Question

The estimated cost per square feet of output is least for:

• A. 75000 square feet output
• B. 100000 square feet output
• C. 125000 square feet output
• D. 150000 square feet output
• E. 175000 square feet output
• A. Decreases up to 125000 sq. ft. and then increases monotonically.
• B. Decreases up to 50000, remains constant between 50000 and 100000, decreases between 100000 and 125000, remains constant between 125000 and 175000 and finally increases between 175000 and 200000.
• C. Remains constant for all levels of monthly output.
• D. Increases up to 50000, remains constant between 50000 and 100000, increases between 100000 and 125000, remains constant between 125000 and 175000 and finally decreases between 175000 and 200000.
• J. Increases up to 100000 and then decreases monotonically.
• A.
• B.
• C.
• D.
• O.

Answer 1

The Correct Option is D

Step 1: Identify all cost components.
Mulchand Textiles has four types of costs: 1. Labour cost (given in table).
2. Material cost (given in table).
3. Electricity cost (geometric progression up to 150000 sq ft, then arithmetic progression beyond that).
4. Administrative cost = fixed Rs. 40000 at all levels.

Step 2: Recall electricity costs.
- At 25000 sq ft: 3800
- At 50000 sq ft: 5700
- Progression is geometric till 150000 sq ft. So ratio $r = \frac{5700}{3800} = 1.5$. Hence, electricity cost values are: \[ 25000 \Rightarrow 3800, \quad 50000 \Rightarrow 5700, \quad 75000 \Rightarrow 8550, \quad 100000 \Rightarrow 12825, \quad 125000 \Rightarrow 19237.5, \quad 150000 \Rightarrow 28856.25 \] Beyond 150000, arithmetic progression begins with a common difference: \[ 38856.5 - 28856.25 = 10000.25 \] So, \[ 175000 \Rightarrow 38856.5, \quad 200000 \Rightarrow 48856.75 \]

Step 3: Compute total costs.
At each output, \[ \text{Total cost} = \text{Labour cost} + \text{Material cost} + \text{Electricity cost} + 40000 \] For example: - At 25000 sq ft: \[ 21500 + 11050 + 3800 + 40000 = 76350 \] - At 50000 sq ft: \[ 41500 + 22000 + 5700 + 40000 = 109200 \] - At 75000 sq ft: \[ 60000 + 33000 + 8550 + 40000 = 141550 \] - At 100000 sq ft: \[ 78000 + 44000 + 12825 + 40000 = 174825 \] - At 125000 sq ft: \[ 95000 + 54750 + 19237.5 + 40000 = 208987.5 \] - At 150000 sq ft: \[ 111000 + 65700 + 28856.25 + 40000 = 245556.25 \] - At 175000 sq ft: \[ 133000 + 76650 + 38856.5 + 40000 = 288506.5 \] - At 200000 sq ft: \[ 160000 + 88000 + 48856.75 + 40000 = 336856.75 \]

Step 4: Calculate cost per square foot.
Divide each total cost by corresponding output: - $76350 / 25000 = 3.054$
- $109200 / 50000 = 2.184$
- $141550 / 75000 \approx 1.887$
- $174825 / 100000 = 1.748$
- $208987.5 / 125000 \approx 1.672$
- $245556.25 / 150000 \approx 1.637$
- $288506.5 / 175000 \approx 1.649$
- $336856.75 / 200000 \approx 1.684$

Step 5: Find the least value.
The smallest cost per square foot is at 150000 sq ft: \[ \approx 1.637 \, \text{Rs. per sq ft} \]

Final Answer: \[ \boxed{\text{D) 150000 square feet output}} \]

Answer 2

Step 1: Define cost components.
At each output level \(Q\) (in sq. ft.), total material cost \(M\) is: \[ M = \text{Spoilage cost} + \text{Usage cost}. \] If spoilage cost per sq. ft. is \(s\) (from graph), then: \[ \text{Total spoilage cost} = s \times Q. \] Hence, \[ \text{Total usage cost} = M - (s \times Q). \]

Step 2: Calculate usage cost per sq. ft.
\[ \text{Usage cost per sq. ft.} = \frac{\text{Total usage cost}}{Q}. \] For example, for \(Q = 25000\): Material cost \(M = 11050\), spoilage cost per sq. ft. \(s = 0.042\). \[ \text{Total spoilage cost} = 25000 \times 0.042 = 1050. \] \[ \text{Usage cost} = 11050 - 1050 = 10000. \] \[ \text{Usage cost per sq. ft.} = \frac{10000}{25000} = 0.40. \]

Step 3: Verify across all outputs.
The same calculation repeated for 50000, 75000, 100000, 125000, 150000, 175000, and 200000 outputs gives usage cost per sq. ft. = 0.40 for all.

Step 4: Conclude.
Since the value remains \emph{constant} at 0.40 for all levels of output, the correct option is: \[ \boxed{\text{Remains constant for all levels of monthly output (Option C)}} \]

Answer 3

Step 1: Use the definition of the plotted quantity.
By the problem statement, the plotted ordinate is \[ \text{Electricity cost per sq. ft.} \;=\; \frac{\text{Total electricity cost (in ₹)}}{\text{Output (in sq. ft.)}}. \] Therefore, for each output level \(Q\), we compute \[ e(Q) \;=\; \frac{\text{Electricity cost at }Q}{Q}. \]

Step 2: Compute \(e(Q)\) for each given output.
From the table (costs read from the question data), the electricity cost values (in ₹) lead to the following per-unit costs: \[ \begin{array}{c|c|c} \text{Output }Q\;(\text{sq. ft.}) & \text{Electricity cost }(₹) & e(Q)=\frac{\text{cost}}{Q}\;(₹/\text{sq. ft.})
\hline 25{,}000 & 3800 & 0.15
50{,}000 & 5700 & 0.11
75{,}000 & 8550 & 0.11
100{,}000 & 12{,}825 & 0.13
125{,}000 & 19{,}237.5 & 0.15
150{,}000 & 28{,}856.25 & 0.19
175{,}000 & 38{,}586.5 & 0.22
200{,}000 & 48{,}856.75 & 0.24 \end{array} \]

Step 3: Infer the qualitative shape.
Reading the sequence of per-unit costs: \[ 0.15 \;\Rightarrow\; 0.11 \;\Rightarrow\; 0.11 \;\Rightarrow\; 0.13 \;\Rightarrow\; 0.15 \;\Rightarrow\; 0.19 \;\Rightarrow\; 0.22 \;\Rightarrow\; 0.24. \] Thus, the curve \emph{first decreases} from \(25{,}000\) to \(50{,}000\)–\(75{,}000\) sq. ft. (down to about \(0.11\)), and then \emph{increases steadily} up to \(200{,}000\) sq. ft. (reaching about \(0.24\)). The only candidate whose shape matches “dip to \(\approx 0.11\) early, then smooth rise to \(\approx 0.24\)” is

Option B. \[ \boxed{\text{Therefore, the correct diagram is }

B.} \]