The average marks obtained by five students A,B,C,D and E is `52(3)/(5)`. If the average marks obtained by A, C and D are `53(2)/(3)` and average marks obtained by B,C and E are `52(2)/(3)`,then the marks obtained by C were

Options

A. 56
B. 53
C. 57
D. 60

Answer

To solve the problem step by step, we will use the information provided to set up equations based on the averages given.

### Step 1: Calculate the total marks of A, B, C, D, and E

The average marks of the five students A, B, C, D, and E is given as \( 52 \frac{3}{5} \).

First, convert the mixed number into an improper fraction:
\[ 52 \frac{3}{5} = \frac{52 \times 5 + 3}{5} = \frac{260 + 3}{5} = \frac{263}{5} \]

Now, to find the total marks of the five students, multiply the average by the number of students (5):
\[ \text{Total marks} = \text{Average} \times \text{Number of students} = \frac{263}{5} \times 5 = 263 \]

This gives us our first equation:
\[ A + B + C + D + E = 263 \quad \text{(Equation 1)} \]

### Step 2: Calculate the total marks of A, C, and D

The average marks of students A, C, and D is given as \( 53 \frac{2}{3} \).

Convert this average into an improper fraction:
\[ 53 \frac{2}{3} = \frac{53 \times 3 + 2}{3} = \frac{159 + 2}{3} = \frac{161}{3} \]

Now, to find the total marks of A, C, and D, multiply the average by the number of students (3):
\[ \text{Total marks of A, C, D} = \frac{161}{3} \times 3 = 161 \]

This gives us our second equation:
\[ A + C + D = 161 \quad \text{(Equation 2)} \]

### Step 3: Calculate the total marks of B, C, and E

The average marks of students B, C, and E is given as \( 52 \frac{2}{3} \).

Convert this average into an improper fraction:
\[ 52 \frac{2}{3} = \frac{52 \times 3 + 2}{3} = \frac{156 + 2}{3} = \frac{158}{3} \]

Now, to find the total marks of B, C, and E, multiply the average by the number of students (3):
\[ \text{Total marks of B, C, E} = \frac{158}{3} \times 3 = 158 \]

This gives us our third equation:
\[ B + C + E = 158 \quad \text{(Equation 3)} \]

### Step 4: Solve for B and E using Equations 1 and 2

Now, we will subtract Equation 2 from Equation 1 to find B and E:
\[ (A + B + C + D + E) - (A + C + D) = 263 - 161 \]
This simplifies to:
\[ B + E = 102 \quad \text{(Equation 4)} \]

### Step 5: Solve for C using Equations 3 and 4

Now, we will subtract Equation 4 from Equation 3 to find C:
\[ (B + C + E) - (B + E) = 158 - 102 \]
This simplifies to:
\[ C = 56 \]

### Conclusion

The marks obtained by student C were \( 56 \).