- $P$ Pa
- $M$ Ma
- $B$ Brother
- $Y$ You
$$P + M + B + Y = 83$$
$$6P = 7M$$
$$M = 3Y$$
Combine and write everything in terms of $B$ and $Y$.
$$21/6 Y + 3Y + B + Y = 83$$
$$45 Y + 6B = 498$$
Sum of even numbers is even, so $Y$ must be even, call it $Y=2n$:
$$90n + 6Y = 498$$
$$15n + Y = 83$$
$$\begin{bmatrix} n \\ B \end{bmatrix} = \begin{bmatrix} n \\ 83 - 15n \end{bmatrix}$$
$$\begin{bmatrix} Y \\ B \end{bmatrix} = \begin{bmatrix} 2n \\ 83 - 15n \end{bmatrix}$$
Insert back in the parents:
$$\begin{bmatrix} P \\ M \\ Y \\ B \end{bmatrix} = \begin{bmatrix} 7n \\ 6n \\ 2n \\ 83 - 15n \end{bmatrix}$$
I think we can assume $B \ge 0$, so $83 - 15n \ge 0$, so $n \le 5$.
Hopefully $B < M$, so $83 - 15n < 6n$, so $n \ge 4$.
So your choices are $n=4$ or $n=5$:
$$\begin{bmatrix} P \\ M \\ Y \\ B \end{bmatrix} = \begin{bmatrix} 28 \\ 24 \\ 8 \\ 23 \end{bmatrix} \text{ or } \begin{bmatrix} 35 \\ 30 \\ 10 \\ 8 \end{bmatrix}$$
Assuming you are humans and your brother isn't adopted, it's probably the second one.
Let's say weight of the persons are $x,y, z, w$ and weight of the person replacing the person of the weight 150 is $a$ kg. After first change average is decreased by $3$.
$\dfrac{x+y+z+w+150}{5}-\dfrac{x+y+z+w+a}{5}=3$
So, $a=135$. Now, $a$ is replaced by a person of weight $(a-30)$kg i.e. $105$ kg.
So, Overall change is the average is, initial average- average after replacing 150 kg person by 105 kg.
$\dfrac{x+y+z+w+150}{5}-\dfrac{x+y+z+w+105}{5}=\dfrac{45}{5}=9$
So, the answer is $9$ kg i.e option B.
Best Answer
Age of each student is reduced by 6 month. So total age reduced is 72 months or 6 years.
It means 2 new students has age 6 years less.
Total age of replaced student = 28 years.
Total age of new students = 28 - 6 = 22 years.
Average age of 2 new students is 11 years.