
The ratio of copper and zinc in an alloy is 9:5. If the weight of copper in the alloy is 48 grams, find the weight of zinc in the alloy.
Answer
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Hint: In order to solve this problem, we need to understand that the ratio of copper and zinc is the same as the ratio of their weight. The ratio is just the smallest possible non factorized value. So If the ratio is given and one of the quantities of ratios is given, we can easily find the fourth quantity by cross multiplication.
Complete step-by-step solution:
In this question, we are given the ratio of the copper to zinc and we need to find the actual weight of the zinc in the alloy.
The ratio of copper to zinc is 9:5.
We can also write the ratio in numerator and denominator form.
Therefore, we can write 9:5 as the $\dfrac{9}{5}$.
As the copper and zinc are in the ratio 9:5 we can also say that the weight of copper and weight zinc are also in the same ratio.
This can be said because the ratio is the simplest fraction of two quantities.
The ratio of the actual weight of the copper and zinc are also in the ratio of 9:5.
Hence, the equation becomes,
$\dfrac{9}{5}=\dfrac{\text{Weight of copper}}{\text{Weight of zinc}}$
Now, in the question, we are given the weight of copper as 48 grams.
So, substituting the values we get,
$\dfrac{9}{5}=\dfrac{48}{\text{Weight of zinc}}$
By cross-multiplying we get,
$9\times \text{weight of zinc}=5\times 48$
Solving this we get,
$\begin{align}
& \text{weight of zinc}=\dfrac{5\times 48}{9} \\
& =\dfrac{80}{3}=26.67
\end{align}$
Therefore, the weight of the zinc is 26.67 grams.
Note: In this question, we can see that the ratio is a unitless quantity, so the ratio of the weights also has to be unitless. Hence as the weight of copper is given in grams, we get the weight of zinc also in grams. And we can see that the answer is round off to two decimal places.
Complete step-by-step solution:
In this question, we are given the ratio of the copper to zinc and we need to find the actual weight of the zinc in the alloy.
The ratio of copper to zinc is 9:5.
We can also write the ratio in numerator and denominator form.
Therefore, we can write 9:5 as the $\dfrac{9}{5}$.
As the copper and zinc are in the ratio 9:5 we can also say that the weight of copper and weight zinc are also in the same ratio.
This can be said because the ratio is the simplest fraction of two quantities.
The ratio of the actual weight of the copper and zinc are also in the ratio of 9:5.
Hence, the equation becomes,
$\dfrac{9}{5}=\dfrac{\text{Weight of copper}}{\text{Weight of zinc}}$
Now, in the question, we are given the weight of copper as 48 grams.
So, substituting the values we get,
$\dfrac{9}{5}=\dfrac{48}{\text{Weight of zinc}}$
By cross-multiplying we get,
$9\times \text{weight of zinc}=5\times 48$
Solving this we get,
$\begin{align}
& \text{weight of zinc}=\dfrac{5\times 48}{9} \\
& =\dfrac{80}{3}=26.67
\end{align}$
Therefore, the weight of the zinc is 26.67 grams.
Note: In this question, we can see that the ratio is a unitless quantity, so the ratio of the weights also has to be unitless. Hence as the weight of copper is given in grams, we get the weight of zinc also in grams. And we can see that the answer is round off to two decimal places.
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