Question

A primary school is having 8 class rooms, each having internal dimensions of 15m × 10m × 4m (height). Only the internal walls of all the class rooms are proposed to be painted. Assume a deduction of 10% internal wall area due to doors, windows etc. The specification suggests two coats of paint application. The spreading rates of the selected paint during base coat and finish coat are 4.5 l/sq.m and 2.5 l/sq.m respectively. The amount of paint (in liters) required for the job will be \underline{\hspace{1cm}}.

Hint:

Always calculate wall area = perimeter × height, deduct 10% for openings, then divide by spreading rate for each coat.

Answer 1

Step 1: Calculate wall area of one classroom.
Perimeter of classroom = \(2 \times (15 + 10) = 50\) m
Height = 4 m
\[ \text{Wall area per classroom} = 50 \times 4 = 200 \; \text{m}^2 \]

Step 2: Total wall area of 8 classrooms.
\[ 200 \times 8 = 1600 \; \text{m}^2 \]

Step 3: Deduction for doors/windows (10%).
\[ \text{Net paintable area} = 1600 \times 0.90 = 1440 \; \text{m}^2 \]

Step 4: Paint requirement for two coats.
- Base coat spreading rate = 4.5 m\(^2\)/liter
\[ \text{Paint required (base coat)} = \frac{1440}{4.5} = 320 \; \text{liters} \] - Finish coat spreading rate = 2.5 m\(^2\)/liter
\[ \text{Paint required (finish coat)} = \frac{1440}{2.5} = 576 \; \text{liters} \]

Step 5: Total paint required.
\[ 320 + 576 = 896 \; \text{liters} \]

Final Answer: \[ \boxed{896 \; \text{liters}} \]