# How to Estimate the Different Types and Sizes of Ceramic Tiles

*Ray is a licensed engineer in the Philippines. He loves to write about mathematics and civil engineering.*

## Introduction About Ceramic Tiles

Ceramic tiles are one of the oldest and widely used types of tiles in the world, especially in comfortable rooms. For so many years, we prefer ceramic tiles because of its durability, dirt resistance, stain resistance, slip resistance and aesthetic properties.

### Types of Ceramic Tiles

The five types of ceramic tiles are porcelain tiles, natural clay tiles, ceramic mosaic tiles, faience mosaic tiles and quarry tiles. Aside from these five main classifications, there are other special-type ceramic tiles. Special-type ceramic tiles include non-slip tile, frost-proof, conductive tile, and gallery tile.

1. **Porcelain Tiles** are types of ceramic tiles used for walls and floors. Precise pressed sand forms porcelain tiles. They have a little absorption rate perfect for outdoors. Although it is a bit pricey, it is still worth the buy.

2. **Natural Clay Tiles** are types of ceramic tiles that consists of pressed dust clay. This type of ceramic tiles has a unique textured image.

3. **Ceramic Mosaic Tiles **are ceramic tiles best for wall and border walls. It is perfect for creating a unique design for walls.

4. **Faience Mosaic Tiles** are the decorative-type of ceramic tiles. It is a term coined for pottery in several parts of the world.

5. **Quarry Tiles** are ceramic tiles made from the extraction of plastic in natural clay. Quarry tiles and bricks are quite similar. Its thickness is usually 13 to 19 millimetres.

It is significant that you also know what type and size of ceramic tile you want to use before estimating the cost. I suggest you canvass for ceramic tiles at nearby tile shops. Also, you have to consider if it will perfectly match the colour of your wall painting. Decide as well if you will place it half or full of the bathroom height. In the following examples, you will have an idea of the cost of putting tiles in a specific room. For terms that are not familiar, a definition of terms section is available on the last part.

## Problem 1: Glazed and Unglazed Ceramic Tiles

A room has the following dimensions shown. Consider that the size of ceramic tiles on the floor is different from that on the wall. Determine the quantity of the following materials:

a. 10 cm x 10 cm glazed wall tiles

b. 25 cm x 25 cm unglazed floor tiles

c. Cement mortar

d. Cement for join filler

Per Square Meter | |
---|---|

Cement Mortar | 0.086 bags |

White Cement Joint Filler | 0.50 kilogram |

Tile Adhesives | 0.11 bags |

**A. Solving for 10 cm x 10 cm Glazed Wall Tiles**

1. Solve the whole wall area of the room. Solve the wall area by multiplying the length by the height.

Area of the first face = 6.00 meters ( 2.00 meters )

Area of the first face = 12.00 square meters

Area of the second face = 3.00 meters ( 2.00 meters )

Area of the second face = 6.00 square meters

Total area = 12.00 square meters + 6.00 square meters

Total area = 18.00 square meters

2. Solve for the area of a piece of glazed tile. Convert centimeters to meters. Given one side of 10.00 cm square tile, divide it by 100.

Area of glazed tile = 0.10 meters ( 0.10 meters )

Area of glazed tile = 0.01 square meters

3. Solve the required number of glazed wall tiles. Divide the total wall area by the area of a piece of glazed tile. The resulting value is the number of pieces of glazed wall tiles.

Pieces = Total area / Area of glazed tile

Pieces = 18.00 square meters / 0.01 square meters

Pieces = 1800__Thus, use 1800 pieces of glazed tiles for the wall.__

**B. Solving for 25 cm x 25 cm Unglazed Floor Tiles**

1. Solve for the whole floor area of the room. Solve floor area by multiplying the length by the height.

Floor Area = 6.00 meters ( 3.00 meters )

Floor Area = 18.00 square meters

2. Solve for the area of a piece of unglazed tile. Convert centimeters to meters. Given one side of 25.00 cm square tile, divide it by 100.

Area of unglazed tile = 0.25 meters ( 0.25 meters )

Area of unglazed tile = 0.0625 square meters

3. Solve the required number of unglazed floor tiles. Divide the total floor area by the area of a piece of unglazed tile.

Pieces = Total area / Area of unglazed floor tile

Pieces = 18.00 square meters / 0.0625 square meters

Pieces = 288__Thus, use 288 pieces of unglazed tiles for the floor.__

**C. Solving for the Cement Mortar**

1. Solve for the total area of the whole wall and floor. The obtained area for the wall and floor are both 18.00 square meters. Solve for the sum of the two area.

Area = 18.00 square meters + 18.00 square meters

Area = 36.00 square meters

2. Refer to the given table 1. In the table are the estimated multipliers for the total area computed. There are 0.086 bags of cement mortar per square meter. Multiply the total area by 0.086.

Cement Mortar: 36.00 square meters ( 0.086 bags/square meter )

Cement Mortar = 3.096 bags

In this case, add 10% allowance for breakage.__Thus, use 3.5 bags of cement mortar.__

**D. Solving for White Cement Joint Filler**

1. Solve for the total area of the whole wall and floor. The obtained area for the wall and floor are both 18.00 square meters. Solve for the sum of the two area.

Total Area = 18.00 square meters + 18.00 square meters

Total Area = 36.00 square meters

2. Refer to the given table above. In the table are the estimated multipliers for the total area computed.

White Cement Joint Filler: 36.00 square meters ( 0.50 kg./square meter )

White Cement Joint Filler = 18 kg

In this case, add 10% allowance for breakage.

Thus, use 20 kilograms of white cement joint filler.

**Summary of Quantities**

1800 pieces 10 cm x 10 cm glazed ceramic wall tiles

288 parts 25 cm x 25 cm unglazed ceramic floor tiles

3.5 bags of cement mortar

20 kg. white cement joint filler

## Problem 2: Mosaic Ceramic Tiles

From the figure given below, determine the following materials:

a. 25 centimeters x 25 centimeters mosaic floor tiles

b. 15 centimeters x 20 centimeters glazed wall tiles

c. Cement mortar and joint filler

d. 20 centimeters internal bead

e. 20 centimeters capping

f. Internal and external corner bead

**A. Solving for 25 cm x 25 cm Mosaic Floor Tiles**

1. Solve the whole floor area of the comfort room. The dimensions are 3.00 meters and 4.00 meters. Multiply the two measurements to get the area.

Floor Area = 4.00 meters ( 3.00 meters )

Floor Area = 12.00 square meters

2. Solve for the area of a piece of mosaic tile. Convert centimeters to meters. Given one side of 25.00 cm square tile, divide it by 100.

Area of mosaic tile = 0.25 meters ( 0.25 meters )

Area of mosaic tile = 0.0625 square meters

3. Solve the required number of mosaic tiles. Divide the total floor area by the area of a piece of mosaic tile.

Pieces = Floor area / Area of mosaic tile

Pieces = 12.00 / 0.0625 = 192__Thus, use 192 pieces of mosaic tiles for the floor.__

**B. Solving for 15 cm x 20 cm Glazed Wall Tiles**

1. Solve for the lateral wall area of the comfort room. Get first the wall perimeter. Do not forget to subtract the horizontal length of the door. Multiply the wall perimeter by the height of the wall tiles.

Wall Perimeter = 2(4.00 meters) + 3.00 meters + (3.00 meters - 0.60 meters)

Wall Perimeter = 13.40 meters

Lateral Wall Area = 13.40 meters ( 2.00 meters)

Lateral Wall Area = 26.8 square meters

2. Solve for the area of a piece of glazed tile. Convert centimeters to meters. Given one side of 15.00 cm square tile, divide it by 100. Multiply 0.15 meter by 0.20 meters.

Area of glazed tile = 0.15 meters ( 0.20 meters )

Area of glazed tile = 0.0.03 square meters

3. Divide the total wall area by the area of a piece of glazed tile.

Pieces = Total area / Area of glazed floor tile

Pieces = 26.8 / 0.03 = 894 pieces

In this case, add 5% allowance for breakage.__Thus, use 939 pieces of glazed tiles for the wall.__

**C. Solving for Cement Mortar and Joint Filler**

1. Solve for the total area of the comfort room including the walls and floor. The obtained area for the wall and floor are 12.00 meters and 26.80 meters, respectively.

Area = 12.00 square meters + 26.80 square meters

Area = 38.80 square meters

2. Refer to the Table 1-1. In the table are the estimated multipliers for the total area computed.

Cement Mortar: 38.80 square meters ( 0.086 bags/square meter )

Cement Mortar = 3.34 bags

White Cement Joint Filler: 38.80 square meters ( 0.50 kg./square meter )

White Cement Joint Filler = 19.4 kilograms

In this case, add 10% allowance for breakage.__Thus, use 3.70 bags of cement mortar and 22 kg. of white cement joint filler.__

**D. Solving for Internal Bead**

1. Solve for the perimeter of the whole comfort room. Place the internal beads along the wall.

Perimeter = 2 (3.00) meters + 2 (4.00) meters

Perimeter = 14.00 meters

2. Solve the total length of the four vertical corners of the comfort room. The height of the wall with tiles is 2.00 meters.

Total Length of vertical corners = 4 ( 2.00 meters )

Total Length of vertical corners = 8.00 meters

3. Solve for the total length containing internal beads. Get the entire perimeter and entire length.

Final Length = Perimeter + Total Length of vertical corners

Final Length = 8.00 meters + 14.00 meters

Final Length = 22.00 meters

4. Divide the total length calculated by the length of the commercial bead or tile. Use a 20-centimeter commercial length for the beads.

Pieces = Final Length / Commercial Length of Bead to Use

Pieces = 22.00 meters / 0.20 meters

Pieces = 110 pieces

Thus, use 110 of 20 centimeters of internal bead.

**E. Solve for Capping**

1. Solve for the perimeter of the inside wall tiles. Place the capping tiles along the wall.

Perimeter = 2 ( 4.00 meters ) + ( 3.00 meters - 0.60 meters )

Perimeter = 10.40 meters

2. Add the length of capping along the door jamb.

Length = ( 2.40 meters )2

Length = 4.80 meters

3. Find the total length for the capping. Add the perimeter and the length of capping along the door jamb.

Total Length = Perimeter + Length of capping along the door jamb

Total Length = 10.40 meters + 4.80 meters

Total Length = 15.20 meters

4. Divide the total length by the length of one cap or tile. Use a 20-centimeter commercial length for the capping tile.

Pieces = Total Length / Commercial length of cap tile

Pieces = 15.20 meters / 0.20 meters

Pieces = 76__Thus, use 76 pieces of 20 cm capping.__

**F. Internal and External Corner Bead**

Use direct computation in computing for the internal and external bead. Thus, use four parts of internal corner beads and four elements of outside corner beads.

**Summary of Quantities**

192 pieces of 25 cm x 25 cm mosaic floor tiles

939 pieces of 15 cm x 20 cm glazed wall tiles

3.7 bags of cement mortar

22 kilograms of white cement joint filler

110 parts of 20 cm internal bead

76 pieces of 20-centimeter lids

4 parts of internal corner beads

4 elements of outside corner beads

## Problem 3: Ordinary Ceramic Floor Tiles

Shown below is the floor area of a big school lobby. It measures 15.00 meters by 30.00 meters. How many 60 cm x 60 cm ceramic floor tiles needed? How much tile adhesive and joint fillers needed?

**A. Solving for 60 cm x 60 cm Ceramic Floor Tiles**

1. Solve the whole floor area of the school lobby. Solve floor area by multiplying the length by the height.

Floor Area = 15.00 meters x 30.00 meters

Floor Area = 450.00 square meters

2. Solve for the area of a piece of ceramic tile. Convert centimeters to meters. Given one side of 60.00 cm square tile, divide it by 100.

Area of ceramic tile = 0.60 meters x 0.60 meters

Area of ceramic tile = 0.36 square meters

3. Divide the total floor area by the area of a piece of mosaic tile. The resulting value is the number of pieces of ceramic floor tiles.

Pieces = Total area / Area of ceramic tile

Pieces = 450.00 square meters / 0.36 square meters

Pieces = 1250

In this case, add 10% allowance for breakage.__Thus, use 1375 pieces of ceramic floor tiles for the lobby.__

**B. Solving for the Tile Adhesive and White Cement Joint Filler**

1. Solve the whole floor area of the school lobby. Multiply the length of the lobby by the width. The width of the lobby is 15.00 meters and length is 30.00 meters.

Area = 15.00 meters ( 30.00 meters )

Area = 450.00 square meters

2. Refer to the Table 1-1. In the table are the estimated multipliers for the total area computed. There are 0.11 bags of tile adhesive per square meter. There are 0.50 kilograms of white cement joint filler per square meter. Multiply the total area by 0.11 and 0.50.

Tile Adhesive: 450.00 square meters x 0.11 bags/square meter

Tile Adhesive = 49.5 bags

White Cement Joint Filler: 450.00 square meters x 0.50 kg./square meter

White Cement Joint Filler = 225 kilograms

In this case, add 10% allowance for breakage.__Thus, use 55 bags of tile adhesive and 248 kg. of white cement joint filler.__

**Summary of Quantities**

1375 pieces of 60 cm x 60 cm ceramic floor tiles

55 bags of tile adhesive

248 kg. of white cement joint filler

## Problem 4: Rectangular Ceramic Tiles

A wall is 5.00 meters high and 8.00 meters long. If the finishing should be glazed ceramic tiles with dimensions of 6.00 cm x 16.50 cm, how many required number of tiles are needed if:

a. the longer side installed vertically.

b. the longer side installed horizontally.

**A. Solving for 6.00 cm x 16.50 cm Glazed Ceramic Wall Tiles Vertically**

1. Designate variables 'a', 'b', 'A', and 'B'. 'A' is the height of the wall, 'B' is the length of the wall, 'a' is the shorter dimension of the tile, and 'b' is the longer dimension of the tile. Convert the dimensions of the tiles from centimeters to meters.

A = 5.00 meters

B = 8.00 meters

a = 6.00 centimeters = 0.06 meters

b = 16.50 centimeters = 0.165 meters

2. In this case, the longer side of the tiles installed vertically. Divide the height of the wall by the longer side of the tile. On the other hand, divide the length of the wall by the shorter side of the tile.

A/b = 5.00 meters / 0.165 meters

A/b = 30.303

A/b = 31 pieces

B/a = 8.00 meters / 0.06 meters

B/a = 133.333

B/a = 134 pieces

3. Get the total number of pieces required by multiplying the two answers from step two. Approximate your answer.

[ A / b ] x [ B / a ] = 31 ( 134 )

[ A / b ] x [ B / a ] = 4154 pieces__Thus, use 4154 pieces of glazed ceramic wall tiles wherein the longer side installed vertically.__

**B. Solving for 6.00 cm x 16.50 cm Glazed Ceramic Wall Tiles Horizontally**

1. Designate variables 'a', 'b', 'A', and 'B'. 'A' is the height of the wall, 'B' is the length of the wall, 'a' is the shorter dimension of the tile, and 'b' is the longer dimension of the tile. Convert the dimensions of the tiles from centimeters to meters.

A = 5.00 meters

B = 8.00 meters

a = 6.00 centimeters = 0.06 meters

b = 16.50 centimeters = 0.165 meters

2. In this case, the longer side of the tiles installed horizontally. Divide the height of the wall by the shorter side of the tile. On the other hand, divide the length of the wall by the longer side of the tile.

A/b = 5.00 meters / 0.06 meters

A/b = 83.333

A/b = 84 pieces

B/a = 8.00 meters / 0.165 meters

B/a = 48.485

B/a = 49 pieces

3. Get the total number of pieces required by multiplying the two answers from step two. Approximate your answer.

[ A / b ] x [ B / a ] = 84 ( 49 )

[ A / b ] x [ B / a ] = 4116 pieces__Thus, use 4116 pieces of glazed ceramic wall tiles wherein the longer side installed horizontally.__

## Definition of Terms

**Glazed Tiles** are glossy-type ceramic tiles. Type of ceramic tiles covered with non-porous liquid glass.

**Unglazed Tiles** are non-shiny ceramic tiles. It has no liquid glass covering.

**Cement Mortar **is a mixed compound that acts as a paste for masonry surfaces, bricks, and stones.

**White Cement Joint Filler** is a cement-based used to seal the corners of the tiles

**Tile Adhesive** is an alternative to cement mortar. Applied mainly on drywall or dry flooring.

**Capping Tiles** give accent to the wall tiles in comfortable rooms. Usually placed at the borderline of wall tiles and non-tiled walls.

**Corner Beads** are thin L-shaped coverings placed in every corner of wall tiles. It secures the edges of tiles.

**Internal Beads** are thin L-shaped coverings placed horizontally on wall tiles.

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**© 2018 Raycaptiosus**