Question

By a reduction of Rs. 2 per kg in the price of sugar, Anita can purchase 2 Kg sugar more for Rs. 224. Find the original price of sugar per Kg. What value of Anita is depicted in the question?

Answer 1

Hint: In this question first we assume the initial price of one Kg of sugar is Rs. \[x\] and weight of the sugar is \[y\] Kg. Then the reduction of Rs. 2 per Kg in the sugar price will give 2 Kg more sugar which we can represent as \[\left( x-2 \right)\] and \[\left( y+2 \right)\] respectively. As we know that in both cases the total price remains the same so on comparing total price in both cases we can obtain the required result.

Complete step by step answer:
Now, we have to find out the total price and total weight of the sugar purchased by Anita if by a reduction of Rs. 2 per kg in the price of sugar, Anita can purchase 2Kg sugar more for Rs. 224.
Let us assume that the price of the sugar per Kg is Rs. \[x\] and the total weight of the sugar purchased by Anita is \[y\] Kg. Now, if the price of sugar gets reduced by Rs. 2 per Kg then it can be represented as Rs. \[\left( x-2 \right)\] which results into purchase of 2 Kg more sugar and it is represented as \[\left( y+2 \right)\] Kg.
Now, we know that in both cases the total price of the sugar remains same and hence,
\[\begin{align}
  & \Rightarrow xy=224\cdots \cdots \cdots \left( i \right) \\
 & \Rightarrow \left( x-2 \right)\left( y+2 \right)=224 \\
 & \Rightarrow xy+2x-2y-4=224 \\
 & \Rightarrow xy+2x-2y=228\cdots \cdots \cdots \left( ii \right) \\
\end{align}\]
Now, by substituting \[\left( i \right)\] in \[\left( ii \right)\] we can write,
\[\begin{align}
  & \Rightarrow 224+2x-2y=228 \\
 & \Rightarrow 2x-2y=4 \\
\end{align}\]
On dividing both sides by 2 we get,
\[\begin{align}
  & \Rightarrow x-y=2 \\
 & \Rightarrow x=2+y\cdots \cdots \cdots \left( iii \right) \\
\end{align}\]
By substituting this value of \[x\] in \[\left( i \right)\] we get,
\[\begin{align}
  & \Rightarrow \left( 2+y \right)y=224 \\
 & \Rightarrow 2y+{{y}^{2}}=224 \\
 & \Rightarrow {{y}^{2}}+2y-224=0 \\
 & \Rightarrow {{y}^{2}}+16y-14y-224=0 \\
 & \Rightarrow y\left( y+16 \right)-14\left( y+16 \right)=0 \\
 & \Rightarrow \left( y+16 \right)\left( y-14 \right)=0 \\
 & \Rightarrow y=-16,14 \\
\end{align}\]
But as \[y\] represents the weight of the sugar so it cannot be negative and hence,
\[\Rightarrow y=14\]
Now, from \[\left( iii \right)\] we can write,
\[\begin{align}
  & \Rightarrow x=2+y \\
 & \Rightarrow x=2+14 \\
 & \Rightarrow x=16 \\
\end{align}\]
Hence, the original price of the sugar is Rs. \[16\] per Kg and the total weight of the sugar purchased by Anita is \[14\] Kg.

Note: In this type of question students have to take care when they solve the quadratic equation \[{{y}^{2}}+2y-224=0\]. One of the students may use \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] to find the roots of quadratic equation \[{{y}^{2}}+2y-224=0\] as follows:
\[\begin{align}
  & \Rightarrow y=\dfrac{-2\pm \sqrt{{{2}^{2}}-4\times 1\times \left( -224 \right)}}{2\times 1} \\
 & \Rightarrow y=\dfrac{-2\pm \sqrt{4+896}}{2} \\
 & \Rightarrow y=\dfrac{-2\pm \sqrt{900}}{2} \\
 & \Rightarrow y=\dfrac{-2\pm 30}{2} \\
 & \Rightarrow y=\dfrac{-2+30}{2},\dfrac{-2-32}{2} \\
 & \Rightarrow y=14,-16 \\
\end{align}\]
Also students have to remember that, as \[y\] represents the weight of the sugar so it cannot be negative.