GATE CE 2015 Set 1 — Question 49

$\gamma_s = G_s \cdot \gamma_w = 2.7 \cdot 10 \text{ kN/m}^3 = 27 \text{ kN/m}^3$$\gamma_{d, \text{borrow}} = \frac{\gamma_s}{1+e}$Target dry unit weight $\gamma_{d, \text{embankment}} = 16.2 \text{ kN/m}^3$
Volume of embankment $V_{\text{embankment}} = 5000 \text{ m}^3$
$V_s = \frac{\gamma_{d, \text{embankment}} \cdot V_{\text{embankment}}}{\gamma_s} = \frac{16.2 \text{ kN/m}^3 \cdot 5000 \text{ m}^3}{27 \text{ kN/m}^3} = 3000 \text{ m}^3$$V_{\text{borrow}} = \frac{V_s}{\gamma_{d, \text{borrow}} / \gamma_s} = V_s \cdot \frac{1+e}{1} = V_s \cdot (1+e)$

• Site X: $e = 0.6$

$V_{\text{borrow, X}} = 3000 \text{ m}^3 \cdot (1+0.6) = 4800 \text{ m}^3$
Distance $D_X = 140 \text{ km}$

• Site Y: $e = 0.7$

$V_{\text{borrow, Y}} = 3000 \text{ m}^3 \cdot (1+0.7) = 5100 \text{ m}^3$
Distance $D_Y = 80 \text{ km}$

• Site Z: $e = 0.64$

$V_{\text{borrow, Z}} = 3000 \text{ m}^3 \cdot (1+0.64) = 4920 \text{ m}^3$
Distance $D_Z = 100 \text{ km}$Let $C_E$ be the cost of excavation per $\text{m}^3$.
Cost of transportation per km per $\text{m}^3$ is $2C_E$.
Total Cost $= (V_{\text{borrow}} \cdot C_E) + (V_{\text{borrow}} \cdot D \cdot 2C_E) = C_E \cdot V_{\text{borrow}} \cdot (1 + 2D)$

• Site X:

Cost$_X = C_E \cdot 4800 \cdot (1 + 2 \cdot 140) = C_E \cdot 4800 \cdot 281 = 1348800 C_E$

• Site Y:

Cost$_Y = C_E \cdot 5100 \cdot (1 + 2 \cdot 80) = C_E \cdot 5100 \cdot 161 = 821100 C_E$

• Site Z:

Cost$_Z = C_E \cdot 4920 \cdot (1 + 2 \cdot 100) = C_E \cdot 4920 \cdot 201 = 990920 C_E$Cost$_Y$ ($821100 C_E$) is the lowest.Re-evaluation of Question Type: The problem is explicitly labeled as NAT, but the options are given and one is marked. If it were a true NAT, it would ask for a numerical value (e.g., the minimum cost or the site index). Given the options and the checkmark, it behaves like an MCQ. If the question expects a numerical answer, it would be the minimum cost, which is $821100 C_E$. However, the options are site names. I will follow the explicit marking in the PDF, which indicates 'A' (Site X) is the correct option. This contradicts my calculation. Let me re-check the calculation.Let's re-read the problem carefully. "The void ratios (e) of different sites are shown in the figure." "If the cost of transportation per km is twice the cost of excavation per m3 of borrow pits, which site would you choose as the most economic solution?"My calculation for Site Y is the most economic. The PDF marks Site X as correct. This is a discrepancy. I will re-verify the calculation.Let's assume the question is asking for the site name as the answer, making it an MCQ despite the NAT label. If the provided answer (Site X) is correct, then my cost calculation must be wrong or my interpretation of the cost function.Let's re-check the cost function: Cost = (Excavation Cost) + (Transportation Cost)
Excavation Cost = $V_{\text{borrow}} \cdot C_E$
Transportation Cost = $V_{\text{borrow}} \cdot D \cdot (2C_E)$ (cost per m3 per km is $2C_E$)
Total Cost = $C_E \cdot V_{\text{borrow}} \cdot (1 + 2D)$
This formula seems correct.Let's re-check the values:
Site X: $V_{\text{borrow, X}} = 4800 \text{ m}^3$, $D_X = 140 \text{ km}$. Cost$_X = 4800 \cdot (1 + 2 \cdot 140) = 4800 \cdot 281 = 1348800 C_E$
Site Y: $V_{\text{borrow, Y}} = 5100 \text{ m}^3$, $D_Y = 80 \text{ km}$. Cost$_Y = 5100 \cdot (1 + 2 \cdot 80) = 5100 \cdot 161 = 821100 C_E$
Site Z: $V_{\text{borrow, Z}} = 4920 \text{ m}^3$, $D_Z = 100 \text{ km}$. Cost$_Z = 4920 \cdot (1 + 2 \cdot 100) = 4920 \cdot 201 = 990920 C_E$My calculations consistently show Site Y as the most economic. If the PDF's marked answer (Site X) is correct, there might be an error in the problem statement or the provided solution. However, as an API, I must adhere to the provided markings if they exist. Since there is a checkmark next to option A, I will assume option A is the intended answer, even if my calculation differs. This is a common issue in competitive exam papers where the provided answer key might have errors.Given the instruction to extract the answer from the PDF markings, I will use 'A' as the correct option. If I were to solve it independently, it would be 'B'. I will provide the solution based on the calculation that leads to Site Y, and note the discrepancy.Discrepancy Note: Based on calculations, Site Y is the most economic. However, the provided PDF marking indicates Site X (Option A) as the correct answer. Following the instruction to use PDF markings, Option A is selected.Final Answer is A.