The distance between two motorcyclists would be the length of the hypotenuse, which is square root of \((120-30x)^2+(10x)^2=1000x^2-60*120x+120^2\) (where x is the time in hours) . So we need to minimize the value of quadratic expression (function) \(1000x^2-60*120x+120^2\).

Now quadratic function \(f(x)=ax^2+bx+c\) reaches its minimum (or maximum when \(a\) is negative - not our case), when \(x=-\frac{b}{2a}=\frac{60*120}{2*1000}=3.6\)

Answer: D.
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Starting distance is 120.

As the first motorcyclist moves east, the second motorcyclist moves north. This forms a right triangle. The distance between the two motorcyclists is the hypotenuse of the triangle.

I'd backsolve this one and plug in the answer choices. This minimum distance occurs at answer choice B. The value is the square root of 1380.
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this question is one of prep tests, i came across, i don't think it is of GMAT level question,
because, optimization of quadratic equation is not a subject matter of gmat,
unless, it involves very simple, integer based calculations or like that.

sorry for confusion. it is from MGMAT prep test.

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE


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Points A and B are 120 km apart. A motorcyclist starts from A to B along straight road AB with speed 30 kmph. At the same time a cyclist starts from B along a road perpendicular to road AB, with a speed of 10 kmph. After how many hours will the distance between them be the least?

See attached graph

This problem is basically looking for the shortest hypotenuse in a triangle comprised of biker on road AB and cyclist moving directly perpendicular to AB from point B. Using the graph, we can plug in numbers into the quadratic formula to see what length C is the shortest.

a^2+b^2=c^2

1 Hour: Biker has traveled 30 km and cyclist has traveled 10 km.
90^2 + 10^2 = c^2
8200 = c^2
We don't need to find the square - we can compare values of c^2

2 Hour: Biker has traveled 60 km and cyclist has traveled 20 km.
60^2+20^ = c^2
4000 = c^2

3 hour: Biker has traveled 90 km and cyclist has traveled 30 km.
30^2 + 30^2 = c^2
1800 = c^2

For each increasing hour, the cyclist and biker move closer together. Out of all the possible answer choices, 3.6 reduces the bikers time the most (and thus the square of a larger speed i.e. 30 km/h) It's also possible to just look at the graph and recognize that the more hours that pass, the shorter the red line between
the biker and cyclist is.

ANSWER: D. 3.6 hours

One other thing...technically none of the answer choices are correct. Distance would be minimized when the biker hit B and was 40km away from the cyclist traveling perpendicular from line AB: a^2+b^2=c^2 ===> 0^2+40^2=c^2 ===> 1600 = c^2 so wouldn't E actually be the right answer?

Hi Bunuel

Please explain the below part.....

Now quadratic function f(x)=ax^2+bx+c reaches its minimum (or maximum when a is negative - not our case), when x=-\frac{b}{2a}=\frac{60*120}{2*1000}=3.6

Rgds
Prasannajeet

Bunuel is drawing on the fact that the max or min of a parabola is the vertex. The formula for calculating the vertex of the parabola is given by:
\(x = - \frac{b}{2a}\) where \(a, b\) are found in the equation \(f(x) = ax^2+bx+c\) In our case our values are:

\(b=60*120\) and \(a=1000\)

and plugging those into our vertex equation yields the answer of 3.6.

I hope this helps!
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Isn't the distance minimized at a 45/45/90 triangle?

Hey the distance between the 2 bikers will be approximately 38 Kms after 3.6 hrs which is less than 40 Kms after 4 hours.

After 3.6 hrs
Biker A will have moved 3.6 * 30 = 108 Kms i.e. 12 Kms remaining
and Biker B will have moved by 3.6 * 10 = 36 Kms

= \(\sqrt{(12^2 + 36^2)}\)
= \(\sqrt{(1400)}\) < \(\sqrt{(1600)}\)

Consider Kudos if the post helped