The following consecutive readings (in m) were taken with a dumpy level and a levelling staff at a common interval of 20 m:
0.385; 1.030; 1.925; 2.825; 3.730; 4.850; 1.045; 2.005; 3.330; 4.580
The dumpy level was shifted after taking the sixth reading. The gradient (in %) of the ground between the first and the last location of the reading is integer).

The correct option is 5 (which corresponds to a gradient of 5%).

Step 1: Understand the readings and shifting of the level
A list of 10 consecutive levelling staff readings is given:
1: 0.385 m
2: 1.030 m
3: 1.925 m
4: 2.825 m
5: 3.730 m
6: 4.850 m
7: 1.045 m
8: 2.005 m
9: 3.330 m
10: 4.580 m
The levelling staff readings were taken at a constant interval of 20 m. The instrument was shifted after taking the sixth reading (4.850 m). This means:
- The first setup of the dumpy level covers readings 1 to 6. Here, the first reading (0.385 m) is the Back Sight (BS) at the first location, and the sixth reading (4.850 m) is the Fore Sight (FS) at the sixth location.
- The instrument is shifted, and the seventh reading (1.045 m) is taken on the same point as the sixth reading. Thus, the seventh reading (1.045 m) is a Back Sight (BS) at the sixth location.
- The second setup covers readings 7 to 10. The last reading (4.580 m) is the Fore Sight (FS) at the tenth (last) location.

Step 2: Determine the elevation difference between the first and last locations
Let us calculate the total rise or fall from the first location to the last location.
For any instrument setup, the difference in elevation between the starting point and the ending point of that setup is given by:
$Elevation\; Difference=Sum\; of\; Back\; Sights-Sum\; of\; Fore\; Sights$
Let us find the sum of the Back Sights (BS) and the sum of the Fore Sights (FS) for both setups:
- Setup 1:
BS₁ = 0.385 m
FS₁ = 4.850 m
- Setup 2:
BS₂ = 1.045 m
FS₂ = 4.580 m
Sum of Back Sights:
$\sum BS=0.385+1.045=1.430m$
Sum of Fore Sights:
$\sum FS=4.850+4.580=9.430m$
The difference in elevation (reduced level) between the first point (location 1) and the last point (location 10) is:
$Difference\; in\; RL=\sum BS-\sum FS=1.430-9.430=-8.000m$
The negative sign indicates a fall of 8.000 m from the first point to the last point. Therefore, the vertical drop (difference in height, h) is 8 m.

Step 3: Calculate the horizontal distance between the first and last locations
There are 10 readings in total. Since the sixth reading (4.850 m) and seventh reading (1.045 m) are taken on the same physical location (change point), the total number of staff stations is 9.
The horizontal distance between consecutive stations is 20 m.
The total horizontal distance (D) between the first station and the ninth (last) station is:
$D=(9-1)\times 20m=8\times 20m=160m$

Step 4: Compute the percentage gradient of the ground
The gradient of the ground (in %) is the ratio of the vertical change in height to the horizontal distance, expressed as a percentage:
$Gradient\; (\%)=\frac{Vertical\; Difference}{Horizontal\; Distance}$
Substituting the calculated values:
$Gradient\; (\%)=\frac{8}{160}\times 100=0.05\times 100=5\%$

Thus, the gradient of the ground between the first and the last location is 5%.