Question

If the cost of x meters of wires is d rupees, then the cost of y meters of wire at the same rate is
A.\[Rs.\left( {\dfrac{{xy}}{d}} \right)\]
B.\[Rs.\left( {xy} \right)\]
C.\[Rs.\left( {yd} \right)\]
D.\[Rs.\left( {\dfrac{{yd}}{x}} \right)\]

Answer 1

Hint: We are given with the cost of x meters of wire and we are asked to find the cost of y meters. So we will use the unitary method here. With the help of which we will be able to find the correct answer.

Complete step-by-step answer:
Given that the cost of x meters of wires is d rupees.
The cost of 1meter of such wire is
 \[
  xmeters \to Rs.d \\
  1meter \to ? \\
   \Rightarrow ? = \dfrac{d}{x} \\
\]
Now we have to find the cost of y meters of wire at the same rate.
So cost of y meters of wire is \[ \Rightarrow y\dfrac{d}{x}\]
This is the cost of y meters of wire \[Rs.\dfrac{{yd}}{x}\] .
So the correct option is D.


Note: Students generally make mistakes in these types of problems. They just focus that the cost of x meters is Rs.d then the cost of y meters is Rs.yd , but this is not the correct answer.Let’s elaborate this concept.
Suppose we say that the cost of 1 chocolate is Rs.5 then the cost of 2 chocolates will be? This we can answer very easily that is Rs.10. But if we are given that the cost of 9 flowers is Rs.20 then what is the cost of 3 such flowers? Now something needs to be calculated! This is the time we need a unitary method.
\[
  9flowers \to Rs.20 \\
  1flowers \to ? \\
   \Rightarrow ? = \dfrac{{20}}{9} = \dfrac{{20}}{9} \\
   \Rightarrow 3 flowers = \dfrac{{20}}{9} \times 3 = \dfrac{{20}}{3} \\
\]
We used the same approach in our problem above. Hope you understood!