Question

An alloy is to contain copper and zinc in the ratio of $9:4$. The zinc required to be melted, with $24$kg of copper is:
A). $10\dfrac{2}{3}$kg
B). $10\dfrac{1}{3}$kg
C). $9\dfrac{2}{3}$kg
D). $9$kg

Answer 1

Hint: First of all, we have to know about the term’s ratio and proportion. In certain situations, the comparison of two quantities by the method of division is very efficient. We can say that the comparison or simplified form of two quantities of the same kind is referred to as ratio. This relation gives us how many times one quantity is equal to the other quantity. In simple words, the ratio is the number, which can be used to express one quantity as a fraction of the other ones.

Complete step-by-step solution:
Proportion is an equation which defines that the two given ratios are equivalent to each other. In other words, the proportion states the equality of the two fractions or the ratios. In proportion, if two sets of given numbers are increasing or decreasing in the same ratio, then the ratios are said to be directly proportional to each other.
For example, the time taken by train to cover $100km$ per hour is equal to the time taken by it to cover the distance of $500km$ for $5$ hours. Such as $100$kmph $=$ $500$km/$5$hrs.
We use the given ratio and assumes the quantity according to that ratio. Now from given data, we will equate the quantities of copper to get the proportional constant. By using the value of the proportional constant, we will find the quantity of zinc.
Given that, the ratio of copper and zinc is $9:4$.
From the above rations assume the quantity of copper as $9x$ and zinc as $4x$.
Given that quantity of copper is $24$kg, then
$9x=24$
Now let’s divide the above equation with $9$ to get the value of $x$, then
$\begin{align}
  & \dfrac{9x}{9}=\dfrac{24}{9} \\
 & x=\dfrac{24}{9}
\end{align}$
Now the quantity of zinc is $4x$, now substituting value of $x$
Quantity $=4x$
       $=4\left( \dfrac{24}{9} \right)$
Now factorizing $4\times 24$ as $2\times 2\times 2\times 2\times 2\times 3$, substituting this value in above equation
Quantity $=\dfrac{2\times 2\times 2\times 2\times 2\times 3}{3\times 3}$
                 $=\dfrac{32}{3}$
If we divide $32$ by $3$ then we will get $10$ as quotient and $2$ as the reminder, hence we can write $\dfrac{32}{3}$ as $10\dfrac{2}{3}$.
Now the quantity of zinc required is $10\dfrac{2}{3}kg$.

Note: We can also solve the problem by simply equating
$\begin{align}
  &\Rightarrow \dfrac{9}{4}=\dfrac{\text{Quantity of copper}}{\text{Quantity of zinc}} \\
 &\Rightarrow \dfrac{9}{4}=\dfrac{24}{\text{Quantity of zinc}} \\
 &\Rightarrow \text{Quantity of zinc}=\dfrac{24\times 4}{9} \\
 & =10\dfrac{2}{3}kg
\end{align}$
From both the methods we get the same answer.