buan15 wrote:

Very difficult to answer under time constraints; any simple way to figure out or hit and trial is the only option?

There's a mechanical way of solving any question like this without relying on hit and trial based on linear algebra

The idea is that we are manipulating the equations by multiplying with a constant, and then adding the 2 equations to get our desired equation.

Our desired equation is one that has the same coefficient for B, S and C

eq 1 - 3B + 7S + C multiply this equation by x

eq 2 - 4B + 10S + C multiply this equation by y

Since we want the coefficients the same after adding, 7x+ 4y (coefficient of B) = 7x + 10y (coefficient of S) = x+y (coefficient of C)

this gives x = -3y/2

Since for simplicity we want x & y to be integers, we can use y = -2, so x = 3

So if we multiply eq 1 by 3 and eq 2 by -2, and add the two equations together we will get B + S + C.

Not sure if my explanation is complicated but in my opinion its a very easy and foolproof way to solve these type of questions.