Yesterday, Manhattan GMAT posted a 700 level GMAT question on our blog. Today, they have followed up with the answer:
This problem requires the use of a special given equation, 5v2 + P = c. We are told that the velocity decreases from 10 meters per second to 5 meters per second. The key to this problem is to recognize that we have a “before” situation and an “after” situation. In the “before” situation, we have a certain velocity and pressure; in the “after” situation, we have a different velocity and pressure. What remains constant between the two situations is the constant c. Thus, we should set up the given equation for both situations.
To distinguish the situations, let’s use subscripts on the variables. P1 and v1 will indicate “before,” while P2 and v2 will indicate “after.”
Before: 5v12 + P1 = c
5(10)2 + P1 = c
500 + P1 = c
After: 5v22 + P2 = c
5(5)2 + P2 = c
125 + P2 = c
Now, we cannot solve for c, but we can set the left sides of these two equations equal to each other, because they are both equal to c.
500 + P1 = 125 + P2
Again, we cannot solve for either pressure, but we do not need to. What we need to find is the increase in pressure – in other words, how much the pressure rises by. As an expression, the increase in pressure is simply P2 – P1. Thus, we rearrange the equation to solve for this difference.
500 – 125 = P2 – P1
375 = P2 – P1
The correct answer is (C).
