In 2010, `60% `of total number of students in a class (Y) were girls. In 2015 in class (Y),the number of girls grew by `5%` and that of boys by 40 respectively, from 2010, as a result of which the ratio of the number of girls to boys in the class (Y) became 21:20. How many students (boys + girls) were in class (Y) in 2010?

Options

A. 250
B. 240
C. 300
D. 200

Answer

To solve the problem step by step, we will break it down into manageable parts.

### Step 1: Define the total number of students in 2010
Let the total number of students in class Y in 2010 be \( X \).

### Step 2: Calculate the number of girls and boys in 2010
Since 60% of the total students were girls:
- Number of girls in 2010 = \( 0.6X \)
- Number of boys in 2010 = \( 0.4X \)

### Step 3: Determine the changes in the number of girls and boys by 2015
According to the problem:
- The number of girls increased by 5% from 2010 to 2015.
- The number of boys increased by 40.

Calculating the number of girls in 2015:
- Increase in girls = \( 5\% \) of \( 0.6X = 0.05 \times 0.6X = 0.03X \)
- Therefore, the number of girls in 2015 = \( 0.6X + 0.03X = 0.63X \)

Calculating the number of boys in 2015:
- Number of boys in 2015 = \( 0.4X + 40 \)

### Step 4: Set up the ratio of girls to boys in 2015
According to the problem, the ratio of girls to boys in 2015 is \( 21:20 \):
\[ \frac{0.63X}{0.4X + 40} = \frac{21}{20} \]

### Step 5: Cross-multiply to solve for \( X \)
Cross-multiplying gives:
\[ 20 \times 0.63X = 21 \times (0.4X + 40) \]
\[ 12.6X = 8.4X + 840 \]

### Step 6: Rearrange the equation
Subtract \( 8.4X \) from both sides:
\[ 12.6X - 8.4X = 840 \]
\[ 4.2X = 840 \]

### Step 7: Solve for \( X \)
Divide both sides by \( 4.2 \):
\[ X = \frac{840}{4.2} = 200 \]

### Conclusion
The total number of students (boys + girls) in class Y in 2010 was \( 200 \).