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

The mean of the remaining 3 boys is 51.

Explanation:

The mean mark of the entire group (
`p_(10)`), dimensionless, is:

`p_(10) = (1)/(10)\cdot \Sigma_(i = 1)^(10) x_(i)` (Eq. 1)

(
`p_(10) = 58`)

`(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) = (1)/(7) \cdot \Sigma_(i = 1)^(7) x_(i)` (Eq. 2)

(
`p_(7) = 61`)

`(1)/(7)\cdot \Sigma_(i=1)^(7) x_(i) = 61` (Eq. 2b)

`p_(3) = (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:

`(1)/(10)\cdot [\Sigma_(i = 1)^(7)x_(i)+\Sigma_(i=8)^(10)x_(i)] = 58`

`(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) = (153)/(3)`

`p_(3) = 51`

The mean of the remaining 3 boys is 51.