The ratio of the amount of work done by (x − 1) labourers in (x + 1) days and by (x + 1) labourers in (x + 2) days is 5 : 6. What is the value of x?

Introduction / Context:
This problem combines the concept of work with algebraic manipulation. Work problems often use the idea that total work done is proportional to the product of the number of workers and the time they work, assuming all workers have the same efficiency. Here, two different groups of labourers work for different numbers of days, and the ratio of the work done by them is given. We are asked to find the value of x that satisfies this condition.

Given Data / Assumptions:
• Group 1 has (x − 1) labourers working for (x + 1) days. • Group 2 has (x + 1) labourers working for (x + 2) days. • Ratio of work done by group 1 to group 2 = 5 : 6. • All labourers are assumed to work at the same constant rate.

Concept / Approach:
When each labourer works at the same rate, the total work is directly proportional to (number of labourers) * (number of days). Let the rate of one labourer be 1 unit of work per day for simplicity. Then work done by each group can be written as the product of the number of labourers and days. We form an equation from the ratio of these products and solve for x. This is a straightforward algebraic equation in x.

Step-by-Step Solution:
Step 1: Work done by group 1 = (x − 1) * (x + 1) = x^2 − 1. Step 2: Work done by group 2 = (x + 1) * (x + 2) = x^2 + 3x + 2. Step 3: Given that the ratio of work is 5 : 6, we write (x^2 − 1) / (x^2 + 3x + 2) = 5 / 6. Step 4: Cross multiply: 6 * (x^2 − 1) = 5 * (x^2 + 3x + 2). Step 5: Expand both sides. Left side: 6x^2 − 6. Right side: 5x^2 + 15x + 10. Step 6: Bring all terms to one side: 6x^2 − 6 − 5x^2 − 15x − 10 = 0. Step 7: Simplify: x^2 − 15x − 16 = 0. Step 8: Factor the quadratic: x^2 − 15x − 16 = 0 can be written as (x − 16)(x + 1) = 0. Step 9: Solutions are x = 16 or x = -1. Since x represents counts of labourers and days, x must be positive and greater than 1, so x = 16.

Verification / Alternative Check:
Substitute x = 16 back into the expressions. Group 1 work: (16 − 1) * (16 + 1) = 15 * 17 = 255. Group 2 work: (16 + 1) * (16 + 2) = 17 * 18 = 306. Now check the ratio 255 : 306. Divide both numbers by 51: 255 / 51 = 5, 306 / 51 = 6. So the ratio is 5 : 6, which confirms the correctness of x = 16.

Why Other Options Are Wrong:
• 15, 17 and 14 do not satisfy the equation x^2 − 15x − 16 = 0. Substituting any of them into the ratio expression will give a ratio different from 5 : 6.

Common Pitfalls:
Students may mistakenly set up the ratio as days to workers or confuse which group corresponds to 5 and which to 6. It is also easy to make errors in expanding and simplifying the algebraic expressions. Carefully writing each step and checking the factorisation of the quadratic equation helps avoid these issues.

Final Answer:
The value of x that satisfies the given condition is 16.