This is a horribly ugly problem, and is nothing like what the GMAT would ask. They tell us the gas used per km, and since we're driving 800 km, we multiply the usage per km by 800 to find the total gas used, and then multiply by $35 to find the cost of the gas used. So the gas costs $70(1000/x + x).
Since the speed is x and the distance is 800, since time = distance/speed, the trip will take 800/x hours, and multiplying this by the hourly wage of $125/hr, the cost of paying the driver will be $100,000/x.
Adding the two costs, the total cost is
70(1000/x + x) + 100,000/x = 170,000/x + 70x ~ 70*(2429/x + x)
which we want to minimize. That's equivalent to minimizing 2429/x + x. Notice now we're adding two things, 2429/x and x, that multiply together to 2429. If ab = 2429 and we want to minimize a+b, we want a and b to be the same, so we want them both to be √2429, so that's the approximate value of x here, and since 50^2 = 2500, either 49 or 50 is correct, depending on whether 49^2 or 50^2 is closer to 2429. Since 49^2 = 7^4 = 2401, 49 is the answer.
The standard way to do this is to use calculus, which is beyond the scope of the GMAT (though so is the solution above imo). To find the minimum value of f(x) = 170,000/x + 70x, we just find the derivative f'(x) and set it equal to 0. The derivative of 170,000x^(-1) is -170,000x^(-2), and the derivative of 70x is just 70, so
f'(x) = -170,000/x^2 + 70
and setting this to zero,
0 = -170,000/x^2 + 70
x^2 = 170,000/70
x^2 ~ 2429
x ~ √2429
Anyway, not a GMAT problem, and not something anyone should feel they need to be able to solve at all, let alone within 2 minutes.