There are 43800 students in 4 schools of a city. If half of the first, two -third of the second, three fourth of the third and four -fifth of the fourth are the same number of students, then find the ratio of number of students of A and D if A,B,C and D be the first, second, third and fourth schools respectively.

Answer

To solve the problem step by step, we will denote the number of students in schools A, B, C, and D as \( A \), \( B \), \( C \), and \( D \) respectively.

### Step 1: Set up the equations based on the problem statement

According to the problem:
- Half of the students in school A: \( \frac{A}{2} \)
- Two-thirds of the students in school B: \( \frac{2B}{3} \)
- Three-fourths of the students in school C: \( \frac{3C}{4} \)
- Four-fifths of the students in school D: \( \frac{4D}{5} \)

We know that these four expressions are equal:
\[ \frac{A}{2} = \frac{2B}{3} = \frac{3C}{4} = \frac{4D}{5} \]

Let’s denote this common value as \( P \):
\[ \frac{A}{2} = P \implies A = 2P \]
\[ \frac{2B}{3} = P \implies B = \frac{3P}{2} \]
\[ \frac{3C}{4} = P \implies C = \frac{4P}{3} \]
\[ \frac{4D}{5} = P \implies D = \frac{5P}{4} \]

### Step 2: Write the total number of students equation

The total number of students in all four schools is given as 43,800:
\[ A + B + C + D = 43800 \]

Substituting the expressions for \( A \), \( B \), \( C \), and \( D \):
\[ 2P + \frac{3P}{2} + \frac{4P}{3} + \frac{5P}{4} = 43800 \]

### Step 3: Find a common denominator and simplify

The least common multiple (LCM) of the denominators (2, 3, 4) is 12. We will convert each term to have a denominator of 12:
\[ 2P = \frac{24P}{12} \]
\[ \frac{3P}{2} = \frac{18P}{12} \]
\[ \frac{4P}{3} = \frac{16P}{12} \]
\[ \frac{5P}{4} = \frac{15P}{12} \]

Now substituting back into the equation:
\[ \frac{24P}{12} + \frac{18P}{12} + \frac{16P}{12} + \frac{15P}{12} = 43800 \]

Combining the fractions:
\[ \frac{24P + 18P + 16P + 15P}{12} = 43800 \]
\[ \frac{73P}{12} = 43800 \]

### Step 4: Solve for \( P \)

To isolate \( P \), multiply both sides by 12:
\[ 73P = 43800 \times 12 \]
\[ 73P = 525600 \]
Now divide by 73:
\[ P = \frac{525600}{73} = 7200 \]

### Step 5: Calculate the number of students in each school

Now we can find \( A \), \( B \), \( C \), and \( D \):
\[ A = 2P = 2 \times 7200 = 14400 \]
\[ B = \frac{3P}{2} = \frac{3 \times 7200}{2} = 10800 \]
\[ C = \frac{4P}{3} = \frac{4 \times 7200}{3} = 9600 \]
\[ D = \frac{5P}{4} = \frac{5 \times 7200}{4} = 9000 \]

### Step 6: Find the ratio of students in A and D

Now we need the ratio of the number of students in school A to school D:
\[ \text{Ratio of A to D} = \frac{A}{D} = \frac{14400}{9000} \]
Simplifying this ratio:
\[ \frac{14400 \div 1800}{9000 \div 1800} = \frac{8}{5} \]

### Final Answer

The ratio of the number of students in school A to school D is:
\[ \text{Ratio of A to D} = 8 : 5 \]

---