need three (3) equations to solve the variables. So, from the given problem, we have,

Equation 1: [1/C] + [1/A] = 1/28 days
Equation 2: [1/B] + [1/C] = 1/34 days
Equation 3: [1/A] + [1/B] = 1/56 days

Now, let's first eliminate variable A (by considering Equations 1 and 3):

1/C + 1/A = 1/28 ==> From Eq. 1
1/A + 1/B = 1/56 ==> From Eq. 3

Since our objective is to eliminate the variable A, we can multiply both sides of either one of the above equations by a value of (-1), so that the value (1/A) will become - (1/A), and thus, can be eliminated when the that equation is combined (added to) the other equation. So, considering Eq. 3,

1/A + 1/B = 1/56
[ 1/A + 1/B = 1/56 ] x (-1)

will give us,

- (1/A) + - (1/B) = - (1/56)

Combining (adding) both Eq. 1 and Eq. 3, we will now get,

[ 1/C + 1/A ] + [ - (1/A) + - (1/B) ] = [ 1/28 ] + [ - (1/56) ]

Simplifying,

1/C + 1/A - 1/A - 1/B = 1/28 - 1/56
1/C - 1/B = (56 - 28) / (28)(56)
1/C - 1/B = 28/1568
1/C - 1/B = 1/56 ==> Let's now call this Equation 4

Next, we can now eliminate variable B (using Equations 4 and 2) to solve for C:

1/C - 1/B = 1/56 ==> From Eq. 4
1/B + 1/C = 1/34 ==> From Eq. 2

Combining (adding) both Eq. 4 and Eq. 2,

[ 1/C - 1/B ] + [ 1/B + 1/C ] = [ 1/56 ] + [ 1/34 ]
1/C - 1/B + 1/B + 1/C = 1/56 + 1/34
1/C + 1/C = (34 + 56) / (56)(34)
2/C = 90/1904
2/C = 45/952
C = 2 (952/45)
C = 1904/45
C = 42.31 days

So, Painter C will be able to finish the painting job in 42.31 days if he is working alone.

Therefore, the correct answer is c. About 42 days.

Q#5: In the roof plan of the "dos-aguas" roofing you designed, the horizontal roofing measurements indicate a total horizontal projected length of 36.5m (from front to rear) and a width of 25.83m (from side to side).

In the roof section/elevation, the slope of the front roof is labeled 28° while the rear roof is labeled 18°.

A spot detail section of the roofing sheet shows a 900mm-wide nominal-width with an effective coverage of 70% due to lapping.

Based on the drawings, what would be the estimated cost for the longspan roofing sheets if the price is Php 323.75 per meter?

a. Approx. Php 484,492.00
b. Approx. Php 524,313.00
c. Approx. Php 208,100.00
d. Approx. Php 610,050.00

Since the roof with two different slopes forms a triangle with two (2) given angles (Angle A = 18° and Angle B = 28°) and one (1) given side (side c = 36.5m) between the angles, the shape forms an "Angle-Side-Angle" ("ASA") problem where we can simply use the Sine Law to find the lengths of the two unknown sloped sides (side a and side b).

Recall that Sine Law is:

[ a / Sin(Angle A) ] = [ b / Sin(Angle B) ] = [ c / Sin(Angle C) ]

where:
Angle A is the interior angle opposite side a
Angle B is the interior angle opposite side b
Angle C is the interior angle opposite side c

First, let's determine the remaining unknown angle (Angle C) of the triangle,

Sum of all Interior Angles in a Triangle = 180°
Angle A + Angle B + Angle C = 180°
Angle C = 180° - Angle B - Angle A
Angle C = 180° - [28°] - [18°]
Angle C = 134°

Next, we'll use Sine Law to determine the unknown length of the front roof (side a):

[ a / Sin(Angle A) ] = [ c / Sin(Angle C) ]
a = Sin(Angle A) x [ c / Sin(Angle C) ]
a = Sin[18°] x [36.5m] / Sin[134°]
a = 15.68m

We'll also use the same principle in determining the remaining unknown length of the rear roof (side b):

[ b / Sin(Angle B) ] = [ c / Sin(Angle C) ]
b = Sin(Angle B) x [ c / Sin(Angle C) ]
b = Sin[28°] x [36.5m] / Sin[134°]
b = 23.82m

All unknowns are now solved. So, we'll now look at the actual roof sheets, where the question states that the 900mm-wide sheet can only effectively cover 70% of its width due to lapping:

Effective Coverage Width of Roof Sheet = 900mm x 0.70
Effective Coverage Width of Roof Sheet = 630mm wide

We can now determine how many of such 900mm-wide roof sheets will cover the entire roof width of [25.83m]. Note that we must use the roof width as shown in the plan (NOT the horizontal projected roof length in the plan) as the roofing sheets are to be laid out with its corrugations parallel to the slope of the roof (for drainage). So,

# of Roof Sheets = [Roof Width] / [Effective Width of 1 Roof Sheet]
# of Roof Sheets = [25.83m] / [0.63m]
# of Roof Sheets = 41 pcs

This means that 41 roof sheets, with an effective width of 630mm (due to lapping), will cover the entire roof width of [25.83m]. Now let's determine the length of each roof sheet by simply getting the total length of the sloped sides (side a and side b) where the longspan roof sheets will be laid,

Total Length of Sloped Sides = [side a] + [side b]
Total Length of Sloped Sides = [15.68m] + [23.82m]
Total Length of Sloped Sides = 39.5m ==> This is the length of each longspan roof sheet

So,

Total Required Length of All Roof Sheets = [Length of Each Roof Sheet] x [# of Roof Sheets]
Total Required Length of All Roof Sheets = [39.5] x [41 pcs]
Total Required Length of All Roof Sheets = 1,619.5m

Since the cost for each longspan roof sheet was given at [Php 323.75 per meter], then,

Total Estimated Cost = [1,619.5m] x [Php 323.75/m]
Total Estimated Cost = Php 524,313.125

Therefore, the correct answer is b. Approx. Php 524,313.00.

Q#6: You are to install an approximately 2.4-m high temporary fence using standard-sized plywood around a triangular corner lot, situated along perpendicular roads, with an area of 429.87sqm. One of the frontage of the lot is measured at 19.32m. How many pieces of plywood would the job require?
a. Approx. 46 pcs
b. Approx. 78 pcs
c. Approx. 131 pcs
d. Approx. 93 pcs

As the question states that the triangular lot is situated along PERPENDICULAR roads (i.e., 2 roads at 90° from each other), then, the triangle formed is a RIGHT triangle.

Since the length of only one (1) side, a road-facing side, was given at [19.32m] (the other two being unknown), and the Area of the triangle was also given as [432.18sqm], then, we can solve the length of the side facing the other road (perpendicular to the given side), using the formula for the Area (A) of a Right Triangle,

A = (1/2) x a x b

where: a and b are the perpendicular sides

[432.18sqm] = (1/2) x [19.32m] x b
b = 2 x [432.18sqm] / [19.32m]
b = 44.50m

Solving the remaining side (the hypotenuse, side c, of the right triangle lot not facing the road), this time using Pythagorean Theorem,

c² = a² + b²
c = √(a² + b²)
c = √([19.32m]² + [44.5m]²)
c = √(2353.5124m²)
c = 48.51m

Now that we know the dimensions of all three (3) sides of the property, we can then determine its perimeter,

Lot Perimeter = [ a ] + [ b ] + [ c ]
Lot Perimeter = [19.32m] + [44.50m] + [48.51m]
Lot Perimeter = 112.33m

Since a standard plywood is 4' wide x 8' long, and the question states that the height of the temporary fence is approximately 2.40m high, then the plywood would be laid out in a vertical orientation (as the 8' length is approx. 2.40m, the same requirement for the fence height). We can now compute the total plywood quantity needed to cover the whole lot perimeter by considering only the width of the plywood (4' or approx. 1.2m),

# of 4' x 8' Plywood Required = [Lot Perimeter] / [Width of 1 Plywood]
# of 4' x 8' Plywood Required = [112.33m] / [~1.2m]
# of 4' x 8' Plywood Required = ~93.6 pcs

Therefore, the correct answer is d. Approx. 93 pcs.

Q#7: You hired 4 workers (A, B, C, and D) to excavate the soil where 54 isolated footings is to be located. Worker A's rate is P422/day and it will take him 6 days to excavate 17 footings. Worker B's rate is P385/day and he can excavate 43 footings in a span of 31 days. Worker C's rate is P395/day and would need 18 days for him to work on 24 footing excavations. Worker D's rate is P418/day wherein he can perform 9 footing excavations within 4 days.

As the one who hired the workers, how much are you expected to pay them in total to work alongside one another on the excavation of the 54 isolated footings of your project?
a. P11,340.00
b. P23,895.00
c. P21,870.00
d. P23,249.00
e. P8,415.00

Let's first determine the work rate of each worker and how long the entire excavation work is going to be, based on the data given in the problem:

Work Rate of A = 17 footings : 6 days = 17 / 6
Work Rate of B = 43 footings : 31 days = 43 / 31
Work Rate of C = 24 footings : 18 days = 24 / 18
Work Rate of D = 9 footings : 4 days = 9 / 4
Total Work = 54 footings : t = 54 / t
t = total time for total work = unknown

Since workers A, B, C, and D will work alongside one another to excavate the entire work, then,

Summation of Work Rate of All Workers = Total Work
[ Work Rate of A ] + [ Work Rate of B ] + [ Work Rate of C ] + [ Work Rate of D ] = Total Work
[ 17 / 6 ] + [ 43 / 31 ] + [ 24 / 18 ] + [ 9 / 4 ] = 54 / t
(104,508 / 13,392) = 54 / t
t = (54) x (13,392 / 104,508)
t = 6.92 days or ~7 days

Now that we have determined the total number of day (t) it will take for the excavation work to be completed by all workers, let's compute the costs/payment to each worker,

Payment Rate of A = P422/day
Payment Rate of B = P385/day
Payment Rate of C = P395/day
Payment Rate of D = P418/day

Cost to Pay All Workers per day = Summation of Payment Rates
Cost to Pay All Workers per day = P1,620/day

Total Payment for Entire Work = [Total # of Days] x [Cost to Pay All Workers per day]
Total Payment for Entire Work = [7 days] x [P1,620/day]
Total Payment for Entire Work = P11,340

Therefore, the correct answer is a. P11,340.00.

Q#8: You designed a circular room having a circumference of 31.42m. Inside that room is a triangular platform, fixed to the floor, with the 3 sides measuring 2.3m, 3.29m, and 2.91m, respectively. Calculate the number of vinyl strips required for the flooring of the room excluding the triangular platform if you specified to use 1.5mm thk x 153mm x 1219mm vinyl strips.

a. 410 pcs
b. 404 pcs
c. 422 pcs
d. 363 pcs

We first need to determine the area of the circular room, recall that,

Area (A) of a circle = π r²

where: r is radius of the circle

However, in my given problem, the measurement of the radius (r) of the room was not given. As there are two unknowns, (A) and (r), we must need another equation. Recall that,

Circumference (C) = 2 π r

From such circumference equation,

r = C / (2 π);

So, substituting (r) in the area equation using the (r) from the circumference equation, we now have,

A = π [ C / (2 π) ]²

Now we can solve the area of the circular room in the given problem,

Gross Area of Circular Room = π x [ C / (2 π) ]²
Gross Area of Circular Room = π x [ [31.42m] / (2 π) ]²
Gross Area of Circular Room = 78.56sqm

Since there is a triangular platform of unequal sides within that gross area of the circular room, which will not be provided with the same floor finish, we need to subtract the area of such platform from the gross area of the room.

To determine the Area (A) of a scalene triangle (a triangle with no equal sides), recall Heron's formula,

A = √ [ p (p - a) (p - b) (p - c) ]

where:
p is the semi-perimeter (i.e., half the perimeter)
a, b, and c are the lengths of the three sides, respectively

Let's first solve for the semi-perimeter (p),

p = (a + b + c) / 2
p = (2.3m + 3.29m + 2.91m) / 2
p = 4.25m

So, solving for the area (A) of the triangular platform,

Area of Platform = √ [ 4.25m x (4.25m - 2.3m) x (4.25m - 3.29m) x (4.25m - 2.91m)]
Area of Platform = √ [10.66104m⁴]
Area of Platform = 3.265sqm

Now that we know the area of both the circular room and the triangular platform, we determine the Net Area of the room (the area only to be provided with the given floor finish),

Net Area of Vinyl Flooring = [Gross Area of Room] - [Area of Platform]
Net Area of Vinyl Flooring = [78.56sqm] - [3.265sqm]
Net Area of Vinyl Flooring = 75.295sqm

Since the material specification for the vinyl flooring is given as [1.5mm thk x 153mm x 1219mm vinyl strips], then,

Area of 1 Vinyl Strip = [0.153m] x [1.219m]
Area of 1 Vinyl Strip = 0.1865sqm

Solving for the quantity of vinyl strips,

# of Vinyl Strips = [Net Area of Vinyl Flooring] / [Area of 1 Vinyl Strip]
# of Vinyl Strips = [75.295sqm] / [0.1865sqm]
# of Vinyl Strips = 403.72 pcs or 404 pcs

Therefore, the correct answer is b. 404 pcs.

Q#9: You are preparing the specifications and costing of the floor finish for your triangularly-shaped lobby. The three sides measure 17.5m, 23.3m, and 18.8m, respectively. Within the lobby is a circular interior landscaped area, which will only have groundcover, having a circumference of 32.2m. The owner wants you to decide and select the most affordable flooring material among his shortlist of finishes:

OPTION	MATERIAL	PRICE	NOTES
A	6mm x 600mm x 600mm Vinyl Tiles	P 2,320.50/box	Each box has 6 pcs. Can order by the piece.
B	12mm x 178mm x 1229mm Wood Planks	P 1,169.00/box	Each box contains 5 pcs. Ordering by box only.
C	30cm x 30cm Granite Tiles	P 97.12/pc	-
D	Natural Cut Stones (Assorted Sizes)	P 102.00/bag	1 bag covers covers an area of 1 square foot.

Which of the 4 materials should you choose as the least expensive in terms of material costs for the floor finish of the lobby?
a. Vinyl Tiles
b. Wood Planks
c. Granite Tiles
d. Natural Cut Stones

Since the lobby is triangularly-shaped, with three (3) different measurements of its sides, then it is a Scalene Triange. We can solve for the floor area of such shape using Heron's formula,

A = √ [ p (p - a) (p - b) (p - c) ]

where:
p is the semi-perimeter (i.e., half the perimeter)
a, b, and c are the lengths of the three sides, respectively

Let's first solve for the semi-perimeter (p),

p = (a + b + c) / 2
p = (17.5m + 23.3m + 18.8m) / 2
p = 29.8m

So, solving for the Gross Area (A) of the triangular lobby,

Gross Area of Lobby = √ [ 29.8m x (29.8m - 17.5m) x (29.8m - 23.3m) x (29.8m - 18.8m)]
Gross Area of Lobby = √ [26207.61m⁴]
Gross Area of Lobby = approx. 161.9sqm

Since their is a circular landscaped area within the lobby area, which will NOT be provided with floor finishing, then, we will be subtractinig the area of such circular landscaping from the gross area of the lobby.

Let's now determine the area of the circular landscaped portion, recall that,

Area (A) of a circle = π r²

where: r is radius of the circle

However, in my given problem, the measurement of the radius (r) of the landscaped portion was not given. As there are two unknowns, (A) and (r), w must need another equation. Recall that,

Circumference (C) = 2 π r

From such circumference equation,

r = C / (2 π);

So, substituting (r) in the area equation using the (r) from the circumference equation, we now have,

A = π [ C / (2 π) ]²

Now we can solve the area of the circular landscape in the given problem,

Area of Circular Landscape = π x [ C / (2 π) ]²
Area of Circular Landscape = π x [ [32.2m] / (2 π) ]²
Area of Circular Landscape = approx. 82.5sqm

Thus, the Net Area of the Lobby (the area that would only be provided with flooring materials) would be,

Net Area of Lobby = [Gross Area of Triangular Lobby] - [Area of Circular Landscape]
Net Area of Lobby = [161.9sqm] - [82.5sqm]
Net Area of Lobby = 79.4sqm

At this point, we can now check the four (4) material options of the owner and decide which one is the most affordable:

OPTION A: VINYL TILES

Area of 1 Vinyl Tile = 0.6m x 0.6m = 0.36sqm
# of Vinyl Tiles = [Net Area of Lobby] / [Area of 1 Vinyl Tile]
# of Vinyl Tiles = [79.4sqm] / [0.36sqm]
# of Vinyl Tiles = 220.55 pcs or 221 pcs

Since the question states that ordering by piece is possible, and 1 box of Vinyl Tiles costs P2,320.50 with 6 pcs included in the box, then,

Price of 1 pc of Vinyl Tile = P2,320.50 / 6 pcs = P386.75/pc

So, the total material cost for Option A (Vinyl Tiles) would be,

Total Cost for OPTION A = P386.75/pc x 221 pcs = P85,471.75

OPTION B: WOOD PLANKS

Area of 1 Wood Plank = 0.178m x 1.229m = 0.218762sqm

Since the question states that, for Wood Planks, we can only order by the box, and 1 box has 5 planks inside, then,

Coverage of 1 box of Wood Planks = [Area of 1 Plank] x [# of Planks in 1 box]
Coverage of 1 box of Wood Planks = 0.218762sqm x 5 pcs
Coverage of 1 box of Wood Planks = 1.09381sqm

Determing the quantity of boxes to cover the net lobby area,

# of Box of Planks = [Net Area of Lobby] / [Coverage of 1 box of Planks]
# of Box of Planks = [79.4sqm] / [1.09381sqm]
# of Box of Planks = 72.59 boxes or 73 boxes

So, the total material cost for Option B (Wood Planks) would be,

Total Cost for OPTION B = P1,169/box x 73 boxes = P85,337.00

OPTION C: GRANITE TILES

Area of 1 Granite Tile = 0.3m x 0.3m = 0.09sqm
# of Granite Tiles = [Net Area of Lobby] / [Area of 1 Granite Tile]
# of Granite Tiles = [79.4sqm] / [0.09sqm]
# of Granite Tiles = 882.22 pcs or 883 pcs

Since the Granite Tiles are ordered by piece, then, the total material cost for Option C (Granite Tiles) would be,

Total Cost for OPTION C = P97.12/pc x 883 pcs = P85,756.96

OPTION D: NATURAL CUT STONES

Since the question states that ordering of Natural Cut Stones is by the bag, where 1 bag covers an area of 1 sq.ft. (which is equivalent to [0.0929 sq.m], then,

# of Bags of Stones = [Net Area of Lobby] / [Coverage of 1 bag of Stones]
# of Bags of Stones = [79.4sqm] / [0.0929sqm/bag]
# of Bags of Stones = 854.68 bags or 855 bags

So, the total material cost for Option D (Natural Cut Stones) would be,

Total Cost for OPTION D = P102/bag x 855 bags = P87,210.00

Summarizing all the computed costs above to compare all the floor finishing material options:

OPTION A: VINYL TILES = P85,471.75
OPTION B: WOOD PLANKS = P85,337.00 ==> Least Expensive
OPTION C: GRANITE TILES = P85,756.96
OPTION D: NATURAL CUT STONES = P87,210.00 ==> Most Expensive

Therefore, the correct answer is b. Wood Planks.

Q#10: You designed a typical 2.3m x 3.1m bathroom with a ceiling height of 2.45m. The window opening (including its frame) in one of the bathroom walls is 11.5% of the floor area of the bathroom. The door opening (including its jambs and header) in another wall measures 703mm wide and 2120mm high. A large mirror, measuring 19.16sq.ft. is also installed flushed in another wall.

There are six (6) typical bathrooms of the same design in your plans. Specifying wall tiles at 2mm tile spacing, how many bags of tile grout will you need to cover all the bathrooms if the grout manufacturer prescribes that the coverage of 1 bag is at 5.85sqm, based on such tile spacing?
a. About 23 bags
b. About 7 bags
c. About 27 bags
d. About 32 bags

Let's first determine the total wall areas of a typical bathroom. Note that the bathroom is rectangular in shape at [2.3m] wide x [3.1m] long/deep, and the height of all walls is given at [2.45m] high.

Area of Wall (Side 1) = [2.3m] x [2.45m] = 5.635sqm
Area of Wall (Side 2) = [3.1m] x [2.45m] = 7.595sqm
Area of Wall (Side 3) = [2.3m] x [2.45m] = 5.635sqm
Area of Wall (Side 4) = [3.1m] x [2.45m] = 7.595sqm
Gross Wall Area (All Sides) = 26.46 sqm. (for 1 Bathroom)

Since the problem states that there are certain features in the walls that will NOT be provided with tile finishes (specifically, the window and door openings as well as the flushed-mirror), then, let's also calculate the areas of such non-tiled portions of the walls,

Area of Window Opening = 15% x [Bathroom Floor Area] = 0.115 x [3.1m x 2.3m] = 0.82sqm
Area of Door Opening = 0.703m x 2.12m = 1.49sqm
Area of Wall Mirror = 19.16sq.ft. x 0.093sqm/sq.ft. = 1.78sqm
Total Area of Non-Tiled Portions= 4.09sqm (for 1 Bathroom)

Now we subtract the total area of non-tiled portions of the walls to the gross wall area of the bathroom,

Net Area of Tiled Walls = [Gross Wall Area] - [Total Area of Non-Tiled Portions]
Net Area of Tiled Walls = [26.46sqm] - [4.09sqm]
Net Area of Tiled Walls = 22.37sqm (for 1 Bathroom)

Since there are six (6) typical bathrooms of the same design, then, the total area of walls to be provided with tile finish would be,

Total Area of Tiled Walls = 22.37sqm x 6 bathrooms
Total Area of Tiled Walls = 134.22sqm (for All Bathrooms)

Computing for the required number of bags of grout needed,

# of Bags of Grout = [Total Area of Tiled Walls] x [Coverage of 1 Bag of Grout]
# of Bags of Grout = [134.22sqm] / [5.85sqm/bag]
# of Bags of Grout = 22.94 bags or ~23 bags

Note that the "2mm tile spacing" (even the unknown tile size) is NOT a factor in the given problem since the given data from the grout manufacturer, i.e., a coverage of [1 bag of grout at 5.85sqm tile area], already considers such tile spacing (and even the unspecified tile sizes).

Therefore, the correct answer is a. About 23 bags.