Points A and B are 120 km apart. A motorcyclist starts from

Question

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?

A. 3 hours
B. 3.4 hours
C. 3.5 hours
D. 3.6 hours
E. None

A. 3 hours
B. 3.4 hours
C. 3.5 hours
D. 3.6 hours
E. None

D, please provide workable method that can solve this in 2mins, i solved but it takes more than 3mins. can any one help.
thanks

Bunuel · Accepted Answer

Untitled.png 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.

gurpreetsingh · Answer

IMO D

Distance traveled by 1st = 30t in t time
distance traveled by 2nd = 10t in t time

now at any instance a right angled triangle is formed whose sides are as follow

1. The distance traveled by 2nd
2. The distance left between the 1st and B
3. Present distance between both the bikers- This is the hypotenuse.

Thus the equation of their distance becomes

d^2 = (10t)^2 + (120-30t)^2

Which is maximum when its differential is 0
=> 20t = 2(120-30t) *3 => t = 3.6

ykaiim · Answer

Great Bunuel.

But, I think if you give some details on this equation, we can better understand.

I tried to solve this by plug-in values. Somehow, 1 hr before there is GC site problem and couldnt post my explanation.

As you move from A, the horizontal distance will reduce as per (120-30t). So, at t=4 hr A will reach at the origin of motoryst B. By refering the properties of triangle, you can figure it out.

1. At t=3.5hr, A will travel 105km (ramaining 15km to reach at origin) and B will travel 35km, so shortest distance = \sqrt{(15^2+35^2)} = \sqrt{(1450)}

2. At t=3.6hr, A will travel 108km (ramaining 12km to reach at origin) and B will travel 36km, so shortest distance = \sqrt{(12^2+36^2)} = \sqrt{(1440)}

So, OA should be D.

SVaidyaraman · Answer

Consider two moving objects, one starting from A and moving at a speed of s1 towards B and the other starting from B and moving perpendicular to AB at a speed of s2. The least distance at any point in time between them is the hypotenuse. The distance between them is given by  \sqrt{(s1*t) ^ 2 + (s2*t) ^ 2}

As the objects move, the length of the hypotenuse will keep changing. The least length of the hypotenuse is given by,
 
 \sqrt{(s1*tmin) ^ 2 + (s2*tmin ) ^ 2}

where  tmin= s1*d / (s1^2 + s2^2) and is the time taken when the least length of the hypotenuse is reached.

and d is the distance between A and B.

For this problem  tmin= 30 * 120 / (30 ^2 + 10 ^ 2) = 3.6 hrs.

Note: tmin  can be derived as done in Gurpreet Singh's post.

WholeLottaLove · Answer

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? :shock:

Attachments

GRAPH.png [ 11.3 KiB | Viewed 47310 times ]

prasannajeet · 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

Kconfused · Answer

I don't think this problems needs calculations of any sort. 
Simply visualize a variable right triangle with A toward B forming the base and perpendicularly away from B as the height. The resulting hypotenuse represents the shortest distance between the bikers at any point. 
The variable, A towards B (Base) diminishes at a much faster rate than does perpendicularly away from B (The height) because of the significant difference in the speeds of the bikers. Now since the hypotenuse (or the distance between the bikers) is the sum of the squares of the sides, it's least when their combination is minimum ( which obviously happens just before 4 hours or closest to 4 hours).

Guess, visualizing this concept takes much lesser time than reading it (say under a minute!!)

himanshujovi · Answer

I dont think differential / integral calculus must be getting tested in GMAT. If yes , then a single differential is just a proof of point of inflection. The equation would need to be differentiated again to confirm that the point was inflection was maxima or minima

KarishmaB · Answer

I would say this question is borderline. GMAT expects you to know how to find the roots of the quadratic using the formula (though it never expects you to actually use it). Also, knowing what the graph of a quadratic looks like is a useful skill on GMAT. We know that when the co-efficient of the x^2 term is positive, it is upward facing so it will have a minimum value. By symmetry, the minimum value will lie right in the center of the roots i.e. it will be the average of the two roots. The two roots are given by 
 \frac{-b + \sqrt{b^2 - 4ac}}{2a} and \frac{-b - \sqrt{b^2 - 4ac}}{2a} . Their average is -b/2a.

So you don't actually need to use differential calculus. I wouldn't recommend my students to ignore this question if they were looking at 750.

GMATGuruNY · Answer

Let M = the motorcyclist and C = the cyclist. After 3.4 hours: M has traveled 102 km of the 120 km between A and B, placing him 18 km from B. C has traveled 34 km from B. Every 1/10 of an hour, M's distance from B decreases by 3 km, while the C's distance from B increases by 1 km, as follows: .....0 hr....3.4 hrs....3.5 hrs....3.6 hrs....3.7 hrs M: 120--->18------> 15 --------> 12 ------> 9 C: 000--->34-------> 35 -------> 36 ------> 37 Since M and C travel along perpendicular roads, the distance between them constitutes the hypotenuse of a right triangle, as follows: (distance)^2 = M^2 + C^2 The colored combinations above imply the following: M^2 + C^2 = 15^2 + 35^2 = 1450 M^2 + C^2 = 12^2 + 36^2 = 1440 M^2 + C^2 = 9^2 + 37^2 = 1450 The results in red are equal and exceed the result in green. Implication: The least distance will be yielded by the values in green, which occur at the 3.6-hour mark. D The following problem is similar but involves much friendlier values: https://gmatclub.com/forum/andrew-start ... 21827.html

samarpan.g28 · Answer

I think the easiest approach will be to go by finding the minima.
Let us consider t hours are taken by motorcyclists and cyclists to keep the least distance between them. Of course, the distance is the hypotenuse between the x and y axis considering their locations on them.
For motorcyclists, its distance from point B is (120-30t) km and for cyclists, its distance from B is 10t km. Let hypotenuse is x km.
Therefore,  (120-30t)^2  +  10t^2  =  x^2 .
Solve the above and we get  1000t^2  - 7200t + 14400 =  x^2 .
The minima is obtained by using -b/2a. Here, it will be 7200/(2*1000)=3.6 hours.

Another way is, see, for all given options A to D, if you put them in the above equation instead of D, you get values of x like 42.something, 38.something, 38.something and finally 37.something for t=3.6. So, we can say option (D) gives us least distance. However, the first method is far easier and faster.

Regor60 · Answer

Just a slight variation that allows T to be determined directly without resorting to roots through completing square:

The car is 120-30T from B at a given time and the cyclist 10T.

So the hypotenuse D is represented by:

D^2 = (120-30T)^2 + (10T)^2

Expanding:

100T^2 - 720T + 1440, dividing out 10 since makes no difference.

Completing the square:

(10T-36)^2 + 144

The lowest value the squared term could be is:

0=10T-36 and T=3.6

Posted from my mobile device

Kinshook · Answer

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?

Let point A be origin (-120,0) and point B be (0,0).

Points A and B are 120 km apart. A motorcyclist starts from A to B along straight road AB with speed 30 kmph.
x = -120 + 30t; where t is time in hours

At the same time a cyclist starts from B along a road perpendicular to road AB, with a speed of 10 kmph. 
y = 10t; where t is time in hours

Motorcyclist coordinates after time t hours = (-120+30t,0)
Cyclist coordinates after time t hours = (0,10t)

Distance between them after time t hours = D =  \sqrt{(-120+3t)^2 + 100t^2}=14400 + 900t^2 - 7200t + 100t^2

D =  \sqrt{14400 + 1000t^2 - 7200t }= \sqrt{1000(t^2 - 2*3.6t + 3.6^2) + 14400 - 12960} = \sqrt{1000(t-3.6)^2 + 1140}

Therefore, at time t = 3.6 hours, distance between them will be least

IMO D

Points A and B are 120 km apart. A motorcyclist starts from

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