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 students who appeared in exam Y only is equal to no of students appeared in exam Y and Z only.
How many students appeared in two exams only?

Options

A. 280
B. 220
C. 340
D. 260

Answer

To solve the problem step by step, we will analyze the information provided and use algebra to find the required values.

### Step 1: Define Variables
1. Let \( A \) be the number of students who appeared in exam Z only.
2. The number of students who did not appear in any exam is also \( A \).
3. Let \( X \) be the number of students who appeared in exam X only.
4. Let \( Y \) be the number of students who appeared in exam Y only.
5. Let \( B \) be the number of students who appeared in exams Y and Z only.
6. Let \( C \) be the number of students who appeared in exams X and Y only.
7. Let \( D \) be the number of students who appeared in exams X and Z only.
8. Let \( E \) be the number of students who appeared in all three exams X, Y, and Z.

### Step 2: Set Up Equations Based on Given Information
1. Total number of students = 1000.
2. Students who appeared in exam Y = 360.
\[ Y + B + C + E = 360 \]
3. The ratio of students who appeared in exams X and Y only (C) to those who appeared in exams Y and Z only (B) is \( 2:3 \).
\[ \frac{C}{B} = \frac{2}{3} \implies C = \frac{2}{3}B \]
4. The number of students who appeared in exams X and Z both (D) is half of those who appeared in only exam Z (A).
\[ D = \frac{1}{2}A \]
5. The number of students who appeared in exam X only (X) is 50% more than those who appeared in exam Y only (Y).
\[ X = 1.5Y \]
6. The number of students who appeared in all three exams is 4% of the total number of students.
\[ E = 0.04 \times 1000 = 40 \]
7. The number of students who appeared in exam Y only (Y) is equal to the number of students who appeared in exam Y and Z only (B).
\[ Y = B \]

### Step 3: Substitute and Solve
1. From \( Y = B \), we can substitute \( B \) in the equation for students in exam Y:
\[ Y + Y + C + E = 360 \implies 2Y + C + 40 = 360 \implies 2Y + C = 320 \]
2. Substitute \( C = \frac{2}{3}Y \) into the equation:
\[ 2Y + \frac{2}{3}Y = 320 \]
Multiply through by 3 to eliminate the fraction:
\[ 6Y + 2Y = 960 \implies 8Y = 960 \implies Y = 120 \]
3. Since \( Y = B \), we have \( B = 120 \).
4. Now substitute \( Y \) back to find \( C \):
\[ C = \frac{2}{3}B = \frac{2}{3} \times 120 = 80 \]
5. Now find \( X \):
\[ X = 1.5Y = 1.5 \times 120 = 180 \]
6. Now find \( D \):
\[ D = \frac{1}{2}A \quad \text{(we will find A later)} \]

### Step 4: Use Total Students to Find A
1. Total students can be expressed as:
\[ X + Y + B + C + D + E + A + A = 1000 \]
Substitute known values:
\[ 180 + 120 + 120 + 80 + D + 40 + 2A = 1000 \]
Substitute \( D = \frac{1}{2}A \):
\[ 180 + 120 + 120 + 80 + \frac{1}{2}A + 40 + 2A = 1000 \]
Combine like terms:
\[ 540 + \frac{5}{2}A = 1000 \implies \frac{5}{2}A = 460 \implies A = \frac{460 \times 2}{5} = 184 \]

### Step 5: Calculate D
1. Now calculate \( D \):
\[ D = \frac{1}{2}A = \frac{1}{2} \times 184 = 92 \]

### Step 6: Calculate Total Students Appeared in Two Exams
1. Students who appeared in two exams only:
\[ C + B + D = 80 + 120 + 92 = 292 \]

### Final Answer
The total number of students who appeared in two exams only is **292**.