---
title: "Road earthwork, rising 1 in 200, falling 1 in 120"
subject: "Estimating and Costing"
chapter: 6
key: "semester_7/estimating_and_costing:road-earthwork-i-rising-falling"
source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/road-earthwork-i-rising-falling
site: BCT Engineering
---

# Road earthwork, rising 1 in 200, falling 1 in 120

Estimating and Costing · Chapter 6: Detailed Estimate

- Asked in: 2076 Chaitra, 10 marks
- Video: [Earthwork Calculation for an Embankment Using the Mean Area Method @CivilConstruction](https://www.youtube.com/watch?v=d8VZscS-xhE) — CivilConstruction
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/estimating_and_costing/road-earthwork-i-rising-falling
- Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6

## Question

Estimate quantities of earthwork of a portion of road from the following data:
(i) Formation width of road is 10 m.
(ii) Side slope in cutting and filling (1:1) and (2:1) (H:V) respectively.

| Distance | 0 | 30 | 60 | 90 | 120 | 150 | 180 |
|---|---|---|---|---|---|---|---|
| R.L. of ground | 102.60 | 103.00 | 102.65 | 102.20 | 101.50 | 101.20 | 100.65 |
| R.L. of formation | 101 | | | | 102.15 | | |
| Gradient | Rising gradient 1 in 200 | | | | Falling gradient 1 in 120 | | |

## Short answer

Using level section formula A=(B+sh)h with B=10m and side slopes 1:1 cutting, 2:1 filling, compute areas from centre heights and then mean-area volumes. Earthwork: cutting 1623.40 m³, filling 137.02 m³.

## Answer

#### Given data and assumptions
- Formation width $B = 10$ m; side slope in cutting $1:1$, in filling $2:1$. Ground is level across the road.
- Formation level at 0 m is 101.00 m, rising 1 in 200 up to 120 m (101.60 m) and falling 1 in 120 from 120 m to 180 m (101.10 m). The printed 102.15 at 120 m does not fit a 1 in 200 gradient, so the stated gradients are used.

#### Method
Level (plain-ground) sections are used. With centre height $h$, formation width $B$ and side slope $s:1$:

$$
A = (B + s\,h)\,h
$$

Volume between two sections = mean area × distance; where a section changes from cutting to filling, the length of each part is found by proportion of the depths and the volume of each part = ½ × end area × length.

| Chainage | GL (m) | FL (m) | Depth of cutting / height of filling (m) |
|---|---:|---:|---:|
| 0+000 | 102.600 | 101.000 | cutting 1.600 |
| 0+030 | 103.000 | 101.150 | cutting 1.850 |
| 0+060 | 102.650 | 101.300 | cutting 1.350 |
| 0+090 | 102.200 | 101.450 | cutting 0.750 |
| 0+120 | 101.500 | 101.600 | filling 0.100 |
| 0+150 | 101.200 | 101.350 | filling 0.150 |
| 0+180 | 100.650 | 101.100 | filling 0.450 |

#### Sectional areas
| Chainage | Centre ht. $h$ (m) | Cutting area (m²) | Filling area (m²) |
|---|---:|---:|---:|
| 0+000 | cut 1.60 | 18.560 | 0.000 |
| 0+030 | cut 1.85 | 21.922 | 0.000 |
| 0+060 | cut 1.35 | 15.323 | 0.000 |
| 0+090 | cut 0.75 | 8.062 | 0.000 |
| 0+120 | fill 0.10 | 0.000 | 1.020 |
| 0+150 | fill 0.15 | 0.000 | 1.545 |
| 0+180 | fill 0.45 | 0.000 | 4.905 |

#### Volumes
| Between | Distance (m) | Cutting (m³) | Filling (m³) |
|---|---:|---:|---:|
| 0+000 – 0+030 | 30.0 | 607.24 | 0.00 |
| 0+030 – 0+060 | 30.0 | 558.68 | 0.00 |
| 0+060 – 0+090 | 30.0 | 350.78 | 0.00 |
| 0+090 – 0+120 | 30.0 | 106.71 | 1.80 |
| 0+120 – 0+150 | 30.0 | 0.00 | 38.47 |
| 0+150 – 0+180 | 30.0 | 0.00 | 96.75 |
| **Total** | 180.0 | **1,623.40** | **137.02** |

**Answer: Earthwork in cutting = 1,623.40 m³; earthwork in filling = 137.02 m³.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Formation level at 0 is 101.00 m, rising 1 in 200 up to 120 m. What is it at 120 m?
   A. 101.60 m
   B. 102.15 m
   C. 101.20 m
   D. 101.00 m
   Answer: A. 120/200 = 0.60 m rise, giving 101.60 m; the printed 102.15 does not fit the gradient. (Key point: Formation levels)

2. For ground level across the road with B = 10 m and cutting slope 1:1, the area at h = 1.6 m is:
   A. (10 + 1.6) x 1.6 = 18.56 m2
   B. 10 x 1.6 = 16 m2
   C. (10 + 2 x 1.6) x 1.6 = 21.12 m2
   D. 1.6 x 1.6 = 2.56 m2
   Answer: A. A = (B + s h) h with s = 1 gives 11.6 x 1.6 = 18.56 m2. (Key point: Area formula)

3. From 120 m to 180 m the formation falls 1 in 120. Formation level at 180 m is:
   A. 101.10 m
   B. 101.60 m
   C. 102.10 m
   D. 101.00 m
   Answer: A. 60/120 = 0.5 m fall; 101.60 - 0.50 = 101.10 m. (Key point: Falling gradient)

4. Which side slope is used for the filling sections?
   A. 2:1
   B. 1:1
   C. 1.5:1
   D. 3:1
   Answer: A. Cutting is 1:1 and filling 2:1 (H:V). (Key point: Slope used)

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