Question

The mean mark of a group of 10 boys is 58. If the mean of 7 of them is 61, what is the mean of the remaining 3 boys

Answer 1

Answer:

The mean of the remaining 3 boys is 51.

Step-by-step explanation:

The mean mark of the entire group ($p_{10}$), dimensionless, is:

$p_{10} = \frac{1}{10}\cdot \Sigma_{i = 1}^{10} x_{i}$ (Eq. 1)

($p_{10} = 58$)

$\frac{1}{10}\cdot \Sigma_{i=1}^{10} x_{i} = 58$ (Eq. 1b)

Where $x_{i}$ is the mark of the i-th boy, dimensionless.

In addition, we know the following mean marks from statement:

$p_{7} = \frac{1}{7} \cdot \Sigma_{i = 1}^{7} x_{i}$ (Eq. 2)

($p_{7} = 61$)

$\frac{1}{7}\cdot \Sigma_{i=1}^{7} x_{i} = 61$ (Eq. 2b)

$p_{3} = \frac{1}{3}\cdot \Sigma_{i=8}^{10}x_{i}$ (Eq. 3)

Where:

$p_{7}$ - Mean mark of the first 7 boys, dimensionless.

$p_{3}$ - Mean mark of the remaining 3 boys, dimensionless.

By applying sum properties in (Eq. 1b) and using (Eq. 2b) and (Eq. 3), we obtain the mean of the remaining 3 boys:

$\frac{1}{10}\cdot [\Sigma_{i = 1}^{7}x_{i}+\Sigma_{i=8}^{10}x_{i}] = 58$

$\frac{1}{10}\cdot [7\cdot (61)+3\cdot p_{3}] = 58$

$7\cdot (61) + 3\cdot p_{3} = 580$

$3\cdot p_{3} = 153$

$p_{3} = \frac{153}{3}$

$p_{3} = 51$

The mean of the remaining 3 boys is 51.