30 labourers, working 7 hours a day can finish a piece of work in 18 days. If the labourers work 6 hours a day, then the number of labourers to finish the same piece of work in 30 days, will be:

### Concept & Work Equivalence Theorem When the total amount of work is identical in two different situations, the total man-hours required remains constant. We can equate the product of Men, Days, and Hours for both scenarios using the foundational Chain Rule formula. $$ M_1 D_1 H_1 = M_2 D_2 H_2 $$ ### Step-by-Step Solution * Scenario 1 data: $M_1 = 30 \text{ labourers}$, $D_1 = 18 \text{ days}$, $H_1 = 7 \text{ hours/day}$. * Scenario 2 data: $M_2 = ? \text{ labourers}$, $D_2 = 30 \text{ days}$, $H_2 = 6 \text{ hours/day}$. * Set up the equivalence equation based on the formula: $30 \times 18 \times 7 = M_2 \times 30 \times 6$. * Divide both sides by $30$ to immediately simplify the equation: $18 \times 7 = M_2 \times 6$. * Divide both sides by $6$: $3 \times 7 = M_2$. * Calculate the final value: $M_2 = 21$. ### Exam Strategy & Shortcut Write down the equation as a fraction to solve for the unknown directly: $M_2 = \frac{30 \times 18 \times 7}{30 \times 6}$. From this vantage point, you can visually cross out the $30$ in the numerator and denominator, leaving $\frac{18 \times 7}{6}$. Since $\frac{18}{6} = 3$, the answer is simply $3 \times 7 = 21$. ### Common Pitfall Multiplying out the entire left side ($30 \times 18 \times 7 = 3780$) before dividing wastes time and increases the probability of basic arithmetic calculation errors. Always cancel factors first. ### Final Answer Therefore, the correct answer is **21**.