Question 1: A lot of 1000 pieces of a job costs the following:
a) Cost of material = Rs. 20000
b) Labour cost = Rs. 15000
c) Cost of tools = Rs. 8000
d) Factory on cost = 150% on labour
e) Office on cost = 30% on factory cost
f) Selling on cost = 20% on cost of production
The selling price of each piece is Rs. 120. Determine whether there is a profit or loss.
Solution:
Given: Material cost = Rs. 20,000
Labour cost = Rs. 15,000
Tool cost = Rs. 8,000
Factory on cost = 150% of Labour cost
Office on cost = 30% of Factory cost
Selling on cost = 20% of Cost of Production
Selling price per piece = Rs. 120
Number of pieces = 1000
Factory on cost = 150% of Labour = 150% of 15,000 = ₹22,500
So, Factory Cost = Material + Labour + Tools + Factory on cost = 20,000 + 15,000 + 8,000 + 22,500 = Rs. 65,500
Office on cost = 30% of Factory Cost = 30% of 65,500 = ₹19,650
So, Total Cost of Production = Factory Cost + Office on cost = 65,500 + 19,650 = Rs. 85,150
Selling on cost = 20% of Cost of Production = 20% of 85,150 = ₹17,030
So, Total Cost (including Selling Cost) = 85,150 + 17,030 = Rs. 102,180
Selling price per piece = Rs. 120
Total Selling Price = 120 × 1000 = Rs. 120,000
Profit = Total Selling Price − Total Cost = 120,000 − 102,180 = Rs. 17,820 (Profit)
Conclusion: There is a Profit of Rs. 17,820 on the lot of 1000 pieces.
Question 2. Calculate the weight of the machine component as shown in figure given below, if the material weighs 7.8 gm/cc. Also determine the cost of material, if its rate is Rs. 60/Kg. All dimensions are in mm.
Solution:
Given Dimensions:
Density: 7.8 gm/cc = 7.8 × 10⁻³ gm/mm³
Material Rate: Rs. 60/kg
1.Cylinder (left):
Diameter = 20 mm
Length = 30 mm
2.Frustum (tapered section):
Large diameter = 40 mm
Small diameter = 20 mm
Height = 25 mm
Volume = (1/3)πh(R₁² + R₁R₂ + R₂²) [where R₁ = 20, R₂ = 10]
=18325.9 mm³
3.Cylinder (middle):
Diameter = 50 mm
Length = 20 mm
Volume = π/4 × 50² × 20 = 39269.9 mm³
Volume = π/4 × d² × h = π/4 × 20² × 30 = 9424.8 mm³
4.Cylinder (right):
Diameter = 40 mm
Length = 40 mm
Volume = π/4 × 40² × 40 = 50265.5 mm³
5.Cone (end part):
Diameter = 40 mm, Angle = 60° → height = (d/2) × cot(30°) = 20 × √3 ≈ 34.64 mm
Volume = (1/3)πr²h = (1/3)π(20²)(34.64) = 14509.9 mm³
So, total volume of the machine component = 9424.8 + 18325.9 + 39269.9 + 50265.5 + 14509.9 = 131796 mm³
Now, total weight of the component = 131796 × 7.8 × 10⁻³ = 1028 gms = 1.028 Kgs (Ans)
And, total cost of the material = 1.028 × 60 Rs = 61.68 Rs. (Ans)
END
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