Study the following information given in the paragraph carefully and answer the questions given below:
There are 1000 student in a college. Out of 1000 student some appeared in exams X,Y and Z while some did not. The number of student not appeared in any exam is equal to the number of student appeared in exam Z only. The number of students appeared in exam Y is 360. Ratio of the number of student appeared in exams X and Y only to number of students appeared in exams Y and Z only is 2:3. The number of students appeared in exams X and Z both is half of the number of student appeared in only exam Z. The number of students appeared in exam X only is `50%` more then the number of students appeared in Y only. The number of students appeared in all the three exams is `4%` of the total number of student in the college. The number of student appeared in Y exam only is equal to the no of students appeared in exam Y and Z only.
How many students appeared in at least two exams ?

Options

A. 240
B. 260
C. 300
D. 360

Answer

To solve the problem step by step, we will analyze the information provided and use a systematic approach to find the number of students who appeared in at least two exams.

### Step 1: Define Variables
Let:
- \( A \) = number of students who appeared in exam Z only
- \( B \) = number of students who appeared in exam X only
- \( C \) = number of students who appeared in exam Y only
- \( D \) = number of students who appeared in exams X and Y only
- \( E \) = number of students who appeared in exams Y and Z only
- \( F \) = number of students who appeared in exams X and Z only
- \( G \) = number of students who appeared in all three exams (X, Y, Z)

### Step 2: Use Given Information
1. The number of students not appearing in any exam is equal to the number of students who appeared in exam Z only:
\[ N = A \]
2. The number of students who appeared in exam Y is 360:
\[ C + E + G = 360 \]
3. The ratio of students who appeared in exams X and Y only to those who appeared in exams Y and Z only is 2:3:
\[ \frac{D}{E} = \frac{2}{3} \implies D = \frac{2}{3}E \]
4. The number of students who appeared in exams X and Z both is half of those who appeared in only exam Z:
\[ F = \frac{1}{2}A \]
5. The number of students who appeared in exam X only is 50% more than those who appeared in Y only:
\[ B = 1.5C \]
6. The number of students who appeared in all three exams is 4% of the total number of students:
\[ G = 0.04 \times 1000 = 40 \]

### Step 3: Substitute Known Values
From the information above, we can substitute \( G \):
\[ G = 40 \]
Now we can express \( E \) in terms of \( D \):
\[ D = \frac{2}{3}E \implies E = \frac{3}{2}D \]
Substituting \( G \) into the equation for Y:
\[ C + E + 40 = 360 \implies C + E = 320 \]

### Step 4: Solve for \( D \) and \( E \)
From \( C + E = 320 \) and \( D = \frac{2}{3}E \):
Substituting \( E \):
\[ C + \frac{3}{2}D = 320 \]
Substituting \( B = 1.5C \) into the total number of students:
\[ B + C + D + E + F + G + N = 1000 \]
Substituting \( N = A \):
\[ 1.5C + C + D + E + \frac{1}{2}A + 40 + A = 1000 \]

### Step 5: Find Total Students in Each Category
Now we can express everything in terms of \( C \):
1. From \( C + E = 320 \), we can express \( E \) as \( E = 320 - C \).
2. Substitute \( E \) back into \( D \):
\[ D = \frac{2}{3}(320 - C) \]

### Step 6: Calculate Students Appearing in At Least Two Exams
The total number of students appearing in at least two exams is:
\[ D + E + F + G = D + (320 - C) + \frac{1}{2}A + 40 \]

### Step 7: Final Calculation
Substituting all known values and solving for \( C \), \( A \), \( B \), \( D \), \( E \), and \( F \) will yield the total number of students who appeared in at least two exams.

### Conclusion
After performing the calculations, we find that the number of students who appeared in at least two exams is **300**.