Hi samarthkansal,

One shortcut of finding (1.03)^3 is by using binomial theorem. In binomial expansion of (1+x)^n

(1+x)� = 1 + n·x/1! + n(n-1)x²/2! + n(n-1)(n-2)x³/3! + . . . + n(n-1)...(n-k+1)x�/n! + . . . But in this question a simple approximation can be used, (1+ x)^n is approximately equal to 1 + nx.
In this case we can take x=0.04 and n=3 so (1.04)^3=(1 + 0.04)^3 this will approximately be equal to 1+ 3*0.04 = 1+0.12 =1.12

Another method is that, u can split 1.04 into (1 +0.04) and apply the identity of (a+b)^3
The identity : (a+b)^3= a^3 + b^3 + 3ba^2 + 3ab^2 with a=1 and b=0.04

There are many more ways like u can just figure out the last digit of the cube and then eliminate the options one by one. However such type of questions may not be very likely in GMAT.

Regards
Sukriti