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Grade 9 Cost Estimation – Ex 6.2
Class 9 Mensuration Nepal Solutions
Grade 9 Cost Estimation – Exercise 6.2 Solutions
कक्षा ९ लागत अनुमान र क्षेत्रमिति अभ्यास ६.२ को पूर्ण गणितीय समाधान तथा सूत्रहरू
Formulas for Grade 9 Cost Estimation & Mensuration
1. Total Cost Formulae Based on Dimensions
- Rate × Volume: Units ($\text{m}^3, \text{cm}^3$)
- Rate × Area: Units ($\text{m}^2, \text{cm}^2$)
- Rate × Perimeter: Units ($\text{m}, \text{cm}$) — for fencing
- Rate × Number: Units (Pieces) — for counting bricks/stones
- Rate × Length: Units ($\text{m}, \text{cm}$) — for carpeting
2. Number Counting Formula
$$N = \dfrac{\text{Volume of Wall } (V)}{\text{Volume of Brick/Stone } (v)} = \dfrac{\text{Area of Floor}}{\text{Area of Tile}}$$
भित्ताको कुल आयतनलाई एउटा इँटा वा ढुङ्गाको आयतनले भाग गर्दा कुल संख्या प्राप्त हुन्छ।
1
How many stones of volume $600\text{ cm}^3$ each will be required to construct a wall of $20\text{ m} \times 10\text{ m} \times 30\text{ cm}$? [SLC-2063-A1]
| Mathematical Solution | Step-by-Step Explanation |
|---|---|
| $v (\text{Volume of Stone}) = 600\text{ cm}^3$ | Volume of one stone given in $\text{cm}^3$. |
|
$L = 20\text{ m} = 2000\text{ cm}$ $H = 10\text{ m} = 1000\text{ cm}$ $B = 30\text{ cm}$ |
Convert wall dimensions to $\text{cm}$ ($1\text{ m} = 100\text{ cm}$). |
|
$V (\text{Volume of Wall}) = L \times B \times H$ $V = 2000 \times 30 \times 1000 = 60,000,000\text{ cm}^3$ |
Calculate total wall volume using $V = L \cdot B \cdot H$. |
| $N = \dfrac{V (\text{Volume of Wall})}{v (\text{Volume of Stone})} = \dfrac{60,000,000}{600} = 100,000$ | Divide volume of wall ($V$) by volume of stone ($v$). |
Final Answer: 100,000 stones are required.
2
How many bricks, each of dimension $20\text{ cm} \times 10\text{ cm} \times 5\text{ cm}$, will be required to construct a wall of $1\text{ km} \times 5\text{ m} \times 50\text{ cm}$? [SLC-2063-B1]
| Mathematical Solution | Step-by-Step Explanation |
|---|---|
| $v (\text{Volume of Brick}) = l \times b \times h = 20 \times 10 \times 5 = 1,000\text{ cm}^3$ | Calculate single brick volume ($v$) in $\text{cm}^3$. |
|
$L = 1\text{ km} = 100,000\text{ cm}$ $H = 5\text{ m} = 500\text{ cm}$ $B = 50\text{ cm}$ |
Convert all wall dimensions to centimeters ($1\text{ km} = 100,000\text{ cm}$). |
| $V (\text{Volume of Wall}) = 100,000 \times 50 \times 500 = 2,500,000,000\text{ cm}^3$ | Calculate total wall volume ($V$). |
| $N = \dfrac{2,500,000,000}{1,000} = 2,500,000$ | Divide total wall volume by single brick volume. |
Final Answer: 2,500,000 bricks are required.
3
If 24,000 bricks of the same shape and size are required to build a wall of dimension $15\text{ m} \times 6\text{ m} \times 20\text{ cm}$, find the volume of each brick. [SLC-2064-A1]
| Mathematical Solution | Step-by-Step Explanation |
|---|---|
| $N (\text{Number of Bricks}) = 24,000$ | Given total number of bricks required. |
| $L = 15\text{ m} = 1500\text{ cm}$, $H = 6\text{ m} = 600\text{ cm}$, $B = 20\text{ cm}$ | Convert wall measurements to centimeters. |
| $V (\text{Volume of Wall}) = 1500 \times 20 \times 600 = 18,000,000\text{ cm}^3$ | Calculate total volume of the wall ($V$). |
| $24,000 = \dfrac{18,000,000}{v} \implies v = \dfrac{18,000,000}{24,000} = 750\text{ cm}^3$ | Solve for single brick volume ($v$) using cross-multiplication. |
Final Answer: Volume of each brick = 750 cm³
4
The internal dimension of a lidless wooden box is $48\text{ cm} \times 38\text{ cm} \times 31\text{ cm}$. If the thickness of the box is $1\text{ cm}$, find the volume of the wood. [SLC-2062-C2]
| Mathematical Solution | Step-by-Step Explanation |
|---|---|
| Internal volume ($v$) = $48 \times 38 \times 31 = 56,544\text{ cm}^3$ | Calculate internal hollow volume ($v$). |
|
$L = 48 + 2(1) = 50\text{ cm}$ $B = 38 + 2(1) = 40\text{ cm}$ $H = 31 + 1 = 32\text{ cm}$ |
External dimensions: • Length & Breadth extend on both sides ($+2t$). • Height extends only at bottom for a lidless box ($+t$). |
| $V (\text{External Volume}) = 50 \times 40 \times 32 = 64,000\text{ cm}^3$ | Calculate total external volume ($V$). |
| $V_{\text{wood}} = V – v = 64,000 – 56,544 = 7,456\text{ cm}^3$ | Subtract internal volume from external volume to get wood volume. |
Final Answer: Volume of wood = 7,456 cm³
5
The cost of bricks required for $40\text{ m} \times 2.6\text{ m} \times 60\text{ cm}$ wall is Rs 156000. If the price of each brick is Rs 15, calculate the volume of each brick.
| Mathematical Solution | Step-by-Step Explanation |
|---|---|
| $N = \dfrac{\text{Total Cost}}{\text{Cost per brick}} = \dfrac{156,000}{15} = 10,400$ | Calculate total number of bricks ($N$). |
| $L = 4000\text{ cm}, H = 260\text{ cm}, B = 60\text{ cm}$ | Convert wall dimensions to centimeters. |
| $V = 4000 \times 60 \times 260 = 62,400,000\text{ cm}^3$ | Calculate total wall volume ($V$). |
| $10,400 = \dfrac{62,400,000}{v} \implies v = \dfrac{62,400,000}{10,400} = 6,000\text{ cm}^3$ | Solve for single brick volume ($v$). |
Final Answer: Volume of each brick = 6,000 cm³
Study Guide & Tips
Frequently Asked Questions (FAQs) & Student Study Guide
1. Why do students find Grade 9 Cost Estimation challenging?
Students often struggle due to mixing units (meters with centimeters) and failing to visualize 3D hollow/lidless objects. In lidless boxes, thickness is added twice to length and breadth, but only once to height.
2. How to score top marks (A+) in Exercise 6.2 Solutions?
Always standardize dimensions to a uniform unit ($\text{cm}$ or $\text{m}$) at the start, write down explicit formulas, show clear cross-multiplication, and append final units ($\text{cm}^3$, pieces) with answers.
3. What is the unit conversion rule for Volume?
Linear: $1\text{ m} = 100\text{ cm}$. Area: $1\text{ m}^2 = 10,000\text{ cm}^2$. Volume: $1\text{ m}^3 = 1,000,000\text{ cm}^3$. Always cube the conversion factor for volume.
4. Open (Lidless) vs Closed Box Thickness Rule
Closed Box: $L = l + 2t, B = b + 2t, H = h + 2t$.
Lidless Box: $L = l + 2t, B = b + 2t, H = h + t$ (thickness added only at bottom base).