Question

An alloy contains \[15\% \] zinc, \[25\% \] brass and rest copper. How much copper is there in 75 kg of that alloy?
A.20 kg
B.30 kg
C.45 kg
D.10 kg

Answer 1

Hint: Here in this question, we have to find how much copper present in 75 kg of alloy. To find this first we have find the percent of copper present in alloy by subtracting the percent of zinc and brass from \[100\% \] and convert the resultant percentage term into fraction form and multiply the fraction with \[75\] using the arithmetic operation we get the amount of copper present in alloy.

Complete step-by-step answer:
Consider the data in given question
An alloy contains
Zinc: \[15\% \]
Brass: \[25\% \]
Now find the percent of copper present in alloy by subtracting the percent of zinc and brass present in \[100\% \] of alloy i.e.,
\[ \Rightarrow 100\% - 15\% - 25\% \]
On simplification, we get
\[ \Rightarrow 60\% \]
\[\therefore \] \[60\% \] of copper present in alloy.
Now, find the amount of copper present in 75 kg of alloy
Simply, we need to calculate the number \[60\% \] of \[75kg\] for this.
Firstly, we have to write the percentage term i.e., \[60\% \] into the fraction.
Remember that "per" means "divided by", and "cent" means 100, so "percent" means "divided by 100" or "out of 100."
Therefore \[60\% \] can be written as:
\[ \Rightarrow \dfrac{{60}}{{100}}\]
When dealing with percent’s the word "of" means "times" or "to multiply".
Multiply the total weight of alloy i.e., \[75kg\] for the above fraction we have
\[ \Rightarrow \dfrac{{60}}{{100}} \times 75\]
\[ \Rightarrow 0.6 \times 75\]
On simplification, we get
\[ \Rightarrow 45\]
Therefore, 45 kg of copper present in 75 kg of alloy.
Hence, option (C) is correct.
So, the correct answer is “Option C”.

Note: Here the question is related to the percentage. By using the specific methods and rules we can convert the number. As we know that the percentage term is written as \[x\% = \dfrac{x}{{100}}\] , here, marked \[x\] as the total value. Using the simple arithmetic operation i.e., multiplication and division to get the required solution.