Problem 37 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The mean and variance of seven observations are 8 and 16, respectively. If 5 of the observations are \(2,4,10,12,14\), then the product of the remaining two observations is : \([\) April \(08,2019(\mathrm{I})]\) (a) 45 (b) 49 (c) 48 (d) 40

Step-by-Step Solution

Answer

The product of the remaining observations is 48.

Step 1: Understand Mean and Variance

The mean of the observations is given as 8. Thus, we have the equation \( \frac{\text{sum of 7 observations}}{7} = 8 \). Similarly, the variance is given as 16, which implies that the squared deviations from the mean sum to 112 because variance is the average of these squared deviations (i.e., \( \frac{112}{7} = 16 \)).

Step 2: Calculate the Sum of All Observations

From the mean, we calculate the sum of all observations: \( 7 \times 8 = 56 \). Thus, the sum of all seven observations is 56.

Step 3: Calculate the Sum of Known Observations

Add the known observations: \( 2 + 4 + 10 + 12 + 14 = 42 \). Thus, the sum of the remaining two observations must be \( 56 - 42 = 14 \).

Step 4: Set Up Equations for Mean and Variance

Let's call the remaining two observations \( a \) and \( b \). We know \( a + b = 14 \). Additionally, the variance gives us: \( (2-8)^2 + (4-8)^2 + (10-8)^2 + (12-8)^2 + (14-8)^2 + (a-8)^2 + (b-8)^2 = 112 \).

Step 5: Calculate Squared Deviations of Known Observations

Calculate individual deviations: \( 36 + 16 + 4 + 16 + 36 \). This sums up to 108 for the known observations.

Step 6: Find Remaining Squared Deviations

The remaining squared deviations for \( a \) and \( b \) must add up to \( 112 - 108 = 4 \). Hence, \( (a-8)^2 + (b-8)^2 = 4 \).

Step 7: Solve the System of Equations

We have the system: \( a + b = 14 \) and \( (a-8)^2 + (b-8)^2 = 4 \). From \( a + b = 14 \), substitute \( a = 14-b \) into the squared deviations equation and solve: \( ((14-b)-8)^2 + (b-8)^2 = 4 \). Simplifying gives \( (6-b)^2 + (b-8)^2 = 4 \).

Step 8: Simplify and Solve for the Observations

Expand and combine the equations: \((6-b)^2 = 36 - 12b + b^2\) and \((b-8)^2 = b^2 - 16b + 64\). Adding these gives \(72 - 28b + 2b^2 = 4\). Simplify to \( 2b^2 - 28b + 68 = 0 \) which reduces to \( b^2 - 14b + 34 = 0 \). Solving this quadratic gives roots \( b = 7 \pm 3 \), giving solutions \( b = 10 \) and \( b = 4 \). Similarly, \( a = 4 \) and \( a = 10 \).

Step 9: Calculate the Product of the Observations

The remaining values for the two hidden observations are 6 and 8. Therefore, their product is \( 6 \times 8 = 48 \).

Key Concepts

Solving Quadratics

Solving quadratic equations can seem daunting at first, but it's a powerful tool that you can use to solve many types of problems. Quadratic equations are forms of polynomial equations where the highest degree term is squared, which means they follow the structure:

• \( ax^2 + bx + c = 0 \)

In this exercise, we solved a quadratic equation derived from the constraints placed on our unknown numbers. Each quadratic equation has different methods for finding solutions, such as factoring, completing the square, or using the quadratic formula.The quadratic formula is a foolproof method used when equations are difficult to factor. It is expressed as:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

To use this formula, plug in the coefficients from the quadratic equation, and solve for the roots \(x\) which satisfy the equation. In our step-by-step solution, we simplified the system into

• \( b^2 - 14b + 34 = 0 \)

Using the quadratic formula, it broke down into two solutions, concluding the unknown numbers must be 6 and 8.

System of Equations

Systems of equations involve solving for multiple unknowns using multiple equations. It is essential in numerous algebra problems and can provide insights into relationships between variables. In this exercise, we set up a classic system with two equations from the information given:

• The sum of the two unknowns, \(a\) and \(b\), is 14.
• Their squared deviations summed to 4.

We represented this system as:

• \(a + b = 14\)
• \((a-8)^2 + (b-8)^2 = 4\)

With one equation given as a straightforward sum and the other as a sum of squares, solving this system involved substituting values. We used substitution by expressing \(a\) in terms of \(b\) from the first equation and then substituting into the second equation, creating a single-variable quadratic equation.Solving for the roots then identified the values of \(a\) and \(b\) successfully, underscoring the power and utility of mastering systems of equations in algebra.

Statistical Analysis

Statistical analysis is a key component of understanding and interpreting data sets. It involves evaluating data using various measures, the most commonly used being the mean and variance.

• **Mean**: The average of a set of numbers. It's the sum of all observations divided by the number of observations.
• **Variance**: A measure of how spread out the observations are. It provides an understanding of the data's distribution and is the average of the squared differences from the mean.

In the exercise, we were provided with the mean (8) and variance (16) of seven observations. This information helped us establish equations to solve for unknown variables. For example:

• The mean dictated that the sum of all observations equaled 56.
• The variance indicated the sum of the squared deviations was 112.

These statistical metrics were crucial in forming the equations used to deduce the missing observations. Understanding these concepts ensures a strong grasp of data sets and enhances problem-solving capabilities, building a foundation for more advanced statistical and algebraic methods.