Final answer:
Using the information provided and solving a system of equations, it is determined that student A initially had 8 pencils.
Step-by-step explanation:
The question is asking for a mathematical solution to determine the initial number of pencils that student A had. We are given that after redistribution, each student A, B, and C has 44 pencils. We start by setting up the initial number of pencils for B and C with A having the least, B the most, and C in between. Let the initial number of pencils A had be x, B had y, and C had z. We need to solve for x using the information given.
For the first allocation, B gives up 3/4 of his pencils, so that leaves B with y - 3/4y. A receives 1/5 of this amount, so A receives (3/4y)*(1/5) = 3/20y. C gives up 48 pencils in the second allocation, leaving C with z - 48. A and B share this equally, so A receives an additional 24 pencils.
After these transactions, A has x + 3/20y + 24 = 44, B has y - 3/4y + 24 = 44, and C has z - 48 = 44.
We now have a system of equations which are:
- x + 3/20y + 24 = 44
- y/4 + 24 = 44 (After simplifying B's pencils)
- z - 48 = 44
Solving for y and z in equations (2) and (3), we get y = 80 and z = 92. Substituting y into equation (1), we get x as the subject,
x + 3/20*80 + 24 = 44
x + 12 + 24 = 44
x + 36 = 44
So, x = 44 - 36 = 8. Therefore, student A initially had 8 pencils.