One of those not-terribly-useful questions which examiners come up with from time to time. You have to tolerate it .. keeping in mind that these poor folk are kept in dark dungeons out in the back of the main building and deserve our empathy.
What is the question on about ?
(a) looking to see if you have some very basic algebraic/arithmetic understanding
(b) if you get past (a) the answer falls out very easily .. if not, then you are up for some heavy simultaneous equations .. all good fun, but quite unnecessary
First off, figure the CG for the load (ignore the asterisks - I still have trouble with formatting in this medium) -
BOW***4200****90*****378000
FUEL****500***100******50000
LOAD***4700****91.1***428000
(if you want to do the calc to 40 decimal places, that's OK, just a bit pointless).
Now, the MAC bit.
Two straightforward ways to approach this problem.
First, a bit more complicated, but relates the solution back to the usual MAC equation -
What you don't have given is the MAC length and LEMAC, so we need to figure a simple way to work them out.
The trick is to be aware of the following -
(a) both the %MAC and FS scales are linear so you can use proportions
(b) for the BOW, 20% = 90 units on the FS scale (doesn't matter what scale they are)
for the fuel load, 30% = 100 units
therefore it follows that 10% = 10 units
without too much strain you can figure out that
length of the MAC (ie 100%MAC) = 100 units (10% = 10 units .. multiple both by 10)
LEMAC = 70 units (if 90 units = 20% and 10 units is equal to 10%, then LEMAC = 90-20)
this gives you the MAC equation at
%MAC = ((FS-LEMAC)/MAC)*100
%MAC = ((FS - 70)/100)*100
which, in this case, comes out to be
%MAC = FS-70 (aren't round numbers great ?)
So the final answer is %MAC = 91.1-70 = 21.1%
The slightly simpler way to get the answer relies on the fact that the scales are linear and that you can use simple proportion (normal back of the prayer wheel sort of stuff)
(a) figure out, as before, that 10% = 10 units
(b) taking the measures from the BOW position as a temporary datum
delta CG = 1.1
delta %MAC = 1.1% (as 10 units = 10%)
So, from the original datum, %MAC = 20 + 1.1 = 21.1%
Not too sure if you would ever find a real world use for this particular exercise but that's the nature of exams, I guess.
For this sort of question, do it, get it right, collect the mark, and move on ....
If the explanation is too rambling and convoluted, ask again for clarification and I'll keep it to the point.