Part 2 - Cost Estimation

Question 1

What is the maximum amount of water that can be contained in the tanks given below. (1 ft³ = 28.317 l)

Solution:

(a)

The tank is a cube shaped tank with length (a) = 6 ft

Volume of the tank (V) = $a^3 = (6)^3 = 216$ ft³

We know, 1 ft³ = 28.317 l

$\therefore 216 \text{ ft}^3 = 216 \times 28.317 = 6116.47 \text{ l}$

Hence, the tank can contain 6116.47 l of water.

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(b)

The tank is a cuboid shaped tank with

Length (l) = 2 m

Breadth (b) = 1.5 m

Height (h) = 2 m

Volume of the tank (V) = $l \times b \times h = 2 \times 1.5 \times 2 = 6$ m³

We know, 1 m³ = 1000 l

$\therefore 6 \text{ m}^3 = 6 \times 1000 = 6000 \text{ l}$

Hence, the tank can contain 6000 l of water.

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(c)

The tank is a cylindrical shaped tank with

Radius (r) = $\frac{1}{2}$ = 0.5 m

Height (h) = 4 m

Volume of the tank (V) = $\pi r^2 h$

$= \frac{22}{7} \times (0.5)^2 \times 4$

$= \frac{22}{7} \times 1$

$= \frac{22}{7} \approx 3.14 \text{ m}^3$

We know, 1 m³ = 1000 l

$\therefore \frac{22}{7} \text{ m}^3 = \frac{22}{7} \times 1000 = 3142.86 \text{ l}$

Hence, the tank can contain 3142.86 l of water.

Question 2

The parking area outside the National Insurance Company Limited Nepal is in geometric shape as shown in the figure. It is planning to pave the area with bricks. A brick occupies the area of 0.222 ft² and the cost of brick per piece is Rs. 16,

1. Find the area of the parking land.
2. How many bricks are needed to pave the whole parking area?
3. If 2 workers can complete the work of paving bricks in 3 days and the wage of a worker per day is Rs. 1200, how much does it cost to pave the bricks including the cost of bricks?

Solution:

The parking land is a trapezium shaped with

Parallel sides: AD = 10 ft and BC = 12 ft

Distance between the parallel sides (height) = 8 ft

(a) Area of the parking land (A) = $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$

$= \frac{1}{2} \times (10 + 12) \times 8$

$= \frac{1}{2} \times 22 \times 8$

$= 88 \text{ ft}^2$

(b) Area occupied by one brick = 0.222 ft²

Number of bricks = $\frac{\text{area of the land}}{\text{area of one brick}}$

$= \frac{88}{0.222}$

$= 396.4 \approx 397 \text{ bricks}$

(rounded up so that the whole area is covered)

(c) Cost of bricks = 397 × 16 = Rs. 6,352

Wage of the workers = 2 workers × 3 days × Rs. 1200 = Rs. 7,200

Total cost to pave the bricks = cost of bricks + wages

$= 6352 + 7200 = \text{Rs. } 13,552$

Hence, the total cost of paving the bricks is Rs. 13,552.

Question 3

The figure given aside is a garden which is on the right side of a newly constructed temple. The management committee of temple has decided to fence around it with barbed wire rounding it 5 times. It can be completed by 3 workers in 2 days. The cost of wire per meter is Rs. 80 and the wage per worker per day is Rs. 1500.

1. Find the perimeter of the land.
2. How much wire is required to fence the garden rounding it 5 times?
3. Find the total cost of fencing the garden rounding it 5 times including the wages to the worker.

Solution:

In the quadrilateral shaped garden ABCD,

AB = 10 m, BC = 11 m, CD = 15 m, DA = 14 m

Number of rounds of wire = 5

Cost of wire per meter = Rs. 80

Number of workers = 3, days = 2, wage per worker per day = Rs. 1500

(a) Perimeter of the garden (P) = AB + BC + CD + DA

$= 10 + 11 + 15 + 14$

$= 50 \text{ m}$

(b) Length of wire required = 5 × perimeter = 5 × 50 = 250 m

(c) Cost of the wire = 250 × 80 = Rs. 20,000

Total wage of the workers = 3 workers × 2 days × Rs. 1500 = Rs. 9,000

Total cost = cost of wire + wages = 20,000 + 9,000 = Rs. 29,000

Hence, the total cost of fencing the garden is Rs. 29,000.

Question 4

There is a cubical shaped room with length 9 ft which has two square shaped windows of length 2 ft each and two windows of size 5 ft × 3 ft each.

1. Find the area of the floor.
2. Find the area of 4 walls excluding the windows and doors.
3. Find the total cost of painting 4 walls including the window and doors at the rate of Rs. 350 per square meter.
4. Another room having length, breadth and height each greater by 1 ft. than that of the first is also to be painted at the rate of Rs. 340 per square ft. Calculate by how much does the total cost increase or decrease while painting the 4 walls including the windows and doors.

Solution:

The room is a cube shaped room with length (a) = 9 ft

So, length (l) = breadth (b) = height (h) = 9 ft

(a) Area of the floor = $l \times b = 9 \times 9 = 81$ ft²

(b) Area of 4 walls = $2h(l + b) = 2 \times 9 \, (9 + 9) = 2 \times 9 \times 18 = 324$ ft²

Area of 2 square windows (A₁) = 2 × (2)² = 8 ft²

Area of 2 windows (A₂) = 2 × (5 × 3) = 30 ft²

Area of 4 walls excluding the windows and doors = 324 − A₁ − A₂

$= 324 - 8 - 30$

$= 286 \text{ ft}^2$

(c) Including the windows and doors, the area of 4 walls = 324 ft²

Total cost of painting 4 walls at the rate of Rs. 350 = 350 × 324 = Rs. 1,13,400

(d) For the second room, length, breadth and height are each greater by 1 ft.

So, it is a cube of length (a) = 9 + 1 = 10 ft

Area of 4 walls (including windows and doors) = $2h(l + b) = 2 \times 10 \, (10 + 10) = 400$ ft²

Cost of painting the second room = 340 × 400 = Rs. 1,36,000

Cost of painting the first room (including windows and doors) = Rs. 1,13,400

Increase in total cost = 1,36,000 − 1,13,400 = Rs. 22,600

Hence, the total cost increases by Rs. 22,600.

Question 5

It requires 60 concrete rings each of diameter 3.5 ft and height 1 ft. to construct a well. The cost of a ring is Rs. 1200. 2 workers can complete the work of keeping rings in 12 days and the wage of 1 worker is Rs. 1500 per day. (1 cubic foot = 28.317 liters)

1. What is the cost of the rings only?
2. How much does it cost to construct the well?
3. Find the maximum amount of water that can be filled in the well.
4. If the water label is seen at the 35^th ring, how much water is there in the well? How many litters of water is needed to fill the well completely?

Solution:

Here, number of rings = 60

Radius of a ring (r) = $\frac{3.5}{2}$ = 1.75 ft

Height of a ring = 1 ft

(a) Cost of one ring = Rs. 1200

Cost of the rings only = 60 × 1200 = Rs. 72,000

(b) Wage of the workers = 2 workers × 12 days × Rs. 1500 = Rs. 36,000

Cost to construct the well = cost of rings + wages = 72,000 + 36,000 = Rs. 1,08,000

(c) The well is a cylinder made of 60 rings, so its height (h) = 60 × 1 = 60 ft

Volume of the well (V) = $\pi r^2 h$

$= \frac{22}{7} \times (1.75)^2 \times 60$

$= 9.625 \times 60$

$= 577.5 \text{ ft}^3$

We know, 1 cubic foot = 28.317 liters

$\therefore 577.5 \text{ ft}^3 = 577.5 \times 28.317 = 16353.10 \text{ l}$

Hence, the maximum amount of water that can be filled in the well is 16353.10 l.

(d) When the water level is at the 35^th ring, the height of water = 35 ft

Volume of water in the well = $\pi r^2 h$

$= \frac{22}{7} \times (1.75)^2 \times 35$

$= 9.625 \times 35$

$= 336.875 \text{ ft}^3$

$= 336.875 \times 28.317 = 9539.29 \text{ l}$

To fill the well completely, the remaining height = 60 − 35 = 25 ft

Volume of water needed = $\pi r^2 h = 9.625 \times 25 = 240.625$ ft³

$= 240.625 \times 28.317 = 6813.78 \text{ l}$

Hence, there is 9539.29 l of water in the well and 6813.78 l more is needed to fill it completely.

Question 6

There are two square based pillars each of height 10 ft placed in the gate of a house. A pyramid of the same base and height 1 ft is placed on the top of each pillar and the length of the base of each pillar is 5 ft.

1. Present the given information in a diagram.
2. Find the slant height of the pyramid.
3. What is the total cost of painting the pillars together with pyramids if the rate of painting is Rs. 94 per square feet?

Solution:

Here, the height of pillar (h₁) = 10 ft

Height of pyramid (h₂) = 1 ft

Since, the base of pillar is square shaped, its length (a) = 5 ft

(a)

(b) We know,

Slant height of pyramid (l) = $\sqrt{(h_2)^2 + \left(\frac{a}{2}\right)^2}$

$= \sqrt{(1)^2 + \left(\frac{5}{2}\right)^2}$

$= \sqrt{1 + 6.25}$

$= \sqrt{7.25} = \frac{\sqrt{29}}{2}$

$= 2.69 \text{ ft}$

(c) Lateral surface area of prism (A₁) = perimeter of base (P) × height (h₁)

$= 4a \times 10$

$= 4 \times 5 \times 10 = 200 \text{ ft}^2$

Lateral surface area of pyramid (A₂) = 2al = 2 × 5 × 2.69 = 26.93 ft²

Total surface area of a pillar with a pyramid = A₁ + A₂ = 200 + 26.93 = 226.93 ft²

Total surface area of two pillars containing pyramids = 2 × 226.93 = 453.85 ft²

(The base area of pillar is not included as it is lying on the ground.)

Rate of painting (R) = Rs. 94 per square feet

Total cost of painting (T) = R × A = 94 × 453.85 = Rs. 42,661.9

Question 7

There are two square based pillars each of height 6 ft in the gate of a house. A square based pyramid of same the base and height 1 ft is placed on the top of each pillar. The length of base of each pillar is 1 ft.

1. Present the given information in a diagram.
2. Write the formula to find out the lateral surface area of a square based pyramid.
3. What is the slant height of the pyramid shaped part?
4. What is the cost of tiling per square feet if the total cost of tiling the pillars together with pyramids is Rs. 2729?

Solution:

Here, the height of pillar (h₁) = 6 ft

Height of pyramid (h₂) = 1 ft

Since, the base of pillar is square shaped, its length (a) = 1 ft

(a)

(b) Lateral surface area of a square based pyramid (A) = $2al$, where a is the length of the base and l is the slant height.

(c) Slant height of pyramid (l) = $\sqrt{(h_2)^2 + \left(\frac{a}{2}\right)^2}$

$= \sqrt{(1)^2 + (0.5)^2}$

$= \sqrt{1.25} = \frac{\sqrt{5}}{2}$

$= 1.12 \text{ ft}$

(d) Lateral surface area of prism (A₁) = 4a × h₁ = 4 × 1 × 6 = 24 ft²

Lateral surface area of pyramid (A₂) = 2al = 2 × 1 × 1.12 = 2.24 ft²

Total surface area of a pillar with a pyramid = A₁ + A₂ = 24 + 2.24 = 26.24 ft²

Total surface area of two pillars containing pyramids = 2 × 26.24 = 52.47 ft²

(The base area of pillar is not included as it is lying on the ground.)

Total cost of tiling = Rs. 2729

Cost of tiling per square feet (R) = $\frac{\text{total cost}}{\text{total area}} = \frac{2729}{52.47}$ = Rs. 52

Hence, the cost of tiling per square feet is Rs. 52.

Question 8

A circular tank with inner diameter 2.80 m and height 3 m is constructed in a nursery to collect rain water. If its upper part is conical shaped with vertical height 0.72 m,

1. write the formula to find the volume of cylinder.
2. what is the maximum amount of water that can be contained in the tank?
3. find the total cost to fill up the whole tank if the cost of water is Rs. 1.85 per liter.

Solution:

Here, radius of the tank (r) = $\frac{2.80}{2}$ = 1.4 m

Height of the cylindrical part (h₁) = 3 m

Height of the conical part (h₂) = 0.72 m

(a) Volume of cylinder (V) = $\pi r^2 h$

(b) Maximum water = volume of cylinder + volume of cone

Volume of cylinder = $\pi r^2 h_1 = \frac{22}{7} \times (1.4)^2 \times 3 = 18.48$ m³

Volume of cone = $\frac{1}{3}\pi r^2 h_2 = \frac{1}{3} \times \frac{22}{7} \times (1.4)^2 \times 0.72 = 1.4784$ m³

Total volume (V) = 18.48 + 1.4784 = 19.9584 m³

We know, 1 m³ = 1000 l

$\therefore 19.9584 \text{ m}^3 = 19.9584 \times 1000 = 19958.4 \text{ l}$

Hence, the maximum amount of water that can be contained is 19958.4 l.

(c) Cost of water = Rs. 1.85 per liter

Total cost to fill the tank = 19958.4 × 1.85 = Rs. 36,923.04

Question 9

A pyramid shaped tent is formed using 8 equal triangular shaped pieces of clothes and the lengths of sides of each triangular piece of clothes are 5 m, 6 m and 6 m,

1. Find the area of a piece of cloth.
2. Find the total area of clothes required for the tent.
3. If the cost of clothes per square meter is Rs. 600, find the total cost of tent construction.

Solution:

Here, the sides of each triangular piece are a = 5 m, b = 6 m and c = 6 m.

(a) Semi-perimeter (s) = $\frac{a + b + c}{2} = \frac{5 + 6 + 6}{2} = \frac{17}{2} = 8.5$ m

By Heron's formula, area of a piece of cloth (A) = $\sqrt{s(s-a)(s-b)(s-c)}$

$= \sqrt{8.5 \, (8.5 - 5)(8.5 - 6)(8.5 - 6)}$

$= \sqrt{8.5 \times 3.5 \times 2.5 \times 2.5}$

$= \sqrt{185.9375} = \frac{5\sqrt{119}}{4}$

$= 13.64 \text{ m}^2$

(b) Total area of clothes required = 8 × area of one piece

$= 8 \times \frac{5\sqrt{119}}{4}$

$= 10\sqrt{119}$

$= 109.09 \text{ m}^2$

(c) Cost of clothes per square meter = Rs. 600

Total cost of tent construction = 600 × 10√119 = 6000√119 = Rs. 65,452.27