Question

If the mean of the numbers $a, b, 8, 5, 10$ is $6$ and their variance is $6.8$, then $ab$ is equal to:

• A. $6$
• B. $7$
• C. $12$
• D. $14$
• E. $25$

Hint:

When given both mean and variance for two unknown variables, always solve for $(a+b)$ and $(a^2+b^2)$ separately. The identity $(a+b)^2 = a^2 + b^2 + 2ab$ will then directly yield the product $ab$.

Answer 1

The Correct Option is C

Concept: The mean ($\mu$) of $n$ numbers is the sum divided by $n$. The variance ($\sigma^2$) is calculated as the mean of the squares minus the square of the mean.
• $\mu = \frac{\sum x_i}{n}$
• $\sigma^2 = \frac{\sum x_i^2}{n} - \mu^2$

Step 1: Use the mean to find the sum of $a$ and $b$.
The mean of $a, b, 8, 5, 10$ is $6$: \[ \frac{a + b + 8 + 5 + 10}{5} = 6 \quad \Rightarrow \quad a + b + 23 = 30 \] \[ a + b = 7 \quad \cdots (1) \]

Step 2: Use the variance to find the sum of squares.
Variance is $6.8$: \[ \frac{a^2 + b^2 + 8^2 + 5^2 + 10^2}{5} - (6)^2 = 6.8 \] \[ \frac{a^2 + b^2 + 64 + 25 + 100}{5} - 36 = 6.8 \] \[ \frac{a^2 + b^2 + 189}{5} = 42.8 \quad \Rightarrow \quad a^2 + b^2 + 189 = 214 \] \[ a^2 + b^2 = 25 \quad \cdots (2) \]

Step 3: Solve for $ab$ using the algebraic identity.
Using $(a + b)^2 = a^2 + b^2 + 2ab$: \[ (7)^2 = 25 + 2ab \quad \Rightarrow \quad 49 = 25 + 2ab \] \[ 2ab = 24 \quad \Rightarrow \quad ab = 12 \]