Math Problem Statement

तीन कक्षा X, Y तथा Z एक बीजगणितीय परीक्षा देती हैं। कक्षा X का औसत 83 है। कक्षा Y का औसत 76 है। कक्षा Z का औसत 85 है। कक्षा X तथा Y का औसत 79 है। कक्षा Y तथा Z का औसत 81 है। तीनों कक्षाओं का औसत क्या होगा?

Solution

The given question is in Hindi, and it involves averages of different classes. Let’s break it down step by step:

Problem Details:

  • There are three classes: X, Y, and Z.
  • 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 combined average score of class X and Y is 79.
  • The combined average score of class Y and Z is 81.

To find:

The average score of all three classes: X, Y, and Z.

Let the number of students in classes X, Y, and Z be denoted by nXn_X, nYn_Y, and nZn_Z, respectively.

Equations based on the dataThe given question is in Hindi, and it involves averages of different classes. Let’s break it down step by step:

Problem Details:

  • There are three classes: X, Y, and Z.
  • 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 combined average score of class X and Y is 79.
  • The combined average score of class Y and Z is 81.

To find:

The average score of all three classes: X, Y, and Z.

Let the number of students in classes X, Y, and Z be denoted by nXn_X, nYn_Y, and nZn_Z, respectively.

Equations based on the

  1. 83nX+76nY=79(nX+nY)83n_X + 76n_Y = 79(n_X + n_Y) (From the average of classes X and Y)
  2. 76nY+85nZ=81(nY+nZ)76n_Y + 85n_Z = 81(n_Y + n_Z) (From the average of classes Y and Z)

We can solve these equations to find the ratio of students in classes nXn_X, nYn_Y, and nZn_Z.

Let me compute this.From solving the equations, we get the following relationships between the number of students in each class:

  • nX=34nYn_X = \frac{3}{4} n_Y
  • nZ=54nYn_Z = \frac{5}{4} n_Y

Now, to find the overall average of classes X, Y, and Z, we will use the formula for the weighted average:

Average=83nX+76nY+85nZnX+nY+nZ\text{Average} = \frac{83n_X + 76n_Y + 85n_Z}{n_X + n_Y + n_Z}

Substitute nX=34nYn_X = \frac{3}{4}n_Y and nZ=54nYn_Z = \frac{5}{4}n_Y into this equation to find the total average. Let's compute this next.The overall average score of all three classes XX, YY, and ZZ is 81.5.

Final Answer:

Option B: 81.5

Would you like more details on this solution, or do you have any questions?

Here are 5 related questions for practice:

  1. What is the formula for weighted average, and how is it used?
  2. How would the solution change if the averages of the classes were different?
  3. How can you solve simultaneous linear equations using substitution?
  4. Why is it important to find the relationships between class sizes in this problem?
  5. Can you explain how averages are affected by the number of participants?

Tip: Always check if the variables can be simplified using ratios before solving a complex equation system!

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Math Problem Analysis

Mathematical Concepts

Averages
Algebra
Linear Equations

Formulas

Weighted Average Formula: \(\text{Average} = \frac{a_1n_1 + a_2n_2 + a_3n_3}{n_1 + n_2 + n_3}\)
Linear equations for class averages

Theorems

Linear system of equations

Suitable Grade Level

Grades 9-10