Directions: Given data shows total male and female doctors in three hospital in a seminar. Read data carefully and answer the question:
In annual seminar of three hospitals, A, B and C some male and female doctors represent their hospitals. Average number of female doctors who represent A and B is 210. Total male doctors in A and B is 810. Number of female doctors is `2/3`rd and `2/5`th of male doctor in A and B repectively. Total female doctor who represent C are `25%` more than total female doctor who represent A and total male doctor who represent C are `33 1/3%` more than female doctor who represent B.
Find average number females in B and C?

Options

A. 250
B. 210
C. 260
D. 240

Answer

To solve the problem step by step, let's break down the information given and calculate the required values.

### Step 1: Define Variables
Let:
- \( M_A \) = Number of male doctors in Hospital A
- \( M_B \) = Number of male doctors in Hospital B
- \( F_A \) = Number of female doctors in Hospital A
- \( F_B \) = Number of female doctors in Hospital B
- \( F_C \) = Number of female doctors in Hospital C
- \( M_C \) = Number of male doctors in Hospital C

### Step 2: Set Up Equations
From the problem, we have the following information:
1. The average number of female doctors in A and B is 210:
\[ \frac{F_A + F_B}{2} = 210 \implies F_A + F_B = 420 \]

2. The total number of male doctors in A and B is 810:
\[ M_A + M_B = 810 \]

3. The number of female doctors in A is \( \frac{2}{3} \) of the male doctors in A:
\[ F_A = \frac{2}{3} M_A \]

4. The number of female doctors in B is \( \frac{2}{5} \) of the male doctors in B:
\[ F_B = \frac{2}{5} M_B \]

### Step 3: Substitute and Solve
Substituting \( F_A \) and \( F_B \) into the equation \( F_A + F_B = 420 \):
\[ \frac{2}{3} M_A + \frac{2}{5} M_B = 420 \]
To eliminate the fractions, we can multiply through by 15 (the least common multiple of 3 and 5):
\[ 10 M_A + 6 M_B = 6300 \quad \text{(Equation 1)} \]

Now we have two equations:
1. \( M_A + M_B = 810 \) (Equation 2)
2. \( 10 M_A + 6 M_B = 6300 \) (Equation 1)

### Step 4: Solve the System of Equations
From Equation 2, we can express \( M_B \) in terms of \( M_A \):
\[ M_B = 810 - M_A \]

Substituting this into Equation 1:
\[ 10 M_A + 6(810 - M_A) = 6300 \]
\[ 10 M_A + 4860 - 6 M_A = 6300 \]
\[ 4 M_A = 6300 - 4860 \]
\[ 4 M_A = 1440 \implies M_A = 360 \]

Now substituting \( M_A \) back to find \( M_B \):
\[ M_B = 810 - 360 = 450 \]

### Step 5: Find Female Doctors in A and B
Now we can find \( F_A \) and \( F_B \):
\[ F_A = \frac{2}{3} M_A = \frac{2}{3} \times 360 = 240 \]
\[ F_B = \frac{2}{5} M_B = \frac{2}{5} \times 450 = 180 \]

### Step 6: Calculate Female Doctors in C
According to the problem, the total female doctors who represent C are 25% more than total female doctors who represent A:
\[ F_C = F_A + 0.25 F_A = 240 + 60 = 300 \]

### Step 7: Calculate Average Number of Females in B and C
Now we can find the average number of female doctors in B and C:
\[ \text{Average} = \frac{F_B + F_C}{2} = \frac{180 + 300}{2} = \frac{480}{2} = 240 \]

### Final Answer
The average number of female doctors in B and C is **240**.

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