The ratio of the amount of work done by (x-1) labours in (x+1) days and that done by (x+1) labours in (x+2) days is 5:6. Then the value of x is

Options

A. 1. 16
B. 2. 15
C. 3. 17
D. 4. 14

Answer

To solve the problem, we need to find the value of \( x \) given the ratio of work done by two groups of laborers. Let's break it down step by step.

### Step 1: Define the Work Done
Let:
- \( m_1 = x - 1 \) (number of labors in the first group)
- \( d_1 = x + 1 \) (number of days worked by the first group)
- \( m_2 = x + 1 \) (number of labors in the second group)
- \( d_2 = x + 2 \) (number of days worked by the second group)

### Step 2: Calculate the Work Done by Each Group
The work done by the first group \( W_1 \) can be calculated as:
\[ W_1 = m_1 \times d_1 = (x - 1)(x + 1) \]

The work done by the second group \( W_2 \) can be calculated as:
\[ W_2 = m_2 \times d_2 = (x + 1)(x + 2) \]

### Step 3: Set Up the Ratio
According to the problem, the ratio of the work done by the two groups is given as:
\[ \frac{W_1}{W_2} = \frac{5}{6} \]

### Step 4: Substitute the Work Done into the Ratio
Substituting \( W_1 \) and \( W_2 \) into the ratio gives us:
\[ \frac{(x - 1)(x + 1)}{(x + 1)(x + 2)} = \frac{5}{6} \]

### Step 5: Cross Multiply to Eliminate the Fraction
Cross multiplying gives us:
\[ 6(x - 1)(x + 1) = 5(x + 1)(x + 2) \]

### Step 6: Expand Both Sides
Expanding both sides:
- Left side:
\[ 6(x^2 - 1) = 6x^2 - 6 \]
- Right side:
\[ 5(x^2 + 3x + 2) = 5x^2 + 15x + 10 \]

### Step 7: Set the Equation to Zero
Setting the equation to zero:
\[ 6x^2 - 6 = 5x^2 + 15x + 10 \]
Rearranging gives:
\[ 6x^2 - 5x^2 - 15x - 6 - 10 = 0 \]
\[ x^2 - 15x - 16 = 0 \]

### Step 8: Factor the Quadratic Equation
We can factor the quadratic equation:
\[ (x - 16)(x + 1) = 0 \]

### Step 9: Solve for \( x \)
Setting each factor to zero gives us:
1. \( x - 16 = 0 \) → \( x = 16 \)
2. \( x + 1 = 0 \) → \( x = -1 \) (not a valid solution since \( x \) must be positive)

Thus, the only valid solution is:
\[ x = 16 \]

### Final Answer
The value of \( x \) is \( 16 \).

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