If the average age decreased by 6 months, calculate the average age of new students 
The average age of 12 students in exam hall is decreased by 6 months, when two of them aged 13 years and 15 years are replaced by two new students. The average age of new students is ?

I am not able to solve this problem and I am facing little ambiguity for solving, how the age will be decreased by six months? Any help appreciated please 
 A: 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.
A: Let $x_1,\ldots,x_{10}$ be the ages of the 10 students that did not move. Let $y_1$ and $y_2$ be the ages of the two new students.
The original average is $$\frac{1}{12}(x_1+\cdots+x_{10} + 13 + 15)$$
and the new average is $$\frac{1}{12}(x_1+\cdots+x_{10} + y_1 + y_2).$$
Since the difference in the averages is $0.5$ years,
$$\frac{1}{12}(x_1+\cdots+x_{10} + 13 + 15)
- \frac{1}{12}(x_1+\cdots+x_{10} + y_1 + y_2) = 0.5.$$
Can you rearrange to compute $\frac{1}{2}(y_1 + y_2)$?
A: Decreasing the average age of the 12 students by 6 months will decrease the total age by $12 \times 6 = 72 \text{ months} = 6 \text{ years}$. Therefore, we choose $A_\text{new} = 11$, which is the solution to 
$$\left(13 - A_\text{new}\right) + \left(15-A_\text{new}\right) = 6$$
A: Let $x_1,\ldots,x_{10}$ be the ages of the 10 students that did not move. Let $y_1$ and $y_2$ be the ages of the two new students.
The original sum of the age students is:
$$old = (x_1+\cdots+x_{10} + 13 + 15)$$
$$new = (x_1+\cdots+x_{10} + y_1 + y_2)$$
Multiply it by 12 to get the sum of months:
$$old = 12 (x_1+\cdots+x_{10} + 13 + 15)$$
$$new = 12 (x_1+\cdots+x_{10} + y_1 + y_2)$$
Compute the average in months:
$$old = \frac{1}{12} 12 (x_1+\cdots+x_{10} + 13 + 15)$$
$$new = \frac{1}{12} 12 (x_1+\cdots+x_{10} + y_1 + y_2)$$
As you can see, we can simplify the formulas to this:
$$old = (x_1+\cdots+x_{10} + 13 + 15)$$
$$new = (x_1+\cdots+x_{10} + y_1 + y_2)$$
We can reduce the formulas further since $$x_1+\cdots+x_{10}$$ are present in both and result and do not affect the result actually so just remove it.
$$old = (x_1+\cdots+x_{10} + 13 + 15)$$
$$new = (x_1+\cdots+x_{10} + y_1 + y_2)$$
Now we have two simple formulas:
$$old = (13 + 15)$$
$$new = (y_1 + y_2)$$
Now let solve the problem. What we know is the the new average is 6 months less than the old average.
$$new = old - 6$$
Technically we have everything we need to solve this:
$$new = (13 + 15) - 6$$
$$new = 28 - 6$$
$$new = 22$$
$$(y_1 + y_2) = 22$$
As we know that 22 is the sum for person y1 and y2, then the average will be:
$$\frac{1}{2} (y_1 + y_2) = \frac{1}{2} 22$$
$$\frac{1}{2} (y_1 + y_2) = 11$$
A: y is the total age of the group.
x is the "average age".
We know the age of 2 students (13 + 15), the rest are not known.
13 + 15 + 10x = y
But if all students have the same age, they each need an additional half a year to compensate for the reduction of the average age of the group.
12 * (x + 0.5) = y
So...
13 + 15 + 10x = 12x + 6
22 = 2x
11 = x
Checking the numbers:
Old situation:
13 + 15 + 10 * 11 = 138
Old average age:
138 / 12 = 11.5
New average age is 11, and the counts are:
12 * 11 = 132
Compensating by adding half a year to each student:
12 * ( 11 + 0.5 ) = 12 * 11.5 = 138
