Final answer:
When variable p, which varies directly with the cube of r and inversely with the square of t, is subjected to r being halved and t being doubled, p reduces to 1/32 of its original value.
Step-by-step explanation:
The student's question requires an understanding of direct and inverse variation. The problem states that p varies directly as r^3 and inversely as t^2. In mathematical terms, this relationship can be expressed as p = k * (r^3) / (t^2), where k is a constant of proportionality.
Now, to find out what happens to p when r is halved and t is doubled, we substitute the new values for r and t into our equation. If r becomes r/2, then r^3 becomes (r/2)^3 = r^3/8. If t is doubled, then t^2 becomes (2t)^2 = 4t^2. Therefore, the new value of p, which we can call p', will be p' = k * (r^3/8) / (4t^2) = p / 32. This shows that p is reduced to 1/32 of its original value.