$$Let\ the\ number\ of\ Vans\ =\ x$$
$$Let\ the\ number\ of\ Cars\ =\ y$$
$$x+y=118\ $$
The question now is How many of them are Vans? (i.e. What is the value of x)
Statement 1:- If 14 more Vans are driven into the garage, there will be twice Vans as Cars in the garage.
$$So,\ y=\frac{\left(x+14\right)}{2}$$
$$x+y=118\ \ \ \left(sub.\ the\ new\ value\ of\ y\right)$$
$$\left(x+14\right)+\frac{\left(x+14\right)}{2}=118+14=132$$
$$\frac{\left(2x+28+x+14\right)}{2}=132$$
$$2x+28+x+14=264$$
$$3x=222$$
$$x=\frac{222}{3}=74$$
$$Thus,\ Statement\ 1\ is\ SUFFICIENT$$

Statement 2:- If x Vans and y Cars are driven into the garage, there will be equal number of Vans and Cars in the garage.
The value of x Vans and y Cars that were driven into the garage is unknown.
$$Hence,\ statement\ 2\ is\ NOT\ SUFFICIENT$$
$$In\ conclusion,\ Statement\ 1\ alone\ is\ SUFFICIENT\ ,\ thereby\ validating\ OPTION\ A\ as\ the\ correct\ answer.$$