Question

How many ounces of iodine worth 30 cents an ounce must be mixed with 50 ounces of iodine worth 18 cents an ounce so that the mixture can be sold for 20 cents an ounce?

Answer 1

Hint: Assume that we need x ounces of iodine whose price is 30 cents an ounce. Now, form a linear equation in one variable (x). Equate the total price of iodine on both sides. In the L.H.S take the sum of product of 30, x and 50, 18. In the R.H.S take the product of (50 + x) ounces of iodine with 20. At last solve the obtained linear equation to get the value of x.

Complete step by step solution:
Here we are mixing certain ounces of iodine whose cost is 30 cents an ounce with 50 ounces of iodine whose cost of each ounce is 18 cents. Now, this mixture of total weight of iodine formed is sold at 20 cents an ounce. We are asked to determine the weight of iodine taken initially whose price was 30 cents an ounce.
Let us assume that x ounce of iodine whose price is 30 cents each ounce is taken initially. So, we have the following relations:
(1) Total price of x ounces of iodine worth 30 cents each ounce = $30x$cents
(2) Total price of 50 ounces of iodine worth 18 cents each ounce = $18\times 50$ cents
(3) Total weight of mixture of iodine formed = $\left( 50+x \right)$ ounces
(4) Total price of (50 + x) ounces of iodine worth 20 cents each ounce = $20\times \left( 50+x \right)$ cents
We need to equate the sum of total price of iodine of the two categories with the total price of the mixture, so we get,
$\begin{align}
  & \Rightarrow 30x+\left( 18\times 50 \right)=20\times \left( 50+x \right) \\
 & \Rightarrow 30x+900=1000+20x \\
\end{align}$
Taking the terms containing the variable x to the L.H.S and the constant terms to the R.H.S we get,
$\begin{align}
  & \Rightarrow 30x-20x=1000-900 \\
 & \Rightarrow 10x=100 \\
 & \therefore x=10 \\
\end{align}$
Hence, 10 ounces of iodine having a price of 30 cents an ounce is required to be mixed.

Note: Here we do not need to convert cents into INR. We can also form two linear equations in two variables to solve this question but it would become confusing that is why we kept only one variable. While forming the linear equation you need to be careful about the arithmetic operation to be performed. You need to understand the question carefully as to perform either addition or subtraction.