Solution of question 2: I m short of time so I would just let you guys imagine the triangle.
Consider the line from B to AC intersecting Ac at E and similarly from C to AB at F.
Their point of intersection as D and join FE.
Now. since area of BDC = 10 and DRC = 5 => ratio is 2.. so BD : DE = 2 as well.
Reason: The line from the opposite vertex divided the triangle into ratio of its bases. Why? because altitude remains same.
Do not progress further unless you grasp this concept because it will be used multiple times.
Similarly if we consider triangle FDB and EDE both have bases BD and DE in ratio 2 so area of FDB and FDE have ratio 2
since area FDB = 8 => area FDE = 4 ...why? read the concept above.
Similarly for triangle ABD and ADE areas are in ratio =2/1
Let area AFD = x and area ADE = y ( our requirement is actually x+y value)
=> since ratio = 2 => 8+x=2y --------------------------------1
similarly for triangle AFD and ADC areas are in ratio 4/5 because area on same bases FD and DC of triangle FDE and EDC have ratio 4/5
=> \(\frac{x}{(y+5)} = \frac{4}{5}\)
=> 5x = 4y+20 ---------------------------------2
using 1 and 2
we get x= 12 and y =10
our requirement = x+y = 22 = answer D
PS: The solution looks complicated but if you try to understand it is very easy question and I think is probable option for 750+ question in GMAT. There is nothing un usual and it only require 1 min of work ..but it took me 10 mins because m out of touch