Assuming you're referring to the gouge posted in the gouge section here, titled "ASTB Prep Test":
28:
Let
C = # of captains
B = # of 1st Lieutenants
A = # of 2nd Lieutenants
The info given states that:
B + A = 30
A + C = 45
C + B = 35
and asks for A+B+C
The easiest way to get it is just to add up all three lines, giving you
2(A+B+C) = 30+45+35 = 110
so A+B+C = 55
Problem 34 is defining an abstract operation *, defined as x*y = y(x+1). The definition tells you exactly what to do.
If -4*j = -6,
then expand out * by the defn:
-6 = -4 *j = j(-3)
So it's clear that j = 2
Problem 46:
The thing here is to remember that the area of a triangle is (1/2) X height X base, regardless of how canted the top point is. So in this case, the area of CED = (1/2) Xheight X base = (1/2) X CD X AD.
Now CD X AD is the area of the rectangle, which we know to be 10.
So the area is 1/2 X 10 = 5.
Note that we skipped computing the value of CD, because we'd end up multiplying by AD anyway.
Problem 47:
This is an equilateral triangle. Draw the altitude from the top vertex. This spilts it into 2 30-60-90 triangles, which the height is a leg of. Use the formula for the ratio of the sides of a 30-60-90 triangle to determine the height, then use the (.5) x height x base formula as above.