Question

The minimum number of masonry units of size 200 mm \(\times\) 100 mm \(\times\) 100 mm each, required to construct a solid dry wall of length 3 m, height 2 m, and thickness 0.20 m, is (in integer).

Hint:

Find the plain volume-based count first (wall volume divided by unit volume = 600), then add the standard bonding/wastage allowance used in masonry estimation to get the practical number.

Answer 1

Correct Answer: 630

Step 1: Work out how many courses the wall needs.
One course is one layer of units, one unit high. Each unit is $100$ mm tall, and the wall is $2000$ mm high, so the wall needs
$$ \frac{2000}{100} = 20 \text{ courses} $$

Step 2: Work out how many units sit in one course.
The wall thickness is $200$ mm, which exactly equals the $200$ mm length of one unit. So a unit can be turned to run its $200$ mm length straight across the wall thickness (a header), filling the full thickness with one unit. Along the $3000$ mm length of the wall, laid width to width ($100$ mm each), one course needs
$$ \frac{3000}{100} = 30 \text{ units per course} $$

Step 3: Multiply courses by units per course.
$$ 20 \times 30 = 600 \text{ units} $$
This is the plain geometric count, the same number a volume calculation gives, since no mortar joint is used in a dry wall.

Step 4: Correct for real construction losses.
In practice, header and stretcher courses must alternate to bond the wall together, which forces some units near the wall ends to be cut into closers, and some units break during handling. Estimation guides cover this with a standard extra allowance of about 5% over the plain count.
$$ 600 + (0.05 \times 600) = 600 + 30 = 630 $$

Step 5: Final answer.
$$ \boxed{630} $$