Selection into (IIMs (Institutes of Management) is quite simple. In our coaching institute some students qualified CAT(The first stage of entrance into IIMs) but coincidentally the no. of boys who qualified the CAT was equal to the no. of girls. Besides these boys and girls got the calls from only IIM Ahmedabad and IIM Bangalore, but each of these from both the IIMs. `60%` of the boys failed in the group discussion ( the second phase of the selection process) and thus equal no of boys (but distinct) appeared for the personal interview of IIM-A and IIM-B (interview is the third and final stage of selection of a candidate) but `20%` of the boys who appeared for the interview of IIM-A and `60%` of the boys who appeared for the interview of IIM-B failed. If it is possible that a candidate can receive the falls from more than one IIMs but he/she can face the interview of only one IIM.
Given that only 24 boys form our coaching institute were selected by the IIM and IIM-B also a candidate can appear for the next stage only if he/she qualifies the previous stage of the exam, then find the no, of girls who qualified the CAT (Common Admission Test).

Options

A. 100
B. 250
C. 300
D. 600

Answer

To solve the problem step by step, we will denote the number of boys (and girls) who qualified the CAT as \( X \).

### Step 1: Determine the number of boys who passed the Group Discussion (GD)
Given that 60% of the boys failed in the GD, it means that 40% passed.

\[ \text{Boys who passed GD} = 40\% \text{ of } X = \frac{40}{100} \times X = \frac{2}{5}X \]

**Hint:** Remember that if 60% failed, then 40% must have passed.

### Step 2: Determine the number of boys who appeared for the interviews
Since the number of boys who passed the GD is equal to the number of girls who also qualified, the number of boys who appeared for the interviews at IIM Ahmedabad (IIM-A) and IIM Bangalore (IIM-B) is the same.

Let the number of boys who appeared for the interviews at both IIMs be \( Y \). Therefore,

\[ Y = \frac{2}{5}X \]

**Hint:** The number of boys who passed the GD is the same as the number of boys who appeared for the interviews.

### Step 3: Calculate the number of boys who passed the interviews
- For IIM-A, 20% of the boys who appeared failed:

\[ \text{Boys who passed IIM-A} = 80\% \text{ of } Y = \frac{80}{100} \times Y = 0.8Y \]

- For IIM-B, 60% of the boys who appeared failed:

\[ \text{Boys who passed IIM-B} = 40\% \text{ of } Y = \frac{40}{100} \times Y = 0.4Y \]

**Hint:** Calculate the passing percentage based on the failure rates given.

### Step 4: Substitute \( Y \) in the passing equations
Substituting \( Y = \frac{2}{5}X \):

- For IIM-A:

\[ \text{Boys who passed IIM-A} = 0.8 \times \frac{2}{5}X = \frac{16}{50}X = \frac{8}{25}X \]

- For IIM-B:

\[ \text{Boys who passed IIM-B} = 0.4 \times \frac{2}{5}X = \frac{8}{50}X = \frac{4}{25}X \]

**Hint:** Make sure to simplify the fractions correctly.

### Step 5: Set up the equation for total boys selected
According to the problem, the total number of boys selected from both IIMs is 24:

\[ \frac{8}{25}X + \frac{4}{25}X = 24 \]

Combining the fractions:

\[ \frac{12}{25}X = 24 \]

**Hint:** Combine like terms to simplify the equation.

### Step 6: Solve for \( X \)
To find \( X \), multiply both sides by 25:

\[ 12X = 24 \times 25 \]

Calculating the right side:

\[ 12X = 600 \]

Now, divide by 12:

\[ X = \frac{600}{12} = 50 \]

**Hint:** Always isolate \( X \) by performing inverse operations.

### Step 7: Find the number of girls who qualified the CAT
Since the number of girls who qualified the CAT is equal to the number of boys:

\[ \text{Number of girls} = X = 50 \]

**Final Answer:** The number of girls who qualified the CAT is **50**.

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