The number of students who take both the subjects mathematics and chemistry is 30. This represents 10% of the enrolment in mathematics and 12% of the enrolment in chemistry. How many students take at least one of these two subjects?

Options

A. 520
B. 490
C. 560
D. 480

Answer

To solve the problem step by step, we will use the information provided about the number of students enrolled in mathematics and chemistry, as well as those taking both subjects.

### Step 1: Identify the known values
We know:
- The number of students taking both subjects (Mathematics and Chemistry) is \( N(M \cap C) = 30 \).
- This represents 10% of the total enrollment in Mathematics.
- This represents 12% of the total enrollment in Chemistry.

### Step 2: Calculate the total enrollment in Mathematics
Let \( N(M) \) be the total enrollment in Mathematics. According to the information given:
\[ N(M \cap C) = 10\% \text{ of } N(M) \]
This can be expressed mathematically as:
\[ 30 = 0.10 \times N(M) \]
To find \( N(M) \), we rearrange the equation:
\[ N(M) = \frac{30}{0.10} = 300 \]

### Step 3: Calculate the total enrollment in Chemistry
Let \( N(C) \) be the total enrollment in Chemistry. According to the information given:
\[ N(M \cap C) = 12\% \text{ of } N(C) \]
This can be expressed mathematically as:
\[ 30 = 0.12 \times N(C) \]
To find \( N(C) \), we rearrange the equation:
\[ N(C) = \frac{30}{0.12} = 250 \]

### Step 4: Calculate the total number of students taking at least one subject
We need to find the number of students taking at least one of the two subjects, which is given by the formula:
\[ N(M \cup C) = N(M) + N(C) - N(M \cap C) \]
Substituting the values we found:
\[ N(M \cup C) = 300 + 250 - 30 \]
Calculating this gives:
\[ N(M \cup C) = 520 \]

### Final Answer
The number of students who take at least one of these two subjects is \( \boxed{520} \).

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