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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. :)
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Jaychoudhary
KarishmaB
Please help with best approach for this question to be solved under 2 min.

Total Cost = Cost of Diesel + Cost of Driver
I want to minimise this cost.

Cost of fuel used = 1/400 ( 1000/x + x) * 800 * 35 = 70*(1000/x + x)
Cost of Driver = 125 * 800/x

We need to minimise 70*(1000/x + x) + 100,000/x = 170,000/x + 70x
Of course differentiation will make it very simple but let's not use it. We will use its fundamentals.

Think of graphs of x and 1/x. x increases consistently while 1/x falls rapidly initially and then it starts falling by smaller and smaller amounts as x increases.

Attachment:
Screenshot 2022-05-05 at 10.31.02.png
Screenshot 2022-05-05 at 10.31.02.png [ 26.29 KiB | Viewed 2553 times ]

Every time x increases by 1 unit, 70x will increase by 70 units. If the increase in x by 1 unit leads to a decrease of more than 70 in 170,000/x, we will get a lower value of the sum.

Let's use the options now.
Put x = 50 in 170,000/x. We get 3400.
Put x = 49 in 170,000/x. We get 346... (something)

So when x goes from 49 to 50, 70x increases by 70 but 170,000/x reduces by something less than 70.
Hence, when x = 49, 170,000/x + 70x will have a lower value than when x is 50.
Now, we can see that if x increases by 1 from 50, 70x will increase by 70 but 170,000/x will decrease by an even smaller amounts. Hence, the minimum value of the cost will be when x = 49 (out of our given options)

Answer (A)

By the way, it is not a GMAT type 2-min problem.
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