Question

A piece of cloth costs Rs. 200. If the piece was 5 m longer and each meter of cloth costs Rs. 2 less the cost of the piece would have remained unchanged. How long is the piece and what is the original rate per meter?

Answer 1

Hint: Here we will assume the length of the piece to be \[x\] meters. Then we will find the rate per meter for the old as well as the new length. We will then form an equation based on the given information and simplify it to get a quadratic equation. We will then use the quadratic formula to get the original price. Using this, we will then find the original rate per meter.

Complete step-by-step answer:
Let us take the length of the original piece as \[x\] meters.
Cost of cloth piece of length \[x = {\rm{Rs}}.200\]
So, the rate per meter \[ = {\rm{Rs}}.\dfrac{{200}}{x}\]……………...\[\left( 1 \right)\]
It is given that the piece was 5 m longer, then
New length \[ = x + 5\] meters
The cost of the piece remains the same as given.
So, the new rate per meter \[ = {\rm{Rs}}.\dfrac{{200}}{{x + 5}}\]……………\[\left( 2 \right)\]
Next, as the difference between the costs of two pieces is 2, so the difference of equation \[\left( 1 \right)\] and \[\left( 2 \right)\] is,
\[\dfrac{{200}}{x} - \dfrac{{200}}{{x + 5}} = 2\]
Taking the L.C.M on LHS, we get
\[ \Rightarrow \dfrac{{200\left( {x + 5} \right) - 200 \times x}}{{x\left( {x + 5} \right)}} = 2\]
On cross multiplication, we get
\[\begin{array}{l} \Rightarrow 200x + 1000 - 200x = 2\left( {x\left( {x + 5} \right)} \right)\\ \Rightarrow 0 + 1000 = 2\left( {{x^2} + 5x} \right)\end{array}\]
Dividing both sides by 2, we get
\[ \Rightarrow \dfrac{{1000}}{2} = \left( {{x^2} + 5x} \right)\]
Simplifying the expression, we get
\[ \Rightarrow 500 = \left( {{x^2} + 5x} \right)\]
\[ \Rightarrow {x^2} + 5x - 500 = 0\]………...\[\left( 3 \right)\]
Now we will solve the above equation using quadratic equation formula which states:
For equation \[a{x^2} + bx + c = 0\] we use formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] to get the value of\[x\].
Substituting \[a = 1,b = 5\] and \[c = - 500\] in the above formula, we get
\[x = \dfrac{{ - 5 \pm \sqrt {{5^2} - 4 \times 1 \times \left( { - 500} \right)} }}{{2 \times 1}}\]
Simplifying the equation, we get
\[\begin{array}{l}x = \dfrac{{ - 5 \pm \sqrt {25 + 2000} }}{2}\\x = \dfrac{{ - 5 \pm \sqrt {2025} }}{2}\end{array}\]
Simplifying the square root, we get
\[x = \dfrac{{ - 5 \pm 45}}{2}\]
So, we get two values of \[x\].
\[x = \dfrac{{ - 5 - 45}}{2} = \dfrac{{ - 50}}{2} = - 25\]
\[x = \dfrac{{ - 5 + 45}}{2} = \dfrac{{40}}{2} = 20\]
As the length of the piece cloth cannot be negative, so the value of \[x\] is 20 meter.
Now we will find the rate per meter by substituting the value of \[x\] in equation \[\left( 1 \right)\], therefore, we get
Rate per meter \[ = {\rm{Rs}}.\dfrac{{200}}{{20}} = {\rm{Rs}}.10\]

So the original length of the cloth is 20 meter and the original rate per meter is Rs.10.

Note:
Here, we have simplified the equation to get a quadratic equation. A quadratic equation is defined as an equation which has the highest degree of 2 and has 2 solutions. Similarly, the cubic equation is defined as an equation which has the highest degree of 3 and has 3 solutions. So, we can say that the number of solutions depends on the highest degree of the equation. It is important to frame the equation carefully as per the given information or else, we will get the wrong answer.