Solving the Cloth Cost Word Problem

The problem asks us to find the original length of a piece of cloth given its total cost and how the cost changes when the length and cost per meter are altered.

Setting up the Equations for Cloth Cost

Let's define the variables:

• Let $L$ be the original length of the piece of cloth in meters.
• Let $C$ be the original cost per meter of the cloth in Rupees.

We are given the total cost of the cloth is Rs. 35. The total cost is the length multiplied by the cost per meter. This gives us our first equation:

Equation 1: $L \times C = 35$

The problem then describes a scenario where the piece were 4 m longer and each meter was to cost Rs. 1 lesser. The new length would be $L+4$ meters, and the new cost per meter would be $C-1$ Rupees. The total cost in this new scenario remains unchanged at Rs. 35. This gives us our second equation:

Equation 2: $(L + 4)(C - 1) = 35$

Solving the System of Equations

We have a system of two equations with two variables, $L$ and $C$. We can use substitution to solve this system.

From Equation 1, we can express $C$ in terms of $L$ (assuming $L$ is not zero):

$C = \frac{35}{L}$

Now, substitute this expression for $C$ into Equation 2:

$(L + 4)\left(\frac{35}{L} - 1\right) = 35$

Next, we expand the left side of the equation:

$L \times \frac{35}{L} - L \times 1 + 4 \times \frac{35}{L} - 4 \times 1 = 35$

Simplify the terms:

$35 - L + \frac{140}{L} - 4 = 35$

Combine the constant terms on the left side:

$31 - L + \frac{140}{L} = 35$

Now, isolate the terms involving $L$ and $\frac{1}{L}$:

$\frac{140}{L} - L = 35 - 31$

$\frac{140}{L} - L = 4$

To eliminate the fraction, multiply the entire equation by $L$ (since $L$ represents length, we know $L > 0$, so we are not multiplying by zero):

$L \times \left(\frac{140}{L}\right) - L \times L = 4 \times L$

$140 - L^2 = 4L$

Rearrange the terms to form a standard quadratic equation $aL^2 + bL + c = 0$:

$L^2 + 4L - 140 = 0$

Solving the Quadratic Equation for Length

We have a quadratic equation $L^2 + 4L - 140 = 0$. We can solve this using the quadratic formula or by factoring. Let's use the quadratic formula $L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1$, $b=4$, and $c=-140$.

Calculate the discriminant, $\Delta = b^2 - 4ac$:

$\Delta = (4)^2 - 4(1)(-140)$

$\Delta = 16 + 560$

$\Delta = 576$

The square root of the discriminant is $\sqrt{576} = 24$.

Now, apply the quadratic formula to find the possible values for $L$:

$L = \frac{-4 \pm 24}{2 \times 1}$

$L = \frac{-4 \pm 24}{2}$

This gives two possible solutions for $L$:

$L_1 = \frac{-4 + 24}{2} = \frac{20}{2} = 10$

$L_2 = \frac{-4 - 24}{2} = \frac{-28}{2} = -14$

Since $L$ represents the length of the cloth, it must be a positive value. Therefore, we discard the negative solution $L_2 = -14$.

The original length of the piece of cloth is $L = 10$ meters.

Verification of the Solution

Let's check if $L=10$ meters satisfies the conditions of the problem.

• Original length $L = 10$ m.
• Original total cost = Rs. 35.
• Original cost per meter $C = \frac{35}{L} = \frac{35}{10} = 3.5$ Rs/m.

Now consider the modified conditions:

• New length = $L + 4 = 10 + 4 = 14$ m.
• New cost per meter = $C - 1 = 3.5 - 1 = 2.5$ Rs/m.
• New total cost = (New length) $\times$ (New cost per meter) = $14 \times 2.5$.

Calculation for new total cost:

The new total cost is Rs. 35, which is the same as the original total cost. This confirms that our calculated length of 10 meters is correct.

The original length of the piece of cloth is 10 meters, which corresponds to Option A.

Revision Table: Key Concepts

Additional Information: Real-World Math Problems

Math word problems like this cloth cost example help develop skills in translating real-world situations into mathematical models. This process involves:

• Identifying the known and unknown quantities.
• Assigning variables to the unknown quantities.
• Formulating equations based on the relationships described in the problem.
• Solving the equations using appropriate algebraic techniques (like substitution, elimination, or factoring).
• Interpreting the results in the context of the original problem, ensuring the solution makes sense (e.g., length cannot be negative).

Practicing these types of problems improves analytical and problem-solving abilities, which are valuable in various academic and everyday situations.