Three maths classes : X, Y and Z, take an algebra test. The average score of class X is 83. The average score of class Y is 76. The average score of class Z is 85. The average score of class X and Y is 79 and average score of class Y and Z is 81. What is the average score of classes X , Y, Z

Answer

To find the average score of classes X, Y, and Z, we can follow these steps:

### Step 1: Understand the given averages
We know the following averages:
- Average score of class X (A_X) = 83
- Average score of class Y (A_Y) = 76
- Average score of class Z (A_Z) = 85
- Average score of classes X and Y (A_XY) = 79
- Average score of classes Y and Z (A_YZ) = 81

### Step 2: Set up equations based on averages
Let the number of students in classes X, Y, and Z be \( n_X \), \( n_Y \), and \( n_Z \) respectively.

From the average scores, we can express the total scores:
- Total score of class X = \( 83 \cdot n_X \)
- Total score of class Y = \( 76 \cdot n_Y \)
- Total score of class Z = \( 85 \cdot n_Z \)

Using the averages of classes X and Y:
\[ A_{XY} = \frac{Total\ Score\ of\ X + Total\ Score\ of\ Y}{n_X + n_Y} = 79 \]
This gives us:
\[ \frac{83n_X + 76n_Y}{n_X + n_Y} = 79 \tag{1} \]

Using the averages of classes Y and Z:
\[ A_{YZ} = \frac{Total\ Score\ of\ Y + Total\ Score\ of\ Z}{n_Y + n_Z} = 81 \]
This gives us:
\[ \frac{76n_Y + 85n_Z}{n_Y + n_Z} = 81 \tag{2} \]

### Step 3: Solve the equations
From equation (1):
\[ 83n_X + 76n_Y = 79(n_X + n_Y) \]
Expanding and rearranging gives:
\[ 83n_X + 76n_Y = 79n_X + 79n_Y \]
\[ 4n_Y = 4n_X \implies n_X = n_Y \tag{3} \]

From equation (2):
\[ 76n_Y + 85n_Z = 81(n_Y + n_Z) \]
Expanding and rearranging gives:
\[ 76n_Y + 85n_Z = 81n_Y + 81n_Z \]
\[ 4n_Z = 5n_Y \implies n_Z = \frac{4}{5}n_Y \tag{4} \]

### Step 4: Substitute values to find averages
Substituting \( n_X = n_Y \) from equation (3) into equation (4):
Let \( n_Y = n \), then \( n_X = n \) and \( n_Z = \frac{4}{5}n \).

Now, we can find the total average score of classes X, Y, and Z:
\[ A_{XYZ} = \frac{Total\ Score\ of\ X + Total\ Score\ of\ Y + Total\ Score\ of\ Z}{n_X + n_Y + n_Z} \]
Substituting the total scores:
\[ A_{XYZ} = \frac{83n + 76n + 85 \cdot \frac{4}{5}n}{n + n + \frac{4}{5}n} \]
Calculating the total scores:
\[ = \frac{83n + 76n + 68n}{2n + \frac{4}{5}n} = \frac{227n}{2n + 0.8n} = \frac{227n}{2.8n} = \frac{227}{2.8} \]
Calculating \( \frac{227}{2.8} \):
\[ = 81.07 \approx 81.1 \]

### Final Answer
The average score of classes X, Y, and Z is approximately **81.1**.

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